Functional Equations

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle first. You know that e to the power, x plus y, equals e to the power x, times e to the power y, right? Now flip it around. Some function f obeys the same rule: f of x plus y equals f of x times f of y. Nobody tells you its formula. Can we still find f? If f is differentiable, is not the zero function, and we know one more number, the slope at zero, yes, we can! These are functional equations, and they come up often in JEE papers. So, let's begin.

0:38The method: let the equation do the work

So, how do we attack such an equation? No formula for f, only a rule. Which tool from this chapter works straight from a definition, with no formula needed? The derivative from first principles, yes! f prime of x is the limit, as h tends to zero, of f of x plus h, minus f of x, all over h. The small dash is called prime, and l i m with the little arrow below it means the limit as h tends to zero. Okay? Now the key move. We do not know f of x plus h, but the equation does! So put y equals h in the equation, and rewrite f of x plus h. Right? But before that, one small job. Put special values, like zeros, and find f of zero, or f of one for some equations. You will soon see why. With me? Then the limit gives f prime of x, and from that, f of x, with f of zero, or f of one, fixing the constant. That's the whole method! Clear?

1:43Watch it work: f(x+y)=f(x)+f(y), f differentiable

Let's watch it on the simplest equation: f of x plus y equals f of x plus f of y, with f differentiable. Move one: put x equals y equals zero. What is f of zero? f of zero equals f of zero plus f of zero. So f of zero equals two times f of zero, which forces f of zero to be zero. Yes! This double arrow means implies. Okay? Move two: put y equals h. f of x plus h equals f of x plus f of h. So f of x plus h, minus f of x, is just f of h. The x has disappeared! Right? Now divide by h, and let h tend to zero. What does f of h over h tend to? f prime of zero, exactly! The arrow means tends to. Here is why: f of zero is zero, so f of h over h is really f of h minus f of zero, over h, the definition of f prime of zero. That is why special values come first. Got it? So f prime of x equals f prime of zero, a constant. Then f of x is f prime of zero times x, plus a constant, and f of zero equals zero makes that constant zero. So, f of x equals f prime of zero times x!

3:09Three standard answers, with their conditions

So that is our first standard answer: f of x plus y equals f of x plus f of y, with f differentiable, gives f of x equals f prime of zero times x. Write it down, okay? The second is our puzzle from the start: f of x plus y equals f of x times f of y. Then f of x equals e to the power f prime of zero times x. An exponential, as we guessed! This sign, three bars with a slash, means not identically equal to, so f is not the zero function. Clear? And the third: f of x times y equals f of x plus f of y, for x greater than zero. Here the special value is one, and f of one is zero. The answer: f of x equals f prime of one times log x. These letters, l n, mean the natural log, read as log x. Remember all three, with their conditions. Got it?

4:06Example 1: f(x+y) = f(x) + f(y) + 2xy with f'(0) = 3

Alright, example one, from the course. f of x plus y equals f of x, plus f of y, plus two x y, for all x and y. And f prime of zero equals three. Find f of x. Pause the video and try it. Done? Okay, let's check. Some students say: this looks like the first standard answer, so f of x is three x. Is that right? No, no, wait! Look at the extra term, two x y. Standard answers fit only the exact forms. Here, use the method. Special values first: x equals y equals zero gives f of zero equals two times f of zero, plus zero, so f of zero is zero. Okay? Now put y equals h. f of x plus h, minus f of x, is f of h plus two x h. Divide by h, and we get f of h over h, plus two x. With me? Now let h tend to zero. What does this whole thing tend to? f prime of zero plus two x, yes! Since f of zero is zero, f of h over h tends to f prime of zero, which is three. So f prime of x equals two x plus three. Perfect! Now from f prime back to f: x squared, with the small raised two, plus three x, plus a constant, capital C, okay? f of zero is zero, so this constant is zero. Never skip this step! So f of x equals x squared plus three x. Not three x, see? That extra two x y is what gave us the x squared. Very good!

5:51Example 2: f(x+y) = f(x) + f(y) + xy(x+y), f'(0) = -1

Now a harder one. f of x plus y equals f of x, plus f of y, plus x y times x plus y, in brackets. And f prime of zero equals minus one. Find f of x. Pause and try. Hint: the same four moves. Done? Okay, let's check. Special values first. Put x equals y equals zero: the extra term becomes zero, so f of zero equals two times f of zero, and f of zero is zero. Clear? Now put y equals h. Tell me, what is f of x plus h, minus f of x? f of h, plus x h times x plus h. Correct! Careful: y becomes h everywhere, also inside the bracket. So it is x plus h, not x plus y. Okay? Divide by h: f of h over h, plus x times x plus h. As h tends to zero, the first part tends to f prime of zero, minus one, and the second tends to x squared. So the limit is minus one plus x squared. Right? So f prime of x equals x squared minus one. Back to f: x cubed over three, minus x, plus a constant, capital C, okay? The small raised three means cubed. And f of zero is zero, so the constant is zero again. So f of x equals x cubed over three, minus x. That's it! Now a topper's habit: put the answer back in. In f of x plus y, minus f of x, minus f of y, the minus x parts cancel, leaving x plus y, whole cubed, minus x cubed, minus y cubed, all over three. That is exactly x y times x plus y. Perfect!

7:41Quick revision: special values, y = h, standard forms

So, quick revision. Special values first: find f of zero, or f of one. Then put y equals h, and use the definition of f prime of x. And use standard answers only for the exact forms, with their conditions. Now go and try the practice problems on this page. All the best!