Help with Taylor Series/Maclaurin Series Question

[student93]

Homework Statement

Problem is attached in this post.

Homework Equations

Problem is attached in this post.

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?

The answer is D. -10!/5!

 

[SammyS]

Homework Statement

Problem is attached in this post.

Homework Equations

Problem is attached in this post.

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?

The answer is D. -10!/5!

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



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

 

[student93]

\displaystyle f^{(10)}(x) = \frac{d^{10}}{dx^{10}}f(x)\ .
It's not \displaystyle \left(f(x)\right)^{10}

I don't understand what you mean?

 

[SammyS]

I don't understand what you mean?

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

 

[student93]

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

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?

 

[SammyS]

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?

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

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

 

[student93]

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

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

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!

 

[SammyS]

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!

What is the 10th derivative of x¹⁰ ?

 

[student93]

what is the 10th derivative of x¹⁰ ?

3,628,800

 

[SammyS]

3,628,800

Yes, that's 10! . Right.

Even if x = 0.

 

[student93]

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

 

[SammyS]

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

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


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

...

 

[student93]

What is the MacLaurin series for $e^{-x^2} \ ?$The 10^th derivative of of the first 5 terms, those with power, 0, 2, 4, 6, 8, are all zero, right?

...

Yes:

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

 

[SammyS]

Yes:

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

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

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

 

[student93]

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

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

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

 

[SammyS]

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

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