Coefficients in Products

Transcript

0:00Introduction

Hello everyone! Okay, a scary-looking question to start. Find the coefficient of x to the power fifty in one plus x, to the power forty one, times one minus x plus x squared, to the power forty. That is eighty one brackets multiplied together! And still, the answer comes out in under a minute, with no expansion at all. How is that possible? Keep that question in mind. Today: coefficients in products. Questions like this come up often in JEE, so let's begin.

0:34Coefficient of x^k in a product

So, the basic idea. When you multiply brackets, every term comes from picking one term from each factor. And the powers of x add. So, to get x to the power k, the powers you pick must add up to k. Okay? Then multiply the coefficients you picked, and add up all such products. That total is the coefficient. Right? Now the most common case. P of x equals the sum of p j times x to the power j: a short polynomial, just a few terms. The big Greek letter sigma means add them all up, and p with a small j below, p j, is the coefficient of x to the power j. And n here is a positive integer. Clear? Now tell me. Suppose you take x to the power j from P of x. Which power of x must come from one plus x to the power n? Yes, exactly, k minus j! Because j plus k minus j is k. Its coefficient there is n C k minus j, with the small n at the top left of the capital C, okay? So, the coefficient of x to the power k is the sum of p j times n C k minus j, over every j. Got it?

1:47Simplify before you expand

Now, a topper's habit: simplify before you expand. First, the same base: one plus x to the power m, times one plus x to the power n, is one plus x to the power m plus n. With me? Next, a quick one for you. What is one plus x, times one minus x? One minus x squared, correct! So, one plus x to the power n, times one minus x to the power n, is one minus x squared, to the power n. We can pair bracket by bracket because both powers are the same n. And notice: this has only even powers of x, so any odd power has coefficient zero. Okay? Next, one minus x, times one plus x plus x squared, is one minus x cubed. Multiply it out once and check: everything in the middle cancels. Clear? And one more. One plus x plus x squared, and so on, up to x to the power k minus one, equals one minus x to the power k, over one minus x. The three dots mean and so on. So a long sum becomes one neat fraction. Right?

3:01Two classic results

Now two classic results. First, x to the power k in one plus x to the power m, times one plus x to the power n. By our rule, take x to the power i from the first, and x to the power k minus i from the second: the sum of m C i times n C k minus i. But the product is just one plus x to the power m plus n! So the sum equals m plus n, C k. This is Vandermonde's identity. Okay? Second, one plus x to the power n, times one plus one over x, to the power n. One plus one over x is the same as x plus one, over x. So its n-th power is one plus x to the power n, over x to the power n. Multiply by the first bracket, and the product becomes one plus x to the power two n, over x to the power n. Got it? Now, the term independent of x is the constant term, with no x left. We divide by x to the power n. So which power of x do we need on top? Yes, exactly, x to the power n! Then the x to the power n cancels. So the term independent of x is the coefficient of x to the power n in one plus x to the power two n, and that is two n C n. With me?

4:27Example 1: x^3 in (1 - x + x^2)(1 + x)^8

Alright, example one. Find the coefficient of x cubed in one minus x plus x squared, times one plus x to the power eight. Pause the video and try it. Done? Okay, let's check. Here P of x is one minus x plus x squared. So p zero is one, p one is minus one, and p two is one. And n is eight. Okay? Now pair them up. The constant one pairs with x cubed from one plus x to the power eight. The minus x term pairs with which power? Yes, exactly, x squared! And the x squared term pairs with plain x. So: one times eight C three, plus minus one times eight C two, plus one times eight C one. The small dot means times. Right? Now, eight C three is fifty six, eight C two is twenty eight, and eight C one is eight. So the answer is fifty six plus twenty eight plus eight, ninety two? No, no, careful! Look at the minus one in front of eight C two; it comes from the minus x. So it is fifty six minus twenty eight plus eight: thirty six. In this topic, marks are usually lost on signs. Clear?

5:50Example 2: x^50 in (1 + x)^41 (1 - x + x^2)^40

Now, remember our scary question from the start? Let's finish it. Find the coefficient of x to the power fifty in one plus x to the power forty one, times one minus x plus x squared, to the power forty. Pause the video and try it. Hint: pair the brackets. Done? Okay, let's check. We want to pair the brackets, but the powers are forty one and forty, not the same. So what do we do? Simple: split off one bracket! One plus x to the power forty one is one plus x, times one plus x to the power forty. Now both big brackets have the same power, forty. With me? And one plus x, times one minus x plus x squared, is one plus x cubed. Multiply it out: the middle terms cancel, just like before. Okay? So the whole product is one plus x, times one plus x cubed, to the power forty. Eighty one brackets have become just this! Right? Now look carefully. One plus x cubed, to the power forty, has only powers that are multiples of three. Then we multiply by one plus x. So which powers of x can appear? Yes, exactly! Times the one, the powers stay at three k. Times the x, they become three k plus one, where k is zero, one, two, and so on. Nothing else. Clear? Now fifty. Fifty is three times sixteen, plus two. Remainder two: so fifty is neither three k nor three k plus one. So the coefficient of x to the power fifty is zero! That is the answer to our opening puzzle, with no expansion at all. That was fast, right?

7:38Quick revision: powers add to k, simplify, Vandermonde

So, quick revision. One: the powers must add up to k; multiply the coefficients, then add. Two: for P of x times one plus x to the power n, the coefficient of x to the power k is the sum of p j times n C k minus j. Three: simplify first and pair brackets, like one plus x to the power n, times one minus x to the power n, which is one minus x squared, to the power n. Four: the classics, Vandermonde, and two n C n for the term independent of x. Now go and try the practice problems on this page. All the best!