Question

Series and derivative problem?

Answer 1

Let E(x) = \[\sum_{n=0}^{\infty} x^n / n!\] .
Prove that E(x) = e^x for all x. Hint: What can you say about d/dx E(x)/ e^x ?

Answer 2

d/dx E(x)/ e^x is 1...

Answer 3

I mean 0

Answer 4

I don't get it, is it not that if we expand the sum, we get
\[1+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+.....\]
And it is e^x
How can we prove definition of e^x?

Answer 5

Wait I am confused...

Answer 6

??

Answer 7

We could find the power series of e^x?

Answer 8

let \[E(x)=\sum_{n=0}^\infty \frac{x^n}{n!}\]Consider \(E(x)/e^x\); take its derivative using the quotient rule giving: $$\frac{E'(x) e^x-E(x)e^x}{(e^x)^2}=e^{-x}\left(E'(x)-E(x)\right)$$now, consider differentiating term-by-term for \(E'(x)\) gives: \[E'(x)=\sum_{n=0}^\infty \frac{nx^{n-1}}{n!}=\sum_{n=1}^\infty\frac{x^{n-1}}{(n-1)!}=\sum_{n=0}^\infty\frac{x^n}{n!}=E(x)\] hence \(E'(x)-E(x)=0\) and the derivative evaluates to \(e^{-x}\cdot0=0\)

Answer 9

... which suggests \(E(x)/e^x=C\) for some constant \(C\), or in other words \(E(x)=Ce^x\). but since clearly we can evaluate \(E(0)=1\) by inspecting the power series, it's clear that \(C=1\) and thus \(E(x)=e^x\) Q.E.D.

Answer 10

How do you evaluate E(0)?

Answer 11

It diverges so it doesn't work, could we solve for E(1) instead? and how would you find that series?

Answer 12

@misspacgirl14 \(E(0)=\sum\limits_{n=0}^\infty\frac{0^n}{n!}=1+0+0+0+\dots=1\) -- every term except that corresponding to \(x^0\) will evaluate to \(0\)

Answer 13

solving for \(E(1)\) would only be useful if you could show \(E(1)=\sum 1/n!=e\)