course Mth 174
I have been extremly busy with everything at school, and finishing up my chemisery course too. I will be doing an assignment a day untill I get caught back up, I also will be doing test 2, I'v been brushing up on my integrals and antiderivatives.
assignment #013Åô•ªgïŒXÜ®šž’òÒÞ€ãݛ⑤w¯
Physics II
04-14-2008
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11:11:46
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 -->
As n gets bigger the number that is being added continues to get smaller, so we are increasing at a decreasing rate so the function will converge to 1 because each time we add a smaller and smaller number to the sum.
Everything you say would be true if the sequence was 1 / n instead of 1 / (3^n + 1); however 1 / n would diverge. As you indicate below, comparison with 1 / 3^n does establish convergence, but you need to spell out the details.
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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11:13:10
With what known series did you compare this series, and how did you show that the comparison was valid?
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RESPONSE -->
this would very close to 1/3^n and I compared the function to that.
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14:30:37
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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I used the ratio test to test out this problem. using (1/(2n!)+1 / 1/(2n) and got a infinate L ergo the function diverges.
You wouldn't have 1/(2n!)+1; rather you would have 1/(2 (n+1)!)/
*&*& 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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14:31:55
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 -->
this is very similar to the first question in the assignment, increasing at a decreasing rate. It would seem to converge to 1.
This series will converge. In fact the series converges absolutely--the absolute value series 1 + .1 + .01 + ... is a geometric series with ratio 1/10 so that it converges to 1 / (1 - 1/10) = 1/ (9/10) = 10/9 = 1.11111 ... . Of course it's obvious from the series itself that the convergent value is 1.11111... .
An alternating series with decreasing absolute values is convergent (e.g., even though the 1/n series diverges, the alternating series (-1)^n * (1/n) converges). So this series is clearly convergent.
More formally,
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 series is actually a geometric series with common ratio -1/10, so it converges to 1 / (1 - (-1/10) ) = 1 / (1.1) = .9090909... .
This result can also be seen from the pattern of 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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14:33:32
What are the first five partial sums of the series?
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RESPONSE -->
1- .1 + .01 - .001 + .0001 = .9091
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14:41:46
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 -->
(-1)^n ((p(-1^n(p-n)))/ n!) x^n
Your signs of grouping are a little ambiguous, but it's clear that you have the right idea. You are also missing a couple of factorial signs:
** 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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14:51:59
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 -->
because using the ratio test I got an R value that converges towards 0 the radius of convergence is infinite.
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14:55:33
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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nx^n / (2n+1) I used this in the ration test and calculated the calculated that the L value would converge towards 0 and thus the radius would be infinite.
Nearly the right expression but this is the coefficient for the nth term; you need to find the limiting value of the ratio of the n+1 term to the nth term. Note that the limiting value of n / (2n + 1), though, would be 1/2.
The limiting value of the correct ratio is 1:
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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15:01:31
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 -->
By using the ratio test we can see that when L= 0 at n= 4 (assuming p &x =2) then the radius of convergence is 4. It will converge when 4>n>-4.
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*&*& As seen in a previous problem 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. *&*&