Question

See attached question..I know this is an exhaustive problem, but I need how ever much help I can get.

Answer 1

a) Let's write the functions as:
\(f(x)=1+x+x^2+..+x^n \text{ and } g(x)=\frac{1-x^{n+1}}{1-x}=\frac{(1-x)(1+x+x^2..+x^n)}{1-x} \implies g(x)=f(x).\)

Answer 2

b) To find p(x), lets first find \(\frac{d}{dx}f(x)\):
\(\frac{d}{dx}f(x)=\frac{d}{dx}{(1+x+x^2+..+x^n)}=0+1+2x+3x^2+..+nx^{n-1}\).
\(p(x)=x\frac{d}{dx}f(x)=x(1+2x+3x^2+..+nx^{n-1})=x+2x^2+3x^3+..+nx^{n}\) \(\implies p(x)=\sum_{i=1}^n ix_i\).

Answer 3

You can find \(q(x)\) in a similar manner.

Answer 4

There's a typo, it should be \(p(x)=\sum_{i=1}^n ix^i\).

Answer 5

c)\[\lim_{x \rightarrow 1}p(x)=\lim_{x \rightarrow 1}(x+2x^2+..+nx^n)=1+2+3+..+n=\sum_{i=1}^{n}i\].