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## #1 2006-05-11 23:39:21

doraeyee_u_v
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### integration

evaluate the limit

lim(x->0)∫(integrate from 2x to x) [(sint t)^m/t^n] dt

where n,m are positive integers.

hope you can read the quation.

## #2 2006-05-12 01:27:42

ganesh
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### Re: integration

doraeyee_u_v's question is

Character is who you are when no one is looking.

## #3 2006-05-12 04:59:25

John E. Franklin
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### Re: integration

There's no x in the equation, so what does it mean?  Maybe x should be t and then it may depend if the sine is in radians or degrees, but the answer is likely zero the the area under the curve from 0 to 0, unless n is way bigger than m.   Forget about n bigger than m, that part might be wrong.  If x is t, then nearly 0 to the power of number 1 or above makes the number get even smaller because a root would make near zero approach a tiny ways toward one (and this is a power, not a root.)
But the numerator is also getting smaller by m power and the sin(t) in radians is approximately t for small t near zero.
Oh maybe I was right, if n is way bigger than m, then perhaps there is a value, like if m is one and n is near infinity.
But still, I think the area is zero.
And ofcourse this all presumes that x and t are one and the same variable.

Last edited by John E. Franklin (2006-05-12 05:09:23)

igloo myrtilles fourmis

## #4 2006-05-12 09:42:34

doraeyee_u_v
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### Re: integration

the question is like what ganesh wrote, this is the question i meant to post but i couldn't type it probably.x shouldn't be same as t, should be independent of t. i think the question want me to do the integration, and then put 2x and x in as the answer, and then limit x->0.

## #5 2006-05-12 13:30:05

George,Y
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### Re: integration

Franklin wrote: but the answer is likely zero the the area under the curve from 0 to 0, unless n is way bigger than m- Right

when n is just 1 larger than m, the limit does exist and is not 0.

before x reached 0, we can treat x as a small amount. 2x is small too.
then the integral can be approximately treated as the area of a trapezoid, with height x=2x-x,
the left base (sinx)^m/x^n, the right base (sin2x)^m/(2x)^n.

Since x and 2x are very small. the left base can be treated as
(x+o(x))^m/x^n= (both*1/xm)= (1+o(x))/x
and the right base
(1+o(x))/2x

Hence the area is
[(1+o(x))/x+(1+o(x))/2x] x/2= 3/4+o(x)

_______________________________________________________
Special Note for o(x)
o(x) is a variable dependent on x which satisfies Limit x->0 {o(x)/x}=0.
Hence o(x) is always much smaller than x, for example x²,x³ or sin²x
o(x)+o(x)=o(x)
o(x) o(x)= atleast o(x) or exactly o(x²)
o(x) anyC =o(x)
o(x) x/x =o(x)
_______________________________________________________

Let's now have a look at the whole limit of 3/4+o(x) - it approaches 3/4 when x gets smaller and smaller, and the limit should be 3/4

when n=1+m, 3/4
when n<1+m, no sense or infinity.
when n>1+m, 0.

Last edited by George,Y (2006-05-12 15:22:16)

X'(y-Xβ)=0

## #6 2006-05-14 10:22:33

doraeyee_u_v
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### Re: integration

is anyone able to do the integration and ignore the limit, i.e. just add a constant in the end.

## #7 2006-05-14 21:11:11

George,Y
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### Re: integration

I admit my method is wrong, because the trapezoid fails to approximate even

So is the answer ln2 instead? It depends on whether (sint)n/tn+1 can be treated as 1/t.

Last edited by George,Y (2006-05-14 21:12:53)

X'(y-Xβ)=0

## #8 2006-05-14 21:34:21

George,Y
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### Re: integration

It took my software 2 minutes to compute n=7 case and 7 minutes to compute n=18 case. Although it didn't return ln2, the numerical approximation are both 0.693147  ! ! !

X'(y-Xβ)=0

## #9 2006-05-14 21:39:50

George,Y
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### Re: integration

After simplification they are both ln2 (log2) !!!

X'(y-Xβ)=0

George,Y
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WEIRD

X'(y-Xβ)=0

## #11 2006-05-14 21:57:20

George,Y
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### Re: integration

The paradox is that the limit sign panetrated through the integral sign and left an incomplete infinity 1/x, and continue to influence following steps.

Ricky, prove why that's valid.

X'(y-Xβ)=0