Manipulating Taylor Series for $e^{2x}$ using $e^x$












0












$begingroup$


I’m a calculus student learning infinite series and sequences for the first time, and I have a question about manipulating the Taylor series for $e^x$ to arrive at one for $e^{2x}$.



My first thought was to say that
$$g(x) = e^x = sum_{n=0}^infty frac{x^n}{n!}$$
And $f(x) = x^2$, so
$$
e^{2x}
= g(f(x))
= left(sum_{n=0}^infty frac{x^n}{n!}right)^2
= sum_{n=0}^infty frac{x^{2n}}{(n!)^2}
$$
However, this isn’t what my textbook got. It said that I can just set $f(x) = e^x$ and $g(x) = 2x$, meaning that $$f(g(x)) = e^{2x} = sum_{n=0}^infty frac{(2x)^n}{n!}$$
I understand this and think it’s a little simpler than what I originally tried out.



So both of these methods make sense to me, but I got different answers for each of them. What am I doing wrong in the first one?






sequences-and-series taylor-expansion







share|cite|improve this question











$endgroup$












  • $begingroup$
    It is true that when you expand the product $(e^x)^2 = (1+x+x^2/2!+x^3/3! + ldots)^2$ then you will arrive at the power-series of $e^{2x}$. You probably didn't compute it correctly. Note how much easier the second method is compare to this one (you need to compute the product of two infinite series). To compute the product of two power-series $sum_{ngeq 0} a_n x^n cdot sum_{ngeq 0} b_n x^n = sum_{ngeq 0} c_n x^n$ see Cauchy product
    $endgroup$
    – Winther
    Jan 15 at 20:11












  • $begingroup$
    Thank you, that makes a lot of sense!!
    $endgroup$
    – elbecker
    Jan 15 at 20:19










  • $begingroup$
    @ElizabethBecker i updated my answer to write out the correct multiplication logic
    $endgroup$
    – gt6989b
    Jan 15 at 20:26






  • 1




    $begingroup$
    Freshman's dream $(a+b)^2 ne a^2 + b^2.$ You need those cross terms. ($sum_limits{n=0}^infty a_n x^n)(sum_limits{n=0}^infty b_n x^n) = sum_limits{n=0}^infty sum_limits{k=0}^n a_{n-k}b_k x^n$
    $endgroup$
    – Doug M
    Jan 15 at 20:30


















0












$begingroup$


I’m a calculus student learning infinite series and sequences for the first time, and I have a question about manipulating the Taylor series for $e^x$ to arrive at one for $e^{2x}$.



My first thought was to say that
$$g(x) = e^x = sum_{n=0}^infty frac{x^n}{n!}$$
And $f(x) = x^2$, so
$$
e^{2x}
= g(f(x))
= left(sum_{n=0}^infty frac{x^n}{n!}right)^2
= sum_{n=0}^infty frac{x^{2n}}{(n!)^2}
$$
However, this isn’t what my textbook got. It said that I can just set $f(x) = e^x$ and $g(x) = 2x$, meaning that $$f(g(x)) = e^{2x} = sum_{n=0}^infty frac{(2x)^n}{n!}$$
I understand this and think it’s a little simpler than what I originally tried out.



So both of these methods make sense to me, but I got different answers for each of them. What am I doing wrong in the first one?






sequences-and-series taylor-expansion







share|cite|improve this question











$endgroup$












  • $begingroup$
    It is true that when you expand the product $(e^x)^2 = (1+x+x^2/2!+x^3/3! + ldots)^2$ then you will arrive at the power-series of $e^{2x}$. You probably didn't compute it correctly. Note how much easier the second method is compare to this one (you need to compute the product of two infinite series). To compute the product of two power-series $sum_{ngeq 0} a_n x^n cdot sum_{ngeq 0} b_n x^n = sum_{ngeq 0} c_n x^n$ see Cauchy product
    $endgroup$
    – Winther
    Jan 15 at 20:11












  • $begingroup$
    Thank you, that makes a lot of sense!!
    $endgroup$
    – elbecker
    Jan 15 at 20:19










  • $begingroup$
    @ElizabethBecker i updated my answer to write out the correct multiplication logic
    $endgroup$
    – gt6989b
    Jan 15 at 20:26






  • 1




    $begingroup$
    Freshman's dream $(a+b)^2 ne a^2 + b^2.$ You need those cross terms. ($sum_limits{n=0}^infty a_n x^n)(sum_limits{n=0}^infty b_n x^n) = sum_limits{n=0}^infty sum_limits{k=0}^n a_{n-k}b_k x^n$
    $endgroup$
    – Doug M
    Jan 15 at 20:30
















0












0








0





$begingroup$


I’m a calculus student learning infinite series and sequences for the first time, and I have a question about manipulating the Taylor series for $e^x$ to arrive at one for $e^{2x}$.



My first thought was to say that
$$g(x) = e^x = sum_{n=0}^infty frac{x^n}{n!}$$
And $f(x) = x^2$, so
$$
e^{2x}
= g(f(x))
= left(sum_{n=0}^infty frac{x^n}{n!}right)^2
= sum_{n=0}^infty frac{x^{2n}}{(n!)^2}
$$
However, this isn’t what my textbook got. It said that I can just set $f(x) = e^x$ and $g(x) = 2x$, meaning that $$f(g(x)) = e^{2x} = sum_{n=0}^infty frac{(2x)^n}{n!}$$
I understand this and think it’s a little simpler than what I originally tried out.



So both of these methods make sense to me, but I got different answers for each of them. What am I doing wrong in the first one?






sequences-and-series taylor-expansion







share|cite|improve this question











$endgroup$




I’m a calculus student learning infinite series and sequences for the first time, and I have a question about manipulating the Taylor series for $e^x$ to arrive at one for $e^{2x}$.



My first thought was to say that
$$g(x) = e^x = sum_{n=0}^infty frac{x^n}{n!}$$
And $f(x) = x^2$, so
$$
e^{2x}
= g(f(x))
= left(sum_{n=0}^infty frac{x^n}{n!}right)^2
= sum_{n=0}^infty frac{x^{2n}}{(n!)^2}
$$
However, this isn’t what my textbook got. It said that I can just set $f(x) = e^x$ and $g(x) = 2x$, meaning that $$f(g(x)) = e^{2x} = sum_{n=0}^infty frac{(2x)^n}{n!}$$
I understand this and think it’s a little simpler than what I originally tried out.



So both of these methods make sense to me, but I got different answers for each of them. What am I doing wrong in the first one?






sequences-and-series taylor-expansion




sequences-and-series taylor-expansion






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 15 at 20:20









gt6989b

35.2k22557




35.2k22557










asked Jan 15 at 20:06









elbeckerelbecker

33




33












  • $begingroup$
    It is true that when you expand the product $(e^x)^2 = (1+x+x^2/2!+x^3/3! + ldots)^2$ then you will arrive at the power-series of $e^{2x}$. You probably didn't compute it correctly. Note how much easier the second method is compare to this one (you need to compute the product of two infinite series). To compute the product of two power-series $sum_{ngeq 0} a_n x^n cdot sum_{ngeq 0} b_n x^n = sum_{ngeq 0} c_n x^n$ see Cauchy product
    $endgroup$
    – Winther
    Jan 15 at 20:11












  • $begingroup$
    Thank you, that makes a lot of sense!!
    $endgroup$
    – elbecker
    Jan 15 at 20:19










  • $begingroup$
    @ElizabethBecker i updated my answer to write out the correct multiplication logic
    $endgroup$
    – gt6989b
    Jan 15 at 20:26






  • 1




    $begingroup$
    Freshman's dream $(a+b)^2 ne a^2 + b^2.$ You need those cross terms. ($sum_limits{n=0}^infty a_n x^n)(sum_limits{n=0}^infty b_n x^n) = sum_limits{n=0}^infty sum_limits{k=0}^n a_{n-k}b_k x^n$
    $endgroup$
    – Doug M
    Jan 15 at 20:30




















  • $begingroup$
    It is true that when you expand the product $(e^x)^2 = (1+x+x^2/2!+x^3/3! + ldots)^2$ then you will arrive at the power-series of $e^{2x}$. You probably didn't compute it correctly. Note how much easier the second method is compare to this one (you need to compute the product of two infinite series). To compute the product of two power-series $sum_{ngeq 0} a_n x^n cdot sum_{ngeq 0} b_n x^n = sum_{ngeq 0} c_n x^n$ see Cauchy product
    $endgroup$
    – Winther
    Jan 15 at 20:11












  • $begingroup$
    Thank you, that makes a lot of sense!!
    $endgroup$
    – elbecker
    Jan 15 at 20:19










  • $begingroup$
    @ElizabethBecker i updated my answer to write out the correct multiplication logic
    $endgroup$
    – gt6989b
    Jan 15 at 20:26






  • 1




    $begingroup$
    Freshman's dream $(a+b)^2 ne a^2 + b^2.$ You need those cross terms. ($sum_limits{n=0}^infty a_n x^n)(sum_limits{n=0}^infty b_n x^n) = sum_limits{n=0}^infty sum_limits{k=0}^n a_{n-k}b_k x^n$
    $endgroup$
    – Doug M
    Jan 15 at 20:30


















$begingroup$
It is true that when you expand the product $(e^x)^2 = (1+x+x^2/2!+x^3/3! + ldots)^2$ then you will arrive at the power-series of $e^{2x}$. You probably didn't compute it correctly. Note how much easier the second method is compare to this one (you need to compute the product of two infinite series). To compute the product of two power-series $sum_{ngeq 0} a_n x^n cdot sum_{ngeq 0} b_n x^n = sum_{ngeq 0} c_n x^n$ see Cauchy product
$endgroup$
– Winther
Jan 15 at 20:11






$begingroup$
It is true that when you expand the product $(e^x)^2 = (1+x+x^2/2!+x^3/3! + ldots)^2$ then you will arrive at the power-series of $e^{2x}$. You probably didn't compute it correctly. Note how much easier the second method is compare to this one (you need to compute the product of two infinite series). To compute the product of two power-series $sum_{ngeq 0} a_n x^n cdot sum_{ngeq 0} b_n x^n = sum_{ngeq 0} c_n x^n$ see Cauchy product
$endgroup$
– Winther
Jan 15 at 20:11














$begingroup$
Thank you, that makes a lot of sense!!
$endgroup$
– elbecker
Jan 15 at 20:19




$begingroup$
Thank you, that makes a lot of sense!!
$endgroup$
– elbecker
Jan 15 at 20:19












$begingroup$
@ElizabethBecker i updated my answer to write out the correct multiplication logic
$endgroup$
– gt6989b
Jan 15 at 20:26




$begingroup$
@ElizabethBecker i updated my answer to write out the correct multiplication logic
$endgroup$
– gt6989b
Jan 15 at 20:26




1




1




$begingroup$
Freshman's dream $(a+b)^2 ne a^2 + b^2.$ You need those cross terms. ($sum_limits{n=0}^infty a_n x^n)(sum_limits{n=0}^infty b_n x^n) = sum_limits{n=0}^infty sum_limits{k=0}^n a_{n-k}b_k x^n$
$endgroup$
– Doug M
Jan 15 at 20:30






$begingroup$
Freshman's dream $(a+b)^2 ne a^2 + b^2.$ You need those cross terms. ($sum_limits{n=0}^infty a_n x^n)(sum_limits{n=0}^infty b_n x^n) = sum_limits{n=0}^infty sum_limits{k=0}^n a_{n-k}b_k x^n$
$endgroup$
– Doug M
Jan 15 at 20:30












2 Answers
2






active

oldest

votes


















6












$begingroup$

Note that
$$
e^y = sum_{k=0}^infty frac{y^k}{k!}
$$
therefore, if $y=2x$ you get
$$
e^{2x}
= sum_{k=0}^infty frac{(2x)^k}{k!}
= sum_{k=0}^infty frac{2^k x^k}{k!}
$$




There are issues with what you did. Even though as you claim
$$
e^{2x}
= left(e^xright)^2
= left(sum_{k=0}^infty frac{x^k}{k!}right)^2,
$$
but squaring that sum is not a simple matter because of the crossing terms, since
$$(x+y)^2 = x^2 + y^2 + 2xy ne x^2 + y^2$$




If you want to continue your method to its logical end, you can use
$$
begin{split}
e^{2x}
&= left(sum_{k=0}^infty frac{x^k}{k!}right)
cdot left(sum_{k=0}^infty frac{x^k}{k!}right)\
&= sum_{n=0}^infty sum_{k=0}^n frac{x^k}{k!} frac{x^{n-k}}{(n-k)!} \
&= sum_{n=0}^infty x^n left( sum_{k=0}^n frac{1}{k!(n-k)!}right) \
&= sum_{n=0}^infty frac{x^n}{n!}
sum_{k=0}^n frac{n!}{k!(n-k)!} \
&= sum_{n=0}^infty frac{x^n}{n!} sum_{k=0}^n binom{n}{k} \
&= sum_{n=0}^infty frac{2^n x^n}{n!},
end{split}
$$
with the last step using the Binomial theorem.






share|cite|improve this answer











$endgroup$





















    4












    $begingroup$

    Your step $$left(sum_{n=0}^infty frac{x^n}{n!}right)^2
    = sum_{n=0}^infty frac{x^{2n}}{(n!)^2}$$
    is akin to claiming that
    $$(a+b)^2=a^2+b^2$$






    share|cite|improve this answer









    $endgroup$














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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      6












      $begingroup$

      Note that
      $$
      e^y = sum_{k=0}^infty frac{y^k}{k!}
      $$
      therefore, if $y=2x$ you get
      $$
      e^{2x}
      = sum_{k=0}^infty frac{(2x)^k}{k!}
      = sum_{k=0}^infty frac{2^k x^k}{k!}
      $$




      There are issues with what you did. Even though as you claim
      $$
      e^{2x}
      = left(e^xright)^2
      = left(sum_{k=0}^infty frac{x^k}{k!}right)^2,
      $$
      but squaring that sum is not a simple matter because of the crossing terms, since
      $$(x+y)^2 = x^2 + y^2 + 2xy ne x^2 + y^2$$




      If you want to continue your method to its logical end, you can use
      $$
      begin{split}
      e^{2x}
      &= left(sum_{k=0}^infty frac{x^k}{k!}right)
      cdot left(sum_{k=0}^infty frac{x^k}{k!}right)\
      &= sum_{n=0}^infty sum_{k=0}^n frac{x^k}{k!} frac{x^{n-k}}{(n-k)!} \
      &= sum_{n=0}^infty x^n left( sum_{k=0}^n frac{1}{k!(n-k)!}right) \
      &= sum_{n=0}^infty frac{x^n}{n!}
      sum_{k=0}^n frac{n!}{k!(n-k)!} \
      &= sum_{n=0}^infty frac{x^n}{n!} sum_{k=0}^n binom{n}{k} \
      &= sum_{n=0}^infty frac{2^n x^n}{n!},
      end{split}
      $$
      with the last step using the Binomial theorem.






      share|cite|improve this answer











      $endgroup$


















        6












        $begingroup$

        Note that
        $$
        e^y = sum_{k=0}^infty frac{y^k}{k!}
        $$
        therefore, if $y=2x$ you get
        $$
        e^{2x}
        = sum_{k=0}^infty frac{(2x)^k}{k!}
        = sum_{k=0}^infty frac{2^k x^k}{k!}
        $$




        There are issues with what you did. Even though as you claim
        $$
        e^{2x}
        = left(e^xright)^2
        = left(sum_{k=0}^infty frac{x^k}{k!}right)^2,
        $$
        but squaring that sum is not a simple matter because of the crossing terms, since
        $$(x+y)^2 = x^2 + y^2 + 2xy ne x^2 + y^2$$




        If you want to continue your method to its logical end, you can use
        $$
        begin{split}
        e^{2x}
        &= left(sum_{k=0}^infty frac{x^k}{k!}right)
        cdot left(sum_{k=0}^infty frac{x^k}{k!}right)\
        &= sum_{n=0}^infty sum_{k=0}^n frac{x^k}{k!} frac{x^{n-k}}{(n-k)!} \
        &= sum_{n=0}^infty x^n left( sum_{k=0}^n frac{1}{k!(n-k)!}right) \
        &= sum_{n=0}^infty frac{x^n}{n!}
        sum_{k=0}^n frac{n!}{k!(n-k)!} \
        &= sum_{n=0}^infty frac{x^n}{n!} sum_{k=0}^n binom{n}{k} \
        &= sum_{n=0}^infty frac{2^n x^n}{n!},
        end{split}
        $$
        with the last step using the Binomial theorem.






        share|cite|improve this answer











        $endgroup$
















          6












          6








          6





          $begingroup$

          Note that
          $$
          e^y = sum_{k=0}^infty frac{y^k}{k!}
          $$
          therefore, if $y=2x$ you get
          $$
          e^{2x}
          = sum_{k=0}^infty frac{(2x)^k}{k!}
          = sum_{k=0}^infty frac{2^k x^k}{k!}
          $$




          There are issues with what you did. Even though as you claim
          $$
          e^{2x}
          = left(e^xright)^2
          = left(sum_{k=0}^infty frac{x^k}{k!}right)^2,
          $$
          but squaring that sum is not a simple matter because of the crossing terms, since
          $$(x+y)^2 = x^2 + y^2 + 2xy ne x^2 + y^2$$




          If you want to continue your method to its logical end, you can use
          $$
          begin{split}
          e^{2x}
          &= left(sum_{k=0}^infty frac{x^k}{k!}right)
          cdot left(sum_{k=0}^infty frac{x^k}{k!}right)\
          &= sum_{n=0}^infty sum_{k=0}^n frac{x^k}{k!} frac{x^{n-k}}{(n-k)!} \
          &= sum_{n=0}^infty x^n left( sum_{k=0}^n frac{1}{k!(n-k)!}right) \
          &= sum_{n=0}^infty frac{x^n}{n!}
          sum_{k=0}^n frac{n!}{k!(n-k)!} \
          &= sum_{n=0}^infty frac{x^n}{n!} sum_{k=0}^n binom{n}{k} \
          &= sum_{n=0}^infty frac{2^n x^n}{n!},
          end{split}
          $$
          with the last step using the Binomial theorem.






          share|cite|improve this answer











          $endgroup$



          Note that
          $$
          e^y = sum_{k=0}^infty frac{y^k}{k!}
          $$
          therefore, if $y=2x$ you get
          $$
          e^{2x}
          = sum_{k=0}^infty frac{(2x)^k}{k!}
          = sum_{k=0}^infty frac{2^k x^k}{k!}
          $$




          There are issues with what you did. Even though as you claim
          $$
          e^{2x}
          = left(e^xright)^2
          = left(sum_{k=0}^infty frac{x^k}{k!}right)^2,
          $$
          but squaring that sum is not a simple matter because of the crossing terms, since
          $$(x+y)^2 = x^2 + y^2 + 2xy ne x^2 + y^2$$




          If you want to continue your method to its logical end, you can use
          $$
          begin{split}
          e^{2x}
          &= left(sum_{k=0}^infty frac{x^k}{k!}right)
          cdot left(sum_{k=0}^infty frac{x^k}{k!}right)\
          &= sum_{n=0}^infty sum_{k=0}^n frac{x^k}{k!} frac{x^{n-k}}{(n-k)!} \
          &= sum_{n=0}^infty x^n left( sum_{k=0}^n frac{1}{k!(n-k)!}right) \
          &= sum_{n=0}^infty frac{x^n}{n!}
          sum_{k=0}^n frac{n!}{k!(n-k)!} \
          &= sum_{n=0}^infty frac{x^n}{n!} sum_{k=0}^n binom{n}{k} \
          &= sum_{n=0}^infty frac{2^n x^n}{n!},
          end{split}
          $$
          with the last step using the Binomial theorem.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Mar 15 at 0:52









          Rócherz

          3,0013821




          3,0013821










          answered Jan 15 at 20:09









          gt6989bgt6989b

          35.2k22557




          35.2k22557























              4












              $begingroup$

              Your step $$left(sum_{n=0}^infty frac{x^n}{n!}right)^2
              = sum_{n=0}^infty frac{x^{2n}}{(n!)^2}$$
              is akin to claiming that
              $$(a+b)^2=a^2+b^2$$






              share|cite|improve this answer









              $endgroup$


















                4












                $begingroup$

                Your step $$left(sum_{n=0}^infty frac{x^n}{n!}right)^2
                = sum_{n=0}^infty frac{x^{2n}}{(n!)^2}$$
                is akin to claiming that
                $$(a+b)^2=a^2+b^2$$






                share|cite|improve this answer









                $endgroup$
















                  4












                  4








                  4





                  $begingroup$

                  Your step $$left(sum_{n=0}^infty frac{x^n}{n!}right)^2
                  = sum_{n=0}^infty frac{x^{2n}}{(n!)^2}$$
                  is akin to claiming that
                  $$(a+b)^2=a^2+b^2$$






                  share|cite|improve this answer









                  $endgroup$



                  Your step $$left(sum_{n=0}^infty frac{x^n}{n!}right)^2
                  = sum_{n=0}^infty frac{x^{2n}}{(n!)^2}$$
                  is akin to claiming that
                  $$(a+b)^2=a^2+b^2$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 15 at 20:26









                  TonyKTonyK

                  43.7k358137




                  43.7k358137






























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