Exam 3 Flashcards
(17 cards)
Form of geometric series and when it converges/diverges
a * r^n-1
|r|< 1 : converges
|r| ≥ 1 : diverges
Sum of geometric series formula
S = a/(1-r)
P-series convergence/divergence
P<1 : diverge
P ≥ 1 : converge
Direct comparison test steps
Determine, as n grows, what pieces will be insignificant (remove additional term such as +2 in the series 1/n+2.)
This will be what you compare to. (1/n in the example)
First compare to determine if the series you’re comparing to is bigger/smaller.
Then determine if bn (bigger function) converges or diverges by using tests like geometric or p-series.
If bn converges, an converges as well.
If bn diverges, an is inconclusive & you have to use the limit comparison test.
If an diverges, bn diverges
An: smaller series
Bn: bigger series
Limit comparison steps
-Divide the series you’re comparing to by the series you’re given.
Ex:
Given: 1/n+10
Series to compare: 1/n
Therefore, (1/n)/(1/n+10)
-rewrite into multiple (KCF)
1/n x (n+10)/1
-cancel terms
(n+10/n)
-evaluate limit
=1 (nth term test)
Therefore, diverge
Alternating series test
If the limit of the series without the alternating piece goes to 0, the series converges
If it doesn’t go to 0, it diverges by the nth term test
Absolute convergence & conditional convergence
By using AST, determine if alternating series converges or diverges.
If the series converges, and, without its alternating piece, converges, then the series is absolutely convergent
If the series converges, and, without its alternating piece, diverges, then the series is conditionally convergent
If the series diverges by failing the AST, then the series diverges by the nth term test
Ratio test
Plug in n+1 for n to get the series an+1
Divide the new series an+1 by the series (multiply an+1 by the reciprocal of the series)
Cancel terms
Evaluate the limit
-if the limit is greater than or equal 0 and less than 1, converge absolutely
-0
Ratio test
Plug in n+1 for n in the series to get the series an+1
Multiply an+1 by the reciprocal of the original series
Cancel terms
Evaluate limit
-L=limit
-0 ≤ L < 1 ; converge absolutely
-L > 1 ; diverges
-L = 1 ; ratio test inconclusive
Root test
In a series that is raised to power n,
Find the limit of the series to the n root
This gets rid of the power n
Evaluate the limit
-L=limit
-0 ≤ L < 1 ; converge absolutely
-L > 1 ; diverges
-L = 1 ; ratio test inconclusive
Radius and Interval of Convergence
Use the ratio test
Set the absolute value limit less than 1
The radius of convergence is the limit with the absolute value bars less than R. R is the radius of convergence.
ex:
1/9|x-4| < 1
|x-4| < 9
The radius of convergence is 9.
Find the interval of convergence by setting the limit between the negative and positive value you got from the radius of convergence. Solve by isolating x in the middle.
ex:
-9 < |x-4| < 9
-5 < x < 13
Test the endpoint values by plugging them in for x in the series.
If div; not ()
If conv; included []
Taylor Polynomial Form
+
Listed out to P2
(f^n(a)/n!)(x-a)^n
f(a)+f’(a)(x-a)+(f’‘(a)/2!)(x-a)^2
Finding Taylor Polynomials
List out the f(a)’s
P0 is “a” (which is the x value given) plugged into f(x).
P1 is “a” (which is the x value given) plugged into the first derivative of f(x).
P2 is “a” (which is the x value given) plugged into the 2nd derivative of f(x).
etc..
Now use the Taylor polynomial form and list out the taylor polynomial.
Taylor Polynomials: Use these polynomials to estimate f(x) where x is a given number.
Plug in whatever x was replaced by into the polynomials and solve each polynomial.
Taylor Polynomials: Use Taylor’s Remainder Theorem to bound the error.
1) Whatever the highest derivative to find the polynomial was, find the next derivative after that.
ex: if the highest polynomial found was P2, find what the next derivative would be. Basically, what you would plug in for P3.
Derivative for P2 was (-2/9)x^-5/3
Derivative for P3 is (10/27)x^-8/13
2) Then plug in the x= number that was given into the new derivative.
(10/27)(8)^-8/13 = .00145
3) Then find the next Taylor polynomial value (the numerator is the value from step 2) and plug in x from what x was replaced by when asked to estimate.
(.00145/3!)(11-8)^3
AST Error Bound
|Rn| ≤ bn+1
The partial sum is only off from the true sum by less than the next term
The answer will be the next term after the partial sum.
Ex: Given the AST: ((-1)^n+1)/n^2, determine a bound on the error R10 if we approximate the sum of the series by the partial sum S10.
|R10| ≤ |a11|
|R10| ≤ |1/11^2|
|R10| ≤ 1/121
Integral Test Error Bound
Set up an integral with the bounds from the value you’re testing to infinity.
ex: For the series: 1/n^4, calculate S5 and estimate the error R5
integral from 5 to infinity: 1/x^4 dx
Solve the integral and plug in the bounds and solve.
integral from 5 to infinity: 1/x^4 dx
=-1/3x^3 from 5 to infinity
=0+1/3(5)^3= 0.0027
