# Homework Help: Find the Maclaurin series for f(x) = (x^2)(e^x)

1. Mar 6, 2008

### ggcheck

Find the Maclaurin series for f(x) = (x^2)(e^x)

the book suggests obtaining the Maclaurin series of f(x) by multiplying the known Maclaurin series for e^x by x^2:

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

= x^2 + x^3 + (x^4)/2! + (x^5)/3! + (x^6)/4! + ... = sum(from n=2, infin) (x^n)/(n-2)!

my question is:

why can we multiply the Maclaurin series for e^x by x^2? wouldn't we have to multiply one Maclaurin series by another, i.e. the Maclaurin series of e^x by the Maclaurin series for x^2?

I hope this questions makes sense. Thanks

Last edited: Mar 6, 2008
2. Mar 6, 2008

### Mathdope

Looks like you missed something.

Edit: you're post was initially blank.

Last edited: Mar 6, 2008
3. Mar 6, 2008

pardon?

4. Mar 6, 2008

### Dick

The Maclaurin series for x^2 IS x^2.

5. Mar 6, 2008

### ggcheck

ugh. I guess I need to study Taylor/Maclaurin series more because I don't understand them very well.

thanks for the help; it would have been nice if my book told me that, though

6. Mar 6, 2008

### Mathdope

Any polynomial in x would be it's own Maclaurin series, just so you know.

7. Mar 6, 2008

### ggcheck

I had no idea... and I read the chapter in my text book on Taylor/Maclaurin series...

8. Mar 6, 2008

### Mathdope

It's pretty clearcut. Just follow the recipe for building a Maclaurin series for a polynomial of the general form, and lo and behold the answer will be the original polynomial.

9. Mar 6, 2008

### ggcheck

I suck a building them

10. Mar 6, 2008

### Mathdope

It only requires finding derivatives and evaluating them at x = 0. When the original function is a polynomial I am hopeful that you find taking a derivative to be a piece of cake!

11. Mar 6, 2008

### ggcheck

but don't you need to notice a pattern to find the equation for the general derivative?

i.e. f^(k)(x) = ?

12. Mar 6, 2008

### Mathdope

Here's an example:

Take the Maclaurin series of f(x) = Ax^3 + Bx^2 + Cx + D (degree 3 polynomial)

f(0) = D
f ' (x) = 3*Ax^2 + 2*Bx + C so f ' (0) = C
f '' (x) = 3*2*Ax + 2*1*B so f ''(0) = 2!B
f ''' (x) = 3*2*1*A so f ''' (0) = 3!A
f ''''(x) = 0 (and all higher derivatives as well are 0)

It's the last bit that makes it easy, the fact that all derivatives that are of order greater than the polynomial degree vanish.

Now if you plug those into the Taylor coefficients out will pop your original polynomial.

13. Mar 6, 2008

### Hurkyl

Staff Emeritus
Why bother finding derivatives? Use facts you know about power series. For example, two convergent power series are equal if and only if they have the same coefficients.

14. Mar 6, 2008

### ggcheck

wow, thank you very much for explaining that to me... I spent at least an hour reading my book and a few websites and I didn't get it until just now (maybe I'm too tired)

would the solution process differ greatly if we were trying to find the Taylor series and not the Maclaurin series (not at a = 0)?

15. Mar 6, 2008

### ggcheck

I'm not following you.

16. Mar 6, 2008

### Hurkyl

Staff Emeritus
I don't think I can effectively answer this question unless you could explain why would you think that you can't.

e^x and x^2 are two functions. Multiplication is defined for any two functions. The expression "e^x" and Maclaurin series for e^x are just two different ways of writing the same function.

17. Mar 6, 2008

### Hurkyl

Staff Emeritus
The Maclaurin series for x^2 is a power series, right?
"x^2" is a power series right?
The expression "x^2" and the Maclaurin series for x^2 both denote the same function, right?
So....

18. Mar 6, 2008

### ggcheck

that part of the question has already been addressed... I did not know that the Maclaurin series of x^2 was x^2

19. Mar 6, 2008

### Hurkyl

Staff Emeritus
This tidbit of knowledge is entirely irrelevant to the problem. You didn't have to know that "x^2" was the Maclaurin series for x^2.

20. Mar 6, 2008

### Mathdope

The process wouldn't differ, i.e. if you wanted to take the Taylor Series of a polynomial about x = a you would evaluate all your derivatives at x = a instead of at x = 0. As an example, if I took the Taylor series of the general 3rd degree polynomial at x = 1 I'd have

f (1) = A + B + C + D
f ' (1) = 3A + 2B + C
f '' (1) = 6A + 2B
f '''(1) = 6A
f ''''(1) = 0

...and then I would plug those in as the Taylor coefficients. The only other bit you would need to include is that now your Taylor series is in powers of (x-1) instead of powers of x.

Last edited: Mar 6, 2008
21. Mar 6, 2008

### ggcheck

ugh how did I get here? I no good with taylor series

22. Mar 6, 2008

### Hurkyl

Staff Emeritus
Of course, you could just rewrite x as (x-1) + 1, and expand the polynomial. Taylor's theorem is probably easier, though... unless you're really good with Pascal's triangle.

23. Mar 6, 2008

### ggcheck

thanks!

24. Mar 6, 2008

### Mathdope

The main point is just that every polynomial is its own Taylor Series, irrespecitve of about what point it is taken. It will always come out to be itself.

25. Mar 6, 2008

### Dick

Sure. If you expand around, say x=1, then the Taylor expansion of x^2 is 1+2(x-1)+(x - 1)^2. And e^x becomes e+e*(x-1)+e*(x-1)^2/2+... Now you have a complicated looking problem. But if you multiply out the expansion of x^2, it's still just x^2. It pays to expand around a point where the expansions become easy.