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

• ggcheck
In summary: So...I don't think I can effectively answer this question unless you could explain why would you think that you can't.
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:
Looks like you missed something.

Edit: you're post was initially blank.

Last edited:
pardon?

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

Dick said:
The Maclaurin series for x^2 IS x^2.
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

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

Mathdope said:
Any polynomial in x would be it's own Maclaurin series, just so you know.
I had no idea... and I read the chapter in my textbook on Taylor/Maclaurin series...

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.

Mathdope said:
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.
I suck a building them

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!

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

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

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.

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

Mathdope said:
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.
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)?

Hurkyl said:
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.
I'm not following you.

ggcheck said:
why can we multiply the Maclaurin series for e^x by x^2?
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.

ggcheck said:
I'm not following you.
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...

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

ggcheck said:
I did not know that the Maclaurin series of x^2 was x^2
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.

ggcheck said:
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)?
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:
ugh how did I get here? I no good with taylor series

Mathdope said:
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) = 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.
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.

thanks!

ggcheck said:
ugh how did I get here? I no good with taylor series
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.

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.

## What is a Maclaurin series?

A Maclaurin series is a type of power series expansion that can represent a function as an infinite sum of powers of x. It is named after the Scottish mathematician Colin Maclaurin.

## How is a Maclaurin series different from a Taylor series?

A Maclaurin series is a special case of a Taylor series, where the center of the series is at x=0. This means that all of the derivatives of the function are evaluated at x=0.

## What is the formula for finding the Maclaurin series of a function?

The formula for finding the Maclaurin series of a function f(x) is given by:
f(x) = f(0) + f'(0)x + (f''(0)/2!)x^2 + (f'''(0)/3!)x^3 + ... + (f^(n)(0)/n!)x^n + ...
where f^(n)(0) represents the nth derivative of f(x) evaluated at x=0.

## What is the Maclaurin series for f(x) = (x^2)(e^x)?

The Maclaurin series for f(x) = (x^2)(e^x) is given by:
f(x) = x^2 + 2x^3 + 3x^4 + 4x^5 + ... + (n+1)x^(n+2) + ...
In other words, the coefficients of the series are (n+1) and the powers of x are increasing by 1 starting from x^2.

## How accurate is the Maclaurin series approximation for f(x) = (x^2)(e^x)?

The accuracy of the Maclaurin series approximation for f(x) = (x^2)(e^x) depends on the value of x. For small values of x, the approximation is very accurate, but as x gets larger, the error in the approximation increases. This is because the series is only an approximation and cannot represent the entire function perfectly. However, by including more terms in the series, we can improve the accuracy of the approximation.

• Calculus and Beyond Homework Help
Replies
6
Views
1K
• Calculus and Beyond Homework Help
Replies
38
Views
2K
• Calculus and Beyond Homework Help
Replies
10
Views
1K
• Calculus and Beyond Homework Help
Replies
6
Views
892
• Calculus and Beyond Homework Help
Replies
1
Views
391
• Calculus and Beyond Homework Help
Replies
7
Views
2K
• Calculus and Beyond Homework Help
Replies
3
Views
381
• Calculus and Beyond Homework Help
Replies
2
Views
1K
• Calculus and Beyond Homework Help
Replies
3
Views
1K
• Calculus and Beyond Homework Help
Replies
16
Views
2K