Series involving trig functions

In summary, the student attempted to solve a part IIB and part IV homework problem, but was having difficulty with parts C and D. They were helped by recognizing the taylor series for e^x and solving for k and a.
  • #1
Noober
20
0

Homework Statement


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

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 [Broken]
http://img485.imageshack.us/img485/6999/dsc07218qf2.jpg [Broken]

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
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.
 
  • #3
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?
 
Last edited:
  • #4
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.
 
  • #5
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...
 
  • #6
Try this approach:

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

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

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

Now you just have to solve [itex]e^x = -2k \cdot e^{-ax/3}[/tex]
 
  • #7
Ok, with that, I got k=-1/2, and a=-3. Right...?
 
  • #8
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!"
 
  • #9
Ok, thanks for the help everyone. Well, actually, just Gib Z, but whatever :).
 

1. What is a series involving trig functions?

A series involving trig functions is a mathematical expression that represents a sum of infinitely many terms, with each term containing a trigonometric function such as sine, cosine, or tangent.

2. How do you determine the convergence of a series involving trig functions?

To determine the convergence of a series involving trig functions, you can use various tests such as the ratio test or the comparison test. These tests analyze the behavior of the terms in the series and determine if the series converges or diverges.

3. Can a series involving trig functions diverge?

Yes, a series involving trig functions can diverge. The convergence of a series depends on the behavior of its terms, and there are series involving trig functions that have terms that do not approach a finite limit, resulting in divergence.

4. What are some common examples of series involving trig functions?

Some common examples of series involving trig functions include the Taylor series for sine, cosine, and tangent, as well as the Maclaurin series for these functions. Other examples include Fourier series and power series with trigonometric terms.

5. How are series involving trig functions used in real-world applications?

Series involving trig functions are used in many real-world applications, such as in physics, engineering, and signal processing. They can be used to model and analyze various phenomena, including oscillations, vibrations, and periodic functions.

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