Question

CALCULUS II: I need to find the Maclaurin series of cos^2 x by using the Maclaurin series of cos x and a half angle formula. Any idea how to do that?

Answer 1

@KingGeorge

Answer 2

A half-angle formula?

Answer 3

Yeah, like cos^2 x = (1 + cos 2x)/2

Answer 4

Right. I was just going to suggest version of that.

Answer 5

Anyways, can you show me what the MacLaurin series for \(\cos(x)\) is?

Answer 6

Sure its\[\cos x = 1 - \frac{ x^2 }{ 2!}+\frac{ x^4 }{ 4!} -\frac{ x^6 }{ 6!}+...+(-1)^n\frac{ x^(2n) }{ (2n)!}\]

the x(2n) is supposed to be x^(2n). For some reason it won't go up to the exponent place.

Answer 7

You need to use { and } instead of ( and ) to get the 2n in the exponent. But yes, that's correct.

Now, if you look at your half-angle formula, you'll notice it has a \(\cos(2x)\). To get the series for that, simply plug in \(2x\) for \(x\) in the series you just wrote out.

Answer 8

You should get \[\cos (2x) = 1 - \frac{ (2x)^2 }{ 2!}+\frac{ (2x)^4 }{ 4!} -\frac{ (2x)^6 }{ 6!}+...+(-1)^n\frac{ (2x)^{2n} }{ (2n)!}\]

Answer 9

But how do you deal with the /2 part of the half angle formula?

Answer 10

We're getting there. When dealing with series, it's best to do everything in simple steps.

Answer 11

Okay, then would be next?

Answer 12

Next, we have the \(1+\cos(2x)\) in the numerator. So we add 1 to the series.\[1+\cos (2x) = 2 - \frac{ 2^2x^2 }{ 2!}+\frac{ 2^4x^4 }{ 4!} -\frac{ 2^6x^6 }{ 6!}+...+(-1)^n\frac{2^{2n}x^{2n} }{ (2n)!}\]

Answer 13

Finally, you divide everything by 2.

Answer 14

And the simplest way to write the series after that, is just to take one off of the power on top of the 2's.\[\frac{1+\cos (2x)}{2} = 1 - \frac{ 2x^2 }{ 2!}+\frac{ 2^3x^4 }{ 4!} -\frac{ 2^5x^6 }{ 6!}+...+(-1)^n\frac{2^{2n-1}x^{2n} }{ (2n)!}\]

Answer 15

Oh!!! okay. Thank you so much!!!

Answer 16

You're welcome.