Assignment 13 Query

course Mth 174

s݂k掼eassignment #013

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VJ]

Physics II

11-09-2007

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11:33:26

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 -->

Looking at this sum, I can compare it to the sum of (1/3)^n. Looking at the limit of (1/3)^n indicates that as the value of n grows, the value of the sum grows toward 1/2 and seems to stay there.

(1/3)^n is a geometric series with a definite limit of 1/2. Just looking at the values of the partial sums can be compelling but it doesn't prove the limit.

However you know that a geometric series with ratio less than 1 converges, so your argument is correct.

Since (1/3)^n converges and 1/(3^n+1) < (1/3)^n, 1/(3^n+1) therefore converges.

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11:38:56

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 with the series (1/3)^n.

To make sure the comparison was valid, I looked at the limit of (1/3)^n and then made sure that 1/(3^n+1) was less than (1/3)^n.

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12:04:02

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 -->

First I used the ratio test to see if the limit of this function was less or more than 1.

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

= lim( 1/(2(n+1)) * (2n)! / 1 )

= lim( (2n)! / ( (2(n+1))(2n)! ) )

= lim( 1 / (2(n + 1)) )

This can be split up into two limits:

lim( 1/2 ) * lim( 1 / (n+1) )

The limit of 1 / (n+1) is equal to 0, and the limit of 1/2 is 1/2. Multiplied together, the overall limit is 0. Therefore this interval converges.

Theorem 9.10 tells us that if the ratio between a_n+1 and a_n is 0, then the radius of convergence is infinite.

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12:21:17

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 wild guess answer is 1 because the sums are oscillating above and below 1 in smaller and smaller increments.

But to prove it, first I need to prove that it converges. To do this I need to make sure that each consecutive sum is less than the previous one:

1 > .1 > .01 > .001

So this seems to be true. Next I need to make sure that the limit of the formula is equal to 0:

lim( (-1)^n * (10)^-n )

This goes to 0 as the value of n goes towards infinity. So this function does converge.

Next the question is the value on which it converges. Using a calculator (and sum estimation), my estimate is the this sum converges on 1 -1/11 = .909.

Note the limitations of graphing calulators when it comes to convergence and divergence. For example, 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 isnt necessary to use a calculator to figure this out. It's easy enough to show that the pattern continues, so the convergent value is .9090909... .

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12:26:39

What are the first five partial sums of the series?

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RESPONSE -->

S_1 = 1 - (1/10) = 0.9

S_2 = 1 - (9/100) = .91

S_3 = 1 - (91/1000) = .909

S_4 = 1 - (909/10000) = .9091

S_5 = 1 - (.09091) = .090909

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12:41:49

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 only thing I could come up with is this:

(p - (n-1)) x^n / n!

** 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^n with a(n) = p ! / (n ! * (p - n) ! )

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13:35:06

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 -->

The first thing I found was the formula:

( n x^n ) / (2n +1)

Next I used the ratio test to find the ratio of the function:

lim( ( ( (n+1) x^(n+1) ) / (2(n+1) +1) ) / ( ( n x^n ) / (2n +1) )

= lim( ( (n+1) x^(n+1) ) / (2(n+1) +1) * (2n+1) / (n x^n)

= lim( (n+1)(2n+1)x / (2n+3)n )

= lim( (2n^2 +3n +1)x / 2n^2 +3n )

= | x | * lim( (2n^2 +3n +1) / 2n^2 )

For large values of n, the two 2n^2 terms will dominate, leaving a value like 2n^2 / 2n^2 = 1. Since the limit is 1, then K must also be equal to 1. Since a = 0, we can ignore it.

K = 1

R = 1/K

R = 1

Good, but you can be a little more rigorous in proving the 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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13:35:46

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 -->

My expression was:

n x^n / (2n+1)

I think I explained how I used it above.

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14:11:48

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 -->

Using the term I found before:

(p - (n-1) ) x^n / n!

Here is my result by using the ratio test:

lim( ( (p - ((n+1)-1) ) x^(n+1) / (n+1)! ) / ( (p - (n-1) ) x^n / n!)

= lim( (p - ((n+1)-1) ) x^(n+1) / (n+1)! ) * ( n! / (p - (n-1) ) )

= lim( (p - n ) x^n x / ( (n+1) n! ) * ( n! / (p - (n-1) ) )

= lim( (p-n)x / ( (p-(n+1))(n+1) )

= x * lim( (p - n) / ( -n^2 + pn + p) )

Since n / n^2 dominates, the function acts like 1/n for large values of n. But it is less than 1/n, so I can't compare it with 1/n to prove that it is divergent. Did I do something wrong with my original formula?

*&*& 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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Very good, but be sure to see my notes.