Question

Consider the function $F(x) = \int_0^x e^{-2t^2} dt$. Express $F(x)$ as a power series in $x$. (hint: $e^x = \sum_{n=1}^\infty \frac{x^n}{n!}$) $\sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{(2n+1)n!}$ $\sum_{n=0}^\infty \frac{(-1)^n 2^n x^{2n+1}}{(2n+1)n!}$ $\sum_{n=0}^\infty \frac{(-1)^n 2^n x^{2n}}{(2n+1)n!}$ $\sum_{n=0}^\infty \frac{(-1)^n x^n}{(2n+1)n!}$

          Consider the function $F(x) = \int_0^x e^{-2t^2} dt$. Express $F(x)$ as a power series in $x$. (hint: $e^x = \sum_{n=1}^\infty \frac{x^n}{n!}$)
$\sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{(2n+1)n!}$
$\sum_{n=0}^\infty \frac{(-1)^n 2^n x^{2n+1}}{(2n+1)n!}$
$\sum_{n=0}^\infty \frac{(-1)^n 2^n x^{2n}}{(2n+1)n!}$
$\sum_{n=0}^\infty \frac{(-1)^n x^n}{(2n+1)n!}$

Consider the function F(x) = ∫0^x e^-2t^2 dt. Express F(x) as a power series in x. (hint: e^x = ∑n=1^∞(x^n)/(n!))
∑n=0^∞((-1)^n x^2n+1)/((2n+1)n!)
∑n=0^∞((-1)^n 2^n x^2n+1)/((2n+1)n!)
∑n=0^∞((-1)^n 2^n x^2n)/((2n+1)n!)
∑n=0^∞((-1)^n x^n)/((2n+1)n!)

Added by Louis A.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Adrian Co

07/28/2021

Step 1

Step 1: Rewrite the given function F(x) as an integral of a power series: F(x) = ∫₀ˣ e^(-2t^2) dt F(x) = ∫₀ˣ ∑_(n=0)^∞ (-2t^2)^n / n! dt F(x) = ∑_(n=0)^∞ ∫₀ˣ (-2t^2)^n / n! dt Show more…

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Consider the function F(x)=int_0^x e^(-2t^(2))dt. Express F(x) as a power series in x. (hint: (:e^(x)=sum_(n=1)^(infty ) (x^(n))/(n!) Consider the function F() e-2t dt. Express F) as a power series in .hint: (-1n2n+1 iM: 2n+1n! -1n2n2n+1 2n+1n! (-1)2nx2n iM: (2n+1n! -1xn 2n+1n!