evaluate this
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is this forum alive??? where are all the smart alecs
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is this even in a level syllabus or sth?
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Originally posted by valentinoo:
is this forum alive??? where are all the smart alecs
I teaching tuition outside mah -
Originally posted by rs rs:
is this even in a level syllabus or sth?
This is called a double integral
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Originally posted by valentinoo:
do the inside first, and thread y like a constant
so, for the inside integral, after integrating, you get
( ln |(1-xy)| )( 1/(-y) ) from 0 to 1
Putting in the values, you get
(-1/y)( ln |1-y| - ln (1) ) = (-1/y)( ln |1-y| )
Our next integral is integrating y from 0 to 1, so 1-y will always be greater or equals to zero => |1-y| = (1-y)
this is a special integral, and I admit I forgot how to do it actually... Found the answer online somewhere, and adapted it here... we can use power series to solve it
since we integrate (-1/y)( ln (1-y) ) from 0 to 1
http://thesaurus.maths.org/mmkb/entry.html?action=entryById&id=2705(-1/y)( ln (1-y) ) = x + x^2/2 + x^3/3 + ...)/x = 1 + x/2 + x^2/3 + ....
Integrating from 0 to 1 we get 1 + 1/2^2 + 1/3^2 + ...
This is the same as summing 1/n^2 from 0 to infinity, and whose answer is pi^2 / 6
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are u sure it's in a level syllabus? im quite sure i've never seen this before...
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Originally posted by rs rs:
are u sure it's in a level syllabus? im quite sure i've never seen this before...
I think is uni first year maths -
oh. okay. (: long way to go.
good luck for your uni anyways!
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i graduated le :D
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no la. i meant valentinoo!
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can you try and evaluate without using sumation 1/n^2 ?
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can
use MatLab
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how? can u show me? thanks eagle! u r good
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I get my answer from here:
http://www.math.umd.edu/~jmr/241/doubleint.html
My Matlab Window:
>> firstint=int('1/(1-x*y)','y',0,1)
firstint =
-1/x*log(1-x)
>> answer=int(firstint,'x',0,1)
answer =
1/6*pi^2 -
LOL.
i hate matlab~
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Originally posted by eagle:
do the inside first, and thread y like a constant
so, for the inside integral, after integrating, you get
( ln |(1-xy)| )( 1/(-y) ) from 0 to 1
Putting in the values, you get
(-1/y)( ln |1-y| - ln (1) ) = (-1/y)( ln |1-y| )
Our next integral is integrating y from 0 to 1, so 1-y will always be greater or equals to zero => |1-y| = (1-y)
this is a special integral, and I admit I forgot how to do it actually... Found the answer online somewhere, and adapted it here... we can use power series to solve it
since we integrate (-1/y)( ln (1-y) ) from 0 to 1
http://thesaurus.maths.org/mmkb/entry.html?action=entryById&id=2705(-1/y)( ln (1-y) ) = x + x^2/2 + x^3/3 + ...)/x = 1 + x/2 + x^2/3 + ....
Integrating from 0 to 1 we get 1 + 1/2^2 + 1/3^2 + ...
This is the same as summing 1/n^2 from 0 to infinity, and whose answer is pi^2 / 6
excellent!!!
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Originally posted by eagle:
I get my answer from here:
http://www.math.umd.edu/~jmr/241/doubleint.html
My Matlab Window:
hey eagle, thanks for that...do u know of any other way beside this 2?
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hi guys any solution? any other method and alternatives?
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can u try doing it using surface integral? contouuur integrral leads to back to summation
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Don't think I know any more methods liao...
I stopped at year 1 engineering maths... And forgotten quite a great deal...
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Originally posted by eagle:
Don't think I know any more methods liao...
I stopped at year 1 engineering maths... And forgotten quite a great deal...
can you get somebody to help me pls? -
erm... I guess you ask your lecturers or friends better liao...
Unless some maths major appear to help in this forum... But I guess you can't expect much for that to happen... The numbers are not a lot even in this country, let alone a forum...
Any reason why there's a need for using surface integrals to solve?