I was wondering if anyone could help me find a Maclaurin series for the function f(x) = sqrt(1+x^2). I found the first, second, third, fourth, fifth, and sixth derivatives and started writing out some terms of the series. So far I have 1 + [(x^2)/2] - [(x^4)/8] + [(x^6)/16] but I'm having trouble writing it in summation notation. Thanks in advance.
Maclaurin Series for f(x) = sqrt(1 + x^2)
- Thread starter Gladius
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Not all are particularly pretty. Just have a good look at the most simplified form of each non-zero derivative.
It looks tempting as far as you have it, but the next term is \(\displaystyle -\frac{5}{128}x^{8}\), and that likely destroys any theories you had up to that point.
Check out the successive derivitives in this form:
\(\displaystyle f' = \frac{x}{f}\)
\(\displaystyle f'' = \frac{1}{f^{3}}\)
\(\displaystyle f''' = \frac{-3x}{f^{5}}\)
\(\displaystyle f^{[IV]} = \frac{3(4x^{2}-1)}{f^{7}}\)
\(\displaystyle f^{[V]} = \frac{-15x(4x^{2}-3)}{f^{9}}\)
\(\displaystyle f^{[VI]} = \frac{45x(1-12x^{2}+8x^{4})}{f^{11}}\)
There are nice enough patterns for some of it.
It looks tempting as far as you have it, but the next term is \(\displaystyle -\frac{5}{128}x^{8}\), and that likely destroys any theories you had up to that point.
Check out the successive derivitives in this form:
\(\displaystyle f' = \frac{x}{f}\)
\(\displaystyle f'' = \frac{1}{f^{3}}\)
\(\displaystyle f''' = \frac{-3x}{f^{5}}\)
\(\displaystyle f^{[IV]} = \frac{3(4x^{2}-1)}{f^{7}}\)
\(\displaystyle f^{[V]} = \frac{-15x(4x^{2}-3)}{f^{9}}\)
\(\displaystyle f^{[VI]} = \frac{45x(1-12x^{2}+8x^{4})}{f^{11}}\)
There are nice enough patterns for some of it.
Dr. Flim-Flam
Junior Member
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- 108
Use binomial series. f(x) = sqrt(1+x^2) = (1+x^2)^(1/2).
Hence Sum as n goes from 0 to infinity [(1/2)/n]x^2n. Note: [(1/2)/n] is a combination term.
Ergo Sum,(0,Inf) [(1/2)/n]x^2n = 1 +(x^2/2)-(x^4/8)+(x^6/16)-(5x^8/128)+(7x^10/256)-(21x^12/1024) +...+
The binomial series always converges when |x| < 1. and converges at both endpoints if k >= 0. k =1/2.
Interval of convergence = [-1,1].
Hence Sum as n goes from 0 to infinity [(1/2)/n]x^2n. Note: [(1/2)/n] is a combination term.
Ergo Sum,(0,Inf) [(1/2)/n]x^2n = 1 +(x^2/2)-(x^4/8)+(x^6/16)-(5x^8/128)+(7x^10/256)-(21x^12/1024) +...+
The binomial series always converges when |x| < 1. and converges at both endpoints if k >= 0. k =1/2.
Interval of convergence = [-1,1].
