Merced College; Don Power

Calculus Lecture - Ch 10

10.1 Sequences

Def: a function whose domain is the set of natural numbers (or, "a set of integers")

(except that it is permissible to designate some other starting point than 1; 0 is frequently used.)

Ex built on f(x)=x²: we might write f(n) = n², n = 1, 2, 3, ...

D: {1, 2, 3, 4, 5, …} term numbers

R: {1, 4, 9, 16, 25, …} terms

Compare the graph of the sequence with the graph of the continuous function

Vocab and notation:

The sequence is usually thought of as the range values (domain is not referred to explicitly)

We use n for an element of the domain, then a_n is the corresp element of the sequence a.

Ex: a₃ = 9 because n = 3,

{a_n} or refer to the entire sequence [w provisions to stress or modify the starting point]

Ex: std notation for our sequence would be {a_n} = {n²}

Task: find first several terms of a series:

Ex: = …{2/1, 3/2, 4/3, …}

Ex: = {1, 0, -1, 0, 1, 0, -1, 0, …}

Task: Find the general term for a given sequence (Exercises 1, 4, 23-31)

Key Question: Does a sequence have a limit (it converges) or not (it diverges)?

Ex: limit of x/(x-1) is 1 as x -> ¥, so converges to 1

Def: [explain graphically] For every epsilon>0, there exists N>0 such that whenever n>N,

abs(a_n-L)<epsilon

Relate to the example series above.

Techniques for continuous functions usually work, including L'Hôpital's Rule

See properties of limits, pg 652

Exactly the same as for limits of continuous functions.

Squeeze Theorem applies

Ex: -1/n £ cos(pn)/n £ 1/n

Thm: if |a_n|->0, then a_n->0

Sequences may be defined recursively

Ex: Newton's Method: Here is an Excel example:

Task: Find 7^(1/3); Equivalent to finding root of f(x) = x^3 - 7
	Newton's method	estimate of	check:
x	x-f(x)/f '(x)	7^(1/3)	ans^3
2	1.916666667	1.916666667	7.041088
1.916667	1.912938458	1.912938458	7.00008
1.912938	1.912931183	1.912931183	7
1.912931	1.912931183
1.912931	1.912931183

Ex: Mechanic's formula for approximating square roots (example on pg 655 in text)

Ex: Fibonacci sequence 1, 1, 2, 3, 5, 8, 13, 21, ...; exercise 46 in text.

10.2 Monotone Sequences

Convenient thm: if a seq is eventually monotonic and bounded, then it converges

Defs

How can you show that a sequence is decreasing [increasing]? For positive-term series:

Show a_n+1 < a_{n [}a_n+1 > a_n

Show a_n+1 - a_n < 0, or [a_n+1 - a_n > 0]

Show a_n+1 / a_n <1, or [a_n+1 / a_n >1]

Show deriv is always negative [deriv is always positive]

Illustration for --Since it's monotonic decreasing, 1st term is an upper bound, and since both num and denon are pos, 0 is a convenient choice for a lower bound (after you find the limit, you can conclude that 1 is the greatest lower bound, but any lower bound is enough for the thm.)

Ex: Show that the sequence {xⁿ / n!} converges to 0, regardless of the choice of x:

Find a_n+1 / a_n = (xⁿ⁺¹ / (n+1)!) / (xⁿ / n!) = (x^.xⁿ / ((n+1)^.n!) / (xⁿ / n!) = x / (n+1)

One thing to notice here is how to factor factorials.

First, it converges because it is bounded and eventually monotone:

The fraction is positive, so it is bounded below by 0

Since x is fixed, eventually n+1 > x, so the fraction < 1

[Later, we'll call this the ratio test, and we will use it a lot]

Second, it converges to 0, because with x fixed, the limit x / (n+1) is 0.

10.3 Infinite Series

Series = sum of terms of a sequence, actually a series is a sequence of partial sums, where a partial sum is a "running total," sum of terms of {a_n} from 1 to n, as n increases from 1 to infinity.

Series converges if the sequence of partial sums converges.

What is the sequence of partial sums for a repeating decimal representation of a fraction?

Ex: 6/11 = 0.54545454...

= 0.54+.0054+.000054+.00000054...

= 0.54 (1+.01+.0001+.000001+...)

=a(1+r+r²+r³+...) or a + ar + ar² + ... Example of a geometric series.

Geometric series

Intro: Induction proofs

Step 1: Prove for n=1 (or initial value);

Assume true for n=k [assuming that there is a highest index for which the proposition is valid]; Step 2: Prove true for n=k+1 [so there is no such highest valid level]

Examples (relevant for calculus):

This was used in calculating Riemann sums

Prove formula for finite geometric series:

Applic: 1/2 +1/4 + 1/8 + …+1/64 = 1/2 * (1 - (1/2)^6) / (1-1/2) = 63/64

Infinite geometric series: Take limit of finite geom. series: Geom series converges to a / (1-r) if |r|<1.

Inf sum of 1/2 +1/4 + 1/8 + … is 1

Violate condition: 1+2+4+8+… = -1

Telescoping series

Ex:

Tasks: Find nth partial sum

Find the infinite sum

Harmonic series Example of a divergent series

Increasing series, but not bounded

a₁+a₂>1/2

a₃+...+a₄>1/2

a₅+...+a₈>1/2

a₉+...+a₁₆.1/2 etc.

Pick a proposed upper bound M; add enough sets of terms, and you will exceed M

10.4 Convergence Tests

Divergence Test: (Theorems 1 and 2)

If the general term u_k does not converge to 0, then the series diverges.

Hence: if series converges, the limit of the general term is 0 [contrapositive]

If the general term u_k does converge to 0, the series may converge or diverge

(you cannot prove convergence by taking the limit of the general term)

Algebraic Properties of Convergent Series:

Sums and differences of convergent series converge

If you factor out a constant from all terms of a convergent series, the result still converges

If you delete a finite number of terms at the beginning of a series, it does not affect the convergence or divergence

Integral Test:

For positive-term series that are eventually decreasing:

Where f(x) is the function that results when k is replaced by x in the general term of the series:

Either:

the series and the integral of the function both diverge, or

they both converge

(Note: if they both converge, they routinely converge to different values.)

p-series

The series (sum of 1/k^p, k=1..¥) converges if p>1 and diverges if p£1

Special cases:

If p is neg, the fraction is gone

If p is 0, you are adding an infinite number of 1's

If p is 1, you get the harmonic series

Thm 6: A series with nonnegative terms converges iff the sequence of partial sums is bounded above.

(This condition is violated by the harmonic series - see the proof at the end of the previous lesson)

Estimates of Sums:

S = S_n + R_n

S_n = a₁ +...+ a_n; R_n = a_n+1 + a_n+2+ ...

For decreasing-term positive-term series,

The first rectangle in R_n (i.e.a_n+1) is pictured twicein the figure below:
the integral of the continuous function beginning at n is above the rectangle,
(hence, the integral from n is greater than R_n)
and the integral beginning at n+1 is below the rectangle

(hence, the integral from n+1 is less than R_n)

Maple evaluation of series, with demonstration of the above discussion of estimates of sums:

10.5 Comparison, Ratio and Root Tests

Comparison Tests

Basic Comparison Test:

Let an £ bn for every n³N

If the bigger series converges, then the smaller series converges

If the smaller series diverges, then the bigger series diverges

Ex: Test Sum of 1 / (2ⁿ+n), Sum of 1 / (sqrt(n) - 1)

Problem: The result is the same if the signs in the denominator are reversed, but this test doesn't show it.

Limit Comparison Test:

Calculate the limit of the ratio of the nth terms for two series, one with a known convergence result.

If the limit is finite and not 0, both series have the same convergence result.

The comparison test is often a p-series or a geometric series.

Ex: The same series above, with the opposite signs in the denominator.

Ratio Test:

Calculate the limit of the ratio of a later term to an earlier term.

Convergence requires that later terms are significantly smaller than earlier terms,...

so this ratio must be less than 1 for the series to converge.

If the ratio is greater than 1, the series diverges

If the ratio is equal to 1, the ratio test fails: it does not tell us anything

We must use a different test

This procedure works well for series that contain factorials

Review factoring of factorials

Caution: The ratio test will routinely fail with series that are comparable to p-series

This test will also work for series that contain negative terms:

Convergence occurs when the absolute value of the ratio is less than 1, etc.

Root Test:

Take the limit of the nth root of the nth term

Same convergence criteria as for root test

Very specialized: it would be used if the entire expression is raised to a power containing the index

10.6 Alternating Series; Conditional Convergence vs Absolute Convergence

Def of alt series

Thm 1: Requirements for convergence:

successive terms decrease in abs value

kth term, in abs value, has limit of 0

Ex: alt harm series

Est of sums: max error: abs(R_n) [ i.e. abs(s-s_n)] £ abs(a_n+1) [a_n+1 is the first discarded term]

Also, the sign of the error is the same as the sign of a_n+1

Two problems:

What is the max possible error?

How many terms are required to get a given level of accuracy?

Ex: Find the max possible error (and the direction of the error) if we use 3 terms to estimate sin(p/6)

Use the series (from the next lesson) sin x = x − x³/3! + x⁵/5! − x⁷/7! ...

Ex: How many terms are needed to estimate sin(p/6) correct to 6 decimal places (i.e. error < 5 x 10^-7)

Use the same series as in the previous example.

Def of abs convergence

Ex: sum((-1)^k/2^2k-1,k=0..infinity) Geometric, with r = -1/4

Def of conditional convergence

Ex: sum((-1)^k/sqrt(k) Alternating p-series with p=1/2

Thm: If a series converges absolutely, then it converges

Ratio Test for absolute convergence

Ex: For the sine function (given above, the general term is (−1)x^2k+1/(2k+1)!, with k = 0..∞

10.7 Maclaurin and Taylor Polynomial Approximations

Example: Find the Maclaurin expansion of f(x)=cos(x)

We assume cos(x)=a₀+a₁x+a₂x²+a₃x³+a₄x⁴+a₅x⁵+... for some set of coefficients a₀, a₁, etc.

Substitute x=0: cos(0)=a₀

\a₀=1, so cos(x)»1

Differentiate: -sin(x)= a₁+2a₂x+3a₃x²+4a₄x³+5a₅x⁴+6a₆x⁵...

Substitute x=0: -sin(0)= a₁

\a₁=0, so cos(x)»1

Differentiate: -cos(x)= 2×1a₂+3×2a₃x+4×3a₄x²+5×4a₅x³+6×5a₆x⁴+7×6a₇x⁵...

Substitute x=0: -cos(0)= 2!a₂

\a₂=-1/2!, so cos(x)»1-x²/2!

Differentiate: sin(x)= 3×2a₃+4×3×2a₄x+5×4×3a₅x²+6×5×4a₆x³+7×6×5a₇x⁴...

Substitute x=0: sin(0)= 3!a₃

\a₃=0, so cos(x)»1-x²/2!

Differentiate: cos(x)= 4×3×2a₄+5×4×3×2a₅x+6×5×4×3a₆x²+7×6×5×4a₇x³...

Substitute x=0: cos(0)= 4!a₄

\a₄=1/4!, so cos(x)»1-x²/2!+x⁴/4!

Using the formula to find the coefficients: , so

	n	f^(n)(x)	f^(n)(0)	an
Original function	0	cos(x)	cos(0)=1	1/0!=1
f¢(x)	1	-sin(x)	-sin(0)=0	0/1!=0
f¢¢(x)	2	-cos(x)	-cos(0)= -1	-1/2!
f^(3)(x)	3	sin(x)	sin(0)=0	0/3!=0
f^(4)(x)	4	cos(x)	cos(0)=1	1/4!

So cos(x) »1+0×x-x²/2!+0×x³+x⁴/4! ...

Upper bound on the error in using a partial sum to estimate the series

1. General formula for Taylor/Maclaurin polynomials:

upper bound on error is |R_n| ≤ M / (n+1)! * |x-x₀|ⁿ⁺¹ where M = |max(f⁽ⁿ⁺¹⁾(x)| on I

In other words, the first discarded term, except for the modification of the numerator

2. Easier, if the series is alternating: |Rn| ≤ |a_n+1|

10.8 Maclaurin and Taylor Series; Power Series

Finding Maclaurin and Taylor Series

By technique of 10.1

Ex: Like X16: cos(x), x₀ = π/6

In 10.9: Build-up from other series by addition, subt, diff, int, mult, div, subst

See the list of series on page 705 (in 10.9)

Finding the radius and interval of convergence

Ex: X42: sum((–1)^k*(x–4)^k/(k+1)², from 0 to ∞

Ex: Like X33: sum((-1)^k-1(x-2)^k/(3^ksqrt(k))

Text mentions that some functions originate as power series, such as Bessel functions.

How? Bessel funcs solve DEs

J₀ solves xy'' + y' + xy = 0 J₀ = sum((−1)^kx^2k / [2^2k(k!)²]) from k=0 to ∞

J₁ solves x²y'' + xy' + (x²−1)y = 0 J₁ = sum((−1)^kx^2k+1 / [k!(k+1)!2^2k+1] from k=0 to ∞

Technique: method of Frobenius: (Complicated: we'll save this for MATH-06)

Let y = a₀ + a₁x + a₂x² + a₃x³ ...

Differentiate repeatedly to calculate the series for y' and y''

Multiply to get the series for the specific terms (e.g. 3xy, –x²y'', etc.)

Substitute into the DE

Use the method of undetermined coefficients: equate the constants, the coefficients of x, etc.

10.9 Convergence of Taylor Series; Computational Methods

Know the series on page 708 or be able to derive them quickly from one another.

Build-up from other series by addition, subt, diff, int, mult, div, subst

Use of the series for ln((1+x) / (1−x)) to find ln(z) when z is greater than 2.

Binomial Series

(1+x)^m =

for |x| < 1 (and you may get convergence at an endpoint)

So (1+x)⁴ =

Notice what happens to the x^m+1 term (here, the x⁵ term) and all later terms:

Therefore, we get convergence for all real numbers if m is a positive integer

10.10 Differentiating and Integrating; Modeling

-- See Stewart's intro page 726, see discussion page 775 on apps (planetary motion, temp distrib in a circular plate, shape of vibrating drumhead)

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