tapia (jat4858) – HW10 – clark – (52990) 1This print-out should have 21 questions.Multiple-choice questions may continue o nthe next column or page – find all choicesbefore answering.001 10.0 pointsDetermine the value of f(2) whenf(x) =x42−2x344+3x546+ . . . .(Hint: differentiate the power series expan-sion of (x2+ 42)−1.)1. f(2) =1502. f(2) =225correct3. f(2) =454. f(2) =1105. f(2) =425Explanation:The g eometr ic series142+ x=14211 + x/42=1421 −x42+x244−x346+ . . .has interval of convergence (−16, 16). But ifwe now restri ct x to the interval (−4, 4) andreplace x by x2we see that142+ x2=1421 −x242+x444−x646+ . . .on the interval (−4, 4). In additio n, in thisinterval the series expansion of the deriva-tive of the left hand side is the term-by-termderivative of the series on the right:−2x(x2+ 42)2=142−2x42+4x344−6x546+ . . ..Consequently, on the interval (−4, 4) thefunction f defined byf(x) =x42−2x344+3x546+ . . .can be identified withf(x) =42x(x2+ 42)2.As x = 2 lies in (−4, 4), we thus see thatf(2) =225.keywords:002 10.0 pointsFind a power series representation for thefunctionf(y) = ln(4 − y) .1. f (y) = ln(4) +∞Xn = 0ynn 4n2. f (y) = −∞Xn = 1ynn4n3. f (y) = ln(4) +∞Xn = 1yn4n4. f (y) =∞Xn = 0ynn 4n5. f (y) = ln(4) −∞Xn = 0yn4n6. f (y) = ln(4) −∞Xn = 1ynn 4ncorrectExplanation:We can either use the known power seriesrepresentationln(1 − x) = −∞Xn = 1xnn,tapia (jat4858) – HW10 – clark – (52 990) 2or the fact thatln(1 − x ) = −Zx011 − sds= −Zx0∞Xn = 0snds= −∞Xn = 0Zx0snds = −∞Xn = 1xnn.For then by properties of logs,f(y) = ln(4)1−14y= l n( 4)+ln1−14y,so thatf(y) = ln(4) −∞Xn = 1ynn 4n.003 10.0 pointsCompare the radius of convergence, R1, ofthe series∞Xn = 0cnxnwith the radius of convergence, R2, of theseries∞Xn = 1n cnxn−1whenlimn → ∞cn+1cn= 2 .1. R1= R2=12correct2. R1= 2R2=123. 2R1= R2=124. R1= R2= 25. R1= 2R2= 26. 2R1= R2= 2Explanation:Whenlimn → ∞cn+1cn= 2 ,the Ratio Test ensures that the series∞Xn = 0cnxnis(i) convergent when |x| <12, and(ii) divergent when |x| >12.On the ot her hand, sincelimn → ∞(n + 1)cn+1ncn= limn → ∞cn+1cn,the Ratio Test ensures also that the series∞Xn = 1n cnxn−1is(i) convergent when |x| <12, and(ii) divergent when |x| >12.Consequently,R1= R2=12.004 10.0 pointsFind a power series representation for thefunctionf(y) =y9y + 1.1. f (y) =∞Xn = 0(−1)n3nyn+12. f (y) =∞Xn = 03nyntapia (jat4858) – HW10 – clark – (52 990) 33. f(y) =∞Xn = 032nyn4. f(y) =∞Xn = 032nyn+15. f(y) =∞Xn = 0(−1)n3nyn6. f(y) =∞Xn = 0(−1)n32nyn+1correctExplanation:After simplification,f(y) =y9y + 1=y1 − (−9y).On the ot her hand,11 − x=∞Xn = 0xn.Thusf(y) = y∞Xn = 0(−9y)n= y∞Xn = 0(−1)n32nyn.Consequently,f(y) =∞Xn = 0(−1)n32nyn+1.keywords:005 10.0 pointsDetermine the interval of convergence forthe power series representation off(x) = tan−1x5centered at the origin obtained by integratingthe power series expansion for 1/(1 + x2).1. interval of cgce. =h−15,152. interval of cgce. =h−15,15i3. interval of cgce. = (−5, 5]4. interval of cgce. = [−5, 5] correct5. interval of cgce. =−15,15i6. interval of cgce. = [−5, 5)Explanation:Since11 − x= 1 + x + x2+ x3+ . . . ,we see t hat11 + x2=11 − (−(x)2)= 1 −x2+ (−x2)2− (x2)3+ . . .=∞Xn = 0(−1)nx2n.NowZx011 + t2dt = tan−1(x) ,whileZx0∞Xn = 0(−1)nx2ndt =∞Xn = 0(−1)n2n + 1x2n+1.Thustan−1(x) =∞Xn = 0(−1)n2n + 1x2n+1,from which it follows thatf(x) = tan−1x5=∞Xn = 0(−1)n2n + 1x52n+1.tapia (jat4858) – HW10 – clark – (52 990) 4To determine the interval of convergence ofthe power series, setan=(−1)n(2n + 1)x52n+1.Thenan+1an=2n + 12n + 3x52,and solimn → ∞an+1an=x52.By the Ratio Test, therefore, the power seriesconverges when |x| < 5 and diverges when|x| > 5.On the ot her hand, at x = 5 the seriesreduces to∞Xn = 0(−1)n2n + 1,which converges by t he Alternating seriesTest, while at x = −5 the series reduces to∞Xn = 0(−1)n+12n + 1,which converges again by the Alternating Se-ries Test. Consequently, the power series rep-resentation for f(x) obtained from the seriesexpansion for 1/(1 − x) hasinterval of convergence = [−5, 5] .keywords:006 10.0 pointsFind a power series representation for thefunctionf(t) = lnr1 − 3t1 + 3t.(Hint: remember properties of logs.)1. f(t) =∞Xn=132n2nt2n2. f(t) = −∞Xn=132n−12n − 1t2n−1correct3. f (t) =∞Xn=132n−12n − 1t2n−14. f (t) = −∞Xn=132n2nt2n5. f (t) =∞Xn=1(−1)n32n−12n − 1t2n−1Explanation:We know thatln(1 + x) = x −x22+x33− . . .=∞Xn=1(−1)n−1nxn,whileln(1 − x) = −x −x22−x33− . . .= −∞Xn=11nxn.Thusln(1 − x ) −ln(1 + x) = −2x +x33+x55+ . . .= −2 ∞Xn=112n − 1x2n−1!.Now by properties of logs,lnr1 − 3t1 + 3t=12ln1 − 3t1 + 3t=12(ln(1 − 3t) −ln(1 + 3t)) .Thusf(t) = −22 ∞Xn=112n − 1(3t)2n−1!,and sof(t) = −∞Xn=132n−12n − 1t2n−1.tapia (jat4858) – HW10 – clark – (52 990) 5007 10.0 pointsExpress the integralI =Ztan−1t4dtas a power series.1. I = C +∞Xn = 0(−1)nt8n+5(2n + 1)(8n + 5)correct2. I = C +∞Xn = 0(−1)nt8n+4(2n + 1)(8n + 5)3. I = C +∞Xn = 0(−1)nt8n+5(8n + 5)4. I = C +∞Xn = 0(−1)nt4n+4(2n + 1)5. I =∞Xn = 0(−1)nt8n+5(2n + 1)(8n + 5)Explanation:We know thattan−1(t) =∞Xn = 0(−1)nt2n+12n + 1.Replacing t wit h t4, we getI =Ztan−1t4dt=Z∞Xn = 0(−1)n(t4)2n+12n + 1dt .Consequently,I = C +∞Xn = 0(−1)nt8n+5(2n + 1)(8n + 5).008 10.0 pointsUse the Taylor series for e−x2to evaluatethe integralI =Z204 e−x2dx .1. I =∞Xk = 0(−1)kk!4 · 22k2. I =∞Xk = 0(−1)k2k + 14 · 22k+13. I =∞Xk = 01k!4 · 22k4. I =nXk = 01k!(2k + 1)4 · 22k+15. I =∞Xk = 0(−1)kk!(2k + 1)4 · 22k+1correctExplanation:The Taylor series for exis given byex= 1 + x +12!x2+ . . . +1n!xn+ . . .and its interval of convergence is (−∞, ∞).Thus we can substitute x → −x2for all val-ues of x, showing thate−x2=∞Xk = 0(−1)kk!x2keverywhere on (−∞, ∞) . ThusI =Z204∞Xk = 0(−1)kk!x2kdx.But we can change the order of summationand int egration on the interval o f convergence,soI = 4∞Xk = 0Z20(−1)kk!x2kdx= 4∞Xk = 0h(−1)kk!(2k +