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Math Help - interesting problem with a function (not defined) - quite intimidating

  1. #1
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    interesting problem with a function (not defined) - quite intimidating

    I found this question to be quite intimidating:

    Suppose a function f has the following properties:
    • f is differentiable everywhere.
    • For all values of x and y, interesting problem with a function (not defined) - quite intimidating-codecogseqn-13.gif.
    • interesting problem with a function (not defined) - quite intimidating-codecogseqn-14.gif.
    • interesting problem with a function (not defined) - quite intimidating-codecogseqn-15.gif.



    a) Show that:
    interesting problem with a function (not defined) - quite intimidating-codecogseqn-17.gif
    b) Show that interesting problem with a function (not defined) - quite intimidating-codecogseqn-18.gif for all values of x by first showing that interesting problem with a function (not defined) - quite intimidating-codecogseqn-19.gif for all x.

    c) Use the definition of the derivative to prove that
    interesting problem with a function (not defined) - quite intimidating-codecogseqn-20.gif for all x.

    I'd assume f would be a constant function. It seems to have similar properties as such a function. Anyway, thanks for help!
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  2. #2
    Super Member ILikeSerena's Avatar
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    Re: interesting problem with a function (not defined) - quite intimidating

    Quote Originally Posted by zachd77 View Post
    I found this question to be quite intimidating:

    Suppose a function f has the following properties:
    • f is differentiable everywhere.
    • For all values of x and y, Click image for larger version. 

Name:	CodeCogsEqn-13.gif 
Views:	3 
Size:	779 Bytes 
ID:	26391.
    • Click image for larger version. 

Name:	CodeCogsEqn-14.gif 
Views:	25 
Size:	469 Bytes 
ID:	26392.
    • Click image for larger version. 

Name:	CodeCogsEqn-15.gif 
Views:	26 
Size:	437 Bytes 
ID:	26393.



    a) Show that:
    Click image for larger version. 

Name:	CodeCogsEqn-17.gif 
Views:	26 
Size:	331 Bytes 
ID:	26394
    b) Show that Click image for larger version. 

Name:	CodeCogsEqn-18.gif 
Views:	25 
Size:	444 Bytes 
ID:	26395 for all values of x by first showing that Click image for larger version. 

Name:	CodeCogsEqn-19.gif 
Views:	41 
Size:	463 Bytes 
ID:	26396 for all x.

    c) Use the definition of the derivative to prove that
    Click image for larger version. 

Name:	CodeCogsEqn-20.gif 
Views:	25 
Size:	531 Bytes 
ID:	26397 for all x.

    I'd assume f would be a constant function. It seems to have similar properties as such a function. Anyway, thanks for help!
    Hi zachd77!

    Have you tried fiddling around with f(x+y)=f(x)f(y)?
    What if you fill in a couple of special values?
    Like x=0 and y=0?
    Or like x=x and y=-x?
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    Re: interesting problem with a function (not defined) - quite intimidating

    yeah... the book I got this from said to try plugging in values.. I'll try it and see what happens.
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    Re: interesting problem with a function (not defined) - quite intimidating

    Ok so I got (a) down so far, but I got stuck on (b) and on (c) I ended up with an indeterminate form. Probably made a mistake or I'm just over thinking this problem a lot.
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    Super Member ILikeSerena's Avatar
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    Re: interesting problem with a function (not defined) - quite intimidating

    So what did you get from filling in x=y=0 respectively y=-x?

    The first should take care of (a), the second in combination with (a) should take care of (b).
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    Re: interesting problem with a function (not defined) - quite intimidating

    Quote Originally Posted by zachd77 View Post
    I found this question to be quite intimidating:

    Suppose a function f has the following properties:
    • f is differentiable everywhere.
    • For all values of x and y, Click image for larger version. 

Name:	CodeCogsEqn-13.gif 
Views:	3 
Size:	779 Bytes 
ID:	26391.
    • Click image for larger version. 

Name:	CodeCogsEqn-14.gif 
Views:	25 
Size:	469 Bytes 
ID:	26392.
    • Click image for larger version. 

Name:	CodeCogsEqn-15.gif 
Views:	26 
Size:	437 Bytes 
ID:	26393.



    a) Show that:
    Click image for larger version. 

Name:	CodeCogsEqn-17.gif 
Views:	26 
Size:	331 Bytes 
ID:	26394
    b) Show that Click image for larger version. 

Name:	CodeCogsEqn-18.gif 
Views:	25 
Size:	444 Bytes 
ID:	26395 for all values of x by first showing that Click image for larger version. 

Name:	CodeCogsEqn-19.gif 
Views:	41 
Size:	463 Bytes 
ID:	26396 for all x.

    c) Use the definition of the derivative to prove that
    Click image for larger version. 

Name:	CodeCogsEqn-20.gif 
Views:	25 
Size:	531 Bytes 
ID:	26397 for all x.

    I'd assume f would be a constant function. It seems to have similar properties as such a function. Anyway, thanks for help!
    a)f(0)=f2(0)=f3(0)=... and so on. It's possible when f(0)=1 only. f(0)=f(x)f(-x)=f(x)f(-x)f(0) => f(x)f(-x)=1
    b)if f(x)<0 then f(x)=f(x/2+x/2)=f2(x/2)<0 it's absurd. As f(x)f(-x)=1 then .


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  7. #7
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    Re: interesting problem with a function (not defined) - quite intimidating

    well since you KNOW f(0) = 1, then:

    1 = f(0) = f(a+(-a)) = f(a)f(-a).

    this means that f(a) ≠ 0 (or else we would get 1 = 0, right?). this works for ANY a.

    note that since f(a)f(-a) = 1, f(a) and f(-a) always have the same sign. so on the interval (-a,a) = (0-a,0+a) f is either entirely positive, or entirely negative, and thus positive (since f(0) = 1 > 0).

    (this is where continuity comes in: suppose for some b > 0, we had f(b) < 0. by continuity, at some point in (0,b), say c, we must have f(c) = 0 (the intermediate value theorem). but there IS no such c. a similar argument works for any b < 0).

    but if f is positive on any interval (-a,a), it must be positive on all of R.

    c) is relatively easy: by definition-

     f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} = \lim_{h \to 0} \frac{f(x)f(h) - f(x)}{h} = \lim_{h \to 0} f(x)\cdot \frac{f(h) - 1}{h}

     = f(x)\cdot \lim_{h \to 0} \frac{f(0)f(h) - f(0)}{h} = f(x)\cdot f'(0), so....

    also: your initial guess that f is constant is wrong: if f was constant, then f'(0) would always be 0 (constant functions are "flat").

    one class of (continuous) functions for which f(x+y) = f(x)f(y) are THESE: f(x) = ax, for a > 0. the trouble with this is, what does ax even MEAN for an irrational x?

    one way of doing this is by imagining a continuous function ax that "connects the dots" where "the dots" are ax for RATIONAL x. it would be helpful if we could produce a specific "a" that we KNEW existed (besides 1, which is boring). that is what this problem is trying to do: figure out what such a function might be like.

    what's coming next is: for what function f(x) = ax (if it exists, and is differentiable) might we have f(x) = f'(x)?
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    Re: interesting problem with a function (not defined) - quite intimidating

    e^x..... that's quite obvious
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    Re: interesting problem with a function (not defined) - quite intimidating

    Since we are given that f is differentiable, it is continuous and it is relatively easy to show that f(x)= e^{ax} (equivalently [/tex]f(x)= r^x[/tex]) are the only continuous functions satisfying f(x+y)= f(x)f(y). There are many NON-continous functions satisfying that but they have to be really discontinuous! In fact, they must be unbounded on any interval.
    Last edited by HallsofIvy; December 29th 2012 at 09:06 AM.
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  10. #10
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    Re: interesting problem with a function (not defined) - quite intimidating

    my point was: how do you define the function f(x) = ex if one doesn't already know what it is? what's so special about "e" that (ex)' = ex, but not for other positive real numbers?

    but, we have something we've learned about f(x) (just ignore e for a bit, pretend you've never heard about it). it equals its own derivative, AND...this derivative (itself) is always positive. that is: f(x) is a strictly increasing function.

    and THAT means: for every f(x) in the range of f, we have an inverse function g:f(x)--->x.

    here's the cool thing about inverse functions:

    suppose we have for a,b in R: [f(b) - f(a)]/(b - a), and f has an inverse on its entire domain (so f-1 is defined for all f(x) in the range of f).

    suppose that f(a) = c, and f(b) = d. we can re-write the fraction above as:

    (c - d)/[f-1(c) - f-1(d)]

    in particular, if b = a+h:

    [f(a+h) - f(a)]/h = (c - d)/[f-1(c) - f-1(d)]

    suppose that we call d-c, k, so that d = c+k.

    then:

    [f(a+h) - f(a)]/h = -k/[f-1(c) - f-1(c+k)] = k/[f-1(c+k) - f-1(c)]

    if f is continuous, then as h-->0, f(a+h) - f(a) -->0, which means that d-c -->0, that is: c+k - c = k -->0.


    so:

    f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} = \lim_{k \to 0} \frac{k}{f^{-1}(c+k) - f^{-1}(c)} = \frac{1}{(f^{-1})'(c)}

    if the derivative of f is never 0 for any of the x's we're considering:

    (f^{-1})'(c) = (f^{-1})'(f(a)) = \frac{1}{f'(a)}

    we can also write this as:

    (f^{-1})'(c) = \frac{1}{f'(f^{-1}(c))}

    which will be more useful for us.

    what does this mean for the PARTICULAR f(x) we've been discussing in this thread? to simplify the notation, let's use g(y) instead of f-1(y).

    it says that:

    g'(y) = \frac{1}{f'(g(y))}

    but, in this case: f'(g(y)) = f(g(y)) = y (since f and g are inverses, and since f'(x) = f(x)). that is:

    g'(y) = 1/y.

    note that since f takes x+y to f(x+y) = f(x)f(y)

    we have g(f(x)f(y)) = g(f(x+y)) = x + y = g(f(x)) + g(f(y))., so writing f(x) = a, f(y) = b:

    g(ab) = g(a) + g(b). holy canoli! we've discovered LOGARITHMS!

    also, since f(0) = 1, g(1) = 0.

    if you know the fundamental theorem of calculus, you know that:

    \int_a^b f'(t)\ dt = f(b) - f(a)

    so for our newly-discovered function g:

    \int_1^x g'(t)\ dt = g(x) - g(1) = g(x)

    that is:

    g(x) = \int_1^x \frac{1}{t}\ dt.

    now, all along we're thinking f(x) = ax for some positive number a. now we have a way of finding a, it is the real number a such that:

    \int_1^a \frac{1}{t}\ dt = 1 <-of course we have to have some means of approximating this integral, which might be tough, but we CAN do it.
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