# Help with Taylor Series/Maclaurin Series Question

1. Apr 28, 2014

### student93

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

Problem is attached in this post.

2. Relevant equations

Problem is attached in this post.

3. The attempt at a solution

I've tried using Maclaurin Series for e^x, and get the term -x^10/5!, however f(0) = 0 which is not the correct answer. Also taking 10 derivatives seems too burdensome of a task etc. Is there any other method to solve this problem?

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2. Apr 28, 2014

### SammyS

Staff Emeritus

$\displaystyle f^{(10)}(x) = \frac{d^{10}}{dx^{10}}f(x)\ .$

It's not $\displaystyle \left(f(x)\right)^{10}$

3. Apr 28, 2014

### student93

I don't understand what you mean?

4. Apr 28, 2014

### SammyS

Staff Emeritus
It's the tenth derivative ... evaluated at x = 0 .

5. Apr 28, 2014

### student93

I know that, but solving for 10 derivatives seems like an impractical way to solve this problem. Is there any other method to solve this problem?

6. Apr 28, 2014

### SammyS

Staff Emeritus
Solving for 10 derivatives is not really that difficult -- using the MacLaurin Series.

What is the 10th derivative of x6, for example?

7. Apr 28, 2014

### student93

10th derivative of x^6 =0, I've tried Maclaurin, and got my 10th powered term as -x^10/5!, but f(0)=0 and the answer is -10!/5!

8. Apr 28, 2014

### SammyS

Staff Emeritus
What is the 10th derivative of x10 ?

9. Apr 28, 2014

### student93

3,628,800

10. Apr 28, 2014

### SammyS

Staff Emeritus
Yes, that's 10! . Right.

Even if x = 0.

11. Apr 28, 2014

### student93

So do I take the derivative of -x^10/5!?

12. Apr 28, 2014

### SammyS

Staff Emeritus
What is the MacLaurin series for $e^{-x^2} \ ?$

The 10th derivative of of the first 5 terms, those with power, 0, 2, 4, 6, 8, are all zero, right?

...

13. Apr 28, 2014

### student93

Yes:

1 - x^2 +x^4/2! - x^6/3! +x^8/4! - x^10/5! ...... etc.

14. Apr 28, 2014

### SammyS

Staff Emeritus
$1 - x^2 +x^4/2! - x^6/3! +x^8/4! - x^{10}/5!+x^{12}/6! \dots$

The 10th derivative of that ... evaluated at x = 0 ?

15. Apr 28, 2014

### student93

I understand it now, thanks for the help. (Since the f^(10)(0)/10! = -1/5! etc.)

16. Apr 28, 2014

### SammyS

Staff Emeritus
Good.

That notation is a bit strange.

If $f(x) = 1 - x^2 +x^4/2! - x^6/3! +x^8/4! - x^{10}/5!+x^{12}/6!\ \dots$

then the 10th derivative is:

$f^{(10)}(x) = - 10!/5!+(12!/2)x^{2}/6!\ \dots$

Then $f^{(10)}(0) = - 10!/5!+(12!/2)0^{2}/6!\ \dots =- 10!/5!$