Question

Use the substitution $(b + x)^r = (b + a)^r (1 + \frac{x - a}{b + a})^r$ in the binomial expansion to find the Taylor series of the function below with the center $a = 4$. $\sqrt{5 + x^2}$ Be sure to include parentheses in your answer. Provide your answer below: $\sum_{n=0}^{\infty} \square \binom{\square}{n} (\square)^n$

Name: use the substitution b xr b ar 1 fracx ab ar in the binomial expansion to find the taylor series of the function below with the center a 4 sqrt5 x2 be sure to include parentheses in your ans 75352
Uploaded: 2020-07-10T17:43:53-08:00
Duration: 55 s
Channel: Thane Stiles
Description: use the substitution b xr b ar 1 fracx ab ar in the binomial expansion to find the taylor series of the function below with the center a 4 sqrt5 x2 be sure to include parentheses in your ans 75352

          Use the substitution $(b + x)^r = (b + a)^r (1 + \frac{x - a}{b + a})^r$ in the binomial expansion to find the Taylor series of the function below with the center $a = 4$.
$\sqrt{5 + x^2}$
Be sure to include parentheses in your answer.
Provide your answer below:
$\sum_{n=0}^{\infty} \square \binom{\square}{n} (\square)^n$

$use the substitution b xr b ar 1 fracx ab ar in the binomial expansion to find the taylor series of the function below with the center a 4 sqrt5 x2 be sure to include parentheses in your ans 75352$

Added by Margaret A.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Thane Stiles

07/10/2020

Step 1

The given substitution is $(b + x)^r = (b + a)^r (1 + \frac{x - a}{b + a})^r$. We need to express $f(x) = \sqrt{5 + x^2}$ in the form $(b+x)^r$. First, let's rewrite $f(x)$ to match the form $(b+x)^r$. We have $f(x) = (5 + x^2)^{1/2}$. The given substitution Show more…

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Use the substitution $(b + x)^r = (b + a)^r (1 + \frac{x - a}{b + a})^r$ in the binomial expansion to find the Taylor series of the function below with the center $a = 4$. $\sqrt{5 + x^2}$ Be sure to include parentheses in your answer. Provide your answer below: $\sum_{n=0}^{\infty} \square \binom{\square}{n} (\square)^n$