Class Notes (1,100,000)
CA (620,000)
UTSG (50,000)
MAT (4,000)
MAT136H1 (900)
Lecture 21

# MAT187H1 Lecture 21: 9.6 Alternating Series and Absolute and Conditional Convergence 10.1 Approximating Functions with PolynomialsPremium

Department
Mathematics
Course Code
MAT136H1
Professor
Anthony Lam
Lecture
21

This preview shows half of the first page. to view the full 3 pages of the document.
MAT187H1Lecture219.6AlternatingSeriesandAbsoluteandConditional
Convergence,10.1ApproximatingFunctionswithPolynomials
Testforconvergence: ..
k=0 k!
(−1)k= 1 − 1 + 1
2! 1
3! +1
4! − .
Since(k+1)!>k! (ie.termsaredecreasing),and
1
k!>1
(k+1)! lim
k→∞
1
k!= lim
k→∞
1
k(k−1)(k−2)... = 0
seriesconvergesbyAlternatingSeriesTest
Note:(inchapter10) e
1
k=0
1
k!=e
1
=1/e
k=0 k!
(−1)k=
ApproximatingSumsofConvergentAlternatingSeries
ConsiderconvergentA.S.
a
1
a
2
+a
3
a
4
+a
5
...(− ) a
k=0 1k+1
k=
DefineRemainderR
n
SS
n
OddtermsofS
n
are>SieSS
odd
<0
EventermsofS
n
are<SieSS
even
>0
TheoremForconvergentseries (whichconvergestoS),then
th
termremainder,(− ) a
k=1 1k+1
k
R
n
=SS
n
hasthesamesignas(n+1)
th
termand0≤|R
n
|<a
n+1
ExampleGiven (− ) ..
k=0 1k1
2k+1 = 1 − 3
1+5
17
1+ . = 4
π
Howmanytermsareneededtoestimate toaccuracyof10
4
?(iefor|R
n
|=10
4
,n=?)
4
π
Foranyn,|R
n
|<a
n+1
=1
2(n+1)+1 =1
2n+3
for <10
4
<>2n+3>10
4
1
2n+3