Question

Find the Taylor series for f(x) centered at the given value of a. (Assume that f has a power series expansion. Do not show that the reminder \(\sf R_n \rightarrow 0\)). Find the associated radius of convergence.

\(\sf \Large f(x)=\sqrt{x} , a= 16\)

Answer 1

Here is what I have so far:

Answer 2

how to find the radius of convergence of this series

Answer 3

@zepdrix @SolomonZelman

Answer 4

ok so ignoring the 4 there
I can easily find a pattern to writ this in summation form
\[4+\sum_{n=0}^{\infty} \frac{1}{2^{3+5n}} (-1)^{n} \frac{1}{(n+1)!} (x-16)^{n+1}\]
you may what to check that though

Answer 5

anyways you can apply ratio test

Answer 6

the numerator can't be one though because I tried finding the 5th term and I'll get \(\sf \Large \frac{-5}{2097152}(x-16)^4\)

Answer 7

you are right
i also didn't account for that 3 on top in that 4th term you have there

Answer 8

I don't want to cheat yet and look how wolfram wrote the sum yet

Answer 9

wolfram can do that? O.o

Answer 10

ok let's looks at what they have...
\[\sum_{n=0}^{\infty} 4^{1-2x} (-16+x)^{n}\left(\begin{matrix}\frac{1}{2} \\ 0\end{matrix}\right)\]

Answer 11

oops that last 0 is suppose to be n

Answer 12

\[\sum_{n=0}^{\infty} 4^{1-2x} (-16+x)^{n}\left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right)\]

Answer 13

i'm not sure how I feel about that fraction in the binomial factor there

Answer 14

\[\sum_{n=0}^{\infty} \frac{f^{(n)}(16)}{n!}(x-16)^{n} \\ \text{ ratio test: } \\ |\frac{f^{(n+1)}(16)}{(n+1)!} (x-16)^{n+1} \div \frac{f^{(n)}(16)}{n!}(x-16)^n| \\ =|\frac{f^{(n+1)}(16)}{f^{(n)}(16)} \cdot (x-16) \frac{n!}{(n+1)!}| \\ = |\frac{f^{(n+1)}(16)}{f^{(n)}(16)} (x-16) \frac{1}{n+1}|\]....

Answer 15

\[|x-16|\lim_{n \rightarrow \infty}| \frac{f^{(n+1)}(16)}{f^{(n)}(16)} \frac{1}{n+1}|<1\]

Answer 16

still thinking

Answer 17

\[\frac{\text{ odd product} (n+1)}{\text{ odd product }(n)}=2n+1 \\ f^{(n)}(16)=\frac{\text{ odd product }(n)}{2^{n} \sqrt{16^{2n-1}}} (-1)^{n+1} \\ \frac{f^{(n+1)}(16)}{f^{n}(16)} =- (2n+1)2^{n-(n+1)} 16^{\frac{2n-1}{2}-\frac{2(n+1)-1}{2}} \\ =-(2n+1)2^{-1} 16^{-1} =-\frac{2n+1}{2(16)}=-\frac{2n+1}{32}\]

testing between two consecutive derivatives we already found...

(-1/256)/(1/8)=-1/32
(3/8192)/(-1/256)=-3/32
seems good

Answer 18

\[|x-16| \lim_{n \rightarrow \infty}|-\frac{2n+1}{32} \cdot \frac{1}{n+1}|<1 \\ \frac{1}{32}|x-16| \lim_{n \rightarrow \infty} |\frac{2n+1}{n+1}|<1 \\ 2 \cdot \frac{1}{32} |x-16|<1\]

Answer 19

hope you can continue from there
:)

Answer 20

okay got it thank you so much...
now i just need to understand it xD