Assignment 13

course Mth 174

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Physics II

03-20-2007

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19:02:56

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 -->

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19:06:22

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 this series to 1/(3^n). Since 3^n grows without bound, the limit of 1/(3^n) as n>>infinity is 0, thus 1/(3^n) converges. For all n, 0<1/(3^n+1)<1/(3^n). Therefore, this is a valid comparison.

unbounded growth of the denominator does not ensure convergence. The p series 1 / x^.99 has an unbounded denominator but diverges.

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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19:22:58

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 -->

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

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

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

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

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

L = lim (n>>infinity) 1 / (4n^2+6n+2) = 0

Since L = 0, the series converges

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19:41:21

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 partial sums of this series are: 1, .9,.91,.909,.9091,.90909,.909091,.9090909,.90909091,.909090909, ...which suggests that it converges to 10/11.

Alternating series test: a(sub n) = 10^(-n) = 1/10^n

a(sub n+1) = 10^(-(n+1)) = 1/10^(n+1)

Therefore, 0

lim (n>>infinity) a(sub n) = lim (n>>infinity) 1/10^n = 0

Therefore, by the alternating series test, the series converges.

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19:41:37

What are the first five partial sums of the series?

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RESPONSE -->

See previous response.

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19:47:31

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 of the series is:

[p(p-1)(p-2)...(p-n+1)]/n! for n>=1

Good, but there is a slightly more elegant way to express the final result:

** 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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20:11:30

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 general term for this series a (sub n) = (n*x^n)/(2n+1)

a(sub n+1) = [(n+1)(x^(n+1))]/(2n+3)

lim (n>>infinity) |a(sub n+1)| / |a (sub n)|

= lim (n>>infinity) {[(n+1)(x^(n+1))]/(2n+3)} / (n*x^n)/(2n+1)

= lim (n>>infinity) [(n+1)(2n+1)*x] / [n(2n+3)]

= x * lim (n>>infinity) [2n^2+3n+1] / [2n^2+3n]

Since the n^2 term will dominate as n becomes very large, this is equivalent to 2n^2/2n^2 for very large n, thus = 1

lim (n>>infinity) = x*1

Therefore, the radius of convergence is 1.

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20:11:41

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 -->

See previous response.

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20:35:46

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 of the series is a(sub n) = [[p(p-1)(p-2)...(p-n+1)x^n] / n!

a(sub n+1) = [p(p-1)(p-2)...(p-n+1)(p-n)x^(n+1)] / (n+1)!

lim (n>>infinity) |a(sub n+1)| / |a(sub n)|

= lim (n>>infinity) |p(p-1)(p-2)...(p-n+1)(p-n)x^(n+1)| / (n+1)! / |p(p-1)(p-2)...(p-n+1)x^n] / n!|

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

= x * lim (n>>infinity) |(p-n)| / |(n+1)|

As n gets very large we get |-n| / |n|, so

lim (n>>infinity) = x * 1, so R=1

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Assignment 13

unbounded growth of the denominator does not ensure convergence. The p series 1 / x^.99 has an unbounded denominator but diverges.

Excellent work. See my notes.