Question

1. Show that the function $f(x) = 1 - \sqrt{1 - x}$ has the following Taylor expansion around $x = 0$: $$ f(x) = \sum_{n=1}^{\infty} \frac{\binom{2n}{n} x^n}{4^n (2n - 1)} $$

Name: 1 show that the function fx 1 sqrt1 x has the following taylor expansion around x 0 fx sumn1infty fracbinom2nn xn4n 2n 1 72549
Uploaded: 2021-07-28T22:17:08-08:00
Duration: 3 min 57 s
Channel: Lucas Finney
Description: 1 show that the function fx 1 sqrt1 x has the following taylor expansion around x 0 fx sumn1infty fracbinom2nn xn4n 2n 1 72549

          1. Show that the function $f(x) = 1 - \sqrt{1 - x}$ has the following Taylor expansion around $x = 0$:
$$ f(x) = \sum_{n=1}^{\infty} \frac{\binom{2n}{n} x^n}{4^n (2n - 1)} $$

$1 show that the function fx 1 sqrt1 x has the following taylor expansion around x 0 fx sumn1infty fracbinom2nn xn4n 2n 1 72549$

Added by Alexander T.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Lucas Finney

07/28/2021

Step 1

Step 1: We use the generalized binomial theorem, which states that for any real number $\alpha$ and $|x| < 1$, $$ (1 + x)^\alpha = \sum_{n=0}^\infty \binom{\alpha}{n} x^n $$ where $\binom{\alpha}{n} = \frac{\alpha(\alpha - 1) \cdots (\alpha - n + 1)}{n!}$ for $n Show more…

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1. Show that the function $f(x) = 1 - \sqrt{1 - x}$ has the following Taylor expansion around $x = 0$: $$ f(x) = \sum_{n=1}^{\infty} \frac{\binom{2n}{n} x^n}{4^n (2n - 1)} $$