Question

I have the function sqrt(1+x)
How can I find the taylor polynomial in the point zero?

Answer 1

you can use that binomial stuff, or just grind it til you find it

Answer 2

\[f(0)=1\\
f'(x)=\frac{1}{2}(1+x)^{-\frac{1}{2}},f'(0)=\frac{1}{2}\] is a start

Answer 3

ohh I see so i just create a poynomial that follows the derivatives

Answer 4

\[c_n=\frac{f^{n}(0)}{n!}\] is the formula to use

Answer 5

So that will be like 1 + x/2 -x^2/8...

Answer 6

it is easiest to find a formula if you do not compute any numbers
for example
\[f''(x)=-\frac{1}{2}\times \frac{1}{2}(1+x)^{-\frac{3}{2}}\]
\[f''(0)=-\frac{1}{2^2}\]
\[c_2=\frac{1}{2^2\times 2!}\]

Answer 7

no i screwed up somewhere, that is wrong

Answer 8

\[f^{(3)}(x)=\frac{3}{2}\times \frac{1}{2}\times \frac{1}{2}(1+x)^{-\frac{5}{2}}\] that is better i flipped by mistake

Answer 9

then
\[c_3=\frac{3}{2^3\times 3!}=\frac{1}{2^4}\]

Answer 10

isnt any of them supposed to be negative?

Answer 11

Like the second derivative

Answer 12

yes it alternates for sure

Answer 13

if you know how to do that generalized binomial thing (probably in your text book) you can write it as
\[\sqrt{x+1}=(1+x)^{\frac{1}{2}}=\sum _{n=0}^{\infty } \binom{\frac{1}{2}}{n} x^n\]

Answer 14

ok :) thanks

Answer 15

yw

Answer 16

is the fourth derivative -15/2^4 ?