Even, Odd and Periodic Functions

Transcript

0:00Introduction

Hello everyone! One small puzzle before we start. Take the integral from minus pi over two to pi over two of sine squared x, over one plus two to the power x, d x. Substitution and integration by parts both get stuck here. And yet, done the right way, it takes about ten seconds. How is that possible? The secret is symmetry, and questions like this come up often in JEE, so let's begin.

0:30Symmetric limits: even and odd functions

So, first idea: symmetric limits. The long S is the integral sign, minus a and a are its limits, and d x says we integrate in x, okay? Split at zero, and fold the left half onto the right by putting minus x for x. So the integral from minus a to a of f of x equals the integral from zero to a of f of x plus f of minus x, in square brackets. With me? Now suppose f of minus x equals f of x for every x. We call such a function even. Then what does the bracket become? Two times f of x, yes, exactly! So for an even function, the integral from minus a to a equals two times the integral from zero to a. Work out one half, and double it. Clear? Now an odd function, where f of minus x equals minus f of x. The bracket becomes zero. So the integral from minus a to a of an odd function is zero, with no working! Got it? But careful, this is a classic trap. Both rules need the limits minus a to a, symmetric about zero. An odd function from zero to a usually does not give zero. So always look at the limits first. Okay?

1:53The 1 + k to the power g(x) trick

Now a special case that comes up often: f is even, g is odd, and k is a positive constant. Right? Look at the integral from minus a to a of f of x, over one plus k to the power g of x, d x. It equals just the integral from zero to a of f of x. The denominator disappears! With me? But why? Replace x by minus x, as in our first rule: f stays the same, and g of x becomes minus g of x. So f of x multiplies two fractions: one over one plus k to the power g, and one over one plus k to the power minus g. What do these add up to? They add up to exactly one, yes! Multiply the top and the bottom of the second fraction by k to the power g: it becomes k to the power g, over k to the power g plus one. Add the two, and the top equals the bottom. Clear?

2:53Limits 0 to 2a: fold at the middle

Now the same folding idea, with limits zero to two a. Fold at the middle point a, and f of x is joined by f of two a minus x. So the integral from zero to two a equals the integral from zero to a of f of x plus f of two a minus x. Okay? If f of two a minus x equals f of x, the answer is two times the integral from zero to a. If it equals minus f of x, the halves cancel, and the answer is zero. Right? Quick question. What is the integral from zero to pi of cos cube x, d x? Here two a is pi, so look at cos cube of pi minus x. Cos of pi minus x is minus cos x, so cos cube of pi minus x is minus cos cube x. The halves cancel, and the integral is zero, with no working. Very good, that is the idea!

3:51Periodic functions: one period, then multiply

Last idea: periodic functions. We say f has period capital T when f of x plus T equals f of x for every x. The graph repeats, okay? So every full period gives the same integral: the integral from a to a plus T equals the integral from zero to T, right? And from zero to n times T, with n a positive integer, there are n full periods, so we get n times one period. Clear? Now take mod sine x: the two vertical bars are the modulus, so the arches below the axis flip up. What is its period? Pi, not two pi, correct! The same arch now repeats every pi. And one arch, the integral from zero to pi, is two. Okay? So from zero to one hundred pi: one hundred periods, one hundred times two, that is two hundred. No need for a hundred pieces! Got it?

4:52Example 1: x2 + x3 cos x and ln((2-x)/(2+x)) from -1 to 1

Now, Example one. Evaluate the integral from minus one to one of x squared plus x cube cos x, d x, and the integral from minus one to one of l n of two minus x over two plus x, d x. The l n you see on screen is the natural log. Pause the video and try it first. Done? Okay, let's check. The limits are minus one to one, symmetric, so our rules apply. Now, x cube cos x is odd. So the whole first integral is zero, right? No, no, careful here! That is the trap. Only the second term is odd: x cube is odd and cos x is even, so their product is odd, and its integral is zero. But x squared is even, so it does not vanish. With me? So only x squared survives. It is even, so we take twice the integral from zero to one of x squared, which is two times one third: two thirds. Clear? Now the second integral. Replace x by minus x in l n of two minus x over two plus x. What do you get? You get l n of two plus x over two minus x: the same fraction, upside down, so its log is minus the original log. So the integrand is odd. Right? Odd function, symmetric limits: the integral is zero. So the answers are two thirds, and zero. Very good, that was the trap to watch for!

6:27Example 2: sin2 x/(1 + 2 to the x), -pi/2 to pi/2, is pi/4

Now Example two, our puzzle from the start. Evaluate the integral from minus pi over two to pi over two of sine squared x, over one plus two to the power x, d x. Pause the video and try it first. Done? Okay, let's check. Compare it with our trick: f of x, over one plus k to the power g of x. So what are f, g and k here? Here f of x is sine squared x, which is even. Then g of x is just x, which is odd, and k is two. The limits are symmetric too, so everything fits perfectly! So the denominator disappears, and we are left with the integral from zero to pi over two of sine squared x, d x. Okay? Now use the double angle formula: sine squared x equals one minus cos two x, over two. Got it? Integrate: x over two, minus sine two x over four. The big square brackets mean the value at the upper limit minus the value at the lower limit. At pi over two we get pi over four; at zero, we get zero. So the answer is pi over four, and our puzzle is solved!

7:46Quick revision: even, odd, 1 + k to the g, 0 to 2a, periods

So, quick revision. Limits minus a to a: an even function gives twice the integral from zero to a, and an odd function gives zero. For f of x over one plus k to the power g of x, with f even and g odd, integrate f from zero to a. Limits zero to two a: compare f of two a minus x with f of x. Period capital T: n periods give n times one period. Now try the practice problems on this page. All the best!