Finding the Coefficients of a Taylor Polynomial: A Tricky Integration Question

In summary, the conversation is about finding the solution to a problem involving the Maclaurin series for e^-4x and sin(5x). The solution is found by multiplying the two series together like polynomials and ignoring powers greater than 3. The conversation also includes a discussion of the number of terms needed for each series to make the final solution a 3rd degree polynomial.
  • #1
izelkay
115
3

Homework Statement


Here's a screenshot of the problem: http://puu.sh/2Bta5


Homework Equations





The Attempt at a Solution


As can be seen by the screenshot, the answer's already given, but I'm not sure how to go about getting it. This one has me stumped since e-4x and sin(5x) are both Maclaurin series that I can easily find. But I'm not sure where to go from there, or if I'm even supposed to use both of the Maclaurin series.
 
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  • #2
izelkay said:

Homework Statement


Here's a screenshot of the problem: http://puu.sh/2Bta5


Homework Equations





The Attempt at a Solution


As can be seen by the screenshot, the answer's already given, but I'm not sure how to go about getting it. This one has me stumped since e-4x and sin(5x) are both Maclaurin series that I can easily find. But I'm not sure where to go from there, or if I'm even supposed to use both of the Maclaurin series.
It looks like you are supposed to write a few terms of the Maclaurin series for e-4x in one area and a few terms for the series for sin(5x) in the other area.
 
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  • #3
izelkay said:

Homework Statement


Here's a screenshot of the problem: http://puu.sh/2Bta5


Homework Equations





The Attempt at a Solution


As can be seen by the screenshot, the answer's already given, but I'm not sure how to go about getting it. This one has me stumped since e-4x and sin(5x) are both Maclaurin series that I can easily find. But I'm not sure where to go from there, or if I'm even supposed to use both of the Maclaurin series.

Using both the Maclaurin series is probably the easiest way to do it. Just multiply them together. Ignore products that will give you powers greater than 3.
 
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  • #4
Dick said:
Using both the Maclaurin series is probably the easiest way to do it. Just multiply them together. Ignore products that will give you powers greater than 3.

Ok, that's what I thought I should do but wasn't sure if that was allowed or not.
Is this correct so far? :

http://puu.sh/2BySD

Mark44 said:
It looks like you are supposed to write a few terms of the Maclaurin series for e-4x in one area and a few terms for the series for sin(5x) in the other area.

I'll try this too, thanks.
 
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  • #5
izelkay said:
Ok, that's what I thought I should do but wasn't sure if that was allowed or not.
Is this correct so far? :

http://puu.sh/2BySD



I'll try this too, thanks.

No, not correct. You don't multiply series term by term. You multiply them like polynomials. Write out each series up to power x^3 and then foil them. Like polynomials. Drop powers bigger than 3.
 
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  • #6
Dick said:
No, not correct. You don't multiply series term by term. You multiply them like polynomials. Write out each series up to power x^3 and then foil them. Like polynomials. Drop powers bigger than 3.

Ah, ok. I think I understand now.
 
  • #7
Never mind, still not getting it. Looking at these series for e^-4x and sin(5x), wouldn't that mean I need 4 terms (n=0 to n=3) for e^-4x to make that 3rd degree, and 2 terms (n=0 to n=1) for sin(5x) to make that 3rd degree?

So for e^-4x I have:
1 - 4x + 8x - (64/6)x3

For sin(5x):
5x + (125/6)x3

FOILing those out and ignoring powers greater than 3, I get:

5x + (125/6)x3 - 20x2 + 40x3

Is there something else I'm doing wrong?
 
  • #8
izelkay said:
For sin(5x):
5x + (125/6)x3

For sin(5x):

5x - (125/6)x3

Not plus.
 
  • #9
Ok, never mind, found my error. That + (125/6)x3 should actually be a - (125/6)x3
 
  • #10
Dick said:
For sin(5x):

5x - (125/6)x3

Not plus.

Just caught it. Thanks for your help, much appreciated.
 

FAQ: Finding the Coefficients of a Taylor Polynomial: A Tricky Integration Question

What is a Taylor Polynomial?

A Taylor Polynomial is a mathematical function used to approximate a more complex function at a specific point. It is a sum of terms that are calculated using the function's derivatives evaluated at that point.

How is a Taylor Polynomial calculated?

A Taylor Polynomial is calculated using a series of derivatives of the function evaluated at the specific point. The more terms included in the polynomial, the more accurate the approximation will be.

What is the purpose of using a Taylor Polynomial?

The main purpose of using a Taylor Polynomial is to approximate a more complex function at a specific point. This allows for simpler calculations and analysis of the function at that point.

What is the difference between a Taylor Polynomial and a Maclaurin Polynomial?

A Taylor Polynomial is centered around a specific point, while a Maclaurin Polynomial is centered around the point x=0. This means that a Maclaurin Polynomial is a special case of a Taylor Polynomial.

Can a Taylor Polynomial be used to find the value of a function at a point outside the interval of convergence?

No, a Taylor Polynomial can only be used to approximate a function within its interval of convergence. If a point is outside this interval, the approximation will not be accurate.

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