Maclaurin Series: cos(2x)/(1+x^2)

[Punkyc7]

F(x)= (cos(2x)/(1+x^2))





Is there anyway to do this without taking a lot of derivatives and looking for a pattern?

 

[SammyS]

So, I take it that you need to find the Maclaurin Series for the function: f(x) = (cos(2x))/(1+x²) .

How many terms do you need, or do you need to write the result using ∑ notation.

If you know the Maclaurin Series for cos(x) and 1/(1+x²) or 1/(1+x), then No, you don't need to take a of derivatives.

 

[Punkyc7]

The first 3 non zero terms,

I know the Maclaurin series for both of them but when I try doing it like that I don't get the right answer. I have only been able to solve it by taking a bunch of derivatives

 

[SammyS]

What is the series for for each?

Can you multiply (a₁+a₂+a₃+...)·(b₁+b₂+b₃+...) ?

It's: a₁b₁+a₁b₂+a₂b₁+a₁b₃+a₂b₂+a₃b₁...

 

[Punkyc7]

((-1)^n *((2x)^2n)/(2n!) and (-1^n)((x^2)^n) since our x is -x^2 in the 1/(1+x^2)

 

[Punkyc7]

Ok so how would you know when to stop the a1 and then move onto the a2 and so on... and yes your allowed to multiply them

 

[SammyS]

((-1)^n *((2x)^2n)/(2n!) and (-1^n)((x^2)^n) since our x is -x^2 in the 1/(1+x^2)

I suppose you mean that:

$$\cos(2x)=\sum_{n=0}^\infty \ (-1)^n\,\frac{(2x)^{2n}}{(2n)!}$$

and

$$\frac{1}{1+x^2}=\sum_{n=0}^\infty \ (-1)^{n}\,x^{2n}$$

Write out the first three terms for each and multiply. Of course group all the terms having the same power of x together.

$$\cos(2x)=1-2x^2+\frac{2x^4}{3}-\dots$$

$$\frac{1}{1+x^2}=1-\frac{1}{x^2}+\frac{1}{x^4}-\dots$$

Multiply them together.

What term(s) do you multiply, then sum, to get the constant term?

What term(s) do you multiply, then sum, to get the linear (x¹) term? -Right, there is none.

What term(s) do you multiply, then sum, to get the squared (x²) term?
…

 

[Punkyc7]

Thank you for your help it makes sense now

 

[SammyS]

Cool! Thanks for the THANKS! We often don't get that!