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Looking for a function, f, which smoothly maps (-1,+1) to (0,\infty), and f(0)=1

10 Answers

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(1 + x)/(1 - x) would do the trick.
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To OP: You may have got your answer, I am adding mine solely to suggest another way to think.

You are smoothly mapping (-1,1) to (0,\infty), to make matters simple let’s think increasing functions only. So f(x)—>0 as x—>-1, and f(x)—>\infty as x—>1.

Now you needn’t worry about f(0)=1 constraint, because in case our function has f(0)=c, we can just replace g(x)=f(x)/c that would satisfy all our requirements of correct limits. So let’s concentrate on the rest.

One readymade function which increases from zero to infinity in some range is tan(x), which takes (0,π/2) to (0,\infty). So if you can take the interval (-1,1) to (0,π/2) by a function h; then you are done; a candidate for f would be tan(h(x)).

Finding a smooth h is almost trivial, since you can fit a line h(x)=ax+b. With the conditions h(-1)=0 and h(1)=π/2, you can solve for a,b; solutions: >!(a=π/4, b=π/4)!<

So our f(x) is tan(ax+b), and c=f(0)=tan(b). So final answer would be g(x)=tan(ax+b)/tan(b).

Happy calculusing!
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I’m not a math expert, but would an exponential function work?

e.g.  y = a^x    f(0) will always be 1, +x will go to infinity and -x will go to zero?
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I’m so glad I’m done with Math in my life.
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You can just start with y = 1/x and shift stuff around as you want. So for this case,

2/(1-x) - 1
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Try this:


f(x) = ae^x + b

Let a = -e/(1-e)

Let b = 1 - a


This fits your description if I am not mistaken
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(x+1)/(e^x + xe^x - 2e^x )

You can put any constant instead of “e”.
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Okay, so the tangent function, tan(x), restricted to (-pi/2, pi/2) has image (-inf,inf). Thus, its module, |tan(x)|, has image \[0,inf). But its domain is still (-pi/2,pi/2), so we have to "stretch" (-1,1) to (-pi/2, pi/2), by multiplying it with pi/2. So,

**f(x)=|tan(pi\*x/2)|.**
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Well, a parabola would go to infinity. You could put the vertex at x = -1 and then the function would be

y = a(x + 1)\^2

Plug in x = 0, y = 1 to find a.
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exp(tan(x 2/pi))

Edit:
I meant
exp(tan(x pi/2))

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