University calculus help

[Eclectic Dreck]

-.- My prof did this question wrong, so if it ends up on the final, which is likely, I won't be able to do it properly:

http://www.wolframalpha.com/input/?i=integral+%28sinx%29%2Fx+from+pi%2F4+to+pi%2F2+%3C+sqrt2%2F2

Wolfram will not show the steps to proving it -.-

Because (sin (x))/x does not have an antiderivative that can be expressed in terms of an elementary function, the only way to come up with a formula for the integral is to express it as a power series, then find the integral of that power series.

As for how to show that the integral is less than sqrt(2)/2... No idea. Sorry.

There is one relatively simple way to demonstrate this.

When integrating, all we are really doing is finding the area under a curve. What we need to determine is if the total area enclosed by the functions y = sin(x)/x, x = pi/4, x = pi/2 is greater than sqrt(2)/2.

This is simple enough and you do not need to resort to calculus to demonstrate this. Indeed, using calculus to demonstrate this is more rigorous or far more difficult for the very reason you mention: there is no anti-derivative and the only way to find an answer is numerical (and a pain without a computer at that!).

Consider for a moment the simpler function f(x) = sin(x). This has an easily computed anti-derivative. In the interval mentioned, the area is exactly sqrt(2)/2.

Dividing this area by any number greater than 1 will obviously result in a reduction. Given that the value in the denominator is linear and further given that we are going to produce an infinite number of rectangles (a riemann sum), you could demonstrate (and I'm not about to actually see if I can find a place to enter proper notation) that while there are certainly infinite values of x that will result in a larger rectagle over the same part of the x-axis, there are relatively more that will cause a reduction. (pi/4 ~ .79 if memory serves and pi/2 ~ 1.57). As such, it seems that one could simply demonstrate that you are reducing the area of sin(x) more often than you are increasing it - a fact born out if you actually graph sin(x)/x! At the moment, given that it is far to late (or early), I cannot decide if it would be easier to prove inductively or deductively, but I suspect this logic could be used in a proof of a sort.

-edit- inductive is probably best. You can demonstrate that (at arbitrary granularity) below a threshold value of x the upper limit of the curve increases. At a certain tipping point it decreases. Because of the linear nature of the denominator (the part that makes induction simple here) you could show that after the tipping point you have a reduction. The Inductive Hypothesis is therefore that the total reduction in the upper limit will be greater than the total increase. Since there is more length available on the reduction side, given the linear nature of the denominator, reducing the granularity of the function would not change this result.

Alternately you could simply resort to summing rectangles to demonstrate your inductive hypothesis and then use a smaller set. As of right now, it seems like either would produce a compelling proof but the latter is more likely rigorous.

-edit 2-

Really thinking about it, a Riemann sum is probably the simplest way in general. It is far more compelling and while my simpler argument makes sense geometrically at least I don't think the argument is sound enough.

 

[Lukeje]

Really thinking about it, a Riemann sum is probably the simplest way in general. It is far more compelling and while my simpler argument makes sense geometrically at least I don't think the argument is sound enough.

Screw it; why bother with Riemann sums when you could instead use Lebesgue integrals?

What the prof tried:

pi/4 < x < pi/2
2/pi < 1/x < 4/pi
2/pi sinx < sinx/x < 4/pi sinx

Then integrate both sides, but you get 2sqrt2/pi -.-

If your professor thought that this would work (and its a method taught on your course) then that would suggest that there was a typo in the question; especially since they're asking for an upper bound (unless of course you have been taught a different way of dealing with such questions).

 

[Eclectic Dreck]

Screw it; why bother with Riemann sums when you could instead use Lebesgue integrals?

Because you could produce a rigorous proof demonstrating that the integral was less than sqrt(2)/2 using a Riemann sum with less work was my logic. You might not get an exact answer, but that was not what the question was being asked. You just need to demonstrate if the inequality is true (or not). Giving an accurate answer is (it would seem) the harder way to go about this.

 

[Brutal Peanut]

*drools on self* Duuuuuuh,....

 

[Jewrean]

If you asked me this 4 years ago I would answer you no problem.

 

[Lukeje]

Screw it; why bother with Riemann sums when you could instead use Lebesgue integrals?

Because you could produce a rigorous proof demonstrating that the integral was less than sqrt(2)/2 using a Riemann sum with less work was my logic. You might not get an exact answer, but that was not what the question was being asked. You just need to demonstrate if the inequality is true (or not). Giving an accurate answer is (it would seem) the harder way to go about this.

It was meant to be a joke; sorry. I've been doing way too much analysis recently.

 

[Eclectic Dreck]

Screw it; why bother with Riemann sums when you could instead use Lebesgue integrals?

Because you could produce a rigorous proof demonstrating that the integral was less than sqrt(2)/2 using a Riemann sum with less work was my logic. You might not get an exact answer, but that was not what the question was being asked. You just need to demonstrate if the inequality is true (or not). Giving an accurate answer is (it would seem) the harder way to go about this.

It was meant to be a joke; sorry. I've been doing way too much analysis recently.

Just as well. After looking into the thing you mentioned (to see if it was something I knew how to do only using a different name), a proof would be fundamentally similar. I don't think you'd get anything for using that form unless you were dealing with some very exotic function or perhaps a series or something.