How do complex number exponents actually work? [duplicate]












3












$begingroup$



This question already has an answer here:




  • Complex Exponent of Complex Numbers

    2 answers




I know Euler's formula and how to take complex exponents, but in it it's $e$ to an imaginary angle, not a number, it seems. From my understanding pi itself is not an angle, but $pi$ radians is. And since cosine can only take in an angle, or at least a representation of one, and the variable is the same everywhere in Euler's formula, the exponent should be an angle and not a number. The question is then raised could I then say $e^{180i}=-1$? Surely this has a single numerical value though, right?



I've been bothered by this for a long time. Am I wrong or how can I take an exponent with a complex number?










share|cite|improve this question











$endgroup$



marked as duplicate by Dietrich Burde, Lord Shark the Unknown, A. Pongrácz, Cesareo, José Carlos Santos 56 mins ago


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.











  • 1




    $begingroup$
    What difference do you see between an "angle" and a "number"?
    $endgroup$
    – Matteo
    14 hours ago










  • $begingroup$
    @Matteo numbers represent angles so you can call them the same... but $frac{pi}{4}$, 45, and 50 can all be called the same. How I'm seeing it, like vectors, we can only talk about certain amounts of basis, but not the basis itself. You can represent a dozen of eggs using the number 1, but also using the number 12, for 12 eggs. Like the dozen of actual eggs themselves, an angle is purely geometric and can't be described as strictly equal to a single number. When we talk about the angle as a number, we really mean the angle is that number amount of "insert arbitrary basis angle here".
    $endgroup$
    – Benjamin Thoburn
    14 hours ago










  • $begingroup$
    Let's put it this way: "cosine can only take in an angle", you say. Does this mean to you what. Cannot you put any number as argument of the cosine function?
    $endgroup$
    – Matteo
    14 hours ago










  • $begingroup$
    I would reason this way (in the first quadrant, then you can work on the others by symmetries). Take a circle of radius one with center in the origin of the axes, then draw a half-line that intersects the circle in the first quadrant. Let $x$ be the length of the arc measured from the point $(1,0)$ (with same unit of measure used for the axes of course). Then the absissa of the intersection point is called $cos (x)$. Does it work? No angles here. Only arc lengths.
    $endgroup$
    – Matteo
    14 hours ago








  • 2




    $begingroup$
    So, do not think of the angle in the exponent as being an angle at all. Think of it as being the ratio of the arc length to the radius length, and it will all make sense.
    $endgroup$
    – Eric Lippert
    9 hours ago
















3












$begingroup$



This question already has an answer here:




  • Complex Exponent of Complex Numbers

    2 answers




I know Euler's formula and how to take complex exponents, but in it it's $e$ to an imaginary angle, not a number, it seems. From my understanding pi itself is not an angle, but $pi$ radians is. And since cosine can only take in an angle, or at least a representation of one, and the variable is the same everywhere in Euler's formula, the exponent should be an angle and not a number. The question is then raised could I then say $e^{180i}=-1$? Surely this has a single numerical value though, right?



I've been bothered by this for a long time. Am I wrong or how can I take an exponent with a complex number?










share|cite|improve this question











$endgroup$



marked as duplicate by Dietrich Burde, Lord Shark the Unknown, A. Pongrácz, Cesareo, José Carlos Santos 56 mins ago


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.











  • 1




    $begingroup$
    What difference do you see between an "angle" and a "number"?
    $endgroup$
    – Matteo
    14 hours ago










  • $begingroup$
    @Matteo numbers represent angles so you can call them the same... but $frac{pi}{4}$, 45, and 50 can all be called the same. How I'm seeing it, like vectors, we can only talk about certain amounts of basis, but not the basis itself. You can represent a dozen of eggs using the number 1, but also using the number 12, for 12 eggs. Like the dozen of actual eggs themselves, an angle is purely geometric and can't be described as strictly equal to a single number. When we talk about the angle as a number, we really mean the angle is that number amount of "insert arbitrary basis angle here".
    $endgroup$
    – Benjamin Thoburn
    14 hours ago










  • $begingroup$
    Let's put it this way: "cosine can only take in an angle", you say. Does this mean to you what. Cannot you put any number as argument of the cosine function?
    $endgroup$
    – Matteo
    14 hours ago










  • $begingroup$
    I would reason this way (in the first quadrant, then you can work on the others by symmetries). Take a circle of radius one with center in the origin of the axes, then draw a half-line that intersects the circle in the first quadrant. Let $x$ be the length of the arc measured from the point $(1,0)$ (with same unit of measure used for the axes of course). Then the absissa of the intersection point is called $cos (x)$. Does it work? No angles here. Only arc lengths.
    $endgroup$
    – Matteo
    14 hours ago








  • 2




    $begingroup$
    So, do not think of the angle in the exponent as being an angle at all. Think of it as being the ratio of the arc length to the radius length, and it will all make sense.
    $endgroup$
    – Eric Lippert
    9 hours ago














3












3








3


1



$begingroup$



This question already has an answer here:




  • Complex Exponent of Complex Numbers

    2 answers




I know Euler's formula and how to take complex exponents, but in it it's $e$ to an imaginary angle, not a number, it seems. From my understanding pi itself is not an angle, but $pi$ radians is. And since cosine can only take in an angle, or at least a representation of one, and the variable is the same everywhere in Euler's formula, the exponent should be an angle and not a number. The question is then raised could I then say $e^{180i}=-1$? Surely this has a single numerical value though, right?



I've been bothered by this for a long time. Am I wrong or how can I take an exponent with a complex number?










share|cite|improve this question











$endgroup$





This question already has an answer here:




  • Complex Exponent of Complex Numbers

    2 answers




I know Euler's formula and how to take complex exponents, but in it it's $e$ to an imaginary angle, not a number, it seems. From my understanding pi itself is not an angle, but $pi$ radians is. And since cosine can only take in an angle, or at least a representation of one, and the variable is the same everywhere in Euler's formula, the exponent should be an angle and not a number. The question is then raised could I then say $e^{180i}=-1$? Surely this has a single numerical value though, right?



I've been bothered by this for a long time. Am I wrong or how can I take an exponent with a complex number?





This question already has an answer here:




  • Complex Exponent of Complex Numbers

    2 answers








complex-numbers






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 2 hours ago







Benjamin Thoburn

















asked 15 hours ago









Benjamin ThoburnBenjamin Thoburn

336111




336111




marked as duplicate by Dietrich Burde, Lord Shark the Unknown, A. Pongrácz, Cesareo, José Carlos Santos 56 mins ago


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.






marked as duplicate by Dietrich Burde, Lord Shark the Unknown, A. Pongrácz, Cesareo, José Carlos Santos 56 mins ago


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.










  • 1




    $begingroup$
    What difference do you see between an "angle" and a "number"?
    $endgroup$
    – Matteo
    14 hours ago










  • $begingroup$
    @Matteo numbers represent angles so you can call them the same... but $frac{pi}{4}$, 45, and 50 can all be called the same. How I'm seeing it, like vectors, we can only talk about certain amounts of basis, but not the basis itself. You can represent a dozen of eggs using the number 1, but also using the number 12, for 12 eggs. Like the dozen of actual eggs themselves, an angle is purely geometric and can't be described as strictly equal to a single number. When we talk about the angle as a number, we really mean the angle is that number amount of "insert arbitrary basis angle here".
    $endgroup$
    – Benjamin Thoburn
    14 hours ago










  • $begingroup$
    Let's put it this way: "cosine can only take in an angle", you say. Does this mean to you what. Cannot you put any number as argument of the cosine function?
    $endgroup$
    – Matteo
    14 hours ago










  • $begingroup$
    I would reason this way (in the first quadrant, then you can work on the others by symmetries). Take a circle of radius one with center in the origin of the axes, then draw a half-line that intersects the circle in the first quadrant. Let $x$ be the length of the arc measured from the point $(1,0)$ (with same unit of measure used for the axes of course). Then the absissa of the intersection point is called $cos (x)$. Does it work? No angles here. Only arc lengths.
    $endgroup$
    – Matteo
    14 hours ago








  • 2




    $begingroup$
    So, do not think of the angle in the exponent as being an angle at all. Think of it as being the ratio of the arc length to the radius length, and it will all make sense.
    $endgroup$
    – Eric Lippert
    9 hours ago














  • 1




    $begingroup$
    What difference do you see between an "angle" and a "number"?
    $endgroup$
    – Matteo
    14 hours ago










  • $begingroup$
    @Matteo numbers represent angles so you can call them the same... but $frac{pi}{4}$, 45, and 50 can all be called the same. How I'm seeing it, like vectors, we can only talk about certain amounts of basis, but not the basis itself. You can represent a dozen of eggs using the number 1, but also using the number 12, for 12 eggs. Like the dozen of actual eggs themselves, an angle is purely geometric and can't be described as strictly equal to a single number. When we talk about the angle as a number, we really mean the angle is that number amount of "insert arbitrary basis angle here".
    $endgroup$
    – Benjamin Thoburn
    14 hours ago










  • $begingroup$
    Let's put it this way: "cosine can only take in an angle", you say. Does this mean to you what. Cannot you put any number as argument of the cosine function?
    $endgroup$
    – Matteo
    14 hours ago










  • $begingroup$
    I would reason this way (in the first quadrant, then you can work on the others by symmetries). Take a circle of radius one with center in the origin of the axes, then draw a half-line that intersects the circle in the first quadrant. Let $x$ be the length of the arc measured from the point $(1,0)$ (with same unit of measure used for the axes of course). Then the absissa of the intersection point is called $cos (x)$. Does it work? No angles here. Only arc lengths.
    $endgroup$
    – Matteo
    14 hours ago








  • 2




    $begingroup$
    So, do not think of the angle in the exponent as being an angle at all. Think of it as being the ratio of the arc length to the radius length, and it will all make sense.
    $endgroup$
    – Eric Lippert
    9 hours ago








1




1




$begingroup$
What difference do you see between an "angle" and a "number"?
$endgroup$
– Matteo
14 hours ago




$begingroup$
What difference do you see between an "angle" and a "number"?
$endgroup$
– Matteo
14 hours ago












$begingroup$
@Matteo numbers represent angles so you can call them the same... but $frac{pi}{4}$, 45, and 50 can all be called the same. How I'm seeing it, like vectors, we can only talk about certain amounts of basis, but not the basis itself. You can represent a dozen of eggs using the number 1, but also using the number 12, for 12 eggs. Like the dozen of actual eggs themselves, an angle is purely geometric and can't be described as strictly equal to a single number. When we talk about the angle as a number, we really mean the angle is that number amount of "insert arbitrary basis angle here".
$endgroup$
– Benjamin Thoburn
14 hours ago




$begingroup$
@Matteo numbers represent angles so you can call them the same... but $frac{pi}{4}$, 45, and 50 can all be called the same. How I'm seeing it, like vectors, we can only talk about certain amounts of basis, but not the basis itself. You can represent a dozen of eggs using the number 1, but also using the number 12, for 12 eggs. Like the dozen of actual eggs themselves, an angle is purely geometric and can't be described as strictly equal to a single number. When we talk about the angle as a number, we really mean the angle is that number amount of "insert arbitrary basis angle here".
$endgroup$
– Benjamin Thoburn
14 hours ago












$begingroup$
Let's put it this way: "cosine can only take in an angle", you say. Does this mean to you what. Cannot you put any number as argument of the cosine function?
$endgroup$
– Matteo
14 hours ago




$begingroup$
Let's put it this way: "cosine can only take in an angle", you say. Does this mean to you what. Cannot you put any number as argument of the cosine function?
$endgroup$
– Matteo
14 hours ago












$begingroup$
I would reason this way (in the first quadrant, then you can work on the others by symmetries). Take a circle of radius one with center in the origin of the axes, then draw a half-line that intersects the circle in the first quadrant. Let $x$ be the length of the arc measured from the point $(1,0)$ (with same unit of measure used for the axes of course). Then the absissa of the intersection point is called $cos (x)$. Does it work? No angles here. Only arc lengths.
$endgroup$
– Matteo
14 hours ago






$begingroup$
I would reason this way (in the first quadrant, then you can work on the others by symmetries). Take a circle of radius one with center in the origin of the axes, then draw a half-line that intersects the circle in the first quadrant. Let $x$ be the length of the arc measured from the point $(1,0)$ (with same unit of measure used for the axes of course). Then the absissa of the intersection point is called $cos (x)$. Does it work? No angles here. Only arc lengths.
$endgroup$
– Matteo
14 hours ago






2




2




$begingroup$
So, do not think of the angle in the exponent as being an angle at all. Think of it as being the ratio of the arc length to the radius length, and it will all make sense.
$endgroup$
– Eric Lippert
9 hours ago




$begingroup$
So, do not think of the angle in the exponent as being an angle at all. Think of it as being the ratio of the arc length to the radius length, and it will all make sense.
$endgroup$
– Eric Lippert
9 hours ago










6 Answers
6






active

oldest

votes


















4












$begingroup$

This is more an answer to the title of your question. A possible (and perhaps the standard) way of properly defining things is as follows. See any reasonable rigorous textbook on analysis for the proofs.



First define the exponential function $exp : mathbb{C} rightarrow mathbb{C}$, by the absolutely convergent series $exp(z) := sum_{n=0}^{infty}{ frac{z^n}{n!}}$. It is easy to prove that $exp$, restricted to the real line, takes real values. A bit more work shows that $exp$, restricted to the real line, is strictly incresing and everywhere $>0$ and that $exp(mathbb{R}) = (0, infty)$. It follows that there is a bijetive function $log{}:(0, infty) rightarrow mathbb{R}$, the inverse of $x mapsto exp(x)$ from $mathbb{R} rightarrow (0, infty)$ The number $e$ is defined as $e: = exp(1)$. For any positive real number $a$ and any complex number $z$, one defines $a^z$ by $a^z := exp(log(a) z)$. Hence, by definition $e^z = exp(z)$ (because $log{(e)}=1$ by definition) and this makes sense for all complex numbers $z$.



One defines the functions $cos, sin : mathbb{C} rightarrow mathbb{R}$ by $cos(x) := frac{1}{2}(e^{ix}+ e^{-ix})$ and $sin(x) := frac{1}{2i}(e^{ix}- e^{-ix})$. It is a (rather non-trivial) Theorem that there is a unique smallest, strictly positive zero $p$ of the function $cos$, restricted to the real line, and then defines the number $pi$ as $ pi := 2p$. It can be shown that the number $pi$ and the functions $sin$ and $cos$ have the "familiar properties". Notice that the identity $e^{ix} = cos(x) + i sin(x)$ for $x in mathbb{R}$ follows directly from the definitions. As does $e^{i pi} = -1$. It always strikes me that people find these two identities so amazing.






share|cite|improve this answer









$endgroup$





















    3












    $begingroup$

    There can be a bunch of definition of number $e$. Let's stick with this definition:
    $$
    frac{d}{dt} e^t = e^t.
    $$



    So we want to find what number can be $z(t)=e^{it} =x(t)+iy(t)$:
    $$
    z'(t)=x'(t) + iy'(t)=ie^{it} = ix(t) - y(t),\
    x'(t) = -y(t),qquad y'(t)=x(t),\
    x''(t) = -y'(t)=-x(t)
    $$



    Finally, $x''(t)+x(t)=0$ and $x(0)=1$, $x'(0)=-y(0)=0$ (since $e^{0i}=e^0=1$), thus $x(t)=cos t$ as the only function that suffice the equation. Then $y=x'(t)=sin t$.



    So, we have found what number is $e^{it}=cos t+isin t$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      I think the OP requires I precise definition of $cos (cdot )$, before, maybe geometrically based?
      $endgroup$
      – Matteo
      14 hours ago










    • $begingroup$
      One technical point. If the function $cos t$ is given in degrees (i.e. $tin[0,360]$) then it doesn't satisfy the differential equation. The derivative of $cos$ in degrees is $frac{180}{pi}$ times the derivative of the usual $cos$.
      $endgroup$
      – Yanko
      14 hours ago










    • $begingroup$
      @Matteo that's true. This is my go-to proof for explaining Euler's formula to people but it didn't help much, although it is quite elegant.
      $endgroup$
      – Benjamin Thoburn
      2 hours ago



















    2












    $begingroup$

    In calculus, the trigonometric functions are not dealing with angles but with real numbers. Implicitly, the numbers are taken to be radians, so that $cospi=-1,sinpi=0,$ which is compatible with Eulers' famous formula



    $$e^{ipi}=cospi+isinpi=-1.$$



    These function are "natural", in the sense that they resemble their derivatives:



    $$(e^{ipi})'=ie^{ipi}=(cos x+isin x)'=i(cos x+isin x).$$





    You can very well define functions assuming arguments in degrees and write for instance $cos_d180=-1,sin_d180=0$. But these functions are a kind of "historical mistake" and do not enjoy the above derivative property, as an extra scaling factor appears.






    share|cite|improve this answer











    $endgroup$





















      2












      $begingroup$

      Angular Units and Trigonometric Functions



      When we talk about arguments to trigonometric functions, there are at least $2$ common angular units: degrees ($360$ to a full rotation) and radians ($2pi$ to a full rotation). A number of calculators also support gradians ($400$ to a full rotation), which are only used in some countries and usually only in certain occupations (e.g. surveying, mining, geology).



      In mathematics, we usually use radians because when angles are measured in radians, we have
      $$
      lim_{xto0}frac{sin(x)}{x}=lim_{xto0}frac{tan(x)}{x}=1tag1
      $$

      That is, for small angles, $sin(x)simtan(x)sim x$. The actual ordering for $|x|ltfracpi2$ is
      $$
      frac{sin(x)}{x}le1lefrac{tan(x)}{x}tag2
      $$

      Furthermore, when $x$ is in radians, we have the nice series
      $$
      sin(x)=sum_{n=0}^infty(-1)^nfrac{x^{2n+1}}{(2n+1)!}tag3
      $$

      and the value of
      $$
      arctan(x)=sum_{n=0}^infty(-1)^nfrac{x^{2n+1}}{2n+1}tag4
      $$

      is in radians.



      Radians are also natural because an arc which subtends $x$ radians on a circle of radius $r$ has length $rx$.



      So when we talk about angles, and don't mention the units, we assume radians.





      The Exponential of Imaginary Numbers



      For $xinmathbb{R}$, we can write
      $$
      e^x=lim_{ntoinfty}left(1+frac xnright)^ntag5
      $$

      so it seems reasonable to write
      $$
      e^{ix}=lim_{ntoinfty}left(1+frac{ix}nright)^ntag6
      $$

      Multiplication by $1+frac{ix}n$ increases the absolute value so minimally, that even when repeated $n$ times, it is insignificant as $ntoinfty$. However, multiplication by $1+frac{ix}n$ rotates a number on the unit circle by a distance of $frac xn$ counter-clockwise along the circle. When this is repeated $n$ times, it rotates a number on the unit circle by a distance of $x$.



      Thus, $e^{ix}$ is a point on the unit circle whose counter-clockwise distance from $1+0i$ is $x$. This is why we use radians when saying
      $$
      e^{ix}=cos(x)+isin(x)tag7
      $$

      To see a more detailed explanation of $(7)$, see this answer.






      share|cite|improve this answer











      $endgroup$





















        0












        $begingroup$

        Just a little note that I hope can be of some use. In order to avoid confusion when dealing with angles and unit of measures of them, one can directly define the trigonometric functions starting with the arc lenght measurement.



        Consider the semicircle of equation
        $$f(x) = sqrt{1-x^2}, xin [-1,1].$$
        It is relatively easy to show that the arc length between point $(y, f(y))$ and point $(1,0)$ can be computed with the (improper) integral
        $$A(y) = int_y^1frac{1}{sqrt{1-x^2}}dx, yin [-1, 1].$$
        Here the unit of measure is clearly the same as the one used for determining abscissae and ordinates.



        $A(y)$ is continuous in $[-1,1]$, differentiable in $(-1,1)$, and strictly decreasing. So it has a well defined inverse function
        $$A^{-1}(x)= cos(x)$$
        with domain $[0, pi]$, where, by definition $A(-1) = pi$. The rest of the cosine function can be defined using appropriate shifts and periodicity.






        share|cite|improve this answer









        $endgroup$





















          0












          $begingroup$

          A number of people are giving fairly complex answers (no pun intended) about exponentiation, but I want to address a much simpler misunderstanding in your answer, which should hopefully clear up some of the issues.




          And since cosine can only take in an angle




          No, the trig functions definitely take numbers. They're related to the circle, in that they're cyclic, and the "width" of their repetition is exactly 2π, which happens to be the circumference of a circle when you measure it in radians. But the argument to trig functions aren't angles except by convention.




          The question is then raised could I then say e180i=−1?




          180 isn't equal to π, any more than it's equal to 50 just because 180 degrees is 50% of circle. 180 degrees is equal to π radians, but since the exponentiation operator also takes numbers, not angles, that equivalence between angle units isn't relevant.






          share|cite|improve this answer








          New contributor




          Xanthir is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.






          $endgroup$




















            6 Answers
            6






            active

            oldest

            votes








            6 Answers
            6






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            4












            $begingroup$

            This is more an answer to the title of your question. A possible (and perhaps the standard) way of properly defining things is as follows. See any reasonable rigorous textbook on analysis for the proofs.



            First define the exponential function $exp : mathbb{C} rightarrow mathbb{C}$, by the absolutely convergent series $exp(z) := sum_{n=0}^{infty}{ frac{z^n}{n!}}$. It is easy to prove that $exp$, restricted to the real line, takes real values. A bit more work shows that $exp$, restricted to the real line, is strictly incresing and everywhere $>0$ and that $exp(mathbb{R}) = (0, infty)$. It follows that there is a bijetive function $log{}:(0, infty) rightarrow mathbb{R}$, the inverse of $x mapsto exp(x)$ from $mathbb{R} rightarrow (0, infty)$ The number $e$ is defined as $e: = exp(1)$. For any positive real number $a$ and any complex number $z$, one defines $a^z$ by $a^z := exp(log(a) z)$. Hence, by definition $e^z = exp(z)$ (because $log{(e)}=1$ by definition) and this makes sense for all complex numbers $z$.



            One defines the functions $cos, sin : mathbb{C} rightarrow mathbb{R}$ by $cos(x) := frac{1}{2}(e^{ix}+ e^{-ix})$ and $sin(x) := frac{1}{2i}(e^{ix}- e^{-ix})$. It is a (rather non-trivial) Theorem that there is a unique smallest, strictly positive zero $p$ of the function $cos$, restricted to the real line, and then defines the number $pi$ as $ pi := 2p$. It can be shown that the number $pi$ and the functions $sin$ and $cos$ have the "familiar properties". Notice that the identity $e^{ix} = cos(x) + i sin(x)$ for $x in mathbb{R}$ follows directly from the definitions. As does $e^{i pi} = -1$. It always strikes me that people find these two identities so amazing.






            share|cite|improve this answer









            $endgroup$


















              4












              $begingroup$

              This is more an answer to the title of your question. A possible (and perhaps the standard) way of properly defining things is as follows. See any reasonable rigorous textbook on analysis for the proofs.



              First define the exponential function $exp : mathbb{C} rightarrow mathbb{C}$, by the absolutely convergent series $exp(z) := sum_{n=0}^{infty}{ frac{z^n}{n!}}$. It is easy to prove that $exp$, restricted to the real line, takes real values. A bit more work shows that $exp$, restricted to the real line, is strictly incresing and everywhere $>0$ and that $exp(mathbb{R}) = (0, infty)$. It follows that there is a bijetive function $log{}:(0, infty) rightarrow mathbb{R}$, the inverse of $x mapsto exp(x)$ from $mathbb{R} rightarrow (0, infty)$ The number $e$ is defined as $e: = exp(1)$. For any positive real number $a$ and any complex number $z$, one defines $a^z$ by $a^z := exp(log(a) z)$. Hence, by definition $e^z = exp(z)$ (because $log{(e)}=1$ by definition) and this makes sense for all complex numbers $z$.



              One defines the functions $cos, sin : mathbb{C} rightarrow mathbb{R}$ by $cos(x) := frac{1}{2}(e^{ix}+ e^{-ix})$ and $sin(x) := frac{1}{2i}(e^{ix}- e^{-ix})$. It is a (rather non-trivial) Theorem that there is a unique smallest, strictly positive zero $p$ of the function $cos$, restricted to the real line, and then defines the number $pi$ as $ pi := 2p$. It can be shown that the number $pi$ and the functions $sin$ and $cos$ have the "familiar properties". Notice that the identity $e^{ix} = cos(x) + i sin(x)$ for $x in mathbb{R}$ follows directly from the definitions. As does $e^{i pi} = -1$. It always strikes me that people find these two identities so amazing.






              share|cite|improve this answer









              $endgroup$
















                4












                4








                4





                $begingroup$

                This is more an answer to the title of your question. A possible (and perhaps the standard) way of properly defining things is as follows. See any reasonable rigorous textbook on analysis for the proofs.



                First define the exponential function $exp : mathbb{C} rightarrow mathbb{C}$, by the absolutely convergent series $exp(z) := sum_{n=0}^{infty}{ frac{z^n}{n!}}$. It is easy to prove that $exp$, restricted to the real line, takes real values. A bit more work shows that $exp$, restricted to the real line, is strictly incresing and everywhere $>0$ and that $exp(mathbb{R}) = (0, infty)$. It follows that there is a bijetive function $log{}:(0, infty) rightarrow mathbb{R}$, the inverse of $x mapsto exp(x)$ from $mathbb{R} rightarrow (0, infty)$ The number $e$ is defined as $e: = exp(1)$. For any positive real number $a$ and any complex number $z$, one defines $a^z$ by $a^z := exp(log(a) z)$. Hence, by definition $e^z = exp(z)$ (because $log{(e)}=1$ by definition) and this makes sense for all complex numbers $z$.



                One defines the functions $cos, sin : mathbb{C} rightarrow mathbb{R}$ by $cos(x) := frac{1}{2}(e^{ix}+ e^{-ix})$ and $sin(x) := frac{1}{2i}(e^{ix}- e^{-ix})$. It is a (rather non-trivial) Theorem that there is a unique smallest, strictly positive zero $p$ of the function $cos$, restricted to the real line, and then defines the number $pi$ as $ pi := 2p$. It can be shown that the number $pi$ and the functions $sin$ and $cos$ have the "familiar properties". Notice that the identity $e^{ix} = cos(x) + i sin(x)$ for $x in mathbb{R}$ follows directly from the definitions. As does $e^{i pi} = -1$. It always strikes me that people find these two identities so amazing.






                share|cite|improve this answer









                $endgroup$



                This is more an answer to the title of your question. A possible (and perhaps the standard) way of properly defining things is as follows. See any reasonable rigorous textbook on analysis for the proofs.



                First define the exponential function $exp : mathbb{C} rightarrow mathbb{C}$, by the absolutely convergent series $exp(z) := sum_{n=0}^{infty}{ frac{z^n}{n!}}$. It is easy to prove that $exp$, restricted to the real line, takes real values. A bit more work shows that $exp$, restricted to the real line, is strictly incresing and everywhere $>0$ and that $exp(mathbb{R}) = (0, infty)$. It follows that there is a bijetive function $log{}:(0, infty) rightarrow mathbb{R}$, the inverse of $x mapsto exp(x)$ from $mathbb{R} rightarrow (0, infty)$ The number $e$ is defined as $e: = exp(1)$. For any positive real number $a$ and any complex number $z$, one defines $a^z$ by $a^z := exp(log(a) z)$. Hence, by definition $e^z = exp(z)$ (because $log{(e)}=1$ by definition) and this makes sense for all complex numbers $z$.



                One defines the functions $cos, sin : mathbb{C} rightarrow mathbb{R}$ by $cos(x) := frac{1}{2}(e^{ix}+ e^{-ix})$ and $sin(x) := frac{1}{2i}(e^{ix}- e^{-ix})$. It is a (rather non-trivial) Theorem that there is a unique smallest, strictly positive zero $p$ of the function $cos$, restricted to the real line, and then defines the number $pi$ as $ pi := 2p$. It can be shown that the number $pi$ and the functions $sin$ and $cos$ have the "familiar properties". Notice that the identity $e^{ix} = cos(x) + i sin(x)$ for $x in mathbb{R}$ follows directly from the definitions. As does $e^{i pi} = -1$. It always strikes me that people find these two identities so amazing.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 14 hours ago









                m.sm.s

                1,292313




                1,292313























                    3












                    $begingroup$

                    There can be a bunch of definition of number $e$. Let's stick with this definition:
                    $$
                    frac{d}{dt} e^t = e^t.
                    $$



                    So we want to find what number can be $z(t)=e^{it} =x(t)+iy(t)$:
                    $$
                    z'(t)=x'(t) + iy'(t)=ie^{it} = ix(t) - y(t),\
                    x'(t) = -y(t),qquad y'(t)=x(t),\
                    x''(t) = -y'(t)=-x(t)
                    $$



                    Finally, $x''(t)+x(t)=0$ and $x(0)=1$, $x'(0)=-y(0)=0$ (since $e^{0i}=e^0=1$), thus $x(t)=cos t$ as the only function that suffice the equation. Then $y=x'(t)=sin t$.



                    So, we have found what number is $e^{it}=cos t+isin t$






                    share|cite|improve this answer









                    $endgroup$













                    • $begingroup$
                      I think the OP requires I precise definition of $cos (cdot )$, before, maybe geometrically based?
                      $endgroup$
                      – Matteo
                      14 hours ago










                    • $begingroup$
                      One technical point. If the function $cos t$ is given in degrees (i.e. $tin[0,360]$) then it doesn't satisfy the differential equation. The derivative of $cos$ in degrees is $frac{180}{pi}$ times the derivative of the usual $cos$.
                      $endgroup$
                      – Yanko
                      14 hours ago










                    • $begingroup$
                      @Matteo that's true. This is my go-to proof for explaining Euler's formula to people but it didn't help much, although it is quite elegant.
                      $endgroup$
                      – Benjamin Thoburn
                      2 hours ago
















                    3












                    $begingroup$

                    There can be a bunch of definition of number $e$. Let's stick with this definition:
                    $$
                    frac{d}{dt} e^t = e^t.
                    $$



                    So we want to find what number can be $z(t)=e^{it} =x(t)+iy(t)$:
                    $$
                    z'(t)=x'(t) + iy'(t)=ie^{it} = ix(t) - y(t),\
                    x'(t) = -y(t),qquad y'(t)=x(t),\
                    x''(t) = -y'(t)=-x(t)
                    $$



                    Finally, $x''(t)+x(t)=0$ and $x(0)=1$, $x'(0)=-y(0)=0$ (since $e^{0i}=e^0=1$), thus $x(t)=cos t$ as the only function that suffice the equation. Then $y=x'(t)=sin t$.



                    So, we have found what number is $e^{it}=cos t+isin t$






                    share|cite|improve this answer









                    $endgroup$













                    • $begingroup$
                      I think the OP requires I precise definition of $cos (cdot )$, before, maybe geometrically based?
                      $endgroup$
                      – Matteo
                      14 hours ago










                    • $begingroup$
                      One technical point. If the function $cos t$ is given in degrees (i.e. $tin[0,360]$) then it doesn't satisfy the differential equation. The derivative of $cos$ in degrees is $frac{180}{pi}$ times the derivative of the usual $cos$.
                      $endgroup$
                      – Yanko
                      14 hours ago










                    • $begingroup$
                      @Matteo that's true. This is my go-to proof for explaining Euler's formula to people but it didn't help much, although it is quite elegant.
                      $endgroup$
                      – Benjamin Thoburn
                      2 hours ago














                    3












                    3








                    3





                    $begingroup$

                    There can be a bunch of definition of number $e$. Let's stick with this definition:
                    $$
                    frac{d}{dt} e^t = e^t.
                    $$



                    So we want to find what number can be $z(t)=e^{it} =x(t)+iy(t)$:
                    $$
                    z'(t)=x'(t) + iy'(t)=ie^{it} = ix(t) - y(t),\
                    x'(t) = -y(t),qquad y'(t)=x(t),\
                    x''(t) = -y'(t)=-x(t)
                    $$



                    Finally, $x''(t)+x(t)=0$ and $x(0)=1$, $x'(0)=-y(0)=0$ (since $e^{0i}=e^0=1$), thus $x(t)=cos t$ as the only function that suffice the equation. Then $y=x'(t)=sin t$.



                    So, we have found what number is $e^{it}=cos t+isin t$






                    share|cite|improve this answer









                    $endgroup$



                    There can be a bunch of definition of number $e$. Let's stick with this definition:
                    $$
                    frac{d}{dt} e^t = e^t.
                    $$



                    So we want to find what number can be $z(t)=e^{it} =x(t)+iy(t)$:
                    $$
                    z'(t)=x'(t) + iy'(t)=ie^{it} = ix(t) - y(t),\
                    x'(t) = -y(t),qquad y'(t)=x(t),\
                    x''(t) = -y'(t)=-x(t)
                    $$



                    Finally, $x''(t)+x(t)=0$ and $x(0)=1$, $x'(0)=-y(0)=0$ (since $e^{0i}=e^0=1$), thus $x(t)=cos t$ as the only function that suffice the equation. Then $y=x'(t)=sin t$.



                    So, we have found what number is $e^{it}=cos t+isin t$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 14 hours ago









                    Vasily MitchVasily Mitch

                    1,62338




                    1,62338












                    • $begingroup$
                      I think the OP requires I precise definition of $cos (cdot )$, before, maybe geometrically based?
                      $endgroup$
                      – Matteo
                      14 hours ago










                    • $begingroup$
                      One technical point. If the function $cos t$ is given in degrees (i.e. $tin[0,360]$) then it doesn't satisfy the differential equation. The derivative of $cos$ in degrees is $frac{180}{pi}$ times the derivative of the usual $cos$.
                      $endgroup$
                      – Yanko
                      14 hours ago










                    • $begingroup$
                      @Matteo that's true. This is my go-to proof for explaining Euler's formula to people but it didn't help much, although it is quite elegant.
                      $endgroup$
                      – Benjamin Thoburn
                      2 hours ago


















                    • $begingroup$
                      I think the OP requires I precise definition of $cos (cdot )$, before, maybe geometrically based?
                      $endgroup$
                      – Matteo
                      14 hours ago










                    • $begingroup$
                      One technical point. If the function $cos t$ is given in degrees (i.e. $tin[0,360]$) then it doesn't satisfy the differential equation. The derivative of $cos$ in degrees is $frac{180}{pi}$ times the derivative of the usual $cos$.
                      $endgroup$
                      – Yanko
                      14 hours ago










                    • $begingroup$
                      @Matteo that's true. This is my go-to proof for explaining Euler's formula to people but it didn't help much, although it is quite elegant.
                      $endgroup$
                      – Benjamin Thoburn
                      2 hours ago
















                    $begingroup$
                    I think the OP requires I precise definition of $cos (cdot )$, before, maybe geometrically based?
                    $endgroup$
                    – Matteo
                    14 hours ago




                    $begingroup$
                    I think the OP requires I precise definition of $cos (cdot )$, before, maybe geometrically based?
                    $endgroup$
                    – Matteo
                    14 hours ago












                    $begingroup$
                    One technical point. If the function $cos t$ is given in degrees (i.e. $tin[0,360]$) then it doesn't satisfy the differential equation. The derivative of $cos$ in degrees is $frac{180}{pi}$ times the derivative of the usual $cos$.
                    $endgroup$
                    – Yanko
                    14 hours ago




                    $begingroup$
                    One technical point. If the function $cos t$ is given in degrees (i.e. $tin[0,360]$) then it doesn't satisfy the differential equation. The derivative of $cos$ in degrees is $frac{180}{pi}$ times the derivative of the usual $cos$.
                    $endgroup$
                    – Yanko
                    14 hours ago












                    $begingroup$
                    @Matteo that's true. This is my go-to proof for explaining Euler's formula to people but it didn't help much, although it is quite elegant.
                    $endgroup$
                    – Benjamin Thoburn
                    2 hours ago




                    $begingroup$
                    @Matteo that's true. This is my go-to proof for explaining Euler's formula to people but it didn't help much, although it is quite elegant.
                    $endgroup$
                    – Benjamin Thoburn
                    2 hours ago











                    2












                    $begingroup$

                    In calculus, the trigonometric functions are not dealing with angles but with real numbers. Implicitly, the numbers are taken to be radians, so that $cospi=-1,sinpi=0,$ which is compatible with Eulers' famous formula



                    $$e^{ipi}=cospi+isinpi=-1.$$



                    These function are "natural", in the sense that they resemble their derivatives:



                    $$(e^{ipi})'=ie^{ipi}=(cos x+isin x)'=i(cos x+isin x).$$





                    You can very well define functions assuming arguments in degrees and write for instance $cos_d180=-1,sin_d180=0$. But these functions are a kind of "historical mistake" and do not enjoy the above derivative property, as an extra scaling factor appears.






                    share|cite|improve this answer











                    $endgroup$


















                      2












                      $begingroup$

                      In calculus, the trigonometric functions are not dealing with angles but with real numbers. Implicitly, the numbers are taken to be radians, so that $cospi=-1,sinpi=0,$ which is compatible with Eulers' famous formula



                      $$e^{ipi}=cospi+isinpi=-1.$$



                      These function are "natural", in the sense that they resemble their derivatives:



                      $$(e^{ipi})'=ie^{ipi}=(cos x+isin x)'=i(cos x+isin x).$$





                      You can very well define functions assuming arguments in degrees and write for instance $cos_d180=-1,sin_d180=0$. But these functions are a kind of "historical mistake" and do not enjoy the above derivative property, as an extra scaling factor appears.






                      share|cite|improve this answer











                      $endgroup$
















                        2












                        2








                        2





                        $begingroup$

                        In calculus, the trigonometric functions are not dealing with angles but with real numbers. Implicitly, the numbers are taken to be radians, so that $cospi=-1,sinpi=0,$ which is compatible with Eulers' famous formula



                        $$e^{ipi}=cospi+isinpi=-1.$$



                        These function are "natural", in the sense that they resemble their derivatives:



                        $$(e^{ipi})'=ie^{ipi}=(cos x+isin x)'=i(cos x+isin x).$$





                        You can very well define functions assuming arguments in degrees and write for instance $cos_d180=-1,sin_d180=0$. But these functions are a kind of "historical mistake" and do not enjoy the above derivative property, as an extra scaling factor appears.






                        share|cite|improve this answer











                        $endgroup$



                        In calculus, the trigonometric functions are not dealing with angles but with real numbers. Implicitly, the numbers are taken to be radians, so that $cospi=-1,sinpi=0,$ which is compatible with Eulers' famous formula



                        $$e^{ipi}=cospi+isinpi=-1.$$



                        These function are "natural", in the sense that they resemble their derivatives:



                        $$(e^{ipi})'=ie^{ipi}=(cos x+isin x)'=i(cos x+isin x).$$





                        You can very well define functions assuming arguments in degrees and write for instance $cos_d180=-1,sin_d180=0$. But these functions are a kind of "historical mistake" and do not enjoy the above derivative property, as an extra scaling factor appears.







                        share|cite|improve this answer














                        share|cite|improve this answer



                        share|cite|improve this answer








                        edited 13 hours ago

























                        answered 13 hours ago









                        Yves DaoustYves Daoust

                        125k671222




                        125k671222























                            2












                            $begingroup$

                            Angular Units and Trigonometric Functions



                            When we talk about arguments to trigonometric functions, there are at least $2$ common angular units: degrees ($360$ to a full rotation) and radians ($2pi$ to a full rotation). A number of calculators also support gradians ($400$ to a full rotation), which are only used in some countries and usually only in certain occupations (e.g. surveying, mining, geology).



                            In mathematics, we usually use radians because when angles are measured in radians, we have
                            $$
                            lim_{xto0}frac{sin(x)}{x}=lim_{xto0}frac{tan(x)}{x}=1tag1
                            $$

                            That is, for small angles, $sin(x)simtan(x)sim x$. The actual ordering for $|x|ltfracpi2$ is
                            $$
                            frac{sin(x)}{x}le1lefrac{tan(x)}{x}tag2
                            $$

                            Furthermore, when $x$ is in radians, we have the nice series
                            $$
                            sin(x)=sum_{n=0}^infty(-1)^nfrac{x^{2n+1}}{(2n+1)!}tag3
                            $$

                            and the value of
                            $$
                            arctan(x)=sum_{n=0}^infty(-1)^nfrac{x^{2n+1}}{2n+1}tag4
                            $$

                            is in radians.



                            Radians are also natural because an arc which subtends $x$ radians on a circle of radius $r$ has length $rx$.



                            So when we talk about angles, and don't mention the units, we assume radians.





                            The Exponential of Imaginary Numbers



                            For $xinmathbb{R}$, we can write
                            $$
                            e^x=lim_{ntoinfty}left(1+frac xnright)^ntag5
                            $$

                            so it seems reasonable to write
                            $$
                            e^{ix}=lim_{ntoinfty}left(1+frac{ix}nright)^ntag6
                            $$

                            Multiplication by $1+frac{ix}n$ increases the absolute value so minimally, that even when repeated $n$ times, it is insignificant as $ntoinfty$. However, multiplication by $1+frac{ix}n$ rotates a number on the unit circle by a distance of $frac xn$ counter-clockwise along the circle. When this is repeated $n$ times, it rotates a number on the unit circle by a distance of $x$.



                            Thus, $e^{ix}$ is a point on the unit circle whose counter-clockwise distance from $1+0i$ is $x$. This is why we use radians when saying
                            $$
                            e^{ix}=cos(x)+isin(x)tag7
                            $$

                            To see a more detailed explanation of $(7)$, see this answer.






                            share|cite|improve this answer











                            $endgroup$


















                              2












                              $begingroup$

                              Angular Units and Trigonometric Functions



                              When we talk about arguments to trigonometric functions, there are at least $2$ common angular units: degrees ($360$ to a full rotation) and radians ($2pi$ to a full rotation). A number of calculators also support gradians ($400$ to a full rotation), which are only used in some countries and usually only in certain occupations (e.g. surveying, mining, geology).



                              In mathematics, we usually use radians because when angles are measured in radians, we have
                              $$
                              lim_{xto0}frac{sin(x)}{x}=lim_{xto0}frac{tan(x)}{x}=1tag1
                              $$

                              That is, for small angles, $sin(x)simtan(x)sim x$. The actual ordering for $|x|ltfracpi2$ is
                              $$
                              frac{sin(x)}{x}le1lefrac{tan(x)}{x}tag2
                              $$

                              Furthermore, when $x$ is in radians, we have the nice series
                              $$
                              sin(x)=sum_{n=0}^infty(-1)^nfrac{x^{2n+1}}{(2n+1)!}tag3
                              $$

                              and the value of
                              $$
                              arctan(x)=sum_{n=0}^infty(-1)^nfrac{x^{2n+1}}{2n+1}tag4
                              $$

                              is in radians.



                              Radians are also natural because an arc which subtends $x$ radians on a circle of radius $r$ has length $rx$.



                              So when we talk about angles, and don't mention the units, we assume radians.





                              The Exponential of Imaginary Numbers



                              For $xinmathbb{R}$, we can write
                              $$
                              e^x=lim_{ntoinfty}left(1+frac xnright)^ntag5
                              $$

                              so it seems reasonable to write
                              $$
                              e^{ix}=lim_{ntoinfty}left(1+frac{ix}nright)^ntag6
                              $$

                              Multiplication by $1+frac{ix}n$ increases the absolute value so minimally, that even when repeated $n$ times, it is insignificant as $ntoinfty$. However, multiplication by $1+frac{ix}n$ rotates a number on the unit circle by a distance of $frac xn$ counter-clockwise along the circle. When this is repeated $n$ times, it rotates a number on the unit circle by a distance of $x$.



                              Thus, $e^{ix}$ is a point on the unit circle whose counter-clockwise distance from $1+0i$ is $x$. This is why we use radians when saying
                              $$
                              e^{ix}=cos(x)+isin(x)tag7
                              $$

                              To see a more detailed explanation of $(7)$, see this answer.






                              share|cite|improve this answer











                              $endgroup$
















                                2












                                2








                                2





                                $begingroup$

                                Angular Units and Trigonometric Functions



                                When we talk about arguments to trigonometric functions, there are at least $2$ common angular units: degrees ($360$ to a full rotation) and radians ($2pi$ to a full rotation). A number of calculators also support gradians ($400$ to a full rotation), which are only used in some countries and usually only in certain occupations (e.g. surveying, mining, geology).



                                In mathematics, we usually use radians because when angles are measured in radians, we have
                                $$
                                lim_{xto0}frac{sin(x)}{x}=lim_{xto0}frac{tan(x)}{x}=1tag1
                                $$

                                That is, for small angles, $sin(x)simtan(x)sim x$. The actual ordering for $|x|ltfracpi2$ is
                                $$
                                frac{sin(x)}{x}le1lefrac{tan(x)}{x}tag2
                                $$

                                Furthermore, when $x$ is in radians, we have the nice series
                                $$
                                sin(x)=sum_{n=0}^infty(-1)^nfrac{x^{2n+1}}{(2n+1)!}tag3
                                $$

                                and the value of
                                $$
                                arctan(x)=sum_{n=0}^infty(-1)^nfrac{x^{2n+1}}{2n+1}tag4
                                $$

                                is in radians.



                                Radians are also natural because an arc which subtends $x$ radians on a circle of radius $r$ has length $rx$.



                                So when we talk about angles, and don't mention the units, we assume radians.





                                The Exponential of Imaginary Numbers



                                For $xinmathbb{R}$, we can write
                                $$
                                e^x=lim_{ntoinfty}left(1+frac xnright)^ntag5
                                $$

                                so it seems reasonable to write
                                $$
                                e^{ix}=lim_{ntoinfty}left(1+frac{ix}nright)^ntag6
                                $$

                                Multiplication by $1+frac{ix}n$ increases the absolute value so minimally, that even when repeated $n$ times, it is insignificant as $ntoinfty$. However, multiplication by $1+frac{ix}n$ rotates a number on the unit circle by a distance of $frac xn$ counter-clockwise along the circle. When this is repeated $n$ times, it rotates a number on the unit circle by a distance of $x$.



                                Thus, $e^{ix}$ is a point on the unit circle whose counter-clockwise distance from $1+0i$ is $x$. This is why we use radians when saying
                                $$
                                e^{ix}=cos(x)+isin(x)tag7
                                $$

                                To see a more detailed explanation of $(7)$, see this answer.






                                share|cite|improve this answer











                                $endgroup$



                                Angular Units and Trigonometric Functions



                                When we talk about arguments to trigonometric functions, there are at least $2$ common angular units: degrees ($360$ to a full rotation) and radians ($2pi$ to a full rotation). A number of calculators also support gradians ($400$ to a full rotation), which are only used in some countries and usually only in certain occupations (e.g. surveying, mining, geology).



                                In mathematics, we usually use radians because when angles are measured in radians, we have
                                $$
                                lim_{xto0}frac{sin(x)}{x}=lim_{xto0}frac{tan(x)}{x}=1tag1
                                $$

                                That is, for small angles, $sin(x)simtan(x)sim x$. The actual ordering for $|x|ltfracpi2$ is
                                $$
                                frac{sin(x)}{x}le1lefrac{tan(x)}{x}tag2
                                $$

                                Furthermore, when $x$ is in radians, we have the nice series
                                $$
                                sin(x)=sum_{n=0}^infty(-1)^nfrac{x^{2n+1}}{(2n+1)!}tag3
                                $$

                                and the value of
                                $$
                                arctan(x)=sum_{n=0}^infty(-1)^nfrac{x^{2n+1}}{2n+1}tag4
                                $$

                                is in radians.



                                Radians are also natural because an arc which subtends $x$ radians on a circle of radius $r$ has length $rx$.



                                So when we talk about angles, and don't mention the units, we assume radians.





                                The Exponential of Imaginary Numbers



                                For $xinmathbb{R}$, we can write
                                $$
                                e^x=lim_{ntoinfty}left(1+frac xnright)^ntag5
                                $$

                                so it seems reasonable to write
                                $$
                                e^{ix}=lim_{ntoinfty}left(1+frac{ix}nright)^ntag6
                                $$

                                Multiplication by $1+frac{ix}n$ increases the absolute value so minimally, that even when repeated $n$ times, it is insignificant as $ntoinfty$. However, multiplication by $1+frac{ix}n$ rotates a number on the unit circle by a distance of $frac xn$ counter-clockwise along the circle. When this is repeated $n$ times, it rotates a number on the unit circle by a distance of $x$.



                                Thus, $e^{ix}$ is a point on the unit circle whose counter-clockwise distance from $1+0i$ is $x$. This is why we use radians when saying
                                $$
                                e^{ix}=cos(x)+isin(x)tag7
                                $$

                                To see a more detailed explanation of $(7)$, see this answer.







                                share|cite|improve this answer














                                share|cite|improve this answer



                                share|cite|improve this answer








                                edited 10 hours ago

























                                answered 13 hours ago









                                robjohnrobjohn

                                266k27305626




                                266k27305626























                                    0












                                    $begingroup$

                                    Just a little note that I hope can be of some use. In order to avoid confusion when dealing with angles and unit of measures of them, one can directly define the trigonometric functions starting with the arc lenght measurement.



                                    Consider the semicircle of equation
                                    $$f(x) = sqrt{1-x^2}, xin [-1,1].$$
                                    It is relatively easy to show that the arc length between point $(y, f(y))$ and point $(1,0)$ can be computed with the (improper) integral
                                    $$A(y) = int_y^1frac{1}{sqrt{1-x^2}}dx, yin [-1, 1].$$
                                    Here the unit of measure is clearly the same as the one used for determining abscissae and ordinates.



                                    $A(y)$ is continuous in $[-1,1]$, differentiable in $(-1,1)$, and strictly decreasing. So it has a well defined inverse function
                                    $$A^{-1}(x)= cos(x)$$
                                    with domain $[0, pi]$, where, by definition $A(-1) = pi$. The rest of the cosine function can be defined using appropriate shifts and periodicity.






                                    share|cite|improve this answer









                                    $endgroup$


















                                      0












                                      $begingroup$

                                      Just a little note that I hope can be of some use. In order to avoid confusion when dealing with angles and unit of measures of them, one can directly define the trigonometric functions starting with the arc lenght measurement.



                                      Consider the semicircle of equation
                                      $$f(x) = sqrt{1-x^2}, xin [-1,1].$$
                                      It is relatively easy to show that the arc length between point $(y, f(y))$ and point $(1,0)$ can be computed with the (improper) integral
                                      $$A(y) = int_y^1frac{1}{sqrt{1-x^2}}dx, yin [-1, 1].$$
                                      Here the unit of measure is clearly the same as the one used for determining abscissae and ordinates.



                                      $A(y)$ is continuous in $[-1,1]$, differentiable in $(-1,1)$, and strictly decreasing. So it has a well defined inverse function
                                      $$A^{-1}(x)= cos(x)$$
                                      with domain $[0, pi]$, where, by definition $A(-1) = pi$. The rest of the cosine function can be defined using appropriate shifts and periodicity.






                                      share|cite|improve this answer









                                      $endgroup$
















                                        0












                                        0








                                        0





                                        $begingroup$

                                        Just a little note that I hope can be of some use. In order to avoid confusion when dealing with angles and unit of measures of them, one can directly define the trigonometric functions starting with the arc lenght measurement.



                                        Consider the semicircle of equation
                                        $$f(x) = sqrt{1-x^2}, xin [-1,1].$$
                                        It is relatively easy to show that the arc length between point $(y, f(y))$ and point $(1,0)$ can be computed with the (improper) integral
                                        $$A(y) = int_y^1frac{1}{sqrt{1-x^2}}dx, yin [-1, 1].$$
                                        Here the unit of measure is clearly the same as the one used for determining abscissae and ordinates.



                                        $A(y)$ is continuous in $[-1,1]$, differentiable in $(-1,1)$, and strictly decreasing. So it has a well defined inverse function
                                        $$A^{-1}(x)= cos(x)$$
                                        with domain $[0, pi]$, where, by definition $A(-1) = pi$. The rest of the cosine function can be defined using appropriate shifts and periodicity.






                                        share|cite|improve this answer









                                        $endgroup$



                                        Just a little note that I hope can be of some use. In order to avoid confusion when dealing with angles and unit of measures of them, one can directly define the trigonometric functions starting with the arc lenght measurement.



                                        Consider the semicircle of equation
                                        $$f(x) = sqrt{1-x^2}, xin [-1,1].$$
                                        It is relatively easy to show that the arc length between point $(y, f(y))$ and point $(1,0)$ can be computed with the (improper) integral
                                        $$A(y) = int_y^1frac{1}{sqrt{1-x^2}}dx, yin [-1, 1].$$
                                        Here the unit of measure is clearly the same as the one used for determining abscissae and ordinates.



                                        $A(y)$ is continuous in $[-1,1]$, differentiable in $(-1,1)$, and strictly decreasing. So it has a well defined inverse function
                                        $$A^{-1}(x)= cos(x)$$
                                        with domain $[0, pi]$, where, by definition $A(-1) = pi$. The rest of the cosine function can be defined using appropriate shifts and periodicity.







                                        share|cite|improve this answer












                                        share|cite|improve this answer



                                        share|cite|improve this answer










                                        answered 10 hours ago









                                        MatteoMatteo

                                        30329




                                        30329























                                            0












                                            $begingroup$

                                            A number of people are giving fairly complex answers (no pun intended) about exponentiation, but I want to address a much simpler misunderstanding in your answer, which should hopefully clear up some of the issues.




                                            And since cosine can only take in an angle




                                            No, the trig functions definitely take numbers. They're related to the circle, in that they're cyclic, and the "width" of their repetition is exactly 2π, which happens to be the circumference of a circle when you measure it in radians. But the argument to trig functions aren't angles except by convention.




                                            The question is then raised could I then say e180i=−1?




                                            180 isn't equal to π, any more than it's equal to 50 just because 180 degrees is 50% of circle. 180 degrees is equal to π radians, but since the exponentiation operator also takes numbers, not angles, that equivalence between angle units isn't relevant.






                                            share|cite|improve this answer








                                            New contributor




                                            Xanthir is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                            Check out our Code of Conduct.






                                            $endgroup$


















                                              0












                                              $begingroup$

                                              A number of people are giving fairly complex answers (no pun intended) about exponentiation, but I want to address a much simpler misunderstanding in your answer, which should hopefully clear up some of the issues.




                                              And since cosine can only take in an angle




                                              No, the trig functions definitely take numbers. They're related to the circle, in that they're cyclic, and the "width" of their repetition is exactly 2π, which happens to be the circumference of a circle when you measure it in radians. But the argument to trig functions aren't angles except by convention.




                                              The question is then raised could I then say e180i=−1?




                                              180 isn't equal to π, any more than it's equal to 50 just because 180 degrees is 50% of circle. 180 degrees is equal to π radians, but since the exponentiation operator also takes numbers, not angles, that equivalence between angle units isn't relevant.






                                              share|cite|improve this answer








                                              New contributor




                                              Xanthir is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                              Check out our Code of Conduct.






                                              $endgroup$
















                                                0












                                                0








                                                0





                                                $begingroup$

                                                A number of people are giving fairly complex answers (no pun intended) about exponentiation, but I want to address a much simpler misunderstanding in your answer, which should hopefully clear up some of the issues.




                                                And since cosine can only take in an angle




                                                No, the trig functions definitely take numbers. They're related to the circle, in that they're cyclic, and the "width" of their repetition is exactly 2π, which happens to be the circumference of a circle when you measure it in radians. But the argument to trig functions aren't angles except by convention.




                                                The question is then raised could I then say e180i=−1?




                                                180 isn't equal to π, any more than it's equal to 50 just because 180 degrees is 50% of circle. 180 degrees is equal to π radians, but since the exponentiation operator also takes numbers, not angles, that equivalence between angle units isn't relevant.






                                                share|cite|improve this answer








                                                New contributor




                                                Xanthir is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                                Check out our Code of Conduct.






                                                $endgroup$



                                                A number of people are giving fairly complex answers (no pun intended) about exponentiation, but I want to address a much simpler misunderstanding in your answer, which should hopefully clear up some of the issues.




                                                And since cosine can only take in an angle




                                                No, the trig functions definitely take numbers. They're related to the circle, in that they're cyclic, and the "width" of their repetition is exactly 2π, which happens to be the circumference of a circle when you measure it in radians. But the argument to trig functions aren't angles except by convention.




                                                The question is then raised could I then say e180i=−1?




                                                180 isn't equal to π, any more than it's equal to 50 just because 180 degrees is 50% of circle. 180 degrees is equal to π radians, but since the exponentiation operator also takes numbers, not angles, that equivalence between angle units isn't relevant.







                                                share|cite|improve this answer








                                                New contributor




                                                Xanthir is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                                Check out our Code of Conduct.









                                                share|cite|improve this answer



                                                share|cite|improve this answer






                                                New contributor




                                                Xanthir is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                                Check out our Code of Conduct.









                                                answered 8 hours ago









                                                XanthirXanthir

                                                1011




                                                1011




                                                New contributor




                                                Xanthir is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                                Check out our Code of Conduct.





                                                New contributor





                                                Xanthir is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                                Check out our Code of Conduct.






                                                Xanthir is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                                Check out our Code of Conduct.















                                                    Popular posts from this blog

                                                    Ponta tanko

                                                    Tantalo (mitologio)

                                                    Erzsébet Schaár