# Thread: Binomial Expansion

1. ## Binomial Expansion

Good Day,

I've been faced with a funny binomial expansion. Attached it with the answer from the book for your reference.

My question is, how do we get to the answer if I expanded (1-2x)^-2 first and then multiplied it with cos x?

2. Originally Posted by dd86
My question is, how do we get to the answer if I expanded (1-2x)^-2 first and then multiplied it with cos x?
Yes,

$(1-2x)^{-2}\cos x=(1+2x+4x^2+8x^3+\ldots)^2(1-x^2/2!+\ldots)=$

$\ldots=1+4x+23x^2/2+30x^3+\ldots$

Fernando Revilla

3. Originally Posted by dd86
Good Day,

I've been faced with a funny binomial expansion. Attached it with the answer from the book for your reference.

My question is, how do we get to the answer if I expanded (1-2x)^-2 first and then multiplied it with cos x?
Your approach is fully correct!... the series expansion of the term $\displaystyle \frac{1}{(1-2 x)^{2}}$ is obtained with the following steps...

$\displaystyle \frac{1}{(1-2 x)^{2}} = \frac{1}{2}\ \frac{d}{d x} \frac{1}{1-2 x} = \frac{1}{2}\ \frac{d}{d x} (1 + 2 x + 4 x^{2} + 8 x^{3} + 16 x^{4} + ...)=$

$\displaystyle = \frac{1}{2}\ (2 + 8 x + 24 x^{2} + 64 x^{3} + ...)= 1 + 4 x + 12 x^{2} + 32 x^{3} + ...$ (1)

Merry Christmas from Italy

$\chi$ $\sigma$

4. Wow OK. I did not come across this in my self-study so far...

So what this means is that when I multiply a binomial expansion with cos x, I have to differentiate the binomial expansion?

What if it was cos 2x or cos3x or cos -x? Do I differentiate that many times for positive integers and integrate that many times for negative ones? What if it was cos ax if a is not an integer?

What about sin ax and tan ax?

It would be great if anyone could just give me the link to read up on this. I'm still a novice so something that's easy to understand would be preferable. But I will post questions if I don't understand

5. I'd also like to know whether I can read up on how to find the rth term in a binomial expansion? I have found myself to be weak in this part for many of the questions I have come across.

Is this covered in a different topic in math?

6. The series expansion of $(1-2x)^{-2}$ can also be found this way:

$\displaystyle \frac{1}{(1-2x)^2} = \frac{1}{1-[1-(1-2x)^2]} = \frac{1}{1-t}$

$\displaystyle \sum_{k \ge 0}t^k = \sum_{k \ge 0}\left[1-(1-2x)^2\right]^k = \sum_{k\ge 0}4^kx^k(1-x)^k.$

@ dd86:
Spoiler:

Originally Posted by dd86
Wow OK. I did not come across this in my self-study so far...
Geometric/Taylor/Maclaurin series, that's what you need for this problem.
So what this means is that when I multiply a binomial expansion with cos x, I have to differentiate the binomial expansion?
No, you replace $\cos{x}$ with its Maclaurin expansion, which is:

$\displaystyle \cos x = \sum^{\infin}_{n=0} \frac{(-1)^n}{(2n)!} x^{2n} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots$

What if it was cos 2x or cos3x or cos -x?
Then you change each of those with their equivalent Maclaurin expansions.
What about sin ax and tan ax?
Same for those too.
It would be great if anyone could just give me the link to read up on this. I'm still a novice so something that's easy to understand would be preferable. But I will post questions if I don't understand
Probably wikipedia can't be bettered in this case.

I'd also like to know whether I can read up on how to find the rth term in a binomial expansion? I have found myself to be weak in this part for many of the questions I have come across.
You only need the factorial definition of the coefficients and the binomial theorem to do that. Do a bit of search to find further explanations.

7. Thank you, The CoffeMachine. Your tip with finding the rth term was very helpful. Had alotta fun in solving the questions I had left out before.

8. Originally Posted by dd86
Wow OK. I did not come across this in my self-study so far...

So what this means is that when I multiply a binomial expansion with cos x, I have to differentiate the binomial expansion?
No, he didn't say that. Chisigma did not use the binomial expansion, he used the derivatives to find the Taylor's series for $(1- 2x^2)^{-2}$. Of course, "a power series is a power series" so both methods give the same result.

What if it was cos 2x or cos3x or cos -x? Do I differentiate that many times for positive integers and integrate that many times for negative ones? What if it was cos ax if a is not an integer?

What about sin ax and tan ax?

It would be great if anyone could just give me the link to read up on this. I'm still a novice so something that's easy to understand would be preferable. But I will post questions if I don't understand