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Thread: Function composition and raising functions to a power.

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

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

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

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

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

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

    for part b, I couldn't figure out how to start...
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  2. #2
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    Re: Functions

    Function composition is denoted by $\displaystyle f\circ g$, 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 $\displaystyle (f\circ g)(x) = x$, which means that g is the (right) inverse of f. This is the relationship between the functions. There are many other relationships, e.g., $\displaystyle (g\circ f)(x) = x$, $\displaystyle f(x) = (g\circ g)(x)$, $\displaystyle g(x)=(f\circ f)(x)$, $\displaystyle f(x)=1-\frac{1}{g(x)}$, $\displaystyle g(x)=f(x+1)+1$, 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, $\displaystyle f\circ g$ is a function (the composition of f and g), while x is a number. So, writing $\displaystyle f\circ g=x$ is not the best. One should write $\displaystyle (f\circ g)(x)=x$ or $\displaystyle f\circ g=\mathrm{id}$ where $\displaystyle \mathrm{id}(x)=x$ is the identity function.

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

    Quote Originally Posted by emakarov View Post
    Function composition is denoted by $\displaystyle f\circ g$, 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 $\displaystyle (f\circ g)(x) = x$, which means that g is the (right) inverse of f. This is the relationship between the functions. There are many other relationships, e.g., $\displaystyle (g\circ f)(x) = x$, $\displaystyle f(x) = (g\circ g)(x)$, $\displaystyle g(x)=(f\circ f)(x)$, $\displaystyle f(x)=1-\frac{1}{g(x)}$, $\displaystyle g(x)=f(x+1)+1$, 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, $\displaystyle f\circ g$ is a function (the composition of f and g), while x is a number. So, writing $\displaystyle f\circ g=x$ is not the best. One should write $\displaystyle (f\circ g)(x)=x$ or $\displaystyle f\circ g=\mathrm{id}$ where $\displaystyle \mathrm{id}(x)=x$ is the identity function.

    Since $\displaystyle f(g(x))=x$, you can cancel 1994 applications $\displaystyle f(g(\dots))$ in $\displaystyle f^{2011}g^{1994}(1/2)$ and get $\displaystyle f^{2011-1994}(1/2)$. In evaluating this, note also that $\displaystyle f^3(x)=x$ (this can be verified directly; it also follows from $\displaystyle g=f^2$ and $\displaystyle f\circ g=\mathrm{id}$).
    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 $\displaystyle f^3=f^{13}=f^{15}=f^{17}= ...$ or for any odd number always

    applies? And is it also true that $\displaystyle f^{2}=f^{60}=f^{90}=f^{100}=f^{102}=...$ always?
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  4. #4
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    Re: Functions

    You are allowed to cross out $\displaystyle f^3$. Therefore, $\displaystyle f^{13}=f^{10}=f^{7}=f^{4}=f$. You are not allowed to cross $\displaystyle f^2$, so, e.g., you can't conclude that $\displaystyle f^4=f^2$.
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  5. #5
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    Re: Functions

    Quote Originally Posted by emakarov View Post
    You are allowed to cross out $\displaystyle f^3$. Therefore, $\displaystyle f^{13}=f^{10}=f^{7}=f^{4}=f$. You are not allowed to cross $\displaystyle f^2$, so, e.g., you can't conclude that $\displaystyle f^4=f^2$.
    I see, so the fact that $\displaystyle f^3$ can be crossed out is applicable on a general basis and not only applicable to this question... Thank you!
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  6. #6
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    Re: Functions

    Of course it's applicable only to this question. The fact that $\displaystyle f^3=\mathrm{id}$ is a very specific property that is true only of some functions. Also, it does not follow from $\displaystyle f^3=\mathrm{id}$ that $\displaystyle f^2=f^{60}$.
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  7. #7
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    Re: Functions

    Quote Originally Posted by emakarov View Post
    Of course it's applicable only to this question. The fact that $\displaystyle f^3=\mathrm{id}$ is a very specific property that is true only of some functions. Also, it does not follow from $\displaystyle f^3=\mathrm{id}$ that $\displaystyle f^2=f^{60}$.
    This is confusing! So how do I know whether $\displaystyle f^2$ or $\displaystyle f^3$ can be cancelled?
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  8. #8
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    Re: Functions

    Quote Originally Posted by Punch View Post
    This is confusing! So how do I know whether $\displaystyle f^2$ or $\displaystyle f^3$ 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 $\displaystyle f$ we have

    $\displaystyle f(f(f(x)))=x$ (*)

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

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

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

    You can repeatedly replace $\displaystyle f(f(f(x)))$ with $\displaystyle x$ or vice versa. This implies, for example, that $\displaystyle x=f^{3k}(x)$ for all natural numbers $\displaystyle k$. All such equalities have to be proved, not just guessed.
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  9. #9
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    Re: Functions

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

    $\displaystyle f(f(f(x)))=x$ (*)

    No, this does not follow from (*).

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

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

    You can repeatedly replace $\displaystyle f(f(f(x)))$ with $\displaystyle x$ or vice versa. This implies, for example, that $\displaystyle x=f^{3k}(x)$ for all natural numbers $\displaystyle k$. 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?
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  10. #10
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    Re: Functions

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