Question

Hi everyone! What is the Interval of Convergence for the Taylor Series of f(x) = cosh(x) at c=1 ? Can anyone walk me through this one? Thanks! :o)

Answer 1

Taylor series for f(x):
\[
f(x)=\sum{f^{(n)}(c)\over n!}(x-c)^n\\
{\rm when}\;f(x)=\cosh x\\
f'=\sinh x\\
f'' =\cosh x\\\vdots\\
f^{(n)}(x)=\cases{\sinh x\;\forall\; n\text{=odd}\\\cosh x\;\forall\; n\text{=even}}
\]

Answer 2

are you going to use the ratio test by any chance?

Answer 3

now, for the above Taylor series expansion,
\[
a_n=\frac{f^{(n)}(1)}{n!}(x-1)^n
\]


yes..

Answer 4

did you already try that?

Answer 5

not really...my professor named cosh(1) as g(x) and sinh(1) as g*(x)...whatever a star is supposed to be...a derivative i guess...but he did it so quick, everyone was confused...by all means continue

Answer 6

the star means the complex conjugate

Answer 7

the only complex conjugate i ever used was like with quotients of radicals and stuff back in algebra...i have never seen this in calculus...whew

Answer 8

\[\sinh x=0.5(e^x-e^{-x})\\
\cosh x=0.5(e^x+e^{-x})\]

Answer 9

remember the Euler identity with the sine and cosine with the complex numbers?
well, if the angles are complex numbers then the regular t-rations map to these hyperbolic t-ratios

Answer 10

also,
cosh(1) = a constant.. not a function of "x"

Answer 11

i have never seen Euler yet...you are explaining things that are more advanced than what I have seen so far

Answer 12

no those are pretty much the fundamental concepts from trig and complex numbers.

Answer 13

and that is why we math-cowboys sing the "calc blues"

Answer 14

hmmm...well i don't remember them then, but i honestly don't ever recal anything about EULER yet

Answer 15

so you have the answer for this then?

Answer 16

can i show you what he did, and maybe you can try and explain his process?

Answer 17

yeah...i have the answer...but i don't care about that...i need to understand the process

Answer 18

sure.
shoot it

Answer 19

not a problem. aint gonna be much different that what I had up there

Answer 20

he wrote down the long Taylor expansion for this...which i will not do to save time...then...

Answer 21

I have it up above for reference.

Answer 22

may be it'd be easier if you posted a picture of that work

Answer 23

the final answer was 1-(1/w)<x<1+(1/w)

Answer 24

i don't even understand where the "w" came from!

Answer 25

\[w=g(x)/g*(x)\]

Answer 26

ok...sure would have been nice if he had mentioned that! lol

Answer 27

did you understand my \(a_n\) above?

Answer 28

not sure...i have to go look again...but in the mean time, can you duplicate his process?...i'm gonna go look again...brb

Answer 29

yeah...i understand that...it's a taylor series

Answer 30

thats what I was going to do.
He gave you my next step that you didnt let me get there

Answer 31

it said you were typing...?

Answer 32

well anyway...if i understand correctly...he named g(x)/g^*(x) = w...then it was w times |(x-1)|<1...then he just kept the w in the equation ...i see it

Answer 33

thanks electro...the missing piece you gave me was what "w" was equal to...i can do the rest...thanks

Answer 34

\[
L=\lim_{n\to\infty}\left|a_{n+1}\over a_n\right|\\
L=\lim_{n\to\infty}\left|\frac{f^{(n+1)}(1)}{(n+1)!}(x-1)^{n+1}\over \frac{f^{(n)}(1)}{n!}(x-1)^n\right|\\
L=\lim_{n\to\infty}\left|{f^{(n+1)}(1)\over f^{(n)}(1)}\times{n!\over (n+1)!}\times (x-1)\right|\\
L=\lim_{n\to\infty}\left|A\times {1\over n+1}\times (x-1)\right|=0\\
\]

Answer 35

you are welcome.

Answer 36

peace out