Every Second or Third Coefficient

Transcript

0:00Introduction

Hello everyone! Okay, a quick puzzle. Put x equals one in one plus x to the power n, and you add all the coefficients. And x equals minus one splits the even ones from the odd ones. But now I want every third coefficient: C zero, plus C three, plus C six, and so on. Which value of x does that? One? Minus one? Neither, right? Today two complex numbers, i and omega, will do it. Sums like this do come up in JEE, so let's begin.

0:36Every second coefficient, with alternating signs

So, we start from one plus x to the power n, with n a positive integer. It equals C zero, plus C one x, plus C two x squared, and so on, the three dots, up to C n x to the power n. Here C r, the capital C with a small r below, means n C r. Okay? Warm up first. I want C zero, minus C two, plus C four, and so on: even ones only, with alternating signs. With x equals minus one, x squared is plus one, so C two stays positive. That fails. So I need a number whose square is minus one. Which number? The letter i, exactly! Its square is minus one. So i to the power r cycles in fours: one, i, minus one, minus i, then back to one. Right? So put x equals i, and look at one plus i to the power n. Even powers give real terms: C zero, minus C two, plus C four, and so on. Odd powers carry an i: C one, minus C three, plus C five, and so on. So the real part gives the first sum, and the imaginary part the second. With me? Now the polar form. One plus i has modulus root two and argument pi over four. So it equals root two, times, cos pi over four, plus i sin pi over four. Clear? By De Moivre's theorem, raise it to the power n. Root two to the power n is two to the power, n over two, and the angle becomes n pi over four. Got it? Compare real and imaginary parts. C zero minus C two plus C four, and so on, is two to the power, n over two, times cos n pi over four. And C one minus C three plus C five, and so on, is the same with sin. One complex number, two sums. Perfect!

2:30Every third coefficient: use omega

Now, for every third, we need omega, the Greek letter like a curly w. Omega is e to the power, two pi i over three, a complex cube root of one. So omega cubed is one, and one plus omega plus omega squared is zero, okay? Now the key question. Take one, plus omega to the power r, plus omega to the power two r. What is it when r is a multiple of three? And otherwise? Very good! For a multiple of three, each term is one, so the total is three. Otherwise the terms are one, omega and omega squared in some order, so the total is zero. It is a filter: it keeps every third r and kills the rest. Clear? So add three expansions: x equals one, omega, and omega squared. Each C r picks up that filter. So one plus one, to the power n, plus one plus omega, to the power n, plus one plus omega squared, to the power n, is three times, C zero plus C three plus C six, and so on. Right? Now many students divide by three and stop: the sum is two to the power n over three. Is that right? No, no, careful! Two to the power n is never a multiple of three, so that is not even a whole number. The two complex terms add up to two cos n pi over three; we prove it in Example two. So the sum is two to the power n, plus two cos n pi over three, all over three. With me? The same idea, shifted by one place, gives the other classes. C one plus C four plus C seven, and so on, has cos of, n minus two, times pi over three. Okay? And C two plus C five plus C eight, and so on, has cos of, n plus two, times pi over three. Same two to the power n, same three below; only the angle moves. Got it?

4:22With an angle: put x = e^(i theta)

One more filter, with an angle, the Greek letter theta. Put x equals e to the power i theta. Then one plus e to the power i theta equals e to the power, i theta over two, times two cos theta over two. Right? Raise both sides to the power n and take real parts. The capital sigma means add over all r. So the sum of C r cos r theta is two to the power n, times cos to the power n of theta over two, times cos n theta over two. Clear? The imaginary parts give the same with sin. Okay? Now put theta equals pi over two in the sine sum. You get C one minus C three plus C five, and so on, again. Everything connects!

5:05Example 1: 8C0 - 8C2 + 8C4 - 8C6 + 8C8

Now, Example one. Find eight C zero, minus eight C two, plus eight C four, minus eight C six, plus eight C eight. Pause the video and try it. Done? Okay, let's check. The numbers below C are even, and the signs alternate. So, the cos formula or the sin formula? The cos one, correct! Here n is eight, so two to the power, eight over two, is two to the power four: sixteen. And cos of eight pi over four is cos two pi, which is one. Answer: sixteen. Got it? Now the long way. The row of eight gives one, twenty eight, seventy, twenty eight and one. One minus twenty eight, plus seventy, minus twenty eight, plus one: sixteen again! Very good!

5:58Example 2: 9C0 + 9C3 + 9C6 + 9C9 using omega

Now Example two, a harder one. Find nine C zero, plus nine C three, plus nine C six, plus nine C nine, from the formula, and check it. Pause the video and try it. Hint: one plus omega plus omega squared is zero. Done? Okay, let's check. First, the proof I promised. One plus omega plus omega squared is zero. So what is one plus omega? Minus omega squared, exactly! And in the same way, one plus omega squared is minus omega. Right? So one plus omega to the power n, plus one plus omega squared to the power n, is minus one to the power n, times, omega to the power two n, plus omega to the power n. And that simplifies to two cos n pi over three. Proved! Clear? Now put n equals nine. We need cos three pi. Is it plus one, or minus one? Minus one! Three pi is a full turn plus pi, so cos three pi equals cos pi. Take plus one, and the answer is not even a whole number. That is the trap. Two to the power nine is five hundred and twelve. Minus two, over three, gives five hundred and ten over three: one hundred and seventy. With me? Now the check. The terms are one, eighty four, eighty four and one. They add up to one hundred and seventy! Perfect! So, quick revision.

7:27Quick revision: filters with i, omega and an angle

One: every second coefficient, alternating, is the real or imaginary part of one plus i to the power n. Two: for every third, add the expansions at one, omega and omega squared, then divide by three. Three: C zero plus C three plus C six, and so on, is two to the power n, plus two cos n pi over three, all over three; with an angle, use one plus e to the power i theta, all to the power n. Now try the practice problems on this page. All the best!