# Thread: Function composition and raising functions to a power.

1. ## Function composition and raising functions to a power.

$f(x)=\frac{1}{1-x}$ (x all real numbers excluding 0 and 1)

$g(x)=1-\frac{1}{x}$ (x all real numbers excluding 0 and 1)

a) Express $fg$ in a similar form and hence describe the relationship between $f$ and $g.$

b) Evaluate $f^{2011}g^{1994}(\frac{1}{2})$

I was able to express $fg$ and $fg=x$, whereby (x all real numbers excluding 0 and 1) but was unable to find the relationship between $f$ and $g$

for part b, I couldn't figure out how to start...

2. ## Re: Functions

Function composition is denoted by $f\circ g$, or f o g in plain text. Some sources may skip the circle, but I would expect them to note this convention explicitly. In any case, it is much better to write a bit more explanations than to let people wonder about notation.

You showed that $(f\circ g)(x) = x$, which means that g is the (right) inverse of f. This is the relationship between the functions. There are many other relationships, e.g., $(g\circ f)(x) = x$, $f(x) = (g\circ g)(x)$, $g(x)=(f\circ f)(x)$, $f(x)=1-\frac{1}{g(x)}$, $g(x)=f(x+1)+1$, etc., but apparently they are not relevant to the question (except maybe the first).

Note also that both sides of equality should have the same type: either functions or real numbers. For example, $f\circ g$ is a function (the composition of f and g), while x is a number. So, writing $f\circ g=x$ is not the best. One should write $(f\circ g)(x)=x$ or $f\circ g=\mathrm{id}$ where $\mathrm{id}(x)=x$ is the identity function.

Since $f(g(x))=x$, you can cancel 1994 applications $f(g(\dots))$ in $f^{2011}g^{1994}(1/2)$ and get $f^{2011-1994}(1/2)$. In evaluating this, note also that $f^3(x)=x$ (this can be verified directly; it also follows from $g=f^2$ and $f\circ g=\mathrm{id}$).

3. ## Re: Functions

Originally Posted by emakarov
Function composition is denoted by $f\circ g$, or f o g in plain text. Some sources may skip the circle, but I would expect them to note this convention explicitly. In any case, it is much better to write a bit more explanations than to let people wonder about notation.

You showed that $(f\circ g)(x) = x$, which means that g is the (right) inverse of f. This is the relationship between the functions. There are many other relationships, e.g., $(g\circ f)(x) = x$, $f(x) = (g\circ g)(x)$, $g(x)=(f\circ f)(x)$, $f(x)=1-\frac{1}{g(x)}$, $g(x)=f(x+1)+1$, etc., but apparently they are not relevant to the question (except maybe the first).

Note also that both sides of equality should have the same type: either functions or real numbers. For example, $f\circ g$ is a function (the composition of f and g), while x is a number. So, writing $f\circ g=x$ is not the best. One should write $(f\circ g)(x)=x$ or $f\circ g=\mathrm{id}$ where $\mathrm{id}(x)=x$ is the identity function.

Since $f(g(x))=x$, you can cancel 1994 applications $f(g(\dots))$ in $f^{2011}g^{1994}(1/2)$ and get $f^{2011-1994}(1/2)$. In evaluating this, note also that $f^3(x)=x$ (this can be verified directly; it also follows from $g=f^2$ and $f\circ g=\mathrm{id}$).
Thank you, one important point I missed was to realise that g is the inverse of f but I have another

doubt, is it true that $f^3=f^{13}=f^{15}=f^{17}= ...$ or for any odd number always

applies? And is it also true that $f^{2}=f^{60}=f^{90}=f^{100}=f^{102}=...$ always?

4. ## Re: Functions

You are allowed to cross out $f^3$. Therefore, $f^{13}=f^{10}=f^{7}=f^{4}=f$. You are not allowed to cross $f^2$, so, e.g., you can't conclude that $f^4=f^2$.

5. ## Re: Functions

Originally Posted by emakarov
You are allowed to cross out $f^3$. Therefore, $f^{13}=f^{10}=f^{7}=f^{4}=f$. You are not allowed to cross $f^2$, so, e.g., you can't conclude that $f^4=f^2$.
I see, so the fact that $f^3$ can be crossed out is applicable on a general basis and not only applicable to this question... Thank you!

6. ## Re: Functions

Of course it's applicable only to this question. The fact that $f^3=\mathrm{id}$ is a very specific property that is true only of some functions. Also, it does not follow from $f^3=\mathrm{id}$ that $f^2=f^{60}$.

7. ## Re: Functions

Originally Posted by emakarov
Of course it's applicable only to this question. The fact that $f^3=\mathrm{id}$ is a very specific property that is true only of some functions. Also, it does not follow from $f^3=\mathrm{id}$ that $f^2=f^{60}$.
This is confusing! So how do I know whether $f^2$ or $f^3$ can be cancelled?

8. ## Re: Functions

Originally Posted by Punch
This is confusing! So how do I know whether $f^2$ or $f^3$ can be cancelled?
You can only use those equalities that follow by the laws of mathematics. In post #2, I stated that for this particular $f$ we have

$f(f(f(x)))=x$ (*)

Originally Posted by Punch
is it true that $f^3=f^{13}=f^{15}=f^{17}= ...$ or for any odd number always
No, this does not follow from (*).

Originally Posted by Punch
And is it also true that $f^{2}=f^{60}=f^{90}=f^{100}=f^{102}=...$ always?
Some of these equalities follow from (*) (e.g., $f^{60}=f^{90}$) and some don't (e.g., $f^{90}=f^{100}$).

Originally Posted by Punch
so the fact that $f^3$ can be crossed out is applicable on a general basis and not only applicable to this question
No, (*) applies only to this particular $f$. Have you checked (*) for this $f$? In doing so, did you use the definition of $f$?

You can repeatedly replace $f(f(f(x)))$ with $x$ or vice versa. This implies, for example, that $x=f^{3k}(x)$ for all natural numbers $k$. All such equalities have to be proved, not just guessed.

9. ## Re: Functions

Originally Posted by emakarov
You can only use those equalities that follow by the laws of mathematics. In post #2, I stated that for this particular $f$ we have

$f(f(f(x)))=x$ (*)

No, this does not follow from (*).

Some of these equalities follow from (*) (e.g., $f^{60}=f^{90}$) and some don't (e.g., $f^{90}=f^{100}$).

No, (*) applies only to this particular $f$. Have you checked (*) for this $f$? In doing so, did you use the definition of $f$?

You can repeatedly replace $f(f(f(x)))$ with $x$ or vice versa. This implies, for example, that $x=f^{3k}(x)$ for all natural numbers $k$. All such equalities have to be proved, not just guessed.
I understand now. So you had tried that f(f(f(x)))=x and that is the reason why we can use f^3=x. Am I right to then say that whether f^2 or f^3 can be cancelled must be tried by showing f^2 or f^3=x before we can use them?

10. ## Re: Functions

Yes. Since $f(x)=\frac{1}{1-x}$, $f(f(x))=\frac{1}{1-\frac{1}{1-x}}=\frac{x-1}{x}=1-\frac{1}{x}=g(x)$. Applying $f$ to both sides we get $f^3(x)=f(g(x))$, and you have already showed that $f(g(x))=x$. As I said in post #2:
Originally Posted by emakarov
In evaluating this, note also that $f^3(x)=x$ (this can be verified directly; it also follows from $g=f^2$ and $f\circ g=\mathrm{id}$).