/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q. 50 In Exercises 49–56 find the Ta... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

In Exercises 49–56 find the Taylor series for the specified function and the given value of x0. Note: These are the same functions and values as in Exercises 41–48.

50.ex,1

Short Answer

Expert verified

The Taylor series for the functionf(x)=exatx=1isPn(x)=∑k=0∞ek!(x−1)k.

Step by step solution

01

Step 1. Given data

We have the functionf(x)=ex

02

Step 2. Table of the taylor series

Any function fwith a derivative of order n, the taylor series at localid="1649409912679" x=1is given by,

Pn(x)=f(1)+f′(1)(x−1)+f′â¶Ä²(1)2!(x−1)2+f′â¶Ä²(1)3!(x−1)3+f′â¶Ä²â€²â¶Ä²(1)4!(x−1)4+…

we can write the general of the Taylor series of the function fis,

Pn(x)=∑k=0∞fk(x0)k!(x−x0)n

So, let us first construct the table of the Taylor series for the function f(x)=exat x=1.

nfn(x)
fn(1)
fn(1)n!
0ex
e1
e
1ex
e1
e
2ex
e1
e2!
.
.
.
.
.
.
.
.
.
.
.
.
kex
e1
ek!
.
.
.
.
.
.
.
.
.
.
.
.
03

Step 3. Taylor series for the f(x)=ex

The Taylor series for the function f(x)=exat x=1is Pn(x)=e+e(x−1)+e2!(x−1)2+e3!(x−1)3+e4!(x−1)4+…

Or we can write this asPn(x)=∑k=0∞ek!(x−1)k

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Study anywhere. Anytime. Across all devices.