Question

Taylor series and radius of convergence.

Answer 1

\[f(x)=\sqrt{x},~~~a = 16\]

Answer 2

I'm not sure how to plug it into Taylor series,

\[f(x) = f(a)+f'(a)(x-a)+\frac{ f''(a) }{ 2! }(x-a)^2+...+\frac{ f^n(a) }{ n! }(x-a)^n\]

\[= 4+\frac{ 1 }{ 2 \times 4 }(x-16)-\frac{ 1 }{ 2 \times 2 \times 4^3 \times 2! }(x-16)^2+\frac{ 3 }{ 2 \times 2 \times 2 \times 4^5 \times 3! }(x-16)^3...\]

Answer 3

I don't know what it looks like as a series

Answer 4

\[f(x) = \sqrt{x},~~~a=16 \implies f(16) = \sqrt{16}=4\]
\[f'(x) = \frac{ 1 }{ 2x^{1/2} }~~~f'(16) = \frac{ 1 }{ 2 \times 4 }\]
\[f''(x) = - \frac{ 1 }{ 2 \times 2x^{3/2} }~~~f'''(16) = -\frac{ 1 }{ 2 \times 2 \times 4^3 }\]
\[f ''' (x) = \frac{ 3 }{ 2 \times 2 \times 2x^{5/2} }~~~f'''(16) = \frac{ 3 }{ 2 \times 2 \times 2 \times 4^5 }\]

Answer 5

@amistre64

Answer 6

Not really sure, where to go from here.

Answer 7

theres a pattern that the derivatives take

Answer 8

\[\frac{1}{2^0}x^{1/2},\frac{1.1}{2^1}x^{-1/2},-\frac{1.1.1}{2^2}x^{-3/2},\frac{1.1.1.3}{2^3}x^{-5/2},-\frac{1.1.1.3.5}{2^4}x^{-7/2}\]

Answer 9

\[\frac{ 1 }{ 4^{2n-1} }(x-16)^n\]

this is the pattern right?

Answer 10

x-4

Answer 11

(x-a) so 16 no?

Answer 12

there seems to be some odd combinatorial from the wolf

Answer 13

pfft, thought i read x=4 :)

Answer 14

Ah lol

Answer 15

So would that just be the taylor series, this one is a bit odd

Answer 16

its a bit odd yes, but i dont think the wolf likes your solution

Answer 17

Mhm yeah I don't think it's complete, we need the other numbers

Answer 18

\[\sum_{n=0}^{\infty}\frac{f^{(n)}(a)}{n!}(x-a)^n\]

Answer 19

so the key is in finding a suitable sequence of coefficients that are equal to the derivatives .. not that you dint know that already

Answer 20

Yeah, I see, but I still feel as if I'm missing something haha

Answer 21

x^(1/2)
1/2^1 x^(-1/2)
-1/2^2 x^(-3/2)
3/2^3 x^(-5/2)
-3.5/2^4 x^(-7/2)
-3.5.7/2^4 x^(-7/2)


(-1)^n prod(2n+1)
----------------
2^n 4^((2n+1)/2)


after some point, we get to something like this ... not that ive adjusted it for n or anything

Answer 22

Mhm yeah I think I see what's going on now.

Now for the radius of convergence, I'll just use ratio test..tedious process

Answer 23

Oh wait, so what you did was use binomial coefficients?

Answer 24

Which gives us the other numbers

Answer 25

That would fill in the confusing gap

Answer 26

\[f(x) = \sum_{n=0}^{\infty} \frac{ 1/2! }{ n!(1/2-n)! }\frac{ 1 }{ 4^{2n-1} }(x-16)^n\]

Answer 27

(-1)^n prod(2n+1)
----------------
2^n 4^((2n+1)/2)

when n starts at 2

Answer 28

water main busted in the front yard ..

Answer 29

Haha, well is what I said look right?

Answer 30

i havent tried playing that much if any with fractional binomials so i cant say for sure, but thats the way the wolf writes it

Answer 31

Link please!

Answer 32

http://www.wolframalpha.com/input/?i=taylor+series+sqrt%28x%29

Answer 33

Ah, nice, thanks for all your help amistre :)!

Answer 34

http://www.wolframalpha.com/input/?i=taylor+series+at+x%3D16+for++sqrt%28x%29

for more specific

Answer 35

youre welcome

Answer 36

Sweet, it also converges at radius 16

Answer 37

:) the wolf is pretty smart :)

Answer 38

I'd say so, hopefully it doesn't take over the world one day...yikes, it did that series in seconds while I've been trying to figure it out for hours.

I really DON'T like this section, it's very frustrating.

Answer 39

its just a hard progression, relies on some serious other prenamed stuff good luck

Answer 40

Thanks again amirstre