Supplemental Instruction
Iowa State University / Leader: / Will
Course: / Math 166
Instructors: / Dr. Johnston
Date: / 4/24/14
1) Tests for Infinite Series:
a) Determine convergence of the following series, specify which test you used:
k=4∞k+1k2-3
b)
n=0∞sinnn2n
2) Tests for Infinite Series:
a)
k=1∞-1nk2-k3/2k3 (Hint:simplify the expression first)
b)
k=1∞41+n2
3) Power Series Convergence:
a) Use the ratio test to determine where the following series converges:
n=1∞2n(x+4)^nn2
b) If the following series converges for -1 < x < 3,
k=1∞akx-1k
Find the radius of convergence; determine where the series below converges:
k=1∞akx+1k
4) Taylor Series:
a) Find a power series for fx= 4x-2 with powers of x-1n using the formula for Taylor series.
(Hint: what does this tell you about the center of the power series?)
5) Remainder Theorem:
Let fx=cos(2x)
a) Find P6x, the Taylor polynomial of degree 6 for f about center c = 0.
b) Make a bound on the error R6(x)=cos2x-P6(x) when 0≤x≤π2
Using the remainder formula:
Rn(x)=Mn+1x-cn+1n+1! where Mn+1=maxfn+1x on 0,π2
6) Remainder Theorem:
a) For the Maclaurin Series for fx=11-3x (which is geometric), the interval of convergence is -13<x<13 . Write out a general nth degree polynomial.
b) When estimating f16 using Pn16 from this power series, determine the degree n such that Rn(x)<0.01 using Rn(x)=f16-Pn16
Hints: simplify f16-Pn16=
k=0∞rk- k=0nrk = k=n+1∞rk < 0.01
7) Extra:
Find a power series for fx=52+3x with powers of (x-1)
1) Tests for Infinite Series:
a) Determine convergence of the following series, specify which test you used:
k=4∞k+1k2-3
Use comparison test:
k+1k2-3 ≥ kk2 = 1k and k=4∞1k diverges (harmonic)
Therefore, our series diverges by the comparison test.
b)
n=0∞sinnn2n
Use comparison test:
sinnn2n≤12n and n=0∞12n converges as a geometric series
Using squeeze theorem principles, the negation of this geometric series is also convergent, so our original series is bounded above and below by converging series. Therefore, our series converges by the comparison test.
2) Tests for Infinite Series:
a)
k=1∞-1nk2-k3/2k3 (Hint:simplify the expression first)
Simplification:
k=1∞-1nk2-k32k232 =k=1∞-1kk2-kk232 = k=1∞-1k1-1k32
Use alternating series test to test for divergence:
limk→∞1-1k32 = 1
Therefore, our series diverges by the alternating series test.
b)
k=1∞41+n2
Use integral test:
1∞41+x2dx = limb→∞1b41+x2dx = 4*(limb→∞tan-1b - tan-11) =π
Therefore, our series converges by the integral test
3) Power Series Convergence:
a) Use the ratio test to determine where the following series converges:
n=1∞2nx+4nn2
limn→∞2n+1x+4n+1n+12*n22nx+4n = x+4limn→∞2*n2n+12
gives us 2 x+4≤1 → -92 x-72
Test endpoints…
n=1∞2n-92+4nn2=n=1∞2n-12nn2= n=1∞-1nn2 converges by AST
n=1∞2n-72+4nn2=n=1∞2n12nn2= n=1∞1n2 converges as a p-series
So our final interval is -92 ≤ x ≤-72
b) If the following series converges for -1 < x < 3,
k=1∞akx-1k this implies that the center is 1, and the radius is 2
Use the ratio test on this series:
limk→∞ak+1x-1n+1akx-1n = x-1limk→∞ak+1ak < 1
Therefore, since the radius of convergence (which is 2) is given byx-c<r
we have that:
x-11limk→∞ak+1ak=2 → limk→∞ak+1ak=12
Find the radius of convergence; determine where the series below converges:
k=1∞akx+1k
Use the ratio test on this series, and apply our new knowledge for the limit:
x+1limk→∞ak+1ak < 1 → 12*x+1 < 1
Thus,
-3< x < 1 so our radius is still 2, and our center is now-1.
4) Taylor Series:
a) Find a power series for fx=14x2 with powers of x-1n using the formula for Taylor series. Where does this series converge?
(Hint: what does this tell you about the center of the power series?)
Immediately, the center c = 1 by definition (look at the formula)
Begin taking powers of f:
fx=14x-2; f'x=-12x-3; f''x=32x-4; f'''x=-6x-5
f1=14; f'1= -12; f''1=32; f'''1= -6;
As a basic pattern, we get fn1=-1nn+1!4
Plug this into the Taylor series formula:
n=0∞fn1n!x-1n = n=0∞-1nn+14x-1n
N-th term test easily shows this series diverges to infinity, BUT since we get to choose the variable x … Let x = 1, then we have a converging series of zero’s.
So, it diverges for all x not equal to 1.
5) Remainder Theorem:
Let fx=cos(2x)
a) Find P6x, the Taylor polynomial of degree 6 for f about center c = 0.
Use the known Maclaurin Series for the function cos(2x):
cos2x=1-2x22!+2x44!-2x66!
b) Make a bound on the error R6(x)=cos2x-P6(x) when 0≤x≤π2
Using the remainder formula:
Rn(x)=Mn+1x-cn+1n+1! where Mn+1=maxfn+1x on 0,π2
fn+1x=f7x=27sinx≤27 sincesinxis bounded by 1
so Mn+1=27
Also need to maximize x-cn+1:
xn+1 for x on the interval 0,π2→π27
Putting these into our remainder formula:
Rn(x)≤27π277!=π77!
6) Remainder Theorem:
a) For the Maclaurin Series for fx=11-3x (which is geometric), the interval of convergence is -13<x<13 . Write out a general nth degree polynomial.
11-3x=1+3x+3x2+3x3…+3xn
b) When estimating f16 using Pn16 from this power series, determine the degree n such that Rn(x)<0.01 using Rn(x)=f16-Pn16
Hints: simplify f16-Pn16=
k=0∞rk- k=0nrk = k=n+1∞rk < 0.01
k=0∞12k - k=0n12k=k=n+1∞12k now convert back to k=0
=k=0∞12k+n+1 =k=0∞12k*12n+1 =12n+1*k=0∞12k = 12n+1*11-12
= 12n < 0.01 → n=7 since 127=1128 < 1100
7) Extra:
Find a power series for fx=52+3x with powers of (x-1)
Derivatives are too time-consuming, so instead transform f(x) to include powers of (x-1):
fx=52+3x = 52+3x-1+3
= 5 5+3(x-1) simplify to look like 1 / 1 – x series…
= 1 1+35(x-1)
Now our function looks like the Taylor Series for a geometric series, substitute for r:
r =-35x-1 → k=0∞rk → k=0∞-35x-1k