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Math Help - Function composition and raising functions to a power.

  1. #1
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    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...
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  2. #2
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    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}).
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    Re: Functions

    Quote Originally Posted by emakarov View Post
    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?
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    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.
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    Re: Functions

    Quote Originally Posted by emakarov View Post
    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!
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    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}.
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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 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?
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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 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 (*)

    Quote Originally Posted by Punch View Post
    is it true that 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 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}).

    Quote Originally Posted by Punch View Post
    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.
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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 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?
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  10. #10
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    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:
    Quote Originally Posted by emakarov View Post
    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}).
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