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## Homework Statement

The problem asks to use the Lagrange form of the remainder in Taylor's Theorem to prove that the Maclaurin series generated by f(x) = xe

^{x}converges to f. From the actual answer, I'm guessing it wants me to use the Remainder Estimation Theorem to accomplish this.

## Homework Equations

The Lagrange form of the remainder of Taylor's Theorem, R

_{n}(x), where a is the center of the Taylor series in an open interval I, c is some number between a and x in the interval I, f

^{(n+1)}is the (n+1)st derivative of f, and n is a positive integer:

[tex]R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}[/tex]

The other relevant equation, I would guess, is the Remainder Estimation Theorem. It states (word-for-word) that "if there are positive constants M and r such that |f

^{(n+1)}(t)| [itex]\leq[/itex] Mr

^{n+1}for all t between a and x, then the remainder R

_{n}(x) in Taylor's Theorem satisfies the inequality

[tex]|R_n(x)| \leq M \frac{r^{n+1}|x-a|^{n+1}}{(n+1)!}[/tex]

If these conditions hold for every n and all other conditions of Taylor's Theorem are satisfied by f, then the series converges to f(x)."

## The Attempt at a Solution

I got the series by multiplying the series for e

^{x}by x. [tex]\sum_{n=0}^{\infty}\frac{x^{n+1}}{n!}[/tex] In my attempt to prove the convergence of this to f, I used the Remainder Estimation Theorem with r = 1, a = 0, and f

^{(n+1)}(t) = the (n+1)st derivative of te

^{t}. |f

^{(n+1)}(t)| [itex]\leq[/itex] M. So, I just needed to find a bound for the (n+1)st derivative. This is where I got confused. I can't find the bound. On the interval [0,x], the bound would be the (n+1)st derivative evaluated at x (the right end of the interval) because xe

^{x}is a positive increasing function. So, M = xe

^{x}+ (n+1)e

^{x}.

However, M = a positive constant, so it can't be dependent on n. On the interval [x,0], a similar thing occurs except this time the bound would be (n+1) because M = 0e

^{0}+ (n+1)e

^{0}= 0 + (n+1) = (n+1). Once again, this is a shifting bound.

Attempting to continue the proof by plugging this into the inequality of the Remainder Estimation Theorem despite the shifting bounds, I get

[tex]|R_n(x)| \leq \frac {xe^x + (n+1)(e^x)}{(n+1)!}|x|^{n+1} = \frac {xe^x|x|^{n+1} + (n+1)e^x|x|^{n+1}}{(n+1)!} = \frac{xe^x|x|^{n+1}}{(n+1)!} + \frac {(n+1)e^x|x|^{n+1}}{(n+1)!} = \frac {xe^x|x|^{n+1}}{(n+1)!} + \frac {e^x|x|^{n+1}}{n!} [/tex]

The n+1 cancels and so as n [itex]\rightarrow\infty[/itex], the right side of this inequality and thus the remainder goes to zero which proves convergence on [0,x] (the convergence on [x,0] follows from the "bound" found earlier through this same method). Assuming there is nothing wrong with this proof, that is the only way I can see how to prove it.

However, the correct answer gives this. |f

^{(n+1)}(t)| = te

^{t}[itex]\leq[/itex] Mr

^{n+1}. r = 1, so M = te

^{t}. On [x,0], M = 0. On [0,x], M = xe

^{x}. Thus, te

^{t}[itex]\leq[/itex] M, and the Remainder Estimation Theorem is satisfied which proves the convergence of the series to f.

And that is why I'm confused. (1) It claims that M = xe

^{x}on [0,x], (2) equates te

^{t}to |f

^{(n+1)}(t)|, and (3) sets M = 0 on [x,0]. There are other similar problems to prove in which a similar thing happens (I have not attempted these again yet, though). The correct answer does all three of these confusing things for h(x) = sin

^{2}(x) and i(x)=cos

^{2}(x). On a third question, the correct answer sets M = 0 on [x,0] for j(x) = sin(x) - x + [itex]\frac {x^3}{6}[/itex]. However, that last one I can find an actual bound for the (n+1)st derivative fine (it isn't M = 0 though).

I thought that there might be something similar going on with all of these problems and that by understanding the one I explained in detail, I could solve the rest.

I'd appreciate any help here. Is there something obvious I'm missing? Because I noticed that https://www.physicsforums.com/showthread.php?t=292812&highlight=remainder+estimation+theorem" [Broken] (but with no attempt made and definitely not as detailed as me) and was asked in another website without getting an answer.

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