My teacher had marked the following solution as a half-done solution.

Is this solution someway incomplete or incorrect ?

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- May 13th 2007, 02:32 PMtotalnewbieCould somebody check this integral ?
My teacher had marked the following solution as a half-done solution.

Is this solution someway incomplete or incorrect ? - May 13th 2007, 03:19 PMCaptainBlack
- May 14th 2007, 01:35 AMtotalnewbie
Then why do I get the negative answer ?

Your and my result also differs a few (0,191 vs. 0,211) considering absolute values. - May 14th 2007, 03:23 AMCaptainBlack
I can't tell you why the answers are different, without explanation

I cannot follow what you think you were doing.

Also given that my answere is cprrect to 10 digits, and yours in supposedly

analyticaly derived being close in wjhatever sense is meaningless. the answers

are different, when they should agree to within the precision of the calculations.

RonL - May 14th 2007, 04:57 AMCaptainBlack
OK with some work I can make some sense of what you have written.

You have:

ln(1+x) = sum_{k=0 to infty} (-1)^k x^{k+1}/(k+1),

so:

(1/x)ln(1+x/5) = sum_{k=0 to infty} (1/x) (-1)^k (x/5)^{k+1}/(k+1),

and you proceed to integrate term by term to get:

integral_{x=0 to 1} (1/x)ln(1+x/5)

.................. = sum_(k=0 to infty) (-1)^k x^{k+1}/[5^{k+1}(k+1)(k+1)] |_0^1.

.................. = sum_(k=0 to infty) (-1)^k/[5^{k+1}(k+1)(k+1)]

lets assume this is correct.

How do you evaluate this:

sum_(k=0 to infty) (-1)^k/[5^{k+1}(k+1)(k+1)]

infinite sum which is the integral you require? - May 14th 2007, 05:30 AMCaptainBlack
OK lets have a go at evaluating this sum:

sum_(k=0 to infty) (-1)^k/[5^{k+1}(k+1)^2]

........... = 1/[5] - 1/[5^2*2^2] + 1/[5^3*3^2] - ..

........... ~= 0.190889

Now the error in this is of the same order as the first ignored term or ~-0.0001,

so your integral is 0.191 correct to 3 significant digits, which agrees with my

calculation.

RonL