Series involving trig functions

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The discussion revolves around solving a homework problem involving Taylor series and trigonometric functions. Participants clarify the correct approach to derive a Taylor series centered at x=π/3 and address errors in substituting values during calculations. One user initially confuses degrees with radians, leading to incorrect results, but later resolves the issue. Assistance is requested for specific parts of the homework, particularly in applying the Maclaurin series and integrating term by term. The conversation highlights the importance of careful variable substitution and understanding series definitions in solving such problems.
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Homework Statement


http://img87.imageshack.us/img87/6784/dsc07215xt5.jpg
http://img72.imageshack.us/img72/9435/dsc07216aa0.jpg

Homework Equations


None.

The Attempt at a Solution


Answers for part I are circled, here's parts II-IV.
http://img485.imageshack.us/img485/8642/dsc07217uq2.jpg
http://img485.imageshack.us/img485/6999/dsc07218qf2.jpg

Any help would be appreciated. This is a very important homework, as you can tell by the title.

If the pics are too small, download them and zoom in, or use the FF extension ImageZoom.
 
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2) Derive a taylor series centered at x= pi/3 and just keep summing terms. You will know you have 4 places accuracy when after summing a certain number of terms for 4 places are still the same, or also when the terms become smaller than 0.0001.

3) You have the right working and series, but it seems you subbed in x=2 instead of x= 0.2.
 
2. I did. I made an Excel thing to find the sums, and it converged to -0.617209760262333. The actual value of cos(63º) is 0.4539905. I can't figure out what's wrong...
Edit: I was doing it in degrees. I got it now.

3. Ah, thanks.

Can anyone help me with Part II B and Part IV?
 
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IV) Let's do a and b first.

A) Just use the definition of a Maclaurin series, not hard to sub in.

B) Using that series for f(x), change the variable to t, integrate term by term.
 
I did that (see above pics), and it looks right. But part C doesn't work. If I set -2ak=1, and 2a²k/3=1, a comes out to be -1/3 and k=3/2. That doesn't work for the third term...
 
Try this approach:

f(x)=6e^{-x/3}, f'(x)=-2e^{-x/3}

kf'(ax) = -2k \cdot e^{-ax/3}

Now for the LHS, you should recognise that's the taylor series for e^x.

Now you just have to solve e^x = -2k \cdot e^{-ax/3}[/tex]
 
Ok, with that, I got k=-1/2, and a=-3. Right...?
 
Aww come on mate you don't need to be spoon fed like this :P Substituting those into the RHS of what I left you to solve, you can see that it is correct = ] but next time i want something more like:

" I got k=-1/2 and a=-3 which works out, yay!"
 
Ok, thanks for the help everyone. Well, actually, just Gib Z, but whatever :).
 

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