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Prove $\dfrac{d}{dx}\left(e^x\right)=e^x$ with Taylor series and power rule.

So far I have

$$1+x+\frac{1}{2}x^2+\frac{1}{3\cdot2}x^3+\cdots+\frac{1}{\infty-1}x^\infty$$

note: Also that is what I got after I took the derivative of the Taylor series, fyi for those who are telling me to take the derivative of the Taylor series. Basically, it goes through an infinite regression.

Thence I applied the power rule upon the Taylor series of $e^x$.

I'm pretty sure I'm supposed to use the binomial theorem to get $(1+\frac{x}{n})^n$; however $n$ in this case is $\infty$, and thus I cannot rewrite the series.

Help s'il vous plaît. Merci.

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    $\begingroup$ By just derivating the taylor series that you wrote with respect to $x$, you'll get the same series again. This means that the derivative of $e^x$ is itself $\endgroup$ – Fareed AF Oct 27 '18 at 20:28
  • $\begingroup$ Do you know the derivative of 2^x? $\endgroup$ – Jacob Wakem Oct 27 '18 at 21:35
  • $\begingroup$ Just use the definition of derivative. $\endgroup$ – Jacob Wakem Oct 27 '18 at 21:40
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You know $$e^x=1+x+\frac{1}{2}x^2+\frac{1}{3\times2}x^3\dots$$ and you want to prove $\frac{d}{dx}e^x=e^x$. To compute the left hand side, differentiate the series from above.

$$\frac{d}{dx}\left(1+x+\frac{1}{2}x^2+\frac{1}{3\times2}x^3\dots\right)$$ What do you get? Do you recognise what you get?


Alternative method:

You also mentioned the formula $e^x=\lim_{n\to\infty}\left((1+\frac{x}{n})^n\right)$. You can also differentiate this with respect to $x$ by moving the derivative into the limit, and you'll get the same result.

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  • $\begingroup$ Look at my note, I already did that but we just get an infinite regress $\endgroup$ – John Rawls Oct 27 '18 at 20:27
  • $\begingroup$ wait actually on the second thought can we say that because e^x = d/d(x) =e^x that is the proof? $\endgroup$ – John Rawls Oct 27 '18 at 20:28
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    $\begingroup$ @JohnRawls We don't get an infinite regress - the successive terms have a generic pattern, so you can see what would happen to the generic pattern, and this allows you to see what happens for every term in the series with just one calculation. You can write the series as $$1+x+\frac{x^2}{2}+\frac{x^3}{3!}+\cdots+\frac{x^k}{k!}+\cdots$$. Then, differentiating the general term reduces the power of $x$ by one, and brings down a $k$ to reduce the $k!$ term. $\endgroup$ – John Doe Oct 27 '18 at 20:32
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    $\begingroup$ I'm not sure what you mean by your second comment, there may be a typo though since d/d(x) isn't actually a function by itself! The point of this question is trying to differentiate $e^x$ and showing you just end up with the same thing you started with. $\endgroup$ – John Doe Oct 27 '18 at 20:34
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$$e^x=\sum_{k=0}^\infty\frac{x^k}{k!}$$ the radius of convergence is $R= \infty$.

the derivative is $$\sum_{k=1}^\infty\frac{kx^{k-1}}{k!}=$$ $$\sum_{k=1}^\infty\frac{x^{k-1}}{(k-1)!}=$$

$$\sum_{K=0}^\infty\frac{x^K}{K!}=e^x$$

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Excluding the infinite term what you have there is the Taylor Series of $e^x$. You don't need to consider an infinite term, as infinite series are determined by the convergence of their partial (finite) sums.

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One way I like to prove it is through a differential equation: $$f'(x)=f(x)$$ $$\frac{f'(x)}{f(x)}=1$$ $$\int\frac{f'(x)}{f(x)}dx=\int dx$$ $$\int\frac{f'(x)}{f(x)}dx=x+c_0$$ Letting $y=f(x)$ gives $$\int\frac{dy}{y}=x+c_0$$ $$\ln|y|=x+c_0$$ $$f(x)=e^{x+c_0}$$ Note: I actually don't know if this proof contains circular logic with the $\int\frac{dy}{y}=\ln|y|$ bit, but I hope it doesn't.

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    $\begingroup$ To remove the circularity, you have to prove that $\int_1^{xy} dt/t = \int_1^xdt/t + \int_1^y dt/t$, thus demonstrating that $\int_1^x dt/t$ is indeed a logarithm function. From that you get $f(x) = e^x$, where $1 = \int_1^e dt/t$. $\endgroup$ – eyeballfrog Oct 27 '18 at 21:12
  • $\begingroup$ @eyeballfrog I just asked a question on this potential proof, and I think your comment (expanded a little) would be a great answer. Here's the link: math.stackexchange.com/q/2973913/583016 $\endgroup$ – clathratus Oct 27 '18 at 21:32
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I suppose that $series$ and $taylor-series$ are studies after function $e^{x}$. But if you can use Taylor-representation it will be easy to differentiate series :

$\displaystyle \frac{d (e^{x})}{dx} = \frac{d}{dx}(\sum_{k=0}^{\infty} \frac{x^{k}}{k!}) = \sum_{k=0}^{\infty}\frac{d(x^{k})}{k!dx} = \sum_{k=0}^{\infty} \frac{x^{k-1}}{(k-1)!} = \sum_{k'=0}^{\infty} \frac{x^{k'}}{k'!}$, where $k' = k-1$.

Adding : $$\lim_{\vartriangle x \to 0} \frac{e^{x+\vartriangle x} - e^{x}}{\vartriangle x} = \frac{e^{x}(e^{\vartriangle x}- 1)}{\vartriangle x}$$ and use Taylor estimation to finish limit.

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As suggested by John Doe since

$$e^x=\sum_{k=0}^{\infty} \frac{x^k}{k!}=\sum_{k=0}^{\infty} a_k(x)$$

and

  • for $k=1 \implies \frac{da_1(x)}{dx}=0$
  • for $k>1 \implies\frac{da_k(x)}{dx}=a_{k-1}(x)$

we have

$$\frac{de^x}{dx}=\frac{d}{dx}\left(1+\sum_{k=1}^{\infty} a_k(x)\right)=0+\frac{d}{dx}\sum_{k=1}^{\infty} a_k(x)=\sum_{k=1}^{\infty} a_{k-1}(x)=\sum_{k=0}^{\infty} a_k(x)=e^x$$

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