Wallis' Formulae and Estimation

Transcript

0:00Introduction

Hello everyone! One small challenge first. Find the integral from zero to pi over two of sine to the power six of x, d x. The usual way, with cosine two x formulas again and again, takes half a page, right? What if you could write the answer in one line? And a second puzzle. Some integrals, like e to the power x squared, cannot be found by any method you know. Yet a question can still ask: is it bigger than one? Today: Wallis' formulae, and estimation. Both come up often in JEE questions. So, let's begin.

0:37Wallis: powers of sine or cosine

First result: the integral from zero to pi over two of sine to the power n of x, d x. The long S is the integral sign, zero and pi over two are its limits, and the small n raised up is the power, a positive whole number. Okay? Now a question. If I change sine to cosine here, does the value change? It does not change at all! Remember the King property: replace x by pi over two minus x. The arrow on screen means replace. The limits stay the same, and sine of pi over two minus x is cosine x. So they are equal. Right? Now the formula. On top: n minus one, times n minus three, and so on. At the bottom: n, times n minus two, and so on. The three dots mean, keep going. Then the cross sign, times, and the factor in the big curly bracket: pi over two if n is even, just one if n is odd. Clear? Both lists go down in twos, and stop at two or one. And careful: this formula is only for the limits zero to pi over two. With me? Quick check. Sine to the power four: do we multiply by pi over two, or not? Yes, we do, since four is even! On top: three, then one. At the bottom: four, then two. Pair them up: three over four, times one over two, times pi over two. The dot means times. Three pi over sixteen, in one line. Got it?

2:17Mixed powers: sin^m x cos^n x from 0 to pi/2

Now, sine and cosine together: the integral from zero to pi over two of sine to the power m of x, times cosine to the power n of x, d x. Okay? On top, two lists, each in square brackets just to group it: m minus one, m minus three, and so on, and n minus one, n minus three, and so on. At the bottom, one list, starting from m plus n and going down in twos. Right? And the last factor. Pi over two comes in only when both powers are even. If even one is odd, multiply by one. Read that word both carefully. Clear? Try sine to the power four, times cosine squared. Where does the bottom list start? At six, yes, exactly! Four plus two. So the bottom is six times four times two, forty eight. On top: three times one for the four, and just one for the two. Both powers are even, so times pi over two. Three over forty eight is one over sixteen, so we get pi over thirty two. Beautiful, right?

3:27Estimation: squeeze the integral

Now the second idea: estimation. We may not find an integral exactly, but we can still trap it. Suppose f of x stays between small m and capital M on the closed interval a to b, shown in square brackets, with a less than b. This sign means less than or equal to. Careful: small m is now a lower bound for f, and capital M an upper bound, not powers. Okay? Now think. If f never goes below m, how small can the integral from a to b be? m times b minus a, correct! That is the integral of the constant m. The same way, on top: capital M times b minus a. So the integral is squeezed between them. Clear? Second rule: comparison. If f of x is less than or equal to g of x on the interval, the integral of f is less than or equal to the integral of g. Right? Third rule. The straight bars mean modulus. The modulus of the integral of f is at most the integral of mod f of x. Why? Inside the integral, positive and negative parts can cancel; with the bars on f, nothing cancels. Got it?

4:45Example 1: sin^6 x and cos^7 x from 0 to pi/2 by Wallis

Example one: our challenge from the start. Evaluate the integral from zero to pi over two of sine to the power six of x, d x, and the same integral of cosine to the power seven of x. Pause the video and try it first. Done? Okay, let's check. Sine to the power six: n is six, which is even. So pi over two comes in at the end. Okay? Pair them up: five over six, times three over four, times one over two, times pi over two. On top, fifteen. At the bottom, forty eight. Fifteen over forty eight is five over sixteen, and times pi over two gives five pi over thirty two. Our half page challenge, done in one line! Now cosine to the power seven. Same pattern, so we multiply by pi over two again at the end, right? No, no, careful here! Seven is odd, so the last factor is just one. Adding a pi over two here is the classic slip. Clear? So: six over seven, times four over five, times two over three. On top, forty eight. At the bottom, one hundred and five. That is sixteen over thirty five, with no pi at all. That's it!

6:05Example 2: e^(x^2) from 0 to 1 lies between 1 and e - 1

Example two, our second puzzle. Show that one is less than the integral from zero to one of e to the power x squared, d x, and that it is less than e minus one. Pause the video and try it first. Done? Okay, let's check. The lower bound first. The round brackets, zero comma one, mean x is strictly between zero and one. There, x squared is positive, so e to the power x squared is greater than e to the power zero, which is one. Right? Now compare, integrating both sides from zero to one. The integral of one is just one. Since e to the power x squared is bigger than one at every point inside, our integral is strictly greater than one. With me? Now the upper bound. The biggest value of e to the power x squared here is e, at x equals one. So capital M is e, and the integral is less than e times one, that is, e. So we are done, right? Oh no, careful here, that is not enough! We need less than e minus one, and e is bigger. That bound is too weak, so we need a sharper comparison. Between zero and one, squaring makes a number smaller, so x squared is less than x. And the exponential function always increases, so e to the power x squared is less than e to the power x. Okay? Integrate from zero to one: the integral of e to the power x gives e minus one. So our integral is less than e minus one. Both bounds proved, and we never needed the exact value. Very good!

7:46Quick revision: Wallis even/odd rule, squeeze, compare

So, quick revision. Wallis, for zero to pi over two only: the lists go down in twos, and pi over two comes in only for even powers. For sine to the m times cosine to the n, the bottom starts at m plus n, and pi over two only if both are even. Estimation: squeeze between m times b minus a and capital M times b minus a, or compare with a bigger function g. Now go and try the practice problems on this page. All the best!