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        riddles >> easy >> Integrate Without Integration
        
(Message started by: rloginunix on Oct 11^th, 2016, 8:40am)

Title: Integrate Without Integration
Post by rloginunix on Oct 11^th, 2016, 8:40am

Without computing the (real) integrals, show that:

2 http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gif http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif(1 - x²) dx = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gifdx / http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif(1 - x²)

where the boundaries of integration are [ -1, 1] ~~[ -a, a ] and a > 0~~

(to be sure, the left is an integral of a square root while the right is an integral of one over (same) square root)

[e]
Correction: a == 1
[/e]

Title: Re: Integrate Without Integration
Post by Michael Dagg on Oct 11^th, 2016, 5:27pm

[hide] It will help to look at the graphs of the integrands, say,
f , g and their difference h = f - g . [/hide]

Title: Re: Integrate Without Integration
Post by dudiobugtron on Oct 11^th, 2016, 6:16pm

Some thoughts:

[hide]sqrt(1-x^2) is the graph of the top half of a circle of radius 1. (Just rearrange x^2 + y^2 = r^2). So the area under the curve is half of Pi*1^2.
The left hand side of rloginunix's equation is therefore equal to Pi.
It remains to prove that the right side of the equation also equals Pi. This is a bit tricky though!
As 1-x^2 approaches 0 when x approaches 1, the value of the function tends to infinity. So I think you can't really calculate the integral for it anyway. So I guess we need to discover an intuitive way of thinking about this shape too, and then defining and approximating the area under it.[/hide]

Title: Re: Integrate Without Integration
Post by rloginunix on Oct 11^th, 2016, 6:34pm

Michael Dagg: Simplify, by drawing less (conclusions). The key idea is there, well done: use its [hide]singular form[/hide].

dudiobugtron: Good job, you are also on the right [hide]path[/hide]. That was a hint, of course. Slow down a bit.

Title: Re: Integrate Without Integration
Post by Michael Dagg on Oct 12^th, 2016, 8:36am

Not necessarily. It is straight forward to rewrite the integrand of the difference to obtain a function that is non-singular and analytic throughout its domain. Once you get that function or one similar but slightly different, you will be convinced that the integrals equate.

Title: Re: Integrate Without Integration
Post by Grimbal on Oct 15^th, 2016, 12:23pm

[hide]On the right side you can replace x = sin y. The integral dissolves in thin air. [/hide]
But that means we compute both integrals and see they are equal. It look like the question was to transform the one into the other.

Title: Re: Integrate Without Integration
Post by rloginunix on Oct 15^th, 2016, 4:28pm

on 10/15/16 at 12:23:38, Grimbal wrote:

[hide] [ ... ] to transform the one [ ... ] [/hide]

Grimbal, as usual, is right on the money. [hide]As a hint, it is not the integral itself that should be transformed[/hide]. (I took the liberty to obscure the quotation of Grimbal's reply for obvious reasons)

Assuming dudiobugtron's left integral observation is successful (I would only add to it that the [hide]relation under that integral is a function since that relation's graph passes the vertical line test[/hide], add to it my earlier "slow down" hint ([hide]when it comes to jumping to the next conclusion[/hide]) and consider Grimbal's fruitful [hide]transformation[/hide] idea.

(if it helps: in the real time trial and error flow I of course did not see that, my perspicacity powers are quite modest, but after the fact I quickly realized that challenge-wise this problem is identical to "Infinitely Fast Ladder" - in the latter problem a psychologically uncomfortable, disturbing or unnatural idea had to be entertained. The current problem, of course, is not as involved)

Title: Re: Integrate Without Integration
Post by Michael Dagg on Oct 18^th, 2016, 8:41am

My first impression was that given all the very nice geometric artwork you've posted here that you were looking for a clever geometric solution. While there are doable geometrical arguments making use of symmetry for that difference, I don't consider any of them clever. In fact, going that route is not as simple as this:

If you make the substitution Grimble suggested (x = cos(theta) will work too), you end up with an integrand of 1. There are numerous ways to interpret the integrand 1. Since a trig substitution was used to transform the integrand, it should come as no surprise that the integrand 1 can be interpretated as the polar equation r = 1 for the unit circle. But the polar intepretation presents a slight problem because the new limits of integration [-pi/2, pi/2] only represent half of the unit circle.

A better interpretation that leads directly to the desired result is to let f(x) = 1 in rectangular coordinates over the interval [-pi/2, pi/2]. Thus the region bounded by f and the x-axis over [-pi/2, pi/2] is a rectangle of height 1 and length pi/2 - (-pi/2) = pi, thus having area pi.

Title: Re: Integrate Without Integration
Post by rloginunix on Oct 18^th, 2016, 7:06pm

Very well.

For reasons that escape me (some one please chime in) the viewership numbers of this problem are unusual: about 180 in the first 48 hours and climbing. For this and the full disclosure reasons I must say that I am not the author of this problem - it was given to me by a friend of mine and in a rather strict and narrow interpretation of "without computing" requirement. In retrospect it may have been better re-worded as "without touching or manipulating", but - you decide (I did try, erroneously, to get a bit too generic initially - as is evident by my corrections of the problem statement)

As Grimbal observed, the above substitution, carried all the way through, amounts to a not allowed computation. Mr. Dagg's latest demonstration uses that substitution as an intermediate step but stops short of the computation. I am not the author of the problem but I would say that it works. Further, the number of curves (or shapes) may be reduced, from two to one.

With a strict and narrow interpretation or looking for, subjectively, "simpler" explanation (fair warning: I am inflicting my way of thinking and my academic baggage on the audience, so take what I am about to say with a grain of salt) if we let f(x) = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif(1 - x²), for now completely disconnected from its parent (left) integral, then what operation or "operator" when applied to f(x) will [hide]push the radical from the numerator into the denominator[/hide]?

Title: Re: Integrate Without Integration
Post by Michael Dagg on Oct 18^th, 2016, 8:01pm

I think I understand your question to mean more than just mere reciprocation but as an equality.

One way is to write

1 - x^2 = 1 - x^2

then factor, say, the LHS as

sqrt(1 - x^2) sqrt(1 - x^2) = 1 - x^2

From this you can write

sqrt(1 - x^2) = (1 - x^2)/sqrt(1 - x^2)

which is what I think you wanted.

Notice you can also obtain

1/sqrt(1 - x^2) = sqrt(1 - x^2)/(1 -x^2).

Title: Re: Integrate Without Integration
Post by rloginunix on Oct 19^th, 2016, 7:19am

Yes, these are all true statements and you are getting there but the transformations undertaken should lead to the f(x)'s curve reuse.

We observe that either of two terms in your third equation (1/http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gifor x²/http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif) look very close to f(x)'s [hide]first derivative[/hide]. What will happen if you compute it and then attempt to find a way to make it look [hide]exactly like the right integrand[/hide]? Simple steps.

Title: Re: Integrate Without Integration
Post by Grimbal on Oct 20^th, 2016, 9:24am

We could compute Int( 2sqrt(1-x^2) +1 )dx and Int( 1/sqrt(1-x^2) +1 )dx and show that they are equal.
We don't compute the original integrals.

Title: Re: Integrate Without Integration
Post by rloginunix on Oct 20^th, 2016, 6:52pm

Very crafty, I must admit. Let us add this solution to the one above with an asterisk of "some integrals computed".

Apparently there exists an explanation that does not involve intermediate substitutions or other integrals computations. If we are willing to reuse the shape of f(x) then what [hide]it's other conveniently valued in the same boundaries thing can be exploited on the right side considering the hint in reply #10[/hide]?

Title: Re: Integrate Without Integration
Post by Michael Dagg on Oct 20^th, 2016, 7:01pm

I get that you seek to establish an inference between the integrands (or some parts of them) based on the premise that the integrals equal pi.

If that is doable, then as matter of equivalence, it should be true that it can be extended to include a rectangle of area pi. After all, that is one interpretation that produces the desired result.

But now that you want to work with f(x) = sqrt(1 - x^2) , f'(x) and the desire to produce 1/f(x) from f'(x), the situation is different. We can produce 1/f(x) from f'(x) exactly but seeking a substitution, say, x = tau(z) and solving f'(tau(z)) - 1/f(x) = 0 isn't the way to proceed.

You mention computing the definite integral of f'(x) over [-1, 1]. As a matter of simplicity, since we obtained f' from f to begin with, this calculation is easy. It turns out that f'(x) is yet another integrand not requiring integration to determine its value. A geometric argument will do since f'(x) is symmetric about the x-axis over [-1, 1]. On the other hand, since f'(x) is an odd function, it should be noted that the definite integral of a function over a symmetric interval is the definite integral of the even part of the function over the interval. Since f'(x) is odd, f'_even(x) = 0 .

Finally, we can transform f'(x) to 1/f(x) with sqrt(1 + [f'(x)]^2), that is, 1/f(x) = sqrt(1 + [f'(x)]^2).

This should come as no surprise since this transform is the integrand for the arc length of f(x) over [-1, 1] which of course is pi.

As for the "curve reuse" I'll let someone else take that.

Title: Re: Integrate Without Integration
Post by rloginunix on Oct 20^th, 2016, 8:10pm

on 10/20/16 at 19:01:40, Michael Dagg wrote:

You mention computing the definite integral of f'(x)

If that is how my comment was understood - I apologize, that was not my intention. When in Reply #10 I said "What will happen if you compute it ..." I thought that it was clear that only f'(x) should be computed, without integration.

"Curve or shape reuse" was just a hinting vehicle, my attempt to make it interesting for every one.

And yes, showing that the [hide]right integral happens to match the arc length of the very same curve square area under which is captured by its left counterpart[/hide] is what secures an explanation.

That is why I shared this problem - it takes quite an effort to [hide]unbuckle oneself from the automatic assumption that both sides stand for square area[/hide] and see just one curve instead of two.

(what still bewilders me is how many people viewed this thread. I always thought that analysis is considered boring and that the folks on this forum are more into "other math")

Title: Re: Integrate Without Integration
Post by SWF on Oct 25^th, 2016, 5:42pm

1/sqrt(1-x^2) - 2*sqrt(1-x^2) = (2x^2 - 1)/sqrt(1-x^2)

Integrating this from -1 to 1 must give zero because the term in the numerator, 2x^2-1=T_2(x), is a Chebyshev polynomial of the first kind. Such polynomials are orthogonal over -1 to 1 with weighting function 1/sqrt(1-x^2). Since T_0=1, this integral is like saying T_2(x) is orthogonal to T_0(x).

Or could substitue x= cos(t), and the integral becomes from 0 to pi of cos(2t), which (without integrating) clearly equals zero.