# Finding a the value of 30th derivative given power series.

• Coderhk
In summary, the conversation is about determining the coefficient of the 30th derivative in a Taylor series formula about the point x = 3. The formula is ∑(ƒ^(n)(a)(x-a)^n)n!, and the question is asking for the kth derivative of (x-a)^2n at x = a. It is clarified that the derivative is non-zero at x = a, and the correct term to use is (x-a)^30/15!, with a coefficient of 30!/15!.

## Homework Statement

The problem is attached as pic

## Homework Equations

∑(ƒ^(n)(a)(x-a)^n)n! (This is the taylor series formula about point x = 3)

## The Attempt at a Solution

So I realized that we should be looking at either the 30th,31st term of the series to determine the coefficient. After we determine the coefficient we divide it by 30! or 31!. Other than this I'm completely lost. I'm not sure weather this series starts with an index of 0 or 1 so I don't know which term I should be looking at. Perhaps, it may even start at 2 for all I know. Please help Thanks.
The answer is E

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What is the ##k##th derivative of ##(x-a)^{2n}## at ##x = a##?

Orodruin said:
What is the ##k##th derivative of ##(x-a)^{2n}## at ##x = a##?
((2n)!/(2n-k)!)*(x-a)^(2n-k)

So for which ##n## is it non-zero?

Orodruin said:
So for which ##n## is it non-zero?
I'm not sure what you mean but wouldn't it be non-zero as long as n is not zero.

No. You are only interested in it being non-zero at ##x=a##.

Well when x = a, the whole expression would become zero. I'm not quite sure what you are getting at though

Coderhk said:
Well when x = a, the whole expression would become zero. I'm not quite sure what you are getting at though
This is not correct. It is a statement that depends on ##n## and ##k##. Let us take a simpler example of ##(x-a)^2##, i.e., ##n = 1##. What are the zeroth (##k = 0##), first (##k = 1##), second (##k = 2##), and third (##k = 3##) derivatives of this expression at ##x = a##?

Orodruin said:
This is not correct. It is a statement that depends on ##n## and ##k##. Let us take a simpler example of ##(x-a)^2##, i.e., ##n = 1##. What are the zeroth (##k = 0##), first (##k = 1##), second (##k = 2##), and third (##k = 3##) derivatives of this expression at ##x = a##?
The zeroth derivative would be the expression itself being (x-a)^2, 1st being 2(x-a), 2nd being 2,3rd being 0. Meaning if k is greater than equal to 2n+1 it will become zero

Coderhk said:
The zeroth derivative would be the expression itself being (x-a)^2, 1st being 2(x-a), 2nd being 2,3rd being 0. Meaning if k is greater than equal to 2n+1 it will become zero
Again, the question asks you explicitly for ##x = a##.

Orodruin said:
Again, the question asks you explicitly for ##x = a##.
If we set x=a everything cancels out and become 0. I still don't get it.

Coderhk said:
If we set x=a everything cancels out and become 0. I still don't get it.
No. You are incorrect. Go back to your expressions for the various derivatives in #9. Which derivative is non-zero at ##x = a##?

Orodruin said:
No. You are incorrect. Go back to your expressions for the various derivatives in #9. Which derivative is non-zero at ##x = a##?
For all less than equal to 2n

Again, incorrect. Insert ##x = a## in the expressions.

Orodruin said:
Again, incorrect. Insert ##x = a## in the expressions.
When we set x= a everything just cancels out

Coderhk said:
When we set x= a everything just cancels out
No it doesn't. You are simply wrong in this statement. These are your expressions:
$$(x-a)^2, \quad 2(x-a), \quad 2, \quad 0.$$
Which of those expressions is non-zero at ##x = a##? Your claim is that they are all zero!

Orodruin said:
No it doesn't. You are simply wrong in this statement. These are your expressions:
$$(x-a)^2, \quad 2(x-a), \quad 2, \quad 0.$$
Which of those expressions is non-zero at ##x = a##? Your claim is that they are all zero!
In this case it is not zero at the 2nd derivative. So in general terms it is not zero at the 2nth derivative

Coderhk said:
In this case it is not zero at the 2nd derivative. So in general terms it is not zero at the 2nth derivative
Right. So in your expression, which term will lead to a non-zero 30th derivative?

Orodruin said:
Right. So in your expression, which term will lead to a non-zero 30th derivative?
((X-a)^30)/15!

Coderhk said:
((X-a)^30)/15!
And it's 30th derivative will have coeefficent 30!/15!

Coderhk said:
And it's 30th derivative will have coeefficent 30!/15!
I get it now. Thank you so much :)

## 1. What is the purpose of finding the value of the 30th derivative in a power series?

The purpose of finding the value of the 30th derivative in a power series is to understand the behavior of a function at a specific point, particularly as the number of derivatives increases. This can provide insight into the convergence or divergence of the series and help solve problems in calculus and differential equations.

## 2. How is the value of the 30th derivative calculated in a power series?

The value of the 30th derivative in a power series can be calculated using the formula for the nth derivative of a power series, which involves taking the coefficient of the nth term and multiplying it by n!. In this case, the coefficient of the 30th term would be used.

## 3. What is the significance of the 30th derivative compared to other derivatives in a power series?

The significance of the 30th derivative lies in its high degree, which can provide more information about the behavior of a function compared to lower derivatives. It can also help determine the convergence or divergence of the series and identify patterns in the coefficients.

## 4. Can the value of the 30th derivative be used to approximate the value of a function at a specific point?

Yes, the value of the 30th derivative can be used to approximate the value of a function at a specific point. This is because as the number of derivatives increases, the power series becomes a better approximation of the function, particularly near the point of expansion.

## 5. Are there any practical applications for finding the value of the 30th derivative in a power series?

Yes, there are practical applications for finding the value of the 30th derivative in a power series. This includes solving problems in physics, engineering, and other fields that involve differential equations and the use of power series to model functions.