/*! 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. 70 Let ∑k=0∞akxk be a power... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Let∑k=0∞akxkbeapowerserieswithapositiveandfiniteradiusofconvergenceÒÏ,andletc≠0.Provethattheseries∑k=0∞ak(cx)khasradiusofconvergenceÒÏc.

Short Answer

Expert verified

The radius of convergence for series∑k=0∞ak(cx)kisÒÏc

Step by step solution

01

Step. 1 Finding the value of ρ for the series ∑k=0∞akxk

Ratio Test for absolute convergence for series ∑k=0∞akxk:

limk→∞ak+1xk+1aKxk=limk→∞ak+1xkxaKxk=ak+1aKx............(EvaluatingtheLimits)

As series ∑k=0∞akxkconverges only when,xak+1aK<1

and therefore

x<akak+1

So, the Radius of convergence =ÒÏ
= akak+1.

02

Step. 2 Finding Radius of Convergence for the Series ∑k=0∞ak(cx)k

Ratio Test for absolute convergence for series ∑k=0∞ak(cx)k:

limk→∞ak+1(cx)k+1aK(cx)k=limk→∞ak+1(cx)k(cx)aK(cx)k=ak+1aKcx............(EvaluatingtheLimits)

As series ∑k=0∞ak(cx)kconverges only when,

ak+1aKcx<1x<ak+1aKcx<ÒÏc

and

therefore

Radius of convergence = ÒÏc.

Hence Proved.

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.