Real Or Infinite?

Calculus Level 3

By the Fundamental Theorem of Calculus , we know that $e^x=\int e^x dx$ . With a bit of unconventional calculus, it follows that:

$e^x-\int e^x dx=0$

$(1-\int dx)e^x=0$

$e^x=(\frac{1}{1-\int dx})(0)$

$e^x=(1+(\int) dx+(\int)^2 dx+(\int)^3 dx...)(0)$

$e^x=(1+\int dx+\iint dx+\iiint dx...)(0)$

$e^x=(1+x+\frac{x^2}{2!}+\frac{x^3}{3!}...\frac{x^n}{n!}...)$

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

By substituting $x=1$ into the equation, we obtain $e=1+1+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}...\frac{1}{n!}...$ , which is a remarkable property in itself. However, can we express any positive real number $a$ as an infinite series of this form? If this is true, what is that series?

\sum_{n=1}^\infty \frac{\ln(a)^n}{an!}

No.

\sum_{n=0}^\infty \frac{\ln(a)^n}{n!}

\sum_{n=1}^\infty \frac{\ln(a)^n}{n!}

1 solution

Andrei Li
Jul 24, 2018

Instead of substituting $x$ with $1$ , we set $x=\ln(a)$ , which gives us:

$e^{\ln(a)}=1+\ln(a)+\frac{\ln(a)^2}{2!}+\frac{\ln(a)^3}{3!}...\frac{\ln(a)^n}{n!}...$

$a=\displaystyle \sum_{n=0}^\infty \frac{ln(a)^n}{n!}$

To test this series, we can substitute $a=2$ for the series' first six terms.

$1+\ln(2)+\frac{ln(2)^2}{2!}+\frac{ln(2)^3}{3!}+\frac{ln(2)^4}{4!}+\frac{ln(2)^5}{5!}\approx1.9998292811061389$ ,

which is less than two ten-thousandths less that $2$ . QED !

Nothing wrong with the solution, but the "proof" in the problem statement at least one issue. It's not at all clear how to define the operator $\int\cdot\,dx$ in such a way that the argument works. Keep in mind that the indefinite integral itself is only defined modulo the constant functions.

Brian Moehring - 2 years, 10 months ago

The book where I had learned about this property did not offer a proof, which means that I, like you, have to wonder on how $0\int^{(n)} dx =\int^{(n)} 0 dx$ . Sorry about any issues this may have raised in your understanding of this problem. I think that it has to do with the definition $f'(x)=\frac{d}{dx}f(x)$ , along with the fact that antiderivatives can be regarded as functions. But who knows?

Andrei Li - 2 years, 10 months ago

One way to fix it is to instead define $\left(\int\cdot\, dx\right)f$ as the antiderivative of $f$ such that $\left(\left(\int\cdot\, dx\right)f\right)(0)=0$ , but then $\left(1 - \int\cdot\, dx\right)e^x = e^0 - 0 = 1.$ Inductively, $\left(\int\cdot\, dx\right)^n(1) = \frac{x^n}{n!}$ so that $\begin{aligned} e^x &= \left(1 - \int\cdot\, dx\right)^{-1}(1) \\ &= \left(1 + \left(\int\cdot\, dx\right) + \left(\int\cdot\, dx\right)^2 + \left(\int\cdot\, dx\right)^3 + \cdots + \left(\int\cdot\, dx\right)^n + \cdots\right)(1) \\ &= 1 + x + \frac{x^2}{2} + \frac{x^3}{3!} + \cdots + \frac{x^n}{n!} + \cdots \end{aligned}$

Brian Moehring - 2 years, 10 months ago

@Brian Moehring – Great solution for the gap!

To avoid this problem altogether, we can expand $e^x$ using Taylor summation:

$f(x)=\sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!}(x^n)$

As $\frac{d}{dx}e^x=e^x$ , and $e^0=1$ , we can conclude that the Taylor series approximation for $e^x$ is:

$\sum_{n=0}^\infty \frac{x^n}{n!}=1+1+\frac{x^2}{2!}+\frac{x^3}{3!}+...+\frac{x^n}{n!}...$

Also, I recently found out that in some special cases, operators can be treated in the manner of the question.

Andrei Li - 2 years, 10 months ago

Real Or Infinite?

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