Assignment 13

course Mth 174

ŕ]܇𗏐Zjassignment #013

د`SB

Physics II

06-22-2009

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19:05:40

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 -->

I compared 1/(3^n + 1) to 1/3^n, which converges as long as n > 1. If n > 1 then 1/(3^n + 1) < 1/3^n

Good. More detail:

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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19:07:10

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 1/(3^n + 1) to 1/3^n. The limit of 1/3^n is 0

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19:15:55

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 -->

ratio test = an+1/ an

= limit as n>infinity of (1/(2n +1!)) / (1/2n!)

= limit as n>infinity of 2n!/ 2n + 1!

= limit as n>infinity of 1 / 2n! = 0

Limit = 0, so R = infinity

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19:25:07

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 -->

an = (-1)^n (n +1)

a(n +1) = (-1)^(n +1) (n +1) +1

In this case an > a(n + 1) so it converges

Good, but a little more detail in analyzing that ratio would be in order.

*&*& 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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19:29:28

What are the first five partial sums of the series?

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RESPONSE -->

S1 = 1

S2= S1 - 0.1 = 1 - 0.1 = .9

S3= S1 - S2 + .01 = 1 - .9 + .01 = .91

S4= S1 - S2 + S3 - .001 = 1 - .9 + .91 - .001 = .909

S5 = S1 - S2 + S3 - S4 + .0001

= 1 - .9 + .91 - .909 + .0001 = .9091

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19:37:11

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 -->

(px^n) ( p -1) / 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^2 with a(n) = p ! / (n ! * (p - n) ! )

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19:49:18

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 -->

an = ((k + 1) (x^ n+1)) . (n + 2)

an + 1 = ((k + 1)(x^ n + 2)) / (n + 3)

ratio test = (an + 1 )/ an so:

limit as n > infinity [((k + 1)(x^ n + 2)) / (n + 3)] / [((k + 1) (x^ n+1)) . (n + 2)]

= limit as n > infinity[( k + 1) (x^n + 2) ( n + 2)]/ [( k + 1) (x ^n+1)(n + 3)]

=limit as n > infinity (x^n+1) (n/3 + 2/3) = infinity

Since the limit = infinity, R = 0

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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19:49:30

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 -->

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19:59:34

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 -->

an = [(px^(n +1) (p - 1))/ (n + 1)]

an+1= [ px^((n + 2) (p-1))/(n + 2)]

ratio = an+ 1/ an so:

limit as n > infinity of [ px^((n + 2) (p-1))/(n + 2)] / [(px^(n +1) (p - 1))/ (n + 1)]

= limit as n > infinity of [px^(n + 2)(p-1)(n+1)]/ [px^(n+1) (p -1) (n + 2)

= limit as n > infinity of [px^(n + 1)(p -1)]/ (n +1) = infinity

The limit = infinity, so R = 0

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