Radius and Interval of Convergence of MacLaurin Series?

Question

What is the MacLaurin series expansion for f(x)= sinh x?

What is the radius of convergence and interval of convergence?

If you can show me HOW you solve this problem, I would really appreciate it =)

Answer 1

MacLaurin series is the Taylor series of the function about x=0, i.e.



f(x) = Σ f^(n) (0)/n! x^n, where f^(n)(0) is the nth order derivative evaluated at x=0.



To solve this, we need the first few derivatives

f(x) = sinh x => f(0) = 0

f'(x) = cosh x => f'(0) = 1

f''(x) = sinh x => f''(0) = 0

f'''(x) = cosh x => f'''(0) = 1

...

f^(n)(x) = cosh x, if n odd

f^(n)(x) = sinh x, if n even



f^(n)(0) = 1, if n is odd,

f^(n)(0) = 0 otherwise



Substituting in, we get the series to be [I use 2n+1 to represent all the odd numbers]



sinh x = Σ x^(2n+1)/(2n+1)!





To figure out the radius of convergence, first try to ratio test:

limit n->∞ |a_(n+1)/a_n| = limit n->∞ (2n+1)!|x|^2/(2(n+1)+1)! = limit n->∞ |x|^2/[(2n+3)(2n+2)] = 0.

Since the value of the ratio is < 1, the series converges for all x, that is, the radius of convergence is infinity.

Answer 2

Radius Of Convergence Maclaurin Series

Answer 3

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You start off with knowing what the general term is, which we will call a[n]. If lim | a[n + 1] / a[n] | n -> infinity is less than 1, the series converges. Once finding the result of |a[n + 1] / a[n]| as n approaches infinity, create an inequality of it being less than 1. Once you obtain the form |x - a| < c Then your radius of convergence will be c. Your interval of convergence is found by solving for x in that inequality. In the above example, -c < x - a < c, or -c + a < x < c + a So the interval of convergence is (-c + a, c + a) All you have to do is find the maclauren series.