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“Functions: Functional Equations, Counting and JEE Advanced Practice” is available to Standard subscribers.

Can you find a function when you know only how it behaves? If f(x + y) = f(x) + f(y) for all x and y, what can f be? In this course we read three ideas that the main chapter only touches: equations whose unknown is a function, the number of solutions of an equation read from a graph (with the fixed points of a function), and the counting of onto functions. Then we solve harder problems and take a JEE Advanced style test.

3Lessons12Problems2Python Labs1Question Set1Mock Test

Read the first lesson freeExtended reading: functional equations

What's inside

Extended reading: functional equations: A functional equation has a function as its unknown. We try special values such as 0 and 1, we replace x by another expression to get a second equation, and we guess a solution and then check it. For $f(x + y) = f(x) + f(y)$ we find $f(0) = 0$

The chapter itself stays free

This is an extra course for Functions and Their Graphs. Its main courses are open to everyone:

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Extended reading: solutions from graphs and fixed points: The number of solutions of g(x) = k is the number of points where the graph of g meets the horizontal line y = k. A modulus inside an equation is handled by putting |x| = t: every positive t gives two values of x. A fixed point of f solves f(x) = x. For an increasing function, f(x) = f⁻¹(x) exactly when f(x) = x.

Extended reading: counting onto functions: The number of functions from m elements to n elements is

n^m

. The number of onto functions is

n^m - \binom{n}{1}(n-1)^m + \binom{n}{2}(n-2)^m - \cdots

, which we get by taking away the functions that miss some element and adding back what was taken twice. Strictly increasing functions from m to n elements number

\binom{n}{m}

, and the functions with

f(f(x)) = x

are counted by choosing which elements are swapped in pairs.