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**doraeyee_u_v****Member**- Registered: 2006-05-02
- Posts: 13

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.

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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 15,405

doraeyee_u_v's question is

Character is who you are when no one is looking.

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**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,585

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 (2006-05-11 07:09:23)*

**igloo** **myrtilles** **fourmis**

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**doraeyee_u_v****Member**- Registered: 2006-05-02
- Posts: 13

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.

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**George,Y****Member**- Registered: 2006-03-12
- Posts: 1,306

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/x[sup]m[/sup])= (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 (2006-05-11 17:22:16)*

**X'(y-Xβ)=0**

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**doraeyee_u_v****Member**- Registered: 2006-05-02
- Posts: 13

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

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**George,Y****Member**- Registered: 2006-03-12
- Posts: 1,306

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

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

*Last edited by George,Y (2006-05-13 23:12:53)*

**X'(y-Xβ)=0**

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**George,Y****Member**- Registered: 2006-03-12
- Posts: 1,306

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**

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**George,Y****Member**- Registered: 2006-03-12
- Posts: 1,306

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

**X'(y-Xβ)=0**

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**George,Y****Member**- Registered: 2006-03-12
- Posts: 1,306

WEIRD

**X'(y-Xβ)=0**

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**George,Y****Member**- Registered: 2006-03-12
- Posts: 1,306

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**

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