Question

can anybody help me solve this problem. i am supposed to convert the limit to definite integral but i m stuck with the integration part. problem is attached in the picture

Answer 1

multiplying a sequence eh .... havent played on that playground yet

Answer 2

This is a weird looking expression. The factor being multiplied does not depend upon n, it is just a constant. So you are being asked to multiply together n copies of a number. If the number is positive and less than 1, then the product of n copies goes to 0 as n increases. If the number is positive and greater than 1, then the product of n copies goes to +infinity as n increases. Does that help?

Of course the problem may be mis-stated, perhaps a typo in the assignment paper.

Good luck!

Answer 3

id assume that either a or r is the ...variable of producation? lol

Answer 4

a is prolly a constant if i see it right

Answer 5

ok, my prior remarks were garbage. I'll give it another try. You are right about roles of r and a.

Answer 6

i m very sorry. i forgot one imp condition
na = 1 always

Answer 7

a is a constant

Answer 8

Supposing a is constant (not depending on n), then the r-th factor would be (1+r^2a^2) to the 1/r-th power, which expands to 1 + (1/r)*(r^2a^2) + positive stuff = 1 + ra^2 + positive, and multiplying those together gives product going to infinity as n increases.

If instead the value of a is to change as n increases, ie na=1, then a is not constant, and get something else -- haven't tried to look at that.

What is the integration approach you mentioned, for calculating this limit?

Answer 9

WolframAlpha can't do it...are you sure that is the correct problem?

Answer 10

\[\lim_{n \rightarrow \infty} \sum_{r=1}^{n}[(1/n).\phi(r/n)] = \int\limits_{0}^{1}\phi(x).dx\]

Answer 11

this is the formula i m supposed to use in order to solve that problem

Answer 12

if you replace a with 1/n and the exponent 1/r with 1/n then I can do it.

But, if WolframAlpha can't do it (the problem as written) then I'm not going to try ;)

Answer 13

OK, thanks Abhi ... Here's a first take on the problem, then.

You want to use the formula you posted 3 messages above, which is just the definition of the Riemann integral of phi(x) over interval (0,1). I'm going to write f(x) instead of phi(x) so I don't have to worry about math-typesetting complexities.

What you're given in the problem is a product of n terms, and the integral is defined as (limit of) a sum of n terms. So take natural-log (log in my notation) of the product to turn it into a sum of n terms. If g(n) is the original product for particular n, then log(g(n)) is sum (r from 1 to n) of log( (1 + r^2/n^2)) ^ (1/r)) terms. (Using a = 1/n)

The exponent 1/r within the log can be brought out in front of the log, and you get each term of the sum equaling (1/r)*log( 1 + r^2/n^2).

But for your limit in definition of Riemann integral, you need 1/n not 1/r, so write the 1/r as 1/n times n/r. Now things are starting to look better. You have 1/n, which is part of the definition of Riemann integral. You have n/r, which is 1/(r/n). And you have, within the log, r^2/n^2, which is (r/n)^2.

Define t = r/n, so t represents a fraction between 1/n and n/n, ie a position within the (0,1) interval. The expression now becomes sum (r = 1 to n, or t = 1/n to n/n) of 1/n * ( log (1+t^2)/t).

So you are looking to integrate log (1+t^2)/t over the interval (0,1).

When you get done, and have a value for the integral, you have to take exp (answer) to invert for the fact that you took logarithms earlier.

Whew! That is a complicated problem!

I'm curious what course you are taking, what university or textbook, poses that problem? It is certainly challenging. Good for you!

Answer 14

@krobe8 : wow.. that was an excellent explanation. i ve gotten as far as u got but i m stuck with that ' integrate log (1+t^2)/t over the interval (0,1) ' part. could you pls help me with this one part and thanks a lot. this problem is from ' Integral Calculus ' by Shanti Narayan

Answer 15

Yes, that integral was itself complicated. I've been meaning to check my work by doing a bit of Gnu-Octave (= free Matlab) computation, but have not gotten around to it. So this is unchecked -- use at your own risk! Or bettter, please check and confirm. I would appreciate it!

To integrate log(1+t^2)/t, the way I proceeded was to set u=t^2, and obtain 1/2*log(1+u)/u also over the interval (0,1). Not really much help. But then, expand log(1+u) in a Taylor series, divide each term by u, integrate term by term, and evaluate at 1 less at 0. The result is 1 - 1/4 + 1/9 - 1/16 ... which (quick check of tables, eg the old edition of CRC math tables, 22nd edition, pg 477 (at end of the "series" section) equals pi^2/12.

So the original limit would be exp(pi^2/24) is my calc. But I'm a very sloppy calculator and would want to check my work again, and also to do a bit of Gnu-Octave computing to be confident, if I were turning it in for an assignment.

My library does not have Narayan's book, unfortunately. But it sounds very good, ie challenging. Should be good for expanding one's mind!

Best wishes,
Ken R.

Answer 16

wow... u r the only one to get the answer among the people i consulted. where can i find more info on that pi^2/8 expansion procedure?. thanks a lot

Answer 17

You can consult "Differential and Integral Calculus", vol 1, by R. Courant, the "new revised" 2nd edition of 1937, pp 440-441, for example, and probably other books.

The method is to consider the even function f(x) = x^2 defined on (-pi,pi) and expand it in a Fourier series. It is even, so you actually get a cosine series -- the sine terms in the Fourier series integrate to zero. The cosine series (for x between -pi and pi) is x^2 = pi^2/3 - 4*(cos(x)/1 - cos(2x)/4 + cos(3x)/9 - cos(4x)/16 + ...). Setting x=0, all the cosines become 1, and you end up with 1 - 1/4 + 1/9 - 1/16 ... = pi^2/12.

I just looked it up. Not my original work!

Answer 18

just use integration by parts

Answer 19

@krobe8 : THANK YOU