Help with Taylor Series/Maclaurin Series Question

In summary: 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!##In summary, the initial problem involves finding the
  • #1
student93
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0

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!
 

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  • #2
student93 said:

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!
attachment.php?attachmentid=69190&d=1398730870.png


[itex]\displaystyle f^{(10)}(x) = \frac{d^{10}}{dx^{10}}f(x)\ .[/itex]



It's not [itex]\displaystyle \left(f(x)\right)^{10}[/itex]
 
  • #3
SammyS said:
attachment.php?attachmentid=69190&d=1398730870.png


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

I don't understand what you mean?
 
  • #4
student93 said:
I don't understand what you mean?

It's the tenth derivative ... evaluated at x = 0 .
 
  • #5
SammyS said:
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?
 
  • #6
student93 said:
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?
 
  • #7
SammyS said:
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!
 
  • #8
student93 said:
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 ?
 
  • #9
sammys said:
what is the 10th derivative of x10 ?

3,628,800
 
  • #10
student93 said:
3,628,800
Yes, that's 10! . Right.

Even if x = 0.
 
  • #11
So do I take the derivative of -x^10/5!?
 
  • #12
student93 said:
So do I take the derivative of -x^10/5!?

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
SammyS said:
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?

...

Yes:

1 - x^2 +x^4/2! - x^6/3! +x^8/4! - x^10/5! ... etc.
 
  • #14
student93 said:
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 ?
 
  • #15
SammyS said:
##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.)
 
  • #16
student93 said:
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!##
 

1. What is a Taylor series/Maclaurin series?

A Taylor series/Maclaurin series is a representation of a function as an infinite sum of terms, where each term is a polynomial of increasing degrees. It is used to approximate a function at a specified point, by using the derivatives of the function at that point.

2. Why are Taylor series/Maclaurin series important?

Taylor series/Maclaurin series are important because they allow us to approximate complicated functions with simpler polynomials, making calculations easier. They also allow us to evaluate functions at points where the function is not defined, by using the Taylor series/Maclaurin series at nearby points.

3. How do you find the Taylor series/Maclaurin series of a function?

The Taylor series/Maclaurin series of a function can be found by using the derivatives of the function at a specified point. The general formula for the Taylor series is: f(x) = f(a) + f'(a)(x-a) + f''(a)(x-a)^2/2! + f'''(a)(x-a)^3/3! + ..., where a is the point at which the Taylor series is being evaluated.

4. What is the difference between a Taylor series and a Maclaurin series?

A Maclaurin series is a special case of a Taylor series, where the point at which the series is being evaluated is a = 0. In other words, a Maclaurin series is a Taylor series centered at x = 0. This results in simpler calculations, as many of the derivatives evaluated at x = 0 will be equal to 0.

5. How accurate is a Taylor series/Maclaurin series approximation?

The accuracy of a Taylor series/Maclaurin series approximation depends on the number of terms used in the series. The more terms included, the more accurate the approximation will be. However, for some functions, the Taylor series/Maclaurin series may not converge for all values of x, leading to a less accurate approximation.

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