Assignment 13

course MTH 174

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Physics II

10-27-2008

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20:30:40

query problem 9.4.4 (9.3.6 3d edition). Using a comparison test determine whether the series sum(1/(3^n+1),n,1,infinity) converges.

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RESPONSE -->

1/3^(n+1) is smaller than 1/3^n so:

Using integral test starting at 1 because f(x) is decreasing and x >= 1:

integral from 1 to infinity of 3^-n dn

integral from 1 to b of 3^-n dn is

limit as b->infinity - 1/(3^b * ln 3) + 1/(3 * ln 3)

as b gets bigger 3^-b goes to 0 so it leaves the result as converging to 1/(3 * ln 3)

Since 1/3^n converges and 1/3^(n+1) is smaller, it must also converge.

Good. FYI:

GOOD STUDENT SOLUTION WITH COMMENT: For large values of n, this series is similar to 1 / 3^n. As n approaches infinity, 1 / 3^n approaches 0. A larger number in the denominator means that the value of the function will be smaller. So, (1 / 3^n) > (1 / (3^n + 1)). We know that since 1 / 3^n converges, so does 1 / (3^n + 1).

COMMENT: We know that 1 / (3^n) converges by the ratio test: The limit of a(n+1) / a(n) is (1 / 3^(n+1) ) / (1 / 3^n) = 1/3, so the series converges.

We can also determine this from an integral test. The integral of b^x, from x = 0 to infinity, converges whenever b is less than 1 (antiderivative is 1 / ln(b) * b^x; as x -> infinity the expression b^x approaches zero, as long as b < 1).

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20:36:57

With what known series did you compare this series, and how did you show that the comparison was valid?

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RESPONSE -->

I compared it to 1/3^n

I showed that 1/3^n converged using the integral test and showed that 1/3^(n+1) is less than 1/3^n

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21:01:46

Query 9.4.10 3d edition 9.3.12). What is the radius of convergence of the series 1 / (2 n) ! and how did you use the ratio test to establish your result?

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RESPONSE -->

Using the ratio test:

lim n ->infinity | 1/(2(n + 1)) ! | / | 1/(2n) ! |

lim n ->infinity (2n) ! /(2(n + 1)) !

Dividing out a portion of the factorial leaves

1/[(2n+2)(2n +1)] = 1/[4n^2 + 6n + 2]

lim n ->infinity 1/[4n^2 + 6n + 2]

As n gets larger the result gets smaller going to 0 which is less than 1 so by the ratio test it converges.

Again good. The following also addresses the radius of convergence:

*&*& The ratio test takes the limit as n -> infinity of a(n+1) / a(n). If the limit is less than 1 then the series converges in much the same way as a geometric series sum(r^n), with r equal to the limiting ratio.

In this case a(n+1) = 1 / (2n + 2)! and a(n) = 1 / (2n) ! so

a(n+1) / a(n)

= 1 / (2n+2) ! / [ 1 / (2n) ! ]

= (2n) ! / (2n + 2) !

= [ 2n * (2n-1) * (2n-2) * (2n - 3) * ... * 1 ] / [ (2n + 2) * (2n + 1) * (2n) * (2n - 1) * ... * 1 ]

= 1 / [ (2n+2) ( 2n+1) ].

As n -> infinity this result approaches zero. Thus the series converges for all values of n, and the radius of convergence is infinite. *&*&

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21:09:29

Query problem 9.4.40 (3d edition 9.3.18) (was 9.2.24) partial sums of 1-.1+.01-.001 ... o what does the series converge?

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RESPONSE -->

The limit as n->infinity of 10^-n is 0 and each successive term is less than the previous and is greater than 0, satisfying the alternating series test.

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21:16:11

What are the first five partial sums of the series?

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RESPONSE -->

1

1 + -0.1 = .9

1 + -0.1 + 0.01 = .91

1 + -0.1 + 0.01 + -0.001 = .909

1 + -0.1+ 0.01 + -0.001 + 0.0001 = .9091

Good. The following includes the convergent value:

This is an alternating series with | a(n) | = .1^n, for n = 0, 1, 2, ... .

Thus limit{n->infinity}(a(n)) = 0.

An alternating series for which | a(n) | -> 0 is convergent.

sum(1/n^.999) diverges and sum(1 / n^1.001) converges, but doing partial sums on your calculator will never reveal this. The calculator is very limited in determining convergence or divergence.

However there is a pattern to the partial sums, which are 1, .9, .91, .909, .9091, .90909, ... . It's easy enough to show that the pattern continues, so the convergent value is .9090909... .

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21:19:16

Query 9.5.6. What is your expression for the general term of the series p x + p(p-1) / 2! * x^2 + p(p-1)(p-2) * x^3 + ?

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RESPONSE -->

The general term is p(p-1)(p-2)...(p-n) * x^n/ n!

I am unsure how to express the p(p-1)(p-2)... part any better.

Good, but see the following for a more economical expression and some important notations:

** The general term is the coefficient of x^n.

In this case you would have the factors of x^2, x^3 etc. go down to p-1, p-2 etc.; the last factor is p minus one greater than the exponent of x. So the factor of x^n would go down to p - n + 1.

This factor would therefore be p ( p - 1) ( p - 2) ... ( p - n + 1).

This expression can be written as p ! / (p-n) !. All the terms of p ! after (p - n + 1) will divide out, leaving you the desired expression p ( p - 1) ( p - 2) ... ( p - n + 1).

The nth term is therefore (p ! / (p - n) !) / n! * x^n, of the form a(n) x^2 with a(n) = p ! / (n ! * (p - n) ! )

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21:26:08

Query 9.5.18. What is the radius of convergence of the series x / 3 + 2 x^2 / 5 + 3 x^2 / 7 + ?

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RESPONSE -->

x / 3 + 2 x^2 / 5 + 3 x^2 / 7 +

The general term is x^n/(2n+1)

|x| lim n->infinity | 2n + 1 | / | 2(n+1) + 1 |

|x| lim n->infinity | 2n + 1 | / | 2n + 3 | = 1 * |x|

K = 1, and R = 1/K so R = 1 so radius of convergence is 1

Your solution is equivalent to the following:

To find the radius of convergence you first find the limit of the ratio | a(n+1) / a(n) | as n -> infinity. The radius of convergence is the reciprocal of this limit.

a(n+1) / a(n) = ( n / (2n+1) ) / (n+1 / (2(n+1) + 1) ) = (2n + 3) / (2n + 1) * n / (n+1).

(2n + 3) / (2n + 1) = ( 1 + 3 / (2n) ) / (1 + 1 / (2n) ), obtained by dividing both numerator and denominator by 2n. In this form we see that as n -> infinity, this expression approaches ( 1 + 0) / ( 1 + 0) = 1.

Similarly n / (n+1) = 1 / (1 + 1/n), which also approaches 1.

Thus (2n + 3) / (2n + 1) * n / (n+1) approaches 1 * 1 = 1, and the limit of a(n+1) / a(n) is therefore 1.

The radius of convergence is the reciprocal of this ratio, which is 1.

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21:29:00

What is your expression for the general term of this series, and how did you use this expression to find the radius of convergence?

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RESPONSE -->

The general term is x^n/(2n+1)

For the general term Cn (x - a)^n

The lim n-> infinity |a n+1| / |a n| = K | x - a |

and K = 1/R

In this case a = 0

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21:38:10

Query 9.5.28 (3d edition 9.4.24). What is the radius of convergence of the series p x + p(p-1) / 2! * x^2 + p(p-1)(p-2) * x^3 + and how did you obtain your result?

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RESPONSE -->

The general term is p(p-1)(p-2)...(p-n)x^n / n!

|x| lim n->infinity

|(p(p-1)(p-2)...(p-n+1) / (n+1)!)| / |(p(p-1)(p-2)...(p-n) / n!)|

The factorials reduce to n+1 and all of the (p - 1)(p-2)... factors reduce to (p -n +1).

The result leaves |x| lim n->infinity |(p - n +1)|/|(n+1)| = |x| * 1 meaning K = 1 and R = 1/K = 1

So the radius of convergence is 1.

*&*& Using the more economical notation:

As seen in 9.4.6 we have

a(n) = p ! / (n ! * (p - n) ! ) so

a(n+1) = p ! / [ (n+1) ! * ( p - (n+1) ) ! ] and

a(n+1) / a(n) = { p ! / [ (n+1) ! * ( p - (n+1) ) ! ] } / {p ! / (n ! * (p - n) ! ) }

= (n ! * ( p - n) !) / {(n+1)! * (p - n - 1) ! }

= (p - n) / (n + 1).

This expression can be written as

(p / n - 1) / (1/n + 1). As n -> infinity both p/n and 1/n approach zero so our limit is -1 / 1 = -1.

Thus the limiting value of | a(n+1) / a(n) | is 1 and the radius of convergence is 1/1 = 1. *&*&

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Your work is excellent. See my notes for comparison, and for a little additional information.