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Inverse functions

  1. Jul 9, 2006 #1
    In one of my older threads, I posted the following:

    log_b (n) = x if and only if b ^ x = n, where b > 0, and b is not equal to one.

    It was said this defines logarithm as the inverse to exponential. I don't really see how that works here, I think it just shows how you write logarithms. Could anyone clear this up?

    Also, I'm wondering something about inverses in general. Say you have a simple example like (x + 1) - 1 = x. This equation says that (x + 1), and (x - 1) are inverses, correct? Is there a way to mathematically show that? Or do you simply conclude this qualitatively? I'm thinking if it's just qualitative, you could say something like this:

    If (x + 1) - 1 = x, then it must be that + 1 and - 1 do opposite things to x. Thus, you can conclude that (x + 1) and (x - 1) are inverse functions, seeing as they do opposite things to x.

    Does that sound right? Is there I way to mathematically show this? I know that if you take (x + 1) and (x - 1) and sub one into the other you get either (x + 1) - 1 = x or (x - 1) + 1 = x, but can you do this backwards, by starting from (x + 1) - 1 = x or (x - 1) + 1 = x?

    Thanks in advance for any help I get.
     
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  3. Jul 9, 2006 #2
    Intuitevely, the inverse of a function f(x), noted f^(-1)(x) is such as :
    f(f^(-1)(x)) = x . The inverse function "cancels" the effect of f(x) over x.
    I would use this definition to prove the inverse.
    The subtle question is : Does the inverse funtion exist ?

    Interesting analysis arises from such questions and a lot of theorems are devoted to such a topic.

    Logarithms and exponentials function can be defined from one another as being inverses , hence Exp(Log(x)) = x

    Of course,this is true only if the conditions you stated are fufilled(i.e. same basis not equal to one)

    For your other question, f(x) = (x+1) has (x-1) as inverse and vice versa. As you mentionned, it "works both ways".
     
    Last edited: Jul 9, 2006
  4. Jul 9, 2006 #3
    This is just how the logarithm is defined. It is defined to be the inverse function of the exponential function.


    As for inverse functions in general, two functions lets call them f(x) and g(x) if and only if

    f(g(x)) = g(f(x)) = x

    If you can show that this relationship holds for two given functions then these functions are inverses of each other.
     
  5. Jul 9, 2006 #4

    arildno

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    Byrgg:
    Since it can be proven that:
    1. The exponential function is strictly increasing
    and
    2. Every strictly increasing function HAS an inverse

    I really don't see your trouble.
     
  6. Jul 10, 2006 #5

    0rthodontist

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    I think I recall from my high school calculus book that log x is defined as [tex]\int_1^x \frac{dy}{y}[/tex], and that ex is defined as the inverse of log x, from which general exponentiation (exponentiation to irrational powers) is defined. I don't know how accurate this is.
     
  7. Jul 10, 2006 #6

    HallsofIvy

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    No, "showing how you write logarithms" wouldn't have anything to do with exponentials, would it? Saying "logbn= x if and only if bx= n" is exactly saying that they are inverses. If f(x)= y, then f-1(y)= x. That's exactly logb[/sup](x)= y, by= x. Another characterization of inverse functions is: f(f-1(x))= x and f-1(f(x))= x. From your equations logb(b^x)= logb(n)= x and [itex]b^{log_b n}= b^x= n[/itex].

    Yes, that is precisely f(f-1(x))= x. Here f-1(x)= x+ 1 and f(x)= x-1. f(f-1(x))= f(x+1)= (x+1)-1= x and f(f-1)(x))= f(x-1)= (x-1)+ 1= x.

    Yes, that's a very good, less technical, way of thinking about inverse functions: whatever f "does", f-1 "undoes".

    Well, once again, f(f-1(x))= x and f-1(f(x))= x are the ways to "mathematically show this". Since applying first one and then the other, whatever one "does" to x, the other "undoes".

    You can see why inverse functions are so important: in order to solve the equation f(x)= b, you must "back out" and "undo" whatever f has "done" to x. If you know the inverse function of f, you only need to apply that function to both sides of the equation: f-1(f(x))= x= f-1(b).

    Of course, solving equations, in general, isn't that easy! That's because most functions don't have inverses. If f(x1) and f(x2) both equal y, what is f-1(y)? Is it x1 or x2? Since there is no way to answer that, only "one to one" functions have inverse. Further, if there is no x such that f(x)= y, then there can be no f-1(y). Only "one to one" and "onto" functions have inverses. Since (2)2= 4 and (-2)2= 4, as simple a function as f(x)= x2 does not have an inverse. f(x)= x2 is neither "one to one" nor "onto". Instead we "modify" the function to : g(x)= x2 if [itex]x\ge 0[/itex] and is undefined for x< 0. We can say that f is "one to one" and "onto" the set of non-negative numbers. Now, if y is a non-negative number, [itex]g^{-1}(y)= \sqrt{y}[/itex] is defined. Of course, if we have the equation f(x)= x2= -4, precisely because f(x)= x2 is not "onto" the set of all real numbers, we have to say there is no solution because f-1(-4) is not defined. Also, because f(x)= x2 is not "one to one", to solve f(x)= x2= 4, we have to re-interpret the equation as g(x)= 4 to get x= 2, then recognize that there are two solutions, 2 and -2.
     
    Last edited: Jul 10, 2006
  8. Jul 10, 2006 #7

    HallsofIvy

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    No, "showing how you write logarithms" wouldn't have anything to do with exponentials, would it? Saying "logbn= x if and only if bx= n" is exactly saying that they are inverses. If f(x)= y, then f-1(y)= x. That's exactly logb[/sup](x)= y, by= x. Another characterization of inverse functions is: f(f-1(x))= x and f-1(f(x))= x. From your equations logb(b^x)= logb(n)= x and [itex]b^{log_b n}= b^x= n[/itex].

    Yes, that is precisely f(f-1(x))= x. Here f-1(x)= x+ 1 and f(x)= x-1. f(f-1(x))= f(x+1)= (x+1)-1= x and f(f-1)(x))= f(x-1)= (x-1)+ 1= x.

    Yes, that's a very good, less technical, way of thinking about inverse functions: whatever f "does", f-1 "undoes".

    Well, once again, f(f-1(x))= x and f-1(f(x))= x are the ways to "mathematically show this". Since applying first one and then the other, whatever one "does" to x, the other "undoes".

    You can see why inverse functions are so important: in order to solve the equation f(x)= b, you must "back out" and "undo" whatever f has "done" to x. If you know the inverse function of f, you only need to apply that function to both sides of the equation: f-1(f(x))= x= f-1(b).

    Of course, solving equations, in general, isn't that easy! That's because most functions don't have inverses. If f(x1) and f(x2) both equal y, what is f-1(y)? Is it x1 or x2? Since there is no way to answer that, only "one to one" functions have inverse. Further, if there is no x such that f(x)= y, then there can be no f-1(y). Only "one to one" and "onto" functions have inverses. Since (2)2= 4 and (-2)2= 4, as simple a function as f(x)= x2 does not have an inverse. f(x)= x2 is neither "one to one" nor "onto". Instead we "modify" the function to : g(x)= x2 if [itex]x\ge 0[/itex] and is undefined for x< 0. We can say that f is "one to one" and "onto" the set of non-negative numbers. Now, if y is a non-negative number, [itex]g^{-1}(y)= \sqrt{y}[/itex] is defined. Of course, if we have the equation f(x)= x2= -4, precisely because f(x)= x2 is not "onto" the set of all real numbers, we have to say there is no solution because f-1(-4) is not defined. Also, because f(x)= x2 is not "one to one", to solve f(x)= x2= 4, we have to re-interpret the equation as g(x)= 4 to get x= 2, then recognize that there are two solutions, 2 and -2.
     
  9. Jul 10, 2006 #8
    Err, HallsofIvy said no to my first question, while everyone else says yes it seems.

    log_b (n) = x if and only if b ^ x = n, where b > 0, and b is not equal to one.

    How does this define logarithm as the inverse of the exponential? log_b (n) = x and b ^ x = n are the same thing...

    Also, is my mathematical explanation here right?(see below)

    (x + 1) - 1 = x

    If you take (x + 1) as the first function of x, you must apply the inverse to it to obtain x, which is x - 1. So if you sub this into x - 1, you get
    (x + 1) - 1 = x, since x - 1 does the opposite thing to x, you just proved that (x - 1) and (x + 1) are inverses.

    Does that sound right?
     
  10. Jul 10, 2006 #9
    Quote : "log_b (n) = x and b ^ x = n are the same thing..."

    They are not the same thing, but they are linked by an equivalence relation.
    It means that if one is true, the other is and vice versa.

    HallsofIvy is right in the sense that logarithm's properties and algebra can be considered separately, without mention of the exponential function.

    But, to rigorusly proove certain properties of to just define the function itself, it is a possibility to use the inverse of that particular function if the properties of the inverse funtion are well defined.

    And I think your last explanation sounds right, just use HallsofIvy's notation/definition to "write it properly".

    Also, his statement gives a damn good explanation of what to be aware of in these situations so listen to the man..... :-)
     
    Last edited: Jul 10, 2006
  11. Jul 10, 2006 #10

    HallsofIvy

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    No, I did not say "no" to your first question! What I said was
    That was in response to your statement
     
  12. Jul 10, 2006 #11
    If I take f(x) = x + 1
    and g(x) = x - 1

    then what you have shown is that g(f(x)) = (x+1)-1 = x
    However you have NOT shown that these two functions are inverses of each other what you have shown is that g is a left inverse for f, or similarly that f is a right inverse for g.

    To show that these two functions are inverses you need to show that
    f(g(x)) = g(f(x)) = x
     
  13. Jul 10, 2006 #12
    Aren't log_b (n) = x if and b ^ x = n two different ways of saying the same thing? If you rewrite log_b (n) = x you get b ^ x = n...
     
  14. Jul 10, 2006 #13
    Also, what is the process to work backwards from (x + 1) - x = x to showing that (x + 1) and (x - 1) are inverses? Or has this already been shown?
     
  15. Jul 10, 2006 #14
    I'm going to assume you mean (x + 1) - 1 = x not (x + 1 ) - x = x.

    let's say we have the function
    f(x) = x + 1
    for simplicity let's also say y = f(x)
    then y = x + 1.


    Now if we want to find the inverse function g(x) = f-1(x)
    Then g(y) = x because I have defined g to be the inverse function of f.

    The typical way to find an inverse function is to start with an initial function like
    y = f(x)

    Now if you can solve for x in terms of y then you have a function in terms of y such as x = g(y) and this function will be the inverse of the original function f, but this is not always possible to do, and actually is very rarely possible to do especially in cases involing several trig functions, and higher order polynomials.

    I'll go through an example with y = x + 1

    Now for sake of ending up with a function in relatively the same form I am going to interchange x and y so that I will end up with a function for y in terms of x, however I am going to denote y by y' to denote that this is not the same y.

    So now I have x = y' + 1
    or y' + 1 = x and subtracting 1 from each side leaves
    y' = x - 1 so I now have found the function that is the inverse of the original, and you can easily show that if I let y = f(x) and y' = g(x) that f(g(x)) = g(f(x)) = x to more rigorously show that these functions are indeed inverses.
     
  16. Jul 10, 2006 #15

    HallsofIvy

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    That true because [itex]b^{log_b(x)}= x[/itex], if [itex]log_b(n)= x[/itex], you get [itex]b^x= n[/itex]. In other words, logb(n)= x is equivalent to bx= n precisely because logb(x) and bx are inverse functions.
     
  17. Jul 10, 2006 #16
    Yeah, I did mean to type (x + 1) - 1 = x, sorry about that.

    When you started the work, you started simply with f(x) = (x + 1). I Wanted to know how you go about it right from (x + 1) - 1 = x.

    Also, what would change if you didn't switch y and x towards the end there?

    As for Hallsof Ivy... I think I'm understanding that better now, those two equations are equivalent because taking the log(base b) of n results in x only if b ^ x = n. This can only be true if the logarithm is the inverse of the exponential, and so therefore this must define them as inverses, right?
     
  18. Jul 10, 2006 #17
    If i hadn't changed y and x, it really wouldn't have made a difference i would have ended up with teh same inverse function, but it would have been in the form of x as a function of y and i would have had x = g(y) = f-1(y) which would still have been the inverse function and if I wanted this to be in the same form as y = f(x) I can just change the variables to y = g(x) and it really doesn't matter as long as you are consistent and understand what you are doing when you do this. And I apologize if this explanation doesn't make much sense because I don't think it does but I'm having trouble trying to explain this in a way that makes some logical sense.

    And I'm not quite sure what you mean by this, I tried to you how to get
    y = x - 1 as the inverse function of y = x + 1 in my last post, so I think I am just not understanding the questiong you're asking here.
     
  19. Jul 10, 2006 #18
    When you started to show the work, you said "let's say we have the function f(x) = x + 1".

    I wanted to know where that came from, I was wondering how you would start at (x + 1) - 1 = x, and then get to the rest.

    Also, how would you work with the resulting function if you didn't switch y and x there? Could you show me how that would work out?
     
  20. Jul 10, 2006 #19

    If I start from there I have (x + 1) - 1 = x + 1 - 1 = x + (1 - 1) = x + 0 = x.

    All that really says is that -1 is the additive invers of 1.

    But I guess you could look at it like this
    give (x + 1) - 1 = x

    Now let u = x + 1 you could say that u = u(x)

    Then we have u - 1 = x

    Now let's say that v = u - 1 but it wouldn't really change this equation if I say that u = x, note that this has nothing to do with the previous function I defined, so I now have v = x - 1 = v(x)

    and so I have v(x) = x -1 so let's substitute u(x) into this giving
    v(u(x)) = u - 1 = u(x) - 1 = (x + 1) - 1 = x saying that v is a left inverse of u ot u is a right inverse of v.

    To me this seems incredibly messy and awkward because I started with an equation, a statement that I can't change and built from this functions, I really don't like this and this process doesn't really seem logical to me. It's preferable to find an inverse the way I stated in one of my previous posts.


    Again I don't really understand your question here. I thought that I tried to explain this in my last post.
     
  21. Jul 10, 2006 #20
    That part there kind of confused me, more specifically when you said "but it wouldn't really change this equation if I say that u = x, note that this has nothing to do with the previous function I defined".

    I didn't really understand that part, could you explain it please?

    Also, why would you assign (x + 1) the value 'u' like you did?

    Now about the reversal of x and y, if you didn't reverse them, it would've ended up like this, right?

    y = x + 1
    y - 1 = x

    What would you have done with this relationship? That is more what I'm asking for this part.
     
    Last edited: Jul 10, 2006
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