Solutions to Taylor and Power Series Problems - Mat104 Solutions to Taylor and Power Series Problems from Old Exams(1(a This is a 0\/0 form We can use

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Mat104 Solutions to Taylor and Power Series Problems from Old Exams(1) (a). This is a 0/0 form. We can use Taylor series to understand the limit.ex=1 +x+x22!+x33!+x44!+· · ·+xnn!+. . .e-x=1-x+x22!-x33!+x44!· · ·+(-1)nxnn!+. . .Thusex-e-x=2x+2x33!+2x55!+. . .From this we find thatex-e-x-2x=2x33!+ higher degree termsAsxapproaches 0, the lowest power ofxwill dominate because the higher degree termsvanish much more rapidly. We can say thatex+e-x-2x2x33!asx0.Next we consider the denominator.xln(1 +x) =x(x-x2/2 +x3/3-x4/4 +. . .) =x2-x32+x43-x54+. . . .Thus the denominatorx2-xln(1 +x) will be dominated by its lowest degree termx32aswe letx0 and soex+e-x-2xx2-xln(1 +x)2x3/3!x3/2=46=23asx0.(1b) Again we have a 0/0 form.In a similar manner we manipulate Taylor series todetermine what power ofxthe numerator and denominator resemble asxapproaches 0.First recall thatcosx= 1-x2/2! +x4/4!-x6/6! +. . .and sinx=x-x3/3! +x5/5!-x7/7! +. . .Then we can easily compute thatcosx2-1 +x4/2=x8/4! plus higher degree termsx2(x-sinx)2=x8/(3!3!) plus higher degree termsThuscosx2-1 +x4/2x2(x-sinx)2x8/4!x8/(3!3!)=3!3!4!=32asx0(2) Rewritentan(1/n) astan(1/n)1/n. This is a 0/0 form and we can use L’Hˆopital’s Rule toshow that the limit is 1.1
2(3) Use the Taylor series for sinxandexto understand how the numerator behaves nearx= 0.(sinx)(ex2)=(x-x3/3! +x5/5!-. . .)(1 +x2+x4/2 +. . .)=(x+x3-x3/3! + higher degree terms )Sosinx·ex2-x= 5x3/6 + higher degree terms.Now for the denominator.ln(1 +x3) =x3-(x3)2/2 + (x3)3/3- · · ·=x3+ higher degree terms.We conclude that the quotient will go to 5/6 asxgoes to 0.(4) Here we use the Taylor series for cosx.cosx= 1-x2/2! +x4/4!-x6/6! +. . .=1-cosx=x2/2!-x4/4! +x6/6!-. . . .So whenxis close to 0, 1-cosxx2/2!. Whennis large, then 1/nwill be close to 0, so1-cos(1/n)1/2n2. Thusn2(1-cos(1/n))1/2 asngoes to infinity.(5) Here it is useful to combine the fractions1sinx-11-e-x=(1-e-x)-sinx(sinx)(1-e-x)Again we use power series to understand how the numerator and denominator behave nearx= 0.1-e-x-sinx=-x2/2 + higher order terms(sinx)(1-e-x)=(x-x3/3! +x5/5! +. . .)(x-x2/2 +x3/3! +. . .)=x2+ higher order terms.So the quotient will behave like-x2/2x2and go to-1/2 asxgoes to 0.(6) Use the Taylor series for cos(x), substitutex3instead ofx. Thus we find thatcos(x3)-1 =-x6/2 + higher order termsSimilarly,sin(x2)-x2=-x63!+ higher order termsand so the quotientcosx3-1sinx2-x2goes to-x6/2-x6/6= 3 asxgoes to 0.

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