Assignment 13

course Mth 174

vЅyY}tIǵassignment #013

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Physics II

07-21-2008

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21:19:36

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 -->

I concluded that 1 / (3^n + 1) <= 1 / (3^n) because the larger the denominator, the lower the value. From n=1 to n=inf, the series converges because sum(1 / x^p), p, 1, infinity) converges when p >= 1. And for all n values from 1 to infinity, they are greater than or equal to 1.

Good work. Your solution is more to the point than the student solution I generally post. For comparison:

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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21:19:43

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

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21:19:56

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 -->

lim(n -> inf) abs[a(n+1)] / abs[a(n)] = L. Converges when L < 1.

a(n) = 1 / (2n)!

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

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

= (2n)(2n - 1)(2n - 2).... / (2n + 1)(2n)(2n - 1)(2n - 2)....

I cancel out (2n)(2n - 1)(2n - 2)..... to get 1 / (2n + 1)

limit as n approaches infinity 1 / (2n + 1) = 0

So the series converges.

Very good. FYI:

*&*& 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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21:20:10

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 -->

I didn't really grasp this part of the section. I know that a series converges if 0 < a(n+1) < a(n) and lim (n -> inf) a(n) = 0. In this series I set a(n) = (-1)^n * 10^(-n) and a(n +1) = -1^(n + 1) * 10^(-n + 1)

Limit as n approaches infinity = 0 so it converges.

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21:20:20

What are the first five partial sums of the series?

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RESPONSE -->

The first five partial sums are 1, .9, .91, .9109, .91091

Good. For comparison and to see the final 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.

This proves convergence.

Its worth looking at the partial sums. There is a pattern to these partial sums:

The first member of the series is 1.

The second member is 1 - .1 = .9.

The third member is 1 - .1 + .01= .91.

The fourth member is 1 - .1 + .01 - .001 = .909.

Continuing this process we see that the members of the sequence 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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21:20:35

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) / 3! * x^3 + ?

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RESPONSE -->

Since each term, after the first term, has a denominator I figured it to be (1 / n!). Each term is also increasing by x^n. The other part I was unsure of, but my guess was (p - n)!

So my general term is: [p * (p - n)! * x^n] / (n + 1)!]

Right idea. See the following for all the details:

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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21:20:45

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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My general ratio for this is [n * x^n] / [1 + 2n]

As n -> infinity, the limit of abs[a(n) + 1] / abs[ a(n)] = zero so raius equals infinity.

The denominator is 1 + 2 n. The numerator is n; the numerator does not include x^n. You didn't work out the details of your limit, but you'll understand 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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21:20:50

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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21:54:31

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 -->

I found the general term to be (p * p! * x^n) / n!

To find the radius I put it in terms of abs(a(n) + 1) / abs(a(n))

a(n) = (p * p! * x^n) / n!

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

abs(a(n) + 1) / abs(a(n)) = abs[p * p! * x^(n + 1) * n!] / abs[p * p! * x^n * (n + 1)!]

limit as n approaches infinty is abs[p * p! * x] / abs[p * p! * (1)!

I know the rules for finding a radius, but I am stuck here and do not know where to go.

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*&*& 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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This looks good. See my notes. Let me know if you have any questions. &#