Chebyshev Expansion (I understanding the provided solution)

[nacho-man]

Hi,
Please refer to attached image.
I did an assignment earlier, and I got this question wrong.
The solutions have been put up, but I struggle to understand how to proceed
from $1*$ to $2*$ and also how the values for $T_4 = 8x^4-8x^2+1$ and $T_2 = 2x^2-1$ were derived.

I would REALLY appreciate if someone could explain this to me, a lot!

My attempt:

$$(\sqrt{1-x^2} T_n')' + \frac{2T_n}{\sqrt{1-x^2}}$$
$$ -\frac{x}{\sqrt{1-x^2}}T_n' + \sqrt{1-x^2}T_n'' + \frac{2T_n}{\sqrt{1-x^2}}$$

But I am unsure where to proceed from here.
I can see that I have the $\frac{2T_n}{\sqrt{1-x^2}}$ term,
but where does the $-n^2$ come from, and how do we deal with $T_n''$ and $T_n'$ ?

Also for

$3*$ I have no clue where to begin. Although, I tried looking at the summation, I tried subbing $n=2$ for example,

$c_2(2-4)T_2 = x^4 + x$ but this doesn't seem helpful.. I have no clue what they do here.

 

[gtoguy97]

The solution for 1* is as follows:We start by applying the product rule to the left side of the equation:$$(\sqrt{1-x^2} T_n')' + \frac{2T_n}{\sqrt{1-x^2}} = (\sqrt{1-x^2})' T_n' + \sqrt{1-x^2} T_n'' + \frac{2T_n}{\sqrt{1-x^2}}$$Now, we can substitute the expression for $(\sqrt{1-x^2})'$ into the equation: $$-\frac{x}{\sqrt{1-x^2}}T_n' + \sqrt{1-x^2}T_n'' + \frac{2T_n}{\sqrt{1-x^2}} = -\frac{x}{\sqrt{1-x^2}}T_n' + \sqrt{1-x^2}T_n'' + \frac{2T_n}{\sqrt{1-x^2}}$$Finally, we multiply both sides of the equation by $\sqrt{1-x^2}$, and arrive at the following equation:$$-xT_n' + (1-x^2)T_n'' + 2T_n = 0$$For 2* we can solve the equation obtained in 1* for $T_n''$:$$T_n'' = \frac{-xT_n' + 2T_n}{1-x^2}$$Now, we substitute this expression for $T_n''$ into the equation. We then obtain the following equation:$$-xT_n' + \frac{-xT_n' + 2T_n}{1-x^2} + 2T_n = 0$$We can simplify this equation and get an expression for $T_n'$:$$T_n' = \frac{2T_n}{1-x^2}$$For 3*, we can use the expression for $T_n'$ obtained in 2* to calculate the