(a) Find the linear Taylor polynomial about $0$ for $(1 + x)^{15}$
For the first question I tried to use the formula: $1 + px + p\frac{(p-1)}{2!} x^2$
This didn't work for me as when I substituted $x$ (part b) I only got 4 decimal digits.
(b) Use the linear Taylor polynomial to find the associated remainders when $x = 0.01$ and $x = 0.1.$ Give these associated remainders to six decimal places.
I'm guessing part (b) will be simply putting the $x$ values into the equation and part (c) will be obvious when part (b) is answered.
(c) Compare the remainders obtained in (b). What conclusions can you draw?
The taylor series is given by: $$(1+x)^{15}=1+15x+R(x),$$ where $R(x)$ is the remainder. The Lagrange form of the remainder is given by: $$R_n(x)=\frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1},$$ where $c$ is some number between $a$ and $x$. Here $a=0$ and $n=1$ so we have: $$R_1(x)=\frac{f^{''}(c)}{2!}x^{2}=\frac{105c^2}{2!}x^{2}.$$
Substitute the appropriate values of $x$ and and the error.