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

[Coderhk]

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

 

[Orodruin]

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

 

[Coderhk]

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

((2n)!/(2n-k)!)*(x-a)^(2n-k)

 

[Orodruin]

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

 

[Coderhk]

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.

 

[Orodruin]

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

 

[Coderhk]

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

 

[Orodruin]

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 $x = a$?

 

[Coderhk]

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

 

[Orodruin]

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 $x = a$.

 

[Coderhk]

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.

 

[Orodruin]

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 $x = a$?

 

[Coderhk]

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

 

[Orodruin]

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

 

[Coderhk]

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

When we set x= a everything just cancels out

 

[Orodruin]

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!

 

[Coderhk]

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

 

[Orodruin]

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?

 

[Coderhk]

Right. So in your expression, which term will lead to a non-zero 30th derivative?

((X-a)^30)/15!

 

[Coderhk]

((X-a)^30)/15!

And it's 30th derivative will have coeefficent 30!/15!

 

[Coderhk]

And it's 30th derivative will have coeefficent 30!/15!

I get it now. Thank you so much :)