Differentiation of a Taylor series

In summary: If so, use parentheses, like this: f'(x) = -\frac{3}{2} \left(\frac{1}{4}+x\right)^{-5/2} \neq... (-\frac{3}{2} \left(\frac{1}{4}+x\right)^{-5/2}eq -\frac{3}{2} \left(\frac{1}{4}+x\right)^{-4/2}
  • #1
cosmictide
37
0

Homework Statement



Hi guys, any help on this question would be hugely appreciated.

The Taylor series about 0 for the function f(x)=(1/4+x)-3/2 is

f(x)=8 - 48x + 240x^2 - 1120x^3 + ...

used differentiation to find the Taylor series about 0 for the function g(x)=(1/4+x)-5/2


The Attempt at a Solution



I tired differentiating each term to obtain 0-48+480x-3360x^2 but that can't be right since the correct answer seems to be g(x)=(1/4+x)^-5/2 to be 32-320x+2240x^2...

I'd be greatful if someone could shed some light on how I go about approaching this.

Thanks in advance. :smile:
 
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  • #2
Is the derivative of (1/4 + x)-3/2 equal to (1/4 + x)-5/2?

I also suggest using etc or LaTeX for readability.
 
  • #3
Orodruin said:
Is the derivative of (1/4 + x)-3/2 equal to (1/4 + x)-5/2?

I also suggest using etc or LaTeX for readability.


Thanks for your reply. :smile:

The derivative of (1/4 + x)-3/2 is -48/(4x+1)5/2 so no they're not equal. I have no idea how to put that right though! Any help would be hugely appreciated.
 
  • #4
Where did the 48 come from? It is correct that it should be a factor, but not 48 ...

Once you get the factor right and have done the differentiation of the Taylor series, you will have a factor times the expression you want on one side and a Taylor series on the other. How would you then solve for the expression you want?
 
  • #5
Orodruin said:
Where did the 48 come from? It is correct that it should be a factor, but not 48 ...

Once you get the factor right and have done the differentiation of the Taylor series, you will have a factor times the expression you want on one side and a Taylor series on the other. How would you then solve for the expression you want?

Okay this time using the chain rule I calculated the derivative to be 3 / 2(1/4+x)5/2...is that any better?

I think once I get the factor right I just use it on the left-hand side to make the equation equal but not sure what it is yet. :cry:
 
  • #6
cosmictide said:
Okay this time using the chain rule I calculated the derivative to be 3 / 2(1/4+x)5/2...is that any better?

I think once I get the factor right I just use it on the left-hand side to make the equation equal but not sure what it is yet. :cry:

Almost correct. The overall sign is not correct.

Yes, set this equal to the derivative of the expansion given at the beginning and then isolate [itex] (1/4+x)^{-5/2} [/itex]
 
  • #7
nrqed said:
Almost correct. The overall sign is not correct.

Yes, set this equal to the derivative of the expansion given at the beginning and then isolate [itex] (1/4+x)^{-5/2} [/itex]

Thanks for the reply. :smile:

So I'm left with:

-3 / 2(1/4+x)5/2= -48x + 240x2 - 1120x3...

How do I isolate it? I've tried multiplying and dividing both sides by 3 and 2 but that's clearly wrong! Any help here would be greatly appreciated.
 
  • #8
cosmictide said:
Thanks for the reply. :smile:

So I'm left with:

-3 / 2(1/4+x)5/2= -48x + 240x2 - 1120x3...

How do I isolate it? I've tried multiplying and dividing both sides by 3 and 2 but that's clearly wrong! Any help here would be greatly appreciated.

You obtained the left side by differentiating the left side of the initial expression, right? So you must also differentiate the right side. You have not done that yet. Do this and then isolate[itex] (1/4+x)^{-5/2} [/itex]. That will work.

And you are welcome :-)
 
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  • #9
nrqed said:
You obtained the left side by differentiating the left side of the initial expression, right? So you must also differentiate the right side. You have not done that yet. Do this and then isolate[itex] (1/4+x)^{-5/2} [/itex]. That will work.

And you are welcome :-)

Thank you so much. :smile:
 
  • #10
cosmictide said:
Thank you so much. :smile:
Good work!
You are very welcome.
 
  • #11
cosmictide said:

Homework Statement



Hi guys, any help on this question would be hugely appreciated.

The Taylor series about 0 for the function f(x)=(1/4+x)-3/2 is

f(x)=8 - 48x + 240x^2 - 1120x^3 + ...

used differentiation to find the Taylor series about 0 for the function g(x)=(1/4+x)-5/2


The Attempt at a Solution



I tired differentiating each term to obtain 0-48+480x-3360x^2 but that can't be right since the correct answer seems to be g(x)=(1/4+x)^-5/2 to be 32-320x+2240x^2...

I'd be greatful if someone could shed some light on how I go about approaching this.

Thanks in advance. :smile:

You wrote\[tex] f(x) = \left(\frac{1}{4}+x\right) - \frac{3}{2}[/tex]
Is that what you meant, or did you want
[tex] f(x) = \left(\frac{1}{4}+x\right) ^{-3/2} \, ?[/tex]
If so, use parentheses, like this: f(x) = (1/4+ x)^(-3/2), or better still, ((1/4)+x)^(-3/2) or maybe [(1/4)+x]^(-3/2).

Anyway, assuming you want the second form above, we have
[tex] f'(x) = -\frac{3}{2} \left(\frac{1}{4}+x\right)^{-5/2} \neq \left(\frac{1}{4}+x\right)^{-5/2} [/tex]
 

FAQ: Differentiation of a Taylor series

What is the Taylor series?

The Taylor series is a mathematical representation of a function as an infinite sum of terms, where each term is dependent on the derivatives of the function evaluated at a specific point. It is used to approximate a function by considering its derivatives at a single point.

What is the purpose of differentiating a Taylor series?

The purpose of differentiating a Taylor series is to find the derivatives of the original function at a specific point. This allows for a more accurate approximation of the function and can be useful in solving problems involving optimization, estimation, and numerical analysis.

What is the process for differentiating a Taylor series?

To differentiate a Taylor series, each term in the series is multiplied by its corresponding power of x and the resulting series is then summed. This results in a new series representing the derivative of the original function at the specific point.

What is the relationship between the Taylor series and the derivative of a function?

The Taylor series provides a way to find the derivatives of a function at a specific point by considering the values of the derivatives at that point. In a sense, it is a generalization of the derivative of a function to higher orders.

What is the importance of Taylor series in mathematics and science?

The Taylor series is an important tool in mathematics and science because it allows for the approximation of complicated functions by simpler ones. It is used in various fields such as physics, engineering, and economics to model and solve real-world problems.

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