Review for Exam 3

Review

Taylor polynomials

The (n^th) degree Taylor polynomial for function f(x) centered at a is a partial sum for the Taylor series of f(x). The polynomial provides an approximation for the function.
	f(x)≈(Underoverscript[Σ, (k=0), n])(((f^((k)))(a))/(k!))(((x-a))^k)=f(a)+f'(a)(x-a)+((f''(a))/2)(((x-a))^2)+…+(((f^((n)))(a))/(n!))(((x-a))^n)
The error in this approximation is called the Taylor remainder (Subscript[R, n])(x), and it satisfies the inequality
	|(Subscript[R, n])(x)|≤(M/((n+1)!))(((x-a))^(n+1))
where M is some number where M≥|(f^((n+1)))(x)| for all values of x of interest.

Questions

Consider f(x)=sin(x).

What is the fourth degree Taylor polynomial for f(x) centered at 0?

We plan to use this polynomial to approximate sin(x) for |x|≤π/2. What is the biggest |(f^((5)))(x)| can be for these values of x?

What is an error estimate for using this approximation using the Taylor remainder error estimate?

What is an approximation for sin(0.6) using its fourth degree Taylor polynomial?

What is an estimate of the error in this approximation?

Exam rules

You can have one page of notes (front and back) and you are allowed to have your favorite calculator.

You must show full work on all problems, especially integrals. (I.e., if you’re using your calculator to evaluate integrals, you’d better be able to show me how you could have gotten your answer without it. Otherwise it’s only good for checking your work....)

Format

Questions will be similar to homework questions and weekly assignment questions.

Topics

Section 11.1: Sequences

Section 11.2: Series

Partial sums.

Geometric series.

Divergence test.

Section 11.3: Integral test

Only applies to series with nonnegative terms (well, also works for series with all negative terms, of course; the point is no changes in sign).

Gives us the behaviors of p-series.

Integral test error estimate.

Section 11.4: Comparison test

Compare with p-series or geometric series.

If the given series is less than a known series, use an appropriate error estimate from the known series.

Section 11.5: Alternating series test

Has to be an alternating series for the test to apply and has to have the terms go to zero.

Alternating series test error estimate.

Section 11.6: Absolute series test

Remove signs and either

Compare to known series like p-series or geometric series, or

Use ratio test or root test

Absolute convergence versus conditional convergence

If removal of the signs produces a convergent series, the original series with the signs converges absolutely.

If removal of the signs produces a divergent series, but the original series converges by the alternating series test, then that convergence is conditional.

Section 11.8: Power series

Center of convergence

Radius of convergence—comes from applying the ratio or root test

Interval of convergence—always must check the endpoints with something other than  the ratio or root test (since it’s indeterminate there).

Section 11.9: Functions as power series (using the geometric series)

Power series representation for (1/(1±x)) (remember that these are only equal for the interval of convergence).

Power series representation for ln(1±x).

Power series representation for (tan^(-1))(x).

Differentiating and integrating power series is allowed (and encouraged!). Differentiation may rough up the interval at the endpoints, whereas integration may smooth things out.

Section 11.10: Taylor series

Taylor series centered at 0 for (e^x), sin(x), cos(x).

If you already have a power series representation for a function, it must be its Taylor series.

Section 11.11: Taylor polynomials

Approximate functions with their Taylor polynomials.

Taylor remainder error estimate.

Studying

Try problems you haven’t worked before from the exercises in your reference textbook.

Look through the weekly homeworks. Where you had problems, check the solutions on Canvas.

Rework problems in Imath.

Finish any worksheets.

Sample questions

Consider the sequence (2/3),-(4/9),(6/27),-(8/81), (10/243),-(12/729),….

Find a formula (Subscript[a, n]) for this sequence.

Table[-2n/(-3)^n,{n,1,6}]

A formula for this sequence is (((-1))^(n+1))((2n)/(3^n)) for n=1,2,3,…

Does this sequence (not the summation of them) converge? If so, to what value?

Compute 
	(lim_(n->∞))|(((-1))^(n+1))((2n)/(3^n))|=(lim_(n->∞))((2n)/(3^n)).
This is an indeterminate limit of the type (∞/∞) so we can use L’Hopital’s rule.
	(lim_(n->∞))((2n)/(3^n))=(lim_(n->∞))(2/((3^n)ln(3)))=0
Since it goes to zero in absolute value, the original must also go to zero.

Analyze each of the given series, if it diverges give reasons why. If it converges, either give the exact value it converges to or give the partial sum whose value approximates the series with error less than 0.001.

(Underoverscript[Σ, (k=0), ∞])2((((4/3)))^k)

(Underoverscript[Σ, (k=0), ∞])2((((4/3)))^k) is a geometric series with r=(4/3)>1. It diverges.

(Underoverscript[Σ, (k=0), ∞])(e^(-k))

(Underoverscript[Σ, (k=0), ∞])(e^(-k)) is a geometric series with a=1 and r=(e^(-1))≈0.367879 which is between -1 and 1. It converges to
	(1/(1-(e^(-1))))=(e/(e-1))≈1.5819767068693265`.

(Underoverscript[Σ, (k=1), ∞])(k/(k+1))

For (Underoverscript[Σ, (k=1), ∞])(k/(k+1)) apply the divergence test by looking at the terms that are being summed.
	(lim_(k->∞))(k/(k+1))=1
Since the individual terms are not approaching 0 as k->∞, the sum of these terms must diverge.

(Underoverscript[Σ, (k=1), ∞])(1/(k^5))

(Underoverscript[Σ, (k=1), ∞])(1/(k^5)) is a p-series with p=5>1. It converges. Approximate it with a partial sum. Use the integral test error estimate to determine which partial sum to use.
	error≤(∫_n^∞)(1/(x^5))d x=((x^(-4))/(-4))(Subscript[|, n]^∞)=(1/(4(n^4)))
Force the error to be small enough by choosing n so that (1/(4(n^4)))<0.001. Solve this inequality of n.
	(1/(4(n^4)))<0.001
	⟶ (1/0.004)<(n^4)
	⟶ n>((1/0.004)^(1/4))≈3.9764
Use the fourth partial sum to approximate this infinite sum.
	(Underoverscript[Σ, (k=1), ∞])(1/(k^5))≈(Underoverscript[Σ, (k=1), 4])(1/(k^5))=1+(1/(2^5))+(1/(3^5))+(1/(4^5))≈1.036

(Underoverscript[Σ, (k=1), ∞])(4/(k·(5^k)))

Compare this to the geometric series  (Underoverscript[Σ, (k=1), ∞])4((((1/5)))^k) since  (4/(k·(5^k)))≤(4/(5^k)). The geometric series converges since r=(1/5) is between -1 and 1. By the comparison test (Underoverscript[Σ, (k=1), ∞])(4/(k·(5^k))) must also converge. Approximate it with a partial sum. Compare the tail with the tail of the geometric series
	error=(Underoverscript[Σ, (k=n+1), ∞])(4/(k·(5^k)))≤(Underoverscript[Σ, (k=n+1), ∞])4((((1/5)))^k)
The tail of the geometric series is also a geometric series with r=1/5 and a=4(((1/5))^(n+1)). The error in the approximation (Underoverscript[Σ, (k=1), n])(4/(k·(5^k))) of (Underoverscript[Σ, (k=1), ∞])(4/(k·(5^k))) is
	error = (Underoverscript[Σ, (k=n+1), ∞])(4/(k·(5^k))) ≤ (Underoverscript[Σ, (k=n+1), ∞])4((((1/5)))^k) =((4(((1/5))^(n+1)))/(1-1/5))=((4(((1/5))^(n+1)))/(4/5))=(((1/5))^n)
Solve the inequality (((1/5))^n)≤0.001 for n.
	(((1/5))^n)≤0.001
	⟶n ln(1/5)≤ln(0.001)
	⟶ n≥ln(0.001)/ln(1/5)≈4.29202967422018`
Use the fifth partial sum.
	(Underoverscript[Σ, (k=1), ∞])(4/(k·(5^k)))≈(Underoverscript[Σ, (k=1), 5])(4/(k·(5^k)))=(4/5)+(4/(2·(5^2)))+(4/(3·(5^3)))+(4/(4·(5^4)))+(4/(5·(5^6)))≈0.8923178666666667`

(Underoverscript[Σ, (k=1), ∞])(n/((n^2)+1))

For large value of n, the one in the denominator becomes insignificant so that (n^2)+1≈(n^2). The limit comparison test tells us that we can compare the convergence of this series with the series
	(Underoverscript[Σ, (k=1), ∞])(n/((n^2)+1))≈(Underoverscript[Σ, (k=1), ∞])(n/(n^2))=(Underoverscript[Σ, (k=1), ∞])(1/n)
The latter is a p-series with p=1. It diverges so that by the limit comparison test (Underoverscript[Σ, (k=1), ∞])(n/((n^2)+1)) diverges.

Alternatively, we can use the integral test on this series.
	(Underoverscript[Σ, (k=1), ∞])(n/((n^2)+1))
	⟶ (∫_1^∞)(x/((x^2)+1))d x
Use the substitution u=(x^2)+1, d u=2x d x or x d x=(1/2)d u to get
	(∫_1^∞)(x/((x^2)+1))d x=(lim_(R->∞))(1/2)(∫_(x=1)^(x=R))(1/u)d u
	=(lim_(R->∞))(1/2)ln(u)(Subscript[|, x=1]^(x=R)) 
	=(lim_(R->∞))(1/2)ln((x^2)+1)(Subscript[|, x=1]^(x=R)) 
	=(lim_(R->∞))(1/2)ln((R^2)+1)-(1/2)ln(2)=∞
Since the improper integral diverges, so too does the original series.

For each series determine if it is absolutely convergent, conditionally convergent, or divergent. Give reasons for your answers.

(Underoverscript[Σ, (k=1), ∞])((((-1))^(k-1))/((k^2)(2^k)))

Use the root test or ratio test with (Subscript[a, k])=((((-1))^(k-1))/((k^2)(2^k))). 
	(lim_(k->∞))((|((((-1))^(k-1))/((k^2)(2^k)))|)^(1/k))=(lim_(k->∞))(1/(((((k^(1/k))))^2)2))=(1/2)<1
or
	(lim_(k->∞))|((((-1))^k)/((((k+1))^2)(2^(k+1))))/((((-1))^(k-1))/((k^2)(2^k)))|=(lim_(k->∞))((k^2)/(((k+1))^2))·(1/2)=(1/2)<1
This series converges absolutely by the root or ratio test.

(Underoverscript[Σ, (k=0), ∞])((((-1))^k)/(Sqrt[k+1]))

Use the absolute convergence test. 
	(Underoverscript[Σ, (k=0), ∞])|((((-1))^k)/(Sqrt[k+1]))|=(Underoverscript[Σ, (k=0), ∞])(1/(Sqrt[k+1]))
By the limit comparison test this compares well with the p-series (Underoverscript[Σ, (k=1), ∞])(1/(k^(1/2))) with p=1/2. Since that diverges, by the limit comparison test (Underoverscript[Σ, (k=0), ∞])(1/(Sqrt[k+1])) diverges. This says the original series cannot be absolutely convergent.

Apply the alternating series test and (Subscript[a, k])=(1/(Sqrt[k+1])):
	(1/(Sqrt[k+1]))>0
(1/(Sqrt[k+1])) gets smaller as k gets larger
(lim_(k->∞))(1/(Sqrt[k+1]))=0
The conditions of the alternating series test are met, (Underoverscript[Σ, (k=0), ∞])((((-1))^k)/(Sqrt[k+1])) converges. Since it doesn’t converge absolutely, it must converge conditionally.

(Underoverscript[Σ, (k=1), ∞])((((-1))^(k+1))/(k^3))

Use the absolute convergence test. 
	(Underoverscript[Σ, (k=1), ∞])|((((-1))^(k+1))/(k^3))|=(Underoverscript[Σ, (k=1), ∞])(1/(k^3))
This is a p-series with p=3>1 that converges. The original series converges absolutely by the absolute convergence test.

Consider the power series (Underoverscript[Σ, (n=1), ∞])((((x-2))^n)/(n (5^n))).

What is its center?

The power series center is 2.

Find its radius of convergence.

To get the radius of convergence apply either the ratio or root test.
	(lim_(n->∞))|(Subscript[a, n+1])/(Subscript[a, n])|=(lim_(n->∞))|((((x-2))^(n+1))/((n+1) (5^(n+1))))/((((x-2))^n)/(n (5^n)))|=(lim_(n->∞))((|x-2|)/5)(n/(n+1))=((|x-2|)/5)
The ratio test implies convergence for |x - 2| < 5. The radius of convergence is 5.

What is its interval of convergence? Be sure to determine convergence at the endpoints of the interval of convergence.

From the ratio test we know it converges for \[LeftBracketingBar]x-2\[RightBracketingBar]<5 or -5<x-2<5 in other words -3<x<7,  and diverges for \[LeftBracketingBar]x-3\[RightBracketingBar]>4. The ratio test is inconclusive for \[LeftBracketingBar]x-3\[RightBracketingBar]=4, in other words x=-3 and x=7. Check these two values of x using other tests.

For x=-3, the power series becomes (Underoverscript[Σ, (n=1), ∞])((((-3-2))^n)/(n (5^n)))=(Underoverscript[Σ, (n=1), ∞])((((-5))^n)/(n (5^n)))=(Underoverscript[Σ, (n=1), ∞])((((-1))^n)/(n )). This is the alternating harmonic series that converges by the alternating series test.

For x=7, the power series becomes (Underoverscript[Σ, (n=1), ∞])((((7-2))^n)/(n (5^n)))=(Underoverscript[Σ, (n=1), ∞])((5^n)/(n (5^n)))=(Underoverscript[Σ, (n=1), ∞])(1/(n )). This is a p-series with p=1. It diverges.

The interval of convergence is [-3,7).

I’m not giving you a formula for function f(x), but I will tell you its value when x=1 and the value of some of its derivatives when x=1:  f(1)=2,f'(1)=-1,f''(1)=(1/2),(f^((3)))(1)=-(3/8).

Using the above information, what is (Subscript[T, 3])(x), the (3^rd) degree Taylor polynomial for f(x) about 1?

The formula for (Subscript[T, 3])(x) is
	(Subscript[T, 3])(x)=(Underoverscript[Σ, (n=0), 3])(((f^((n)))(1))/(n!))(((x-1))^n)=f(1)+f'(1)(x-1)+((f''(1))/2)(((x-1))^2)+((f'''(1))/(6!))(((x-1))^3)
	=2-(x-1)+(1/4)(((x-1))^2)-(1/16)(((x-1))^3)

Using your Taylor polynomial approximation for f(x), what is an approximation for f(0)?

(Subscript[T, 3])(0)=2-(0-1)+(1/4)(((0-1))^2)-(1/16)(((0-1))^3)=(53/16)=3.3125

An extra piece of information about this function is that |(f^((4)))(x)|≤3 for all x. Use that information to estimate the error in approximating f(0) with (Subscript[T, 3])(0).

The error in approximating f(x) with (Subscript[T, 3])(x) is the remainder (Subscript[R, 3])(x). The remainder satisfies the inequality
	|(Subscript[R, 3])(x)|≤((M|x-1((|)^4))/(4!))
where M≥|(f^((4)))(x)|. In this case from the given information we can take M=3. That means
	(Subscript[T, 3])(0)=2-(0-1)+(1/4)(((0-1))^2)-(1/16)(((0-1))^3)=3.3125`
approximates f(0) with error (Subscript[R, 3])(0) that satisfies
	|(Subscript[R, 3])(0)|≤((3|0-1((|)^4))/(4!))=(3/24)=(1/8)=0.125