# Basic Taylor Polynomial Question involving e^(-x)^2

1. Nov 15, 2014

### RJLiberator

1. The problem statement, all variables and given/known data
Consider:

F(x) =$\int_0^x e^{-x^2} \, dx$
Find the Taylor polynomial p3(x) for the function F(x) centered at a = 0.

2. Relevant equations
Tabulated Taylor polynomial value for standard e^x

3. The attempt at a solution

I started out by using the tabulated value for Taylor polynomial e^x and replaced all x with -x^2.
Going to the degree of three I received:

e^(-x)^2 = 1 - (x^2)/1! + x^4/2! - x^6/3!

Is this all there is to it? The question simply states to find the Taylor Polynomial p3(x) for the function, nothing else. I don't need to find the sum or the remainder per the problems question. Correct?

If I were to go at the sum, would I integrate each value?

Thank you for any help, I think I am just getting used to the wording here more than anything.

2. Nov 15, 2014

### gopher_p

The Taylor polynomial $p_3$ of degree $3$ for $F$ centered at $0$ should look something like $$p_3(x)=F(0)+F'(0)x+\frac{F''(0)}{2}x^2+\frac{F'''(0)}{6}x^3,$$ yes? If you remember the Fundamental Theorem of Calculus and what is says about computing $F'(x)$, it shouldn't be too hard to get the coefficients, right? So you can just do the problem "properly" without worrying whether your methods work.

I'm not saying your way won't work, though you'd theoretically only need the degree $2$ Taylor polynomial for $f(x)=e^{-x^2}$ to extract the degree $3$ polynomial for $F$. But the question you'd want to ask is whether the derivative of the degree $n$ Taylor polynomial of a function $g$ really is the degree $n-1$ Taylor polynomial for the function $g'$ and vice -versa; i.e. is the (anti)derivative of the Taylor polynomial the Taylor polynomial of the (anti)derivative? I don't think that question is too incredibly difficult to answer for those who understand what it is asking.

3. Nov 17, 2014

### RJLiberator

Hm. Bare with me. I am in the early process of understanding Taylor Series wordage:
So what you are saying is that I should take the derivative of e^(-x)^2 a few times to acquire the coefficients? If I do this I answer:
f(0) = 1
f'(0) = 1
f''(0) = 2
f'''(0) = 4
Which would mean my original equation is incorrect. I am not sure what you mean by doing the problem properly without worrying.

I guess my first question would be: Is my answer correct? All I did was input -x^2 to the e^x common taylor series polynomial expansion for the first 3 degrees.

Then my second question would be how to proceed with this problem 'properly' ? I will review the fundamental theorem of calculus.

Question 3: Out or curiosity, why would I need to only go to degree 2 for this problem? Doesn't it state to go to degree 3?

4. Nov 17, 2014

### Staff: Mentor

1. Yes, what you did looks fine.
2. Integrate the series you found.
3. If you have terms up to degree 2, when you integrate, what will be the highest degree you get?

5. Nov 17, 2014

### RJLiberator

Mark44, you are shedding light on this topic!

1. Thank you for your confirmation.
2. Hm, ok. I am going to mess around with a few things. I know my coefficients then, now I got to find where to plug them in and integrate.
3. Ah.. integrating would bring it to degree three.

6. Nov 17, 2014

### gopher_p

The way that I was suggesting you do the problem is as follows;

For $F(x)=\int_0^xe^{-t^2}\ \mathrm{d}t$,

$F(0)=\int_0^0e^{-t^2}\ \mathrm{d}t=0$

$F'(x)=e^{-x^2}\Rightarrow F'(0)=1$, where the derivative is given by the FTC.

$F''(x)=-2xe^{-x^2}\Rightarrow F''(0)=0$

$F'''(x)=-2e^{-x^2}+4x^2e^{-x^2}\Rightarrow F'''(0)=-2$

Then the Taylor polynomial of degree three centered at $a=0$ is given by $$p_3(x)=F(0)+F'(0)x+\frac{F''(0)}{2}x^2+\frac{F'''(0)}{6}x^3=x-\frac{1}{3}x^3$$

So we're just (a) appealing to the definition of the Taylor polynomial and (b) computing the coefficients directly. I guess that's what I meant by "properly".

7. Nov 18, 2014

### RJLiberator

OK, so with your help, I've been working through this problem.

Question 1: So for F(0) you made the bounds 0 to 0. I don't understand why. I see that the answer is clearly 0 when the bounds are both 0 to 0, but why do this?

Other then this part, everything you wrote now makes complete sense to me.

Question 2: In my original solution where I just inputted -x^2 to the e^x tabulated value i had NOT integrated yet. Did I HAVE to integrate after I replaced those values? If so, the two answers match up nicely. I would imagine so, by the definition of the Taylor Series, correct?

8. Nov 18, 2014

### gopher_p

One of the "skills" that you learned in you preparation for calculus was that, given a formula for $F(x)$, you compute $F(a)$ by plugging in $a$ for $x$. So given $F(x)=\int_0^xe^{-t^2}\ \mathrm{d}t$, get $F(0)$ by plugging in $0$ for $x$; $F(0)=\int_0^0e^{-t^2}\ \mathrm{d}t$. This is why it is important to use a different variable of integration - $t$ here - inside the integral.

I'm confused by the question. It's clear that the degree two Taylor polynomial $\tilde p_2$ for $f(x)=e^{-x^2}$ is (a) not a degree three polynomial and (b) not a Taylor polynomial for the function $F$, right? So something must be done if we are to extract $p_3$ from $\tilde p_2$.

The real question, in my mind, is whether it is true that $p_3(x)=\int_0^x\tilde p_2(t)\ \mathrm{d}t$. In this case it is. It works here because the lower limit of integration, $0$, is the same as the center of the Taylor polynomials. You need to be careful in a more general setting.

9. Nov 18, 2014

### PeroK

You should note that where you wrote:

F(x) =$\int_0^x e^{-x^2} \, dx$

gopher_p wrote:

F(x) =$\int_0^x e^{-t^2} \, dt$

I think you need to understand the difference and why the second one correctly defines F as a function of x, with t as the dummy variable.

This may also be the reason why you did not understand the substitution $x = 0$

10. Nov 18, 2014

### RJLiberator

Guys, thanks for your help here.

You were a help to me and my classmates.

Also, thanks for pointing out my subtle mistakes that cause errors in conceptual understanding. I am starting to understand this problem due to this.

I see why the dummy variable matters.

As always, excellent help.