How to Calculate Remainder for Convergent Series in Calculus 2?

[GreenPrint]

Determine how many terms of the convergent series must be summed to be sure that the remainder is less than 10^-4.

Ʃ[n=1,∞] cos(k)/k^(3/2)

I'm not sure how to solve this problem. I'm only aware of remainder formulas for the integral test and for alternating series. I'm not sure that this particular problem pertains to the remainder from the integral test sense we are not given the number of terms and it's what we are solving for and not exactly sure what I'm suppose to do.

 

[Mark44]

Determine how many terms of the convergent series must be summed to be sure that the remainder is less than 10^-4.

Ʃ[n=1,∞] cos(k)/k^(3/2)

The summation index should match the one used in the function being summed.

I'm not sure how to solve this problem. I'm only aware of remainder formulas for the integral test and for alternating series. I'm not sure that this particular problem pertains to the remainder from the integral test sense we are not given the number of terms and it's what we are solving for and not exactly sure what I'm suppose to do.

Your book should have a theorem about the remainder of a Taylor series when you approximate a Taylor series by its first n terms. That's what you need to use.

 

[GreenPrint]

Convergence of Taylor Series

Let f have derivatives of all orders on an open interval I containing a. The Taylor series for f centered at a converges to f for all x in I if and only if lim n->inf R_n(x)=0 for all x in I, where

R_n(x) = ( f(c)^(n+1) (x-a)^(n+1) )/(n+1)!

is the remainder at x (with c between x and a)

I'm not sure how to solve this problem because I don't know what function the series represents. My book has examples in which it finds the remainder term of the summation from k=0 to infinity of x^k/k! and such were the the function is actually known and proves that the interval of convergence is -infinity to positive infinity and such

but none of this stuff were you have to find how many terms are needed to get to a certain range and I'm not sure how to do this type of problem with this type of series.

 

[Mark44]

Scratch that - I was thinking we were dealing with a Taylor series, but obviously we're not. Let me think about it a bit...

 

[Mark44]

I'm stumped. According to Wolframalpha, the series converges to about .210049. I calculated the partial sums up to 300 terms in Excel, and the partial sums still don't seem to be settling down very quickly.

Checking in several of my calculus books, I don't see any error estimate theorems other than the ones you already cited.

 

[GreenPrint]

Ya I'm stumped to.

 

[romsofia]

Have you learned about the Lagrange error bound?

 

[GreenPrint]

Nope I have not.

 

[romsofia]

I believe you can you use it here.

Here is a thread that shows an example on PF.

https://www.physicsforums.com/showthread.php?t=73523

 

[Mark44]

I believe you can you use it here.

Here is a thread that shows an example on PF.

https://www.physicsforums.com/showthread.php?t=73523

As far as I can tell, that won't be helpful here - the series isn't a Taylor series, which is what the thread in the link is working with.

 

[romsofia]

As far as I can tell, that won't be helpful here - the series isn't a Taylor series, which is what the thread in the link is working with.

I thought you could use the error bound formula for power series, but I maybe wrong.

EDIT:

I'm wrong, it's for taylor only. My friend is about to post a technique he knows of.

 

[ChowPuppy]

You could do this just by plugging in numbers like sooo, although this isn't very elegant:
cos(1)/(3/2) = 0.3602
cos(2)/(9/4) = -0.18495
cos(5)/(3/2)^5 = 0.03735
cos(10)/(3/2)^10 = -0.0145
cos(20)/(3/2)^20 = 0.000122
Were getting close..
cos(21)/(3/2)^21 = -0.0001098
cos(22)/(3/2)^22 = 0.000133
cos(23)/(3/2)^23 = -0.0000475
cos(24)/(3/2)^24 = 0.000025

So you need 23 terms before you can be sure.

 

[Mark44]

You could do this just by plugging in numbers like sooo, although this isn't very elegant:
cos(1)/(3/2) = 0.3602
cos(2)/(9/4) = -0.18495
cos(5)/(3/2)^5 = 0.03735
cos(10)/(3/2)^10 = -0.0145
cos(20)/(3/2)^20 = 0.000122
Were getting close..
cos(21)/(3/2)^21 = -0.0001098
cos(22)/(3/2)^22 = 0.000133
cos(23)/(3/2)^23 = -0.0000475
cos(24)/(3/2)^24 = 0.000025

So you need 23 terms before you can be sure.

This is incorrect - the general term in the series is cos(k)/k^(3/2), not cos(k)/(3/2)^(k).

 

[Dick]

|cos(k)/k^(3/2)|<=1/k^(3/2). It's bounded by a convergent p-series. Can't you use the integral test to get an error estimate as well?