Series, Equations and Inequalities

Transcript

0:00Introduction

Hello everyone! Let me start with a puzzle. Take tan inverse of two, add tan inverse of two over four, then tan inverse of two over nine, and keep going forever, with n squared at the bottom each time. Infinitely many positive angles, all added up. So the total must be infinite, right? Hmm, not so fast. The total is a neat, finite angle, and you will find it yourself today. Series, equations and inequalities with inverse functions come up often in JEE, so let's begin.

0:36Telescoping: write each term as a difference

So, here is the key tool for series. Tan inverse of a minus b, over one plus a times b, equals tan inverse a, minus tan inverse b. The small minus one up top means inverse, not a power, okay? It needs a times b to be greater than minus one; that is the greater than sign you see in the bracket. So, see the numerator as a difference, and the denominator as one plus a product. Right? Now try it on a typical term, one over one plus n plus n squared. So a minus b must be one, and a times b must be n plus n squared. Which two expressions do that? Yes, exactly: n plus one, and n! They differ by one, and their product is n plus n squared. So one over one plus n plus n squared equals the bracket n plus one, minus n, over one plus n times the bracket n plus one. Clear? Both numbers are positive, so their product is certainly greater than minus one. So each term is tan inverse of n plus one, minus tan inverse n. Got it? Now watch what happens when we add. For n equal to one, tan inverse two minus tan inverse one. For n equal to two, tan inverse three minus tan inverse two. The three dots stand for the terms in between, up to the last term, tan inverse of capital N plus one, minus tan inverse of capital N, okay? See it? Every middle piece cancels with its neighbour, like a telescope folding shut. Right? So only two pieces survive: the last, tan inverse of capital N plus one, and the first, minus tan inverse one, where tan inverse one is pi over four, with the Greek letter pi. The big Greek letter sigma means add up, from n equal to one up to capital N, okay? So the sum equals tan inverse of capital N plus one, minus pi over four. That is telescoping. With me?

2:37Equations: take tan, solve, then check

Next, equations, where the unknown x sits inside inverse functions. Step one is to take tan of both sides, or sine, or cos, whichever fits. Okay? Now the inverse functions are gone, so step two is to solve the ordinary equation that is left. Right? So, once we have the roots, are we done? Not yet, and students lose marks here. Step three is to check every root in the original equation. Why? Different angles can have the same tan, so taking a function of both sides can add roots. Clear?

3:14Inequalities: follow the direction

Now, inequalities. The key word here is monotonicity. Sine inverse, the dark blue curve, and tan inverse, the solid green curve, always rise. So a bigger input gives a bigger angle, and the inequality keeps its direction. Okay? Now cos inverse. Let me ask you this: if cos inverse x is greater than cos inverse y, which is bigger, x or y? Yes, exactly, y is bigger! Cos inverse, the orange curve, and cot inverse, the dashed blue curve, always fall. So a bigger input gives a smaller angle, and the inequality flips. On screen: cos inverse x greater than cos inverse y implies x is less than y; the double arrow means implies. This is a common trap. Right? And keep x inside the domain. For sine inverse and cos inverse, x lies from minus one to one; the square brackets include the end points. Got it?

4:13Example 1: solve tan inverse 2x + tan inverse 3x = pi/4

Now the first example. Solve tan inverse two x, plus tan inverse three x, equals pi over four. Pause the video and try it first. Done? Okay, let's check. Take tan of both sides. Tan of a sum is the sum of the tans, over one minus their product. So we get two x plus three x, over one minus two x times three x; the dot means times. That is five x over one minus six x squared, and it equals one, since tan of pi over four is one. Okay? Cross multiply: six x squared, plus five x, minus one, equals zero. It factorises as the bracket six x minus one, times the bracket x plus one; the double arrow means implies. So x is one sixth, or x is minus one. Right? So, two answers? Let me ask you this: does x equal to minus one really satisfy the original equation? It does not, and this is the trap! With x equal to minus one, we get tan inverse of minus two, plus tan inverse of minus three. Both angles are negative, so their sum is less than zero; that is the less than sign on screen. It can never be pi over four. Taking tan added this extra root. With me? Now x equal to one sixth. Two x is one third, and three x is one half. And tan inverse of one third, plus tan inverse of one half, is exactly pi over four. So the only answer is x equal to one sixth. Got it?

5:54Example 2: infinite sum of tan inverse 2 over n squared

Now the harder one, and it is our puzzle from the start! Find the sum, from n equal to one to infinity, of tan inverse of two over n squared; the sideways eight is the infinity sign. Pause the video and try it first. Hint: write the two as a difference. Done? Okay, let's check. We need a minus b equal to two, and one plus a times b equal to n squared, so a times b is n squared minus one. Which two expressions do that? Yes, exactly: n plus one, and n minus one! They differ by two, and their product is n squared minus one. So two over n squared equals the bracket n plus one, minus the bracket n minus one, over one plus the bracket n squared minus one. Okay? Here n squared minus one is never negative, so it is greater than minus one. So each term is tan inverse of n plus one, minus tan inverse of n minus one. Right? Now careful, this telescope is a little different: each piece cancels with a piece two terms later. So when we add up to capital N, which pieces survive? Two from the end, and two from the start! The sum up to capital N is tan inverse of capital N plus one, plus tan inverse of capital N, minus tan inverse one, minus tan inverse zero. Got it? Now let capital N grow. What does tan inverse of a huge number approach? It approaches pi over two, exactly! The arrow means approaches: tan inverse stays below pi over two, but gets closer and closer as the input grows. So both tan inverse terms with capital N approach pi over two. Clear? So the sum is pi over two, plus pi over two, minus pi over four, minus zero, and that gives three pi over four. Infinitely many positive angles, and the total is just three pi over four. Our puzzle is solved! With me?

8:01Quick revision: series, equations and inequalities

So, quick revision. For series, write each term as tan inverse a, minus tan inverse b, and let the middle cancel. For equations, take tan, sine or cos of both sides, solve, then check every root. And for inequalities, sine inverse and tan inverse keep the direction, while cos inverse and cot inverse flip it. Now go and try the practice problems on this page. All the best!