Binomial Series — Question 5

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Question 5

For (1+x)1/3=∑n=0∞cnxn(1+x)^{1/3}=\sum_{n=0}^{\infty}c_nx^n, derive a recurrence for cn+1c_{n+1} from cnc_n. Use it to compute through x4x^4, then state the convergence interval.

Original worksheet page 1: question and worked solution for 4-18-005
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Question 5 – Solution

Step 1: Derive the recurrence.

Since cn=(1/3n)c_n=\binom{1/3}{n}, cn+1cn=(1/3)−nn+1,cn+1=cn1/3−nn+1,c0=1.\frac{c_{n+1}}{c_n}=\frac{(1/3)-n}{n+1},\qquad \boxed{c_{n+1}=c_n\frac{1/3-n}{n+1}},\quad c_0=1.

Step 2: Generate coefficients.

c1=13,c2=−19,c3=581,c4=−10243.c_1=\frac13,\qquad c_2=-\frac19,\qquad c_3=\frac5{81},\qquad c_4=-\frac{10}{243}. Therefore (1+x)1/3=1+x3−x29+5x381−10x4243+⋯.\boxed{(1+x)^{1/3}=1+\frac{x}{3}-\frac{x^2}{9}+\frac{5x^3}{81}-\frac{10x^4}{243}+\cdots}.

Step 3: State convergence.

The radius is 11. Since the coefficient magnitudes are order n−4/3n^{-4/3}, both endpoint series converge absolutely.

Conclusion.

The interval is [−1,1]\boxed{[-1,1]}.

Original worksheet page 2: question and worked solution for 4-18-005

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