Question

I'm trying (unsuccessfully) to find the first several coefficients in a Taylor series that looks like cos(x) by using substitution. I'll post my approach below.

Answer 1

So the formula looks like this: \[3 \cos(9x^{2})\]and we know that cos(x) expands to \[1-\frac{ x^{2} }{ 2! }+\frac{ x^{4} }{ 4! }-\frac{ x^{6} }{ 6! }+\frac{ x^{8} }{ 8! }+ ...\] In sigma notation, we get \[\sum_{n=0}^{\infty}(-1)^{n}\frac{ x^{2n} }{ (2n)! }\]It seems as though I should be able to substitute and get \[3\sum_{n=0}^{\infty}(-1)^{n}\frac{ (9x^{2})^{2n} }{ (2n)! }\]Is that right? My first coefficient, c_0 =3. Next, I think my second coefficient would be \[\frac{9^{2}}{2!}\]. Only that's not coming up right. :/

Answer 2

Oh, actually that's (81*3)/2!.

Answer 3

If you already know the expansion for cos(x), then the expansion of cos(9x^2) can be obtained by substituting all x (in the first expansion) by 9x^2, giving:
\(\huge 1-\frac{(9x^2)^2}{2!}+\frac{(9x^2)^4}{4!}-\frac{(9x^2)^6}{6!}+...\)
Finally multiply the expansion throughout by the "external" factor 3.

Answer 4

So that gives me \[c_{0}=3\]\[c_{2}=\frac{9^2*3}{2!}=\frac{243}{2!}\] and so on. I tried submitting that answer with no luck. It's saying that the function is given as \[\sum_{n=0}^{\infty}c_{n}x^{n}\]

Answer 5

I think WeBWorK is going to be my undoing. :/

Answer 6

I do not know what the "3" in your expressions mean. If x=3, then x^2=9!

Answer 7

I'm leaving the x out for simplicity in writing, since they're just asking for \[c_{n}\] where n=0,2,4,6 and 8.

Answer 8

It's the 3 that I multiply each expression by after I've evaluated the sum at n.

Answer 9

The usual notation for c2 is for x^2, c4 for x^4, etc.
So the new expansion has non-zero coefficients c0, c4, c8, etc because the powers of x are x^0, x^4, x^8, etc.

Answer 10

Maybe this will help. Here's a shot of what I've been trying to do:

Answer 11

Ok, just got the answer back from my prof, and now your answer makes more sense. The coefficient for x^2 is zero (which is why we start at x^4.) I've been thinking too much like a programmer.

Answer 12

Thanks for your help!

Answer 13

You're welcome! :)
and
Welcome on board (the boat of programmers)!

Answer 14

Thanks. I've actually been on that boat since the mid 90s. I'm self-taught, though, so I didn't have any calculus requirements when I started out.