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?
Yes,
$\displaystyle (1-2x)^{-2}\cos x=(1+2x+4x^2+8x^3+\ldots)^2(1-x^2/2!+\ldots)=$
$\displaystyle \ldots=1+4x+23x^2/2+30x^3+\ldots$
Fernando Revilla
Your approach is fully correct!... the series expansion of the term $\displaystyle \displaystyle \frac{1}{(1-2 x)^{2}}$ is obtained with the following steps...
$\displaystyle \displaystyle \frac{1}{(1-2 x)^{2}} = \frac{1}{2}\ \frac{d}{d x} \frac{1}{1-2 x} = \frac{1}{2}\ \frac{d}{d x} (1 + 2 x + 4 x^{2} + 8 x^{3} + 16 x^{4} + ...)=$
$\displaystyle \displaystyle = \frac{1}{2}\ (2 + 8 x + 24 x^{2} + 64 x^{3} + ...)= 1 + 4 x + 12 x^{2} + 32 x^{3} + ...$ (1)
Merry Christmas from Italy
$\displaystyle \chi$ $\displaystyle \sigma$
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
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?
Will post a sample question about this soon.
The series expansion of $\displaystyle (1-2x)^{-2}$ can also be found this way:
$\displaystyle \displaystyle \frac{1}{(1-2x)^2} = \frac{1}{1-[1-(1-2x)^2]} = \frac{1}{1-t} $
$\displaystyle \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:
No, he didn't say that. Chisigma did not use the binomial expansion, he used the derivatives to find the Taylor's series for $\displaystyle (1- 2x^2)^{-2}$. 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