How do complex number exponents actually work?












3












$begingroup$


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









$endgroup$








  • 1




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










  • $begingroup$
    Possible duplicate of Complex Exponent of Complex Numbers
    $endgroup$
    – Dietrich Burde
    5 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
    5 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
    5 hours ago






  • 1




    $begingroup$
    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
    12 mins ago
















3












$begingroup$


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









$endgroup$








  • 1




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










  • $begingroup$
    Possible duplicate of Complex Exponent of Complex Numbers
    $endgroup$
    – Dietrich Burde
    5 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
    5 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
    5 hours ago






  • 1




    $begingroup$
    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
    12 mins ago














3












3








3





$begingroup$


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









$endgroup$




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?







complex-numbers






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 6 hours ago









Benjamin ThoburnBenjamin Thoburn

336111




336111








  • 1




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










  • $begingroup$
    Possible duplicate of Complex Exponent of Complex Numbers
    $endgroup$
    – Dietrich Burde
    5 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
    5 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
    5 hours ago






  • 1




    $begingroup$
    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
    12 mins ago














  • 1




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










  • $begingroup$
    Possible duplicate of Complex Exponent of Complex Numbers
    $endgroup$
    – Dietrich Burde
    5 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
    5 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
    5 hours ago






  • 1




    $begingroup$
    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
    12 mins ago








1




1




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




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












$begingroup$
Possible duplicate of Complex Exponent of Complex Numbers
$endgroup$
– Dietrich Burde
5 hours ago




$begingroup$
Possible duplicate of Complex Exponent of Complex Numbers
$endgroup$
– Dietrich Burde
5 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
5 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
5 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
5 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
5 hours ago




1




1




$begingroup$
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
12 mins 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
12 mins ago










5 Answers
5






active

oldest

votes


















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
    5 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
    5 hours ago



















3












$begingroup$

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









$endgroup$





















    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









        $endgroup$













          Your Answer





          StackExchange.ifUsing("editor", function () {
          return StackExchange.using("mathjaxEditing", function () {
          StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
          StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
          });
          });
          }, "mathjax-editing");

          StackExchange.ready(function() {
          var channelOptions = {
          tags: "".split(" "),
          id: "69"
          };
          initTagRenderer("".split(" "), "".split(" "), channelOptions);

          StackExchange.using("externalEditor", function() {
          // Have to fire editor after snippets, if snippets enabled
          if (StackExchange.settings.snippets.snippetsEnabled) {
          StackExchange.using("snippets", function() {
          createEditor();
          });
          }
          else {
          createEditor();
          }
          });

          function createEditor() {
          StackExchange.prepareEditor({
          heartbeatType: 'answer',
          autoActivateHeartbeat: false,
          convertImagesToLinks: true,
          noModals: true,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: 10,
          bindNavPrevention: true,
          postfix: "",
          imageUploader: {
          brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
          contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
          allowUrls: true
          },
          noCode: true, onDemand: true,
          discardSelector: ".discard-answer"
          ,immediatelyShowMarkdownHelp:true
          });


          }
          });














          draft saved

          draft discarded


















          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3078651%2fhow-do-complex-number-exponents-actually-work%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          5 Answers
          5






          active

          oldest

          votes








          5 Answers
          5






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          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
            5 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
            5 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
            5 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
            5 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 5 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
            5 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
            5 hours ago


















          • $begingroup$
            I think the OP requires I precise definition of $cos (cdot )$, before, maybe geometrically based?
            $endgroup$
            – Matteo
            5 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
            5 hours ago
















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




          $begingroup$
          I think the OP requires I precise definition of $cos (cdot )$, before, maybe geometrically based?
          $endgroup$
          – Matteo
          5 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
          5 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
          5 hours ago











          3












          $begingroup$

          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












            $begingroup$

            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












              3








              3





              $begingroup$

              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 5 hours ago









              m.sm.s

              1,272313




              1,272313























                  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 4 hours ago

























                      answered 4 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 1 hour ago

























                              answered 4 hours ago









                              robjohnrobjohn

                              266k27304626




                              266k27304626























                                  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 1 hour ago









                                      MatteoMatteo

                                      30329




                                      30329






























                                          draft saved

                                          draft discarded




















































                                          Thanks for contributing an answer to Mathematics Stack Exchange!


                                          • Please be sure to answer the question. Provide details and share your research!

                                          But avoid



                                          • Asking for help, clarification, or responding to other answers.

                                          • Making statements based on opinion; back them up with references or personal experience.


                                          Use MathJax to format equations. MathJax reference.


                                          To learn more, see our tips on writing great answers.




                                          draft saved


                                          draft discarded














                                          StackExchange.ready(
                                          function () {
                                          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3078651%2fhow-do-complex-number-exponents-actually-work%23new-answer', 'question_page');
                                          }
                                          );

                                          Post as a guest















                                          Required, but never shown





















































                                          Required, but never shown














                                          Required, but never shown












                                          Required, but never shown







                                          Required, but never shown

































                                          Required, but never shown














                                          Required, but never shown












                                          Required, but never shown







                                          Required, but never shown







                                          Popular posts from this blog

                                          404 Error Contact Form 7 ajax form submitting

                                          How to know if a Active Directory user can login interactively

                                          TypeError: fit_transform() missing 1 required positional argument: 'X'