1 | initial version |
Hello, @Arnab! I suppose the following is the answer you're looking for. First define $f$:
f(x) = x^2 + 1
Now compose simply by writing:
f2(x) = f(f(x))
f3(x) = f(f(f(x)))
As you might guess, $f2$ is the composition $f\circ f$ with itself, while $f3$ is the composition $f\circ f\circ f$.
The same mechanism works for different functions. Let's define
f(x) = x^2 + 1
g(t) = t^3
Then you can write
fg(t) = f(g(t))
gf(x) = g(f(x))
The first one will give you $t^6+1$, while the second will give you $(x^2+1)^3$.
I hope this helps!
2 | No.2 Revision |
Hello, @Arnab! I suppose the following is the answer you're looking for. First define $f$:
f(x) = x^2 + 1
Now compose simply by writing:
f2(x) = f(f(x))
f3(x) = f(f(f(x)))
As you might guess, $f2$ is the composition $f\circ f$ ($f$ with itself, itself), while $f3$ is the composition $f\circ f\circ f$.f$ (three times $f$).
The same mechanism works for different functions. Let's define
f(x) = x^2 + 1
g(t) = t^3
Then you can write
fg(t) = f(g(t))
gf(x) = g(f(x))
The first one will give you $t^6+1$, while the second will give you $(x^2+1)^3$.
I hope this helps!