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I'm trying to understand the following question

Give a new derivation of the formula $$ \frac {\mathrm d}{\mathrm dx} e^x = e^x $$ by differentiating the power series $$ 1 + x + \frac{x^2}2 + \frac{x^3}{3!} + \frac{x^4}{4!} + \cdots + \frac{x^k}{k!} + \cdots $$ and seeing what you get.

Do I differentiate $e^x$ and compare it to the series of $e^x$? Stuck here.

Mohsen Shahriari
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dodgevipert56
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  • Differentiate term by term. – Bernard Aug 17 '20 at 18:30
  • And wouldn't that produce 1 + x + x^2/2 + x^3/3! + x^4/4! + … + x^k/k! + … again an be equal to E^x? Is that what the question is about? – dodgevipert56 Aug 17 '20 at 18:33
  • @dodgevipert56 Yes, that's right. Note that $e$ is invariably written in lower case, just like we write $\pi$ instead of $\Pi$ for the circle constant. – Théophile Aug 17 '20 at 18:37
  • The solution asked is circular, since to find the series given for $e^x$, we have to know $\frac{d{e^x}}{dx}=e^x$. – MasB Aug 17 '20 at 19:11

2 Answers2

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We have the Maclaurin-Taylor series expansion for the exponential function $$e^x=\sum_{k=0}^\infty \frac{x^k}{k!} = 1 + x{} + \frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\frac{x^5}{5!}+...$$ If we differentiate implicitly, then $$\frac{d(e^x)}{dx} =\frac{d}{dx}\sum_{k=0}^\infty \frac{x^k}{k!} = \frac{d}{dx} \bigg( 1 + x{} + \frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\frac{x^5}{5!}+... \bigg)$$ And we know by the linearity of the differential operator, i.e. $(f(x)+g(x))' = f'(x)+g'(x)$, we get $$\frac{d(e^x)}{dx} =\sum_{k=0}^\infty \frac{d}{dx} \bigg( \frac{x^k}{k!} \bigg) = \sum_{k=0}^\infty \frac{1}{k!} \frac{d}{dx} \bigg( x^k \bigg)$$ applying the constant rule for derivatives, i.e. $(cf(x))=cf'(x)$ and $(c)'=0$ ($c$ constant). Furthermore, knowing the power rule, $(x^n)' = nx^{n-1}$ and using the recursivity of the factorial, i.e. $k! = k(k-1)!$, for $k > 0$, $$\frac{d(e^x)}{dx} = \sum_{k=0}^\infty \frac{1}{k!} kx^{k-1} = \frac{1}{0!} 0 \cdot x^{0-1} + \sum_{k=1}^\infty \frac{1}{k!} kx^{k-1}$$ So $$\frac{d(e^x)}{dx}=\sum_{k=1}^\infty \frac{1}{k(k-1)!} kx^{k-1} = \sum_{k=1}^\infty \frac{1}{(k-1)!} x^{k-1}$$ We will now proceed a change of index $k-1=k'$, $$\frac{d(e^x)}{dx}=\sum_{k'=0}^\infty \frac{1}{k'!} x^{k'}=\sum_{k=0}^\infty \frac{x^{k}}{k!} $$ (since the terms of the sum are infinite)

Which is exactly $e^x$. Thus, $$\frac{d}{dx} \bigg( e^x \bigg) = e^x$$

Darius Chitu
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The power series of $e^x$ is given by $$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots = \sum_{k = 0}^{\infty} \frac{x^{k}}{k!}$$

Differentiating term-by-term, we see $\frac{d}{dx} \frac{x^k}{k!} = k \cdot \frac{x^{k - 1}}{k!} = \frac{x^{k - 1}}{(k - 1)!}$.

In other words, each term gets "shifted" down by one. If the sum were finite, this would mean that the last term would vanish. But since the sum is infinite, there is no "last term". Thus, we conclude $\frac{d}{dx} e^x = e^x$.

Raoul
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Ekesh Kumar
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