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It is easy to prove:

$$e^{\ln\left(1+x\right)}=1+x$$

in algebraic way. However, how about prove it by Taylor expansion?

Know that: $$\ln\left(1+x\right)=x-\dfrac{x^2}{2}+\dfrac{x^3}{3}-\dfrac{x^4}{4}+\cdots\\e^z=1+z+\dfrac{z^2}{2}+\dfrac{z^3}{6}+\dfrac{z^4}{24}+\cdots$$ Then, $$e^{\ln\left(1+x\right)}=e^{x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\dots}=e^x\cdot e^{-\frac{x^2}{2}}\cdot e^\frac{x^3}{3}\cdot e^{-\frac{x^4}{4}}\cdots\\=\left(1+x+\dfrac{1}{2}x^2+\dfrac{1}{6}x^3+\cdots\right)\left[1-\dfrac{x^2}{2}+\dfrac{1}{2}\left(\dfrac{x^2}{2}\right)^2-\dfrac{1}{6}\left(\dfrac{x^2}{2}\right)^3+\cdots\right]\left[1+\dfrac{x^3}{6}+\dfrac{1}{2}\left(\dfrac{x^3}{6}\right)^2+\dfrac{1}{6}\left(\dfrac{x^3}{6}\right)^3+\cdots\right]\left[1+\dfrac{x^4}{24}+\dfrac{1}{2}\left(\dfrac{x^4}{24}\right)^2+\dfrac{1}{6}\left(\dfrac{x^4}{24}\right)^3+\cdots\right]$$ Till here, I am stuck. This expansion is very big that I can't really simplify. Though I find the coefficient of $x^0,x^1,x^2,x^3$ are $1,1,0,0$ respectively. However, I can't prove the higher degree coefficient. If there are someone can answer or giving tips, that's appreciated. Thank you!

MafPrivate
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    See https://math.stackexchange.com/questions/3325684/combinatorial-argument-for-exponential-and-logarithmic-function-being-inverse – Angina Seng Sep 15 '19 at 11:46
  • From the power series definitions it is easier to show (for $|x|$ small enough) $\exp'(x)=\exp(x), \log'(1+x)= 1/(1+x)$ $ \implies (\log(\exp(x)))'= 1, \log(\exp(x))=x+\log(\exp(0))=x \implies \exp(\log(1+x)) =1+x$ – reuns Sep 15 '19 at 13:03
  • @reuns this prove is quite algebraic, this is easy to see. – MafPrivate Sep 15 '19 at 13:19
  • It is based on power series coefficients : the derivative has a concrete meaning on power series coefficients and $f' = 1 \implies f = x+c$ is a trivial power series identity, same for $(f(g))' = f'(g) g'$ – reuns Sep 15 '19 at 13:20
  • Let $f(x)=\exp (\log x)$ with $|x|<1$ (so the power series for $\log x$ converges). If a power series converges on an open interval then it can be differentiated term-wise on that interval to get the derivative of the series. So $(\exp z)'=\exp z$ and $(\log x)'=1-x+x^2-...=1/(1+x). $ So by the chain rule, $f'(x)=f(x)/(1+x),$ so $f(x)=(1+x)f'(x).$ Differentiating $that$ gives $f'(x)=(1+x)f''(x)+f'(x).$ So $f''(x)=0$ for $|x|<1$. So $f(x)=A+Bx$ with constants $A, B.$ We compute $A=f(0)=1.$ And from $f'(x)=f(x)/(1+x)$ we have $ B=f'(0)=f(0)/(1+0)=f(0)=1.$ – DanielWainfleet Sep 21 '19 at 10:44
  • If you define $e^z$ as the power series then before you can say $\exp (\log (1+x))=\prod_{n=1}^{\infty} \exp ((-1)^{n+1}x^n/n),$ you must prove that $\exp (u) \exp (v)=\exp (u+v).$ – DanielWainfleet Sep 21 '19 at 10:58

1 Answers1

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Staying with $$e^{\ln\left(1+x\right)}=e^{x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}}$$ let $y=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}$ and use $$e^y=1+y+\frac 12 y^2+\frac 16 y^3+\frac 1{24} y^4$$ So, replacing $y$ and expanding $$y=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+O(x^5)$$ $$y^2=x^2-x^3+\frac{11 x^4}{12}-\frac{5 x^5}{6}+\frac{13 x^6}{36}-\frac{x^7}{6}+\frac{x^8}{16}$$ $$y^3=x^3-\frac{3 x^4}{2}+\frac{7 x^5}{4}-\frac{15 x^6}{8}+\frac{4 x^7}{3}-\frac{41 x^8}{48}+\frac{205 x^9}{432}-\frac{17 x^{10}}{96}+\frac{x^{11}}{16}-\frac{x^{12}}{64}$$ $$y^4=x^4-2 x^5+\frac{17 x^6}{6}-\frac{7 x^7}{2}+\frac{155 x^8}{48}-\frac{31 x^9}{12}+\frac{49 x^{10}}{27}-\frac{223 x^{11}}{216}+\frac{1355 x^{12}}{2592}-\frac{97 x^{13}}{432}+\frac{7 x^{14}}{96}-\frac{x^{15}}{48}+\frac{x^{16}}{256}$$ So, all of the above gives $$e^{x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}}=1+x-\frac{5 x^5}{24}-\frac{x^6}{72}-\frac{x^7}{144}+\cdots$$

Edit

What could be interesting is to define $$y_n=\sum_{k=1}^n (-1)^{k-1} \frac{x^k} k$$ to show that the expansion $$\exp\left(\sum_{k=0}^p \frac{y_n^k} {k!} \right)=1+x +a_{p+1} x^{p+1}+O(x^{p+2})$$ Starting at $p=2$, the sequence for the $a$ coefficients is $$\left\{-\frac{1}{2},\frac{5}{24},-\frac{5}{24},\frac{119}{720},-\frac{103}{720},\frac{5039}{40320},-\frac{40321}{362880},\frac{362879}{3628800},-\frac{329891}{3628 800},\frac{39916799}{479001600}\right\}$$

Claude Leibovici
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