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    I suppose Autograph isn't really good enough. It doesn't show anything.
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    (Original post by mrsmann)
    I suppose Autograph isn't really good enough. It doesn't show anything.
    Well, should it? A lot of those points (I don't want to say "most") won't be defined.
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    (Original post by mrsmann)
    I suppose Autograph isn't really good enough. It doesn't show anything.
    it's only as shocking as the fact y=x! doesn't show up.

    i didn't want to use the factorial example... can anyone name a function that is only real for all positive and integral negative values? it's on the tip of my.. brain.
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    I still haven't got an explanation as to why it crosses at e.

    Pleaaasee.
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    (Original post by thomas795135)
    I still haven't got an explanation as to why it crosses at e.

    Pleaaasee.
    Because x=y=e satisfies that equation.

    Don't see what's hard to understand.
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    (Original post by Chewwy)
    can anyone name a function that is only real for all positive and integral negative values? it's on the tip of my.. brain.
    1+i sin(pi x)?
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    (Original post by thomas795135)
    I still haven't got an explanation as to why it crosses at e.

    Pleaaasee.
    Because

    \displaystyle \frac{\ln t}{t} has a maximum at t = e.

    No seriously. The equation can be rewritten as

    \displaystyle \frac{\ln x}{x} = \frac{\ln y}{y}

    For all functions f(t) that are not injective, but for which (usually) there are (at most of the points) either two or zero t-values corresponding to each f-value, the graph of f(x) = f(y) will display the same kind of behaviour as the one you described: There is one set of points where x = y (which is obviously a line) included in the graph (because obviously, if x = y then f(x) = f(y)), but then there's another set of points(which incidentally forms a curve as long as f(x) is reasonably continuous), corresponding to when x and y different but for which the function f(t) nonetheless yields the same value.

    Now, can you sorta see that any point x = y = wherever the function has a maximum or a minimum (in your case, x = y = e) sorta belongs to both categories? This is why the lines cross there)

    A perhaps simpler example is f(t) = t^2 for instance. Can you see why the graphs of

    x^2 = y^2 (i.e. f(x) = f(y))

    are two lines that cross at the origin? They cross at the origin because t = 0 is the minimum of f(t).

    is two lines crossing at the origin.
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    (Original post by generalebriety)
    Because x=y=e satisfies that equation.

    Don't see what's hard to understand.
    Well, x=y=3 also satisfies that equation, but the lines don't cross at 3, do they?
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    (Original post by ukgea)
    Now, can you sorta see that any point x = y = wherever the function has a maximum or a minimum (in your case, x = y = e) sorta belongs to both categories? This is why the lines cross there)
    A better way to think of it is perhaps:

    Say you're walking along the latter curve (the one where in general x \neq y), by taking a small x-value and a large y-value so that f(x) = f(y), and then continually increase x and decrease y while still f(x) = f(y).

    Essentially, if you see it in the graph of f(t), you're basically taking two points on the graph, one t = x and one t = y, at the same height, and then moving the left one to the right and the right one two the left while still maintaining the same height. You can continue to do this, and it's clear that the points will pass each other at the maximum/minimum of the graph, and then continue in opposite directions. Which means that on this curve, x = y (i.e. when the points pass each other) will be at the maximum/minimum of f(t).
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    (Original post by DFranklin)
    Well, x=y=3 also satisfies that equation, but the lines don't cross at 3, do they?
    Oh, I see. Must have misread the question, I see what's going on now. :p: Sorry.
 
 
 
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