Question

Can someone help me find the sigma notation for this Taylor series and calculate the radius of convergence?

Answer 1

The original function it gives me is \[f(x) = \sqrt{1-x ^{2}}\]I am supposed to find the first four nonzero terms and find the radius of convergence. So far I have the first four nonzero terms as \[f(x) = 1-(1/2)x^2-(1/8)x^4 - (1/16)x^6\] I think I may be able to use the ratio test but I'm not quite sure how to get this into sigma notation.

Answer 2

looks like the denominator is a power of 2

Answer 3

first four non zero terms \[1-\frac{x^2}{2}-\frac{x^4}{2^3}-\frac{x^6}{2^4}\] hmm not that clear, especially since the next one does not fit the pattern

Answer 4

next one is \[-\frac{5x^8}{2^7}\]

Answer 5

Yes the 2^3 confused me. The closest thing I got that almost works is a factorial, but that also off by a bit

Answer 6

i will make a guess that the radius of convergence is the same as the domain of \(\sqrt{1-x^2}\) namely \(|x|<1\)

Answer 7

but checking it without an formula is anyone's guess

Answer 8

these terms are all simplified. Originally they would be \[\frac{ 1 }{ \ 0!} -\frac{ 1 }{ \ 2!} - \frac{ 3 }{ \ 4!} - \frac{ 45 }{ \ 6!}\] i don't know if this helps at all?

Answer 9

oh crap it is one of those binomial ones \[\left(1-x^2\right)^{\frac{1}{2}}\]

Answer 10

Does that make a difference with how the radius of convergence is found?

Answer 11

no, binomial \((1+x)^n\) is good for \(|x|<1\) so you have \(|x^2|<1\) same as \(|x|<1\)

Answer 12

ok so then the radius is just |x| < 1? Is it just a matter of knowing the radius for \[(1+x)^{n} \] ?

Answer 13

sorry, r = |1|