Taylor series integration of cosx -1 / x

In summary: So my original answer was wrong?In summary, the student attempted to solve the integration of ((cosX)-1)/x dx using the Taylor Series approach. They split the integral into two parts and obtained a solution that was different from the one in the solutions manual. However, it was pointed out that their approach was invalid as they had made a mistake in integrating the n=0 term of their series. The correct approach would be to write the Maclaurin series of (cos(x)-1) and use that to solve the integral.
  • #1
Feodalherren
605
6

Homework Statement



∫((cosX)-1)/x dx

Homework Equations


Taylor Series


The Attempt at a Solution



My approach was basically to to split the integral into two more manageable parts which gave me

∫(cosX/x)dx - ∫(1/x)dx

The solutions manual did it completely differently and instead changed the index of the sum to start from n=1 instead of n=0 which "ate up" the 1 on top. I guess my question is, did I do it wrong or is my solution correct, only not as pretty?

I got the same sum as they did, except mine starts at 0 and I have an extra natural log.
 
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  • #2
Feodalherren said:

Homework Statement



∫((cosX)-1)/x dx

Homework Equations


Taylor Series

The Attempt at a Solution



My approach was basically to to split the integral into two more manageable parts which gave me

∫(cosX/x)dx - ∫(1/x)dx

The solutions manual did it completely differently and instead changed the index of the sum to start from n=1 instead of n=0 which "ate up" the 1 on top. I guess my question is, did I do it wrong or is my solution correct, only not as pretty?

I got the same sum as they did, except mine starts at 0 and I have an extra natural log.

You should really post what you actually did. The answer shouldn't have any natural logs at all. The Taylor series of cos(x) already has a 1 in it.
 
  • #3
I solved the two integrals: ∫(cosX/x)dx - ∫(1/x)dx.

The first one is exactly the same answer as the book got, except they shifted their index. I solved it by using the infinite sum definition of cosine. The integral of 1/x is ln x, that's where the log came from.
 
  • #4
Feodalherren said:
I solved the two integrals: ∫(cosX/x)dx - ∫(1/x)dx.

The first one is exactly the same answer as the book got, except they shifted their index. I solved it by using the infinite sum definition of cosine. The integral of 1/x is ln x, that's where the log came from.

What did you do with the 1 in the series expansion of cos(x)? What do you think the series expansion of cos(x) is? These are the questions I'm trying to get at.
 
  • #5
Feodalherren said:
I solved the two integrals: ∫(cosX/x)dx - ∫(1/x)dx.

The first one is exactly the same answer as the book got, except they shifted their index. I solved it by using the infinite sum definition of cosine. The integral of 1/x is ln x, that's where the log came from.

The integral of cos(x)/x is non-elementary, but it is expressible in terms of already-defined non-elementary functions. Is that what you did?
 
  • #6
Here's a picture of what I did. The light is kinda iffy but hopefully you'll see what's going on.

photo_2.jpg
 
  • #7
What you did is invalid. You have ##\frac{\cos x}{x} = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n-1}}{(2n)!}##. That's fine, but your integration of that isn't fine. Look at the n=0 term. You integrated that incorrectly.
 
  • #8
Hmm I'm not sure I understand what you mean. What term exactly are you referring to?
 
  • #9
The n=0 term of your sum, of course. You computed ##\int \frac{x^{2n-1}}{(2n!)}\,dx## as ##\frac{x^{2n}}{(2n)(2n!)}##. That is not valid when n=0.
 
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  • #10
Feodalherren, you're making this much harder than it needs to be. Do not split (cos(x) -1)/x into two terms. Write the Maclaurin series expansion of cos(x) - 1, preferably without the summation sign so that you can get the first few terms right.
 
  • #11
Ohh now I got it. Thanks.
 

FAQ: Taylor series integration of cosx -1 / x

What is Taylor series integration?

Taylor series integration is a method used to approximate the integral of a function using its Taylor series. This involves expanding the function into an infinite sum of terms and then integrating each term individually. By truncating the series at a certain point, we can get a good approximation of the integral.

What is the Taylor series for cos(x)?

The Taylor series for cos(x) is 1 - x^2/2! + x^4/4! - x^6/6! + ..., where n! represents the factorial of n.

How do you integrate using the Taylor series for cos(x)?

To integrate using the Taylor series for cos(x), we first need to find the Taylor series for the function we want to integrate. Then, we integrate each term of the series separately. Finally, we plug in the upper and lower limits of integration and subtract the results to get an approximation of the integral.

What is the purpose of dividing by x in the Taylor series integration of cosx -1 / x?

Dividing by x in the Taylor series integration of cosx -1 / x is necessary because the Taylor series for cos(x) is centered at x = 0. By dividing by x, we are essentially shifting the center of the series to the value of x we are integrating around, which allows us to get a more accurate approximation of the integral.

What are the limitations of using Taylor series integration?

One limitation of using Taylor series integration is that it only works for well-behaved, smooth functions. If the function is not smooth or has discontinuities, the Taylor series may not converge and the approximation will not be accurate. Additionally, the higher order terms in the series may become increasingly difficult to calculate, leading to a less accurate result.

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