Question

Prove e^x = sum from n=0 to infinity of (x^n)/(n!)

Answer 1

I'm going to start this proof then as my question. My question is basically why we need some constant term c in bounding the Remainder of the Taylor series.

Answer 2

ask*

Answer 3

Let f(x) = e^x, and let \[T _{n} = \sum_{k=0}^{n}\frac{ x^k }{ k! }\]

Answer 4

We can define the Remainder of the Taylor series as \[R _{n} = f(x) - T _{n}\]

Answer 5

In other words, \[f(x) = T _{n} + R _{n}\]

Answer 6

To show that e^x = sum from n=0 to infinity of (x^n)/(n!), it will suffice to show that \[\lim_{n \rightarrow \infty}f(x) = \lim_{n \rightarrow \infty} [T _{n} + R]\]

Answer 7

And
\[\lim_{n \rightarrow \infty}f(x) = f(x) = \lim_{n \rightarrow \infty}T _{n} + \lim_{n \rightarrow \infty}R _{n}\]

Answer 8

Where the limit as n approaches infinity of Tn is the limit of the partial sums of the Taylor series.

Answer 9

If \[\lim_{n \rightarrow \infty}R _{n} = 0\]
Then f(x) = limit as n approaches infinity of Tn, which means e^x = sum from n=0 to infinity of (x^n)/(n!)

Answer 10

So, we will first show that the limit as n approaches infinity of the Remainder of the Taylor series is zero.

Answer 11

By Taylor's Inequality we have
\[\left| R _{n} \right| \le \frac{ e^c \left| x \right|^{n+1} }{ (n + 1)! }\]

Answer 12

Where c is some constant such that 0 < c < x. This is the part where I need help. Why does c have to be a constant between zero and x?

Answer 13

I missed a summation symbol!

Answer 14

The reason why 0 < c < x is because in the general statement of Taylor's remainder theorem, it states that there exists a value c in between x and the point you are centering your Taylor expansion around (in your case you are expanding the series around the point 0).

Answer 15

By Taylor's Inequality,
\[\left| R _{n} \right|=\sum_{k=n+1}^{\infty}\frac{ e^{c}\left| x \right|^{k} }{ k! }\]

Answer 16

So, I think I am struggling to understand (or remember) what it means to expand the series around the point 0.

Answer 17

Okay, I am looking at WolframAlpha and it is talking about Taylor Series and expanding around a point, "a".

Answer 18

Does expanding around a point mean that is the starting point?

Answer 19

Is it a horizontal shift?

Answer 20

Ahh I see, Taylor's remainder theorem states

\[|R_n| = \sum_{k=n+1}^{\infty} \frac{ e^c|x-x_0|^k }{ (n+1)! } = \frac{ e^c}{(n+1)!}(x-x_0)^{n+1}\]

So in our case, x_0 is 0, mainly due to convenience.

Answer 21

I understand how that works mathematically, however, conceptually I am still unsure what "expand around a point means".

Answer 22

@ganeshie8 Do you know what "expand around a point" means conceptually in relation to a Taylor Series Expansion?

Answer 23

http://math.stackexchange.com/questions/43559/whats-the-difference-between-taylor-series-around-different-points

Answer 24

Okay, so the point chosen will affect the radius of convergence for the series.

Answer 25

When you expand the taylor series around an arbitrary point, you are basically forming a better and better approximation to your original function (at that specific location) by higher degree polynomials. And as that higher degree polynomial goes towards an infinite degree (taking n to infinity), your function (at that specific point) actually becomes the taylors series expansion

Answer 26

Okay.

Answer 27

I think I can finish the proof from here.

Answer 28

Cool : )

Answer 29

So,
\[\left| R _{n} \right|\le \sum_{k=n+1}^{\infty}\frac{ e ^{c}\left| x \right|^{k} }{ k! } = e ^{c}\sum_{k=n+1}^{\infty}\frac{ \left| x \right|^{k} }{ k! } = e ^{c}*(0) = 0\]

Answer 30

Therefore,
\[\left| R _{n} \right|\le0 \]
Which implies that the Remainder is equal to zero by the squeeze theorem. Thus f(x) = limit as n approaches infinity of the Partial Sum of the Taylor Series, and so e^x = the sum from n=0 to infinity of (x^n)/n!

Answer 31

It seems obvious but I'd still be wary of the use of (without explicit proof)

\[\lim_{k \rightarrow \infty} \sum_{k=n+1}^{\infty} \frac{|x|^k}{k!} = 0\]

All else looks good. Nice job : )

Answer 32

Thanks for your help domu! Great point, I'll go ahead and prove that to myself.

Answer 33

By the Ratio Test,
\[\frac{ \frac{ \left| x \right|^{k+1} }{ (k+1)! } }{ \frac{ \left| x \right|^{k} }{ (k)! } }=\frac{ \left| x \right|^{k+1} k! }{ \left| x \right|^k (k+1)! }=\frac{ \left| x \right| }{ k+1 } = 0\]

Answer 34

This is true for all real values of x since k is approaching infinity. Thus, by the Ratio Test, this series converges to zero and has an infinite radius of convergence.

Answer 35

So at this point I think I'm done. Feel free to ask me any questions or give me any comments.

Answer 36

define
\[ e = \lim_{n \to \infty } \left( 1 + \frac 1 n\right)^n \]
and expand it nominally show that
\[ \lim_{n\to \infty } \left | \left( 1 + \frac 1 n\right)^n - \sum_{k=1}^\infty \frac 1 {k!} \right |<\epsilon \]