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Taylor Series

  • Thread starter JG89
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Homework Statement




Find the first four non-vanishing terms of the Taylor series for xcotx in the neighborhood of x = 0.


Homework Equations





The Attempt at a Solution



LaTex isn't working for me so hopefully this doesn't look too messy.

cotx = cosx/sinx. I know what the Taylor series is for cosx and sinx about x = 0.

cosx = 1 - x^2/2! + x^4/4! + ..

sinx = x - x^3/3! + x^5/5! +...

so cotx = (1 - x^2/2! + x^4/4! + ..) / (x - x^3/3! + x^5/5! +...)

and so xcotx = (x - x^3/2! + x^5/4! + ..) / (x - x^3/3! + x^5/5! +...)

And so the first non-vanishing terms are:

1: x / (x - x^3/3! + x^5/5! +...)

2: -x^3/2! / (x - x^3/3! + x^5/5! +...)

and so on?

Is this right?
 

Answers and Replies

  • #2
Dick
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No, those aren't 'terms', those are still quotients. First take your expansion for x*cot(x) and cancel an x in the numerator and the denominator. Now the denominator has the form 1-C, where C is an infinite series. Use 1/(1-C)=1+C+C^2+C^3+... To expand the infinite series, if you look at the form of your product it's a pretty good guess that the first four nonvanishing terms are going to be multiples of 1, x^2, x^4 and x^6. So you only need to keep terms of those powers.
 
  • #3
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I cancelled out an x in the numerator with an x in the denominator and so now I have

xcotx = (1 - x^2/2! + x^4/4! - x^6/6! + ...) / (1 - x^2/3! + x^4/5! - x^6/7! + ...)

The infinite series in denominator resembles the infinite series 1 - x^2 + x^4 - x^6 + ... = 1/(1+x^2). I just can't figure out how to get rid of the coefficients 1/3!, 1/5! and so on
 
  • #4
Dick
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You don't get rid of them. You use them to calculate the first four terms. Just truncate the expansions until you've accounted for all the the terms with exponent less than or equal to six.
 
  • #5
726
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Do you mean like this:

So for example, since I have:

xcotx = (1 - x^2/2! + x^4/4! - x^6/6! + ...) / (1 - x^2/3! + x^4/5! - x^6/7! + ...)

Then the first term will be:

1 / (1 - x^2/3! + x^4/5! - x^6/7! + ...)

= (3!5!7!) / (3!5!7! - 5!7!x^2 + 3!7!x^4 - 3!5!x^6)

When I do this with the other terms, they will all have a similar denominator which I can make into a single denominator and then just collect like terms in the numerator?
 
Last edited:
  • #6
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Wait, that doesn't quite make sense to me.

*sighs* I'm gonna catch some sleep then work on this again tomorrow. Take it easy
 
  • #7
Dick
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Here's a simpler example. (1+x)/(1-x+x^2/2). That's (1+x)/(1-C), where C is x-x^2/2, right? 1/(1-C)=1+C+C^2+C^3+... (geometric series expansion). Suppose I want the first three terms, a+bx+cx^2+dx^3. I've got (1+x)*(1+(x-x^2/2)+(x-x^2/2)^2+(x-x^2/2)^3+.... Now you think, wait, I don't need the C^4 term or anything larger because they don't contribute to a, b, c, d. Then you think, in the C^3 part, the only term that counts is x^3. In C^2, I don't need the (x^2/2)^2 term. Just multiply it out being selective about what terms you keep.
 
  • #8
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Ah, I get what you're saying now.

So for this example, (1+x) / (1 - x + x^2/2) = (1+x)(1/(1-c) where c = x - x^2/2

Now, 1/(1-c) = 1 + c^2 + c^3 + .... I will only take it up to c^3. So we have

(1+x)(1 + c + c^2 + c^3) = 1 + c^2 + c^3 + x + xc + xc^2 + xc^3

After expanding the c^2's and c^3's and collecting like terms only up to the third degree, I end up with

1 + x + (3x^2)/2 + (x^3)/2

Is this right?
 
  • #9
Dick
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Right. Now can you pull the same trick with the original problem?
 
  • #10
726
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So xcotx = (1 - x^2/2! + x^4/4!) / (1 - x^2/3! + x^4/5!) = (1 - x^2/2! + x^4/4!) (1/(1-c))

Where c = x^2/3! - x^4/5!

Now, 1/(1-c) = 1 + c + c^2 + ... I will only take it up to c^2 because I see that after c^3, the lowest exponent on the x's is greater than 6. So I have

(1 - x^2/2! + x^4/4!)(1 + c + c^2)

After multiplying out and collecting like terms up to the sixth degree, I end up with

1 - (x^2)/3 - (x^4)/45 - (x^6)/180

I think this is right because I plugged in 0.1 into xcotx, and into my above expression, and the answers were the same up until the fifth decimal place or something like that
 
  • #11
Dick
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They are ok up to the fifth place because the first problem is with your x^6. Firstly, there is ONE term from c^3 that can contribute. You may have also missed or added a thing or two wrong. But you've got the idea.
 
  • #12
726
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Thanks for the help, this is a great technique :)
 
  • #13
726
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Dick, I have a question.

We used the expansion 1/(1-c) = 1 + c + c^2 + c^3 + ... but this expansion is only valid for |c| < 1, so say for example that c = x - (x^2)/2, then does that mean that the Taylor series is only a good approximation for x satisfying |x - (x^2)/2| < 1?
 
  • #14
Dick
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True. But what you are doing is expanding a taylor series around x=0. You don't have to worry about large values of x. The full taylor series may wind up converging even if |c|>1.
 

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