Proof that $e^{ab} = left(e^aright)^b$, using the series for $e^x$












2












$begingroup$


I want to prove that



$$e^{ab}=left(e^{a}right)^{b}$$where



$$a,b in mathbb{R}$$ using the infinite series for $$e^{x}$$



My attempt:



$$e^{ab}=sum_{n=0}^{infty}frac{(ab)^{n}}{n!}=sum_{n=0}^{infty}frac{(a)^{n}(b)^{n}}{n!}=sum_{n=0}^{infty}frac{(a)^{n}}{n!}b^{n}=1+(a)b+left(frac{a^{2}}{2}right)b^{2}+left(frac{a^{3}}{6}right)b^{3}+cdots$$



At this point I don't know how to proceed. I don't see how I could get b out of the series and have it as a power that the series is raised to. I thought that it resembled the geometric series, but that did not seem to lead anywhere.






sequences-and-series proof-writing exponential-function







share|cite|improve this question











$endgroup$












  • $begingroup$
    Please see math.meta.stackexchange.com/questions/5020
    $endgroup$
    – Lord Shark the Unknown
    Dec 30 '18 at 17:23






  • 2




    $begingroup$
    Unless I'm blind, isn't $n^{mk} = (n^m)^k$ true always, whether or not $n=e$?
    $endgroup$
    – Zachary Hunter
    Dec 30 '18 at 17:26






  • 2




    $begingroup$
    He’s specifically trying to show it for $e$ using the series expansion.
    $endgroup$
    – KM101
    Dec 30 '18 at 17:31










  • $begingroup$
    You are complicating things for yourself by trying to show this using series. This is a general property. See for example math.stackexchange.com/questions/91869/… or math.stackexchange.com/questions/614162/laws-of-indices or math.stackexchange.com/questions/2086109/…
    $endgroup$
    – Winther
    Dec 30 '18 at 17:47


















2












$begingroup$


I want to prove that



$$e^{ab}=left(e^{a}right)^{b}$$where



$$a,b in mathbb{R}$$ using the infinite series for $$e^{x}$$



My attempt:



$$e^{ab}=sum_{n=0}^{infty}frac{(ab)^{n}}{n!}=sum_{n=0}^{infty}frac{(a)^{n}(b)^{n}}{n!}=sum_{n=0}^{infty}frac{(a)^{n}}{n!}b^{n}=1+(a)b+left(frac{a^{2}}{2}right)b^{2}+left(frac{a^{3}}{6}right)b^{3}+cdots$$



At this point I don't know how to proceed. I don't see how I could get b out of the series and have it as a power that the series is raised to. I thought that it resembled the geometric series, but that did not seem to lead anywhere.






sequences-and-series proof-writing exponential-function







share|cite|improve this question











$endgroup$












  • $begingroup$
    Please see math.meta.stackexchange.com/questions/5020
    $endgroup$
    – Lord Shark the Unknown
    Dec 30 '18 at 17:23






  • 2




    $begingroup$
    Unless I'm blind, isn't $n^{mk} = (n^m)^k$ true always, whether or not $n=e$?
    $endgroup$
    – Zachary Hunter
    Dec 30 '18 at 17:26






  • 2




    $begingroup$
    He’s specifically trying to show it for $e$ using the series expansion.
    $endgroup$
    – KM101
    Dec 30 '18 at 17:31










  • $begingroup$
    You are complicating things for yourself by trying to show this using series. This is a general property. See for example math.stackexchange.com/questions/91869/… or math.stackexchange.com/questions/614162/laws-of-indices or math.stackexchange.com/questions/2086109/…
    $endgroup$
    – Winther
    Dec 30 '18 at 17:47
















2












2








2


1



$begingroup$


I want to prove that



$$e^{ab}=left(e^{a}right)^{b}$$where



$$a,b in mathbb{R}$$ using the infinite series for $$e^{x}$$



My attempt:



$$e^{ab}=sum_{n=0}^{infty}frac{(ab)^{n}}{n!}=sum_{n=0}^{infty}frac{(a)^{n}(b)^{n}}{n!}=sum_{n=0}^{infty}frac{(a)^{n}}{n!}b^{n}=1+(a)b+left(frac{a^{2}}{2}right)b^{2}+left(frac{a^{3}}{6}right)b^{3}+cdots$$



At this point I don't know how to proceed. I don't see how I could get b out of the series and have it as a power that the series is raised to. I thought that it resembled the geometric series, but that did not seem to lead anywhere.






sequences-and-series proof-writing exponential-function







share|cite|improve this question











$endgroup$




I want to prove that



$$e^{ab}=left(e^{a}right)^{b}$$where



$$a,b in mathbb{R}$$ using the infinite series for $$e^{x}$$



My attempt:



$$e^{ab}=sum_{n=0}^{infty}frac{(ab)^{n}}{n!}=sum_{n=0}^{infty}frac{(a)^{n}(b)^{n}}{n!}=sum_{n=0}^{infty}frac{(a)^{n}}{n!}b^{n}=1+(a)b+left(frac{a^{2}}{2}right)b^{2}+left(frac{a^{3}}{6}right)b^{3}+cdots$$



At this point I don't know how to proceed. I don't see how I could get b out of the series and have it as a power that the series is raised to. I thought that it resembled the geometric series, but that did not seem to lead anywhere.






sequences-and-series proof-writing exponential-function




sequences-and-series proof-writing exponential-function






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 30 '18 at 17:41









Blue

47.8k870152




47.8k870152










asked Dec 30 '18 at 17:22









bhjkbhjk

112




112












  • $begingroup$
    Please see math.meta.stackexchange.com/questions/5020
    $endgroup$
    – Lord Shark the Unknown
    Dec 30 '18 at 17:23






  • 2




    $begingroup$
    Unless I'm blind, isn't $n^{mk} = (n^m)^k$ true always, whether or not $n=e$?
    $endgroup$
    – Zachary Hunter
    Dec 30 '18 at 17:26






  • 2




    $begingroup$
    He’s specifically trying to show it for $e$ using the series expansion.
    $endgroup$
    – KM101
    Dec 30 '18 at 17:31










  • $begingroup$
    You are complicating things for yourself by trying to show this using series. This is a general property. See for example math.stackexchange.com/questions/91869/… or math.stackexchange.com/questions/614162/laws-of-indices or math.stackexchange.com/questions/2086109/…
    $endgroup$
    – Winther
    Dec 30 '18 at 17:47




















  • $begingroup$
    Please see math.meta.stackexchange.com/questions/5020
    $endgroup$
    – Lord Shark the Unknown
    Dec 30 '18 at 17:23






  • 2




    $begingroup$
    Unless I'm blind, isn't $n^{mk} = (n^m)^k$ true always, whether or not $n=e$?
    $endgroup$
    – Zachary Hunter
    Dec 30 '18 at 17:26






  • 2




    $begingroup$
    He’s specifically trying to show it for $e$ using the series expansion.
    $endgroup$
    – KM101
    Dec 30 '18 at 17:31










  • $begingroup$
    You are complicating things for yourself by trying to show this using series. This is a general property. See for example math.stackexchange.com/questions/91869/… or math.stackexchange.com/questions/614162/laws-of-indices or math.stackexchange.com/questions/2086109/…
    $endgroup$
    – Winther
    Dec 30 '18 at 17:47


















$begingroup$
Please see math.meta.stackexchange.com/questions/5020
$endgroup$
– Lord Shark the Unknown
Dec 30 '18 at 17:23




$begingroup$
Please see math.meta.stackexchange.com/questions/5020
$endgroup$
– Lord Shark the Unknown
Dec 30 '18 at 17:23




2




2




$begingroup$
Unless I'm blind, isn't $n^{mk} = (n^m)^k$ true always, whether or not $n=e$?
$endgroup$
– Zachary Hunter
Dec 30 '18 at 17:26




$begingroup$
Unless I'm blind, isn't $n^{mk} = (n^m)^k$ true always, whether or not $n=e$?
$endgroup$
– Zachary Hunter
Dec 30 '18 at 17:26




2




2




$begingroup$
He’s specifically trying to show it for $e$ using the series expansion.
$endgroup$
– KM101
Dec 30 '18 at 17:31




$begingroup$
He’s specifically trying to show it for $e$ using the series expansion.
$endgroup$
– KM101
Dec 30 '18 at 17:31












$begingroup$
You are complicating things for yourself by trying to show this using series. This is a general property. See for example math.stackexchange.com/questions/91869/… or math.stackexchange.com/questions/614162/laws-of-indices or math.stackexchange.com/questions/2086109/…
$endgroup$
– Winther
Dec 30 '18 at 17:47






$begingroup$
You are complicating things for yourself by trying to show this using series. This is a general property. See for example math.stackexchange.com/questions/91869/… or math.stackexchange.com/questions/614162/laws-of-indices or math.stackexchange.com/questions/2086109/…
$endgroup$
– Winther
Dec 30 '18 at 17:47












2 Answers
2






active

oldest

votes


















5












$begingroup$

The problem with trying to prove this is that you need a rigorous definition for $x^y$ where $x$ can be any positive number and $y$ can be any real number. Now, we already have a rigorous definition for $exp(y)=e^y$ for any $yinBbb{R}$ as $sum_{n=0}^infty frac{y^n}{n!}$. Therefore, I usually people write $x^y$ in terms of the exponential function:



$$x^y=exp(yln x)$$



Thus, the easiest way to prove $(e^a)^b$ is to just use this definition of $x^y$:



$$(e^a)^b=exp(bln e^a)=exp(ba)=e^{ab}$$



I know that's not really the series solution you wanted, but I still think it is helpful to see this kind of proof.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    How would you go about proving then that ln((e^a)^b) = b*ln(e^a)?
    $endgroup$
    – bhjk
    Dec 30 '18 at 17:46










  • $begingroup$
    @bhjk: That follows directly from the definition of $x^y$ presented here.
    $endgroup$
    – Henning Makholm
    Dec 30 '18 at 17:48








  • 2




    $begingroup$
    Note that this definition doesn't really work "where $x$ and $y$ can be any real number", but only where $x$ is a positive real (and $y$ is an arbitrary real). (Irrational powers of negative reals are usually not considered well-defined at all).
    $endgroup$
    – Henning Makholm
    Dec 30 '18 at 17:51





















0












$begingroup$

If we assume the "meaning" of $e^a$ and if $bin mathbb{Q}$, $b=frac{p}{q}$, then we know what does it mean to $textit{raise a number by the power $p$ and taking the $q-th$ root of that.}$ That is to say, $(e^a)^p=e^a...e^a=e^{ap}$, and $(e^{ap})^{frac{1}{q}}$ is a number which gives $e^{ap}$ when multiplied $q$-times, and we are done.



Now if the number $b$ is not a rational, then consider the following number (can be defined in this way, uses l.u.b. property of $mathbb{R}$) $$(e^a)^b=sup{(e^a)^x,,|,, xin mathbb{Q}text{ and }x<b}=sup{e^{ax},,|,, xin mathbb{Q}text{ and }x<b}=e^{ab}.$$






share|cite|improve this answer











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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    5












    $begingroup$

    The problem with trying to prove this is that you need a rigorous definition for $x^y$ where $x$ can be any positive number and $y$ can be any real number. Now, we already have a rigorous definition for $exp(y)=e^y$ for any $yinBbb{R}$ as $sum_{n=0}^infty frac{y^n}{n!}$. Therefore, I usually people write $x^y$ in terms of the exponential function:



    $$x^y=exp(yln x)$$



    Thus, the easiest way to prove $(e^a)^b$ is to just use this definition of $x^y$:



    $$(e^a)^b=exp(bln e^a)=exp(ba)=e^{ab}$$



    I know that's not really the series solution you wanted, but I still think it is helpful to see this kind of proof.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      How would you go about proving then that ln((e^a)^b) = b*ln(e^a)?
      $endgroup$
      – bhjk
      Dec 30 '18 at 17:46










    • $begingroup$
      @bhjk: That follows directly from the definition of $x^y$ presented here.
      $endgroup$
      – Henning Makholm
      Dec 30 '18 at 17:48








    • 2




      $begingroup$
      Note that this definition doesn't really work "where $x$ and $y$ can be any real number", but only where $x$ is a positive real (and $y$ is an arbitrary real). (Irrational powers of negative reals are usually not considered well-defined at all).
      $endgroup$
      – Henning Makholm
      Dec 30 '18 at 17:51


















    5












    $begingroup$

    The problem with trying to prove this is that you need a rigorous definition for $x^y$ where $x$ can be any positive number and $y$ can be any real number. Now, we already have a rigorous definition for $exp(y)=e^y$ for any $yinBbb{R}$ as $sum_{n=0}^infty frac{y^n}{n!}$. Therefore, I usually people write $x^y$ in terms of the exponential function:



    $$x^y=exp(yln x)$$



    Thus, the easiest way to prove $(e^a)^b$ is to just use this definition of $x^y$:



    $$(e^a)^b=exp(bln e^a)=exp(ba)=e^{ab}$$



    I know that's not really the series solution you wanted, but I still think it is helpful to see this kind of proof.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      How would you go about proving then that ln((e^a)^b) = b*ln(e^a)?
      $endgroup$
      – bhjk
      Dec 30 '18 at 17:46










    • $begingroup$
      @bhjk: That follows directly from the definition of $x^y$ presented here.
      $endgroup$
      – Henning Makholm
      Dec 30 '18 at 17:48








    • 2




      $begingroup$
      Note that this definition doesn't really work "where $x$ and $y$ can be any real number", but only where $x$ is a positive real (and $y$ is an arbitrary real). (Irrational powers of negative reals are usually not considered well-defined at all).
      $endgroup$
      – Henning Makholm
      Dec 30 '18 at 17:51
















    5












    5








    5





    $begingroup$

    The problem with trying to prove this is that you need a rigorous definition for $x^y$ where $x$ can be any positive number and $y$ can be any real number. Now, we already have a rigorous definition for $exp(y)=e^y$ for any $yinBbb{R}$ as $sum_{n=0}^infty frac{y^n}{n!}$. Therefore, I usually people write $x^y$ in terms of the exponential function:



    $$x^y=exp(yln x)$$



    Thus, the easiest way to prove $(e^a)^b$ is to just use this definition of $x^y$:



    $$(e^a)^b=exp(bln e^a)=exp(ba)=e^{ab}$$



    I know that's not really the series solution you wanted, but I still think it is helpful to see this kind of proof.






    share|cite|improve this answer











    $endgroup$



    The problem with trying to prove this is that you need a rigorous definition for $x^y$ where $x$ can be any positive number and $y$ can be any real number. Now, we already have a rigorous definition for $exp(y)=e^y$ for any $yinBbb{R}$ as $sum_{n=0}^infty frac{y^n}{n!}$. Therefore, I usually people write $x^y$ in terms of the exponential function:



    $$x^y=exp(yln x)$$



    Thus, the easiest way to prove $(e^a)^b$ is to just use this definition of $x^y$:



    $$(e^a)^b=exp(bln e^a)=exp(ba)=e^{ab}$$



    I know that's not really the series solution you wanted, but I still think it is helpful to see this kind of proof.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 30 '18 at 18:25

























    answered Dec 30 '18 at 17:39









    Noble MushtakNoble Mushtak

    15.2k1735




    15.2k1735












    • $begingroup$
      How would you go about proving then that ln((e^a)^b) = b*ln(e^a)?
      $endgroup$
      – bhjk
      Dec 30 '18 at 17:46










    • $begingroup$
      @bhjk: That follows directly from the definition of $x^y$ presented here.
      $endgroup$
      – Henning Makholm
      Dec 30 '18 at 17:48








    • 2




      $begingroup$
      Note that this definition doesn't really work "where $x$ and $y$ can be any real number", but only where $x$ is a positive real (and $y$ is an arbitrary real). (Irrational powers of negative reals are usually not considered well-defined at all).
      $endgroup$
      – Henning Makholm
      Dec 30 '18 at 17:51




















    • $begingroup$
      How would you go about proving then that ln((e^a)^b) = b*ln(e^a)?
      $endgroup$
      – bhjk
      Dec 30 '18 at 17:46










    • $begingroup$
      @bhjk: That follows directly from the definition of $x^y$ presented here.
      $endgroup$
      – Henning Makholm
      Dec 30 '18 at 17:48








    • 2




      $begingroup$
      Note that this definition doesn't really work "where $x$ and $y$ can be any real number", but only where $x$ is a positive real (and $y$ is an arbitrary real). (Irrational powers of negative reals are usually not considered well-defined at all).
      $endgroup$
      – Henning Makholm
      Dec 30 '18 at 17:51


















    $begingroup$
    How would you go about proving then that ln((e^a)^b) = b*ln(e^a)?
    $endgroup$
    – bhjk
    Dec 30 '18 at 17:46




    $begingroup$
    How would you go about proving then that ln((e^a)^b) = b*ln(e^a)?
    $endgroup$
    – bhjk
    Dec 30 '18 at 17:46












    $begingroup$
    @bhjk: That follows directly from the definition of $x^y$ presented here.
    $endgroup$
    – Henning Makholm
    Dec 30 '18 at 17:48






    $begingroup$
    @bhjk: That follows directly from the definition of $x^y$ presented here.
    $endgroup$
    – Henning Makholm
    Dec 30 '18 at 17:48






    2




    2




    $begingroup$
    Note that this definition doesn't really work "where $x$ and $y$ can be any real number", but only where $x$ is a positive real (and $y$ is an arbitrary real). (Irrational powers of negative reals are usually not considered well-defined at all).
    $endgroup$
    – Henning Makholm
    Dec 30 '18 at 17:51






    $begingroup$
    Note that this definition doesn't really work "where $x$ and $y$ can be any real number", but only where $x$ is a positive real (and $y$ is an arbitrary real). (Irrational powers of negative reals are usually not considered well-defined at all).
    $endgroup$
    – Henning Makholm
    Dec 30 '18 at 17:51













    0












    $begingroup$

    If we assume the "meaning" of $e^a$ and if $bin mathbb{Q}$, $b=frac{p}{q}$, then we know what does it mean to $textit{raise a number by the power $p$ and taking the $q-th$ root of that.}$ That is to say, $(e^a)^p=e^a...e^a=e^{ap}$, and $(e^{ap})^{frac{1}{q}}$ is a number which gives $e^{ap}$ when multiplied $q$-times, and we are done.



    Now if the number $b$ is not a rational, then consider the following number (can be defined in this way, uses l.u.b. property of $mathbb{R}$) $$(e^a)^b=sup{(e^a)^x,,|,, xin mathbb{Q}text{ and }x<b}=sup{e^{ax},,|,, xin mathbb{Q}text{ and }x<b}=e^{ab}.$$






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      If we assume the "meaning" of $e^a$ and if $bin mathbb{Q}$, $b=frac{p}{q}$, then we know what does it mean to $textit{raise a number by the power $p$ and taking the $q-th$ root of that.}$ That is to say, $(e^a)^p=e^a...e^a=e^{ap}$, and $(e^{ap})^{frac{1}{q}}$ is a number which gives $e^{ap}$ when multiplied $q$-times, and we are done.



      Now if the number $b$ is not a rational, then consider the following number (can be defined in this way, uses l.u.b. property of $mathbb{R}$) $$(e^a)^b=sup{(e^a)^x,,|,, xin mathbb{Q}text{ and }x<b}=sup{e^{ax},,|,, xin mathbb{Q}text{ and }x<b}=e^{ab}.$$






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        If we assume the "meaning" of $e^a$ and if $bin mathbb{Q}$, $b=frac{p}{q}$, then we know what does it mean to $textit{raise a number by the power $p$ and taking the $q-th$ root of that.}$ That is to say, $(e^a)^p=e^a...e^a=e^{ap}$, and $(e^{ap})^{frac{1}{q}}$ is a number which gives $e^{ap}$ when multiplied $q$-times, and we are done.



        Now if the number $b$ is not a rational, then consider the following number (can be defined in this way, uses l.u.b. property of $mathbb{R}$) $$(e^a)^b=sup{(e^a)^x,,|,, xin mathbb{Q}text{ and }x<b}=sup{e^{ax},,|,, xin mathbb{Q}text{ and }x<b}=e^{ab}.$$






        share|cite|improve this answer











        $endgroup$



        If we assume the "meaning" of $e^a$ and if $bin mathbb{Q}$, $b=frac{p}{q}$, then we know what does it mean to $textit{raise a number by the power $p$ and taking the $q-th$ root of that.}$ That is to say, $(e^a)^p=e^a...e^a=e^{ap}$, and $(e^{ap})^{frac{1}{q}}$ is a number which gives $e^{ap}$ when multiplied $q$-times, and we are done.



        Now if the number $b$ is not a rational, then consider the following number (can be defined in this way, uses l.u.b. property of $mathbb{R}$) $$(e^a)^b=sup{(e^a)^x,,|,, xin mathbb{Q}text{ and }x<b}=sup{e^{ax},,|,, xin mathbb{Q}text{ and }x<b}=e^{ab}.$$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 30 '18 at 18:21

























        answered Dec 30 '18 at 18:14









        DèöDèö

        16113




        16113






























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