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January 10th, 2015, 06:04 PM   #1
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Additional formula for tangent function

Suppose we wish to find a real-valued, differentiable function F(x) that satisfies the functional
equation F(x+y) = (F(x) + F(y))/ (1-F(x)*F(y))

Show that F necessarily satisfies F(0) = 0. Hint: Use the above to get an expression for F(0+0) and then use the fact that we seek F to be real-valued.

Set a = F'(0). Show that F must satisfy the diff eq dF/dx = a(1+F(x)^2)

Hint: Differentiate the above WRT to y then set y = 0
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January 19th, 2015, 10:00 AM   #2
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I doubt that you can find F(0) without being told that 0 lies in the domain of F. What have you managed to do so far?
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January 19th, 2015, 11:12 AM   #3
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It is normal to assume such things in this type of question. This is pretty easy, because you are told what to do!

$$\begin{align*} &\text{If $f(0)$ exists, then let } & x = y &= 0 \\ &\text{so that} & f(0) &= {2f(0) \over 1 - f^2(0)} \\ && \implies -f(0) \left(1 + f^2(0)\right) &= 0 \\ &&\implies 1 + f^2(0) &= 0 \qquad \text{or} \qquad f(0) = 0 \\[8pt] &\text{if $f$ is a real valued function} & f^2(0) &\ne -1 \implies f(0) = 0 \\[12pt] &\text{holding $x$ fixed, we have} & f'(x+y) &= {f'(y)\big(1-f(x)f(y)\big)+f'(y)f(x)\big(f(x) + f(y)\big) \over \big(1 - f(x)f(y)\big)^2} \qquad \text{using the quotient rule} \\ &\text{now setting $y=0$} & f'(x) &= f'(0) + f'(0)f^2(x) \\ &\text{writing $a = f'(0)$, we get} & f'(x) &= a\left(1+f^2(x)\right) \end{align*} $$

Now we can see that if $f'(0) = 0$ we have $f(x) = 0$ for all $x$.
If $f'(0) \gt 0$ then $f(x)$ is strictly increasing.
If $f'(0) \lt 0$ then $f(x)$ is strictly decreasing.

Last edited by v8archie; January 19th, 2015 at 11:16 AM.
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January 21st, 2015, 03:03 PM   #4
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Why should one assume 0 is in the domain, but not $\pi/2$? Unless F(x) is identically zero, it isn't differentiable everywhere and there are values of x and y for which F doesn't satisfy the functional equation.
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January 23rd, 2015, 04:21 AM   #5
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I think if you are asked to "show that F(0)= 0" you can reasonably assume that F(0) exists!
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January 23rd, 2015, 07:19 AM   #6
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Quote:
Originally Posted by skipjack View Post
Why should one assume 0 is in the domain, but not $\pi/2$? Unless F(x) is identically zero, it isn't differentiable everywhere and there are values of x and y for which F doesn't satisfy the functional equation.
Yes, I should have said that we must assume that $f(x)$ is differentiable, not necessarily everywhere, but our findings only hold where it is differentiable.

We don't need to assume anything about the function at $\frac\pi2$ because we don't use any properties there.

There are an infinite number of functions that are solutions to the functional equation for which $\frac\pi2$ is not special at all.

Last edited by v8archie; January 23rd, 2015 at 07:43 AM.
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January 23rd, 2015, 07:48 AM   #7
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Quote:
Originally Posted by Country Boy View Post
I think if you are asked to "show that F(0)= 0" you can reasonably assume that F(0) exists!
There may be solutions to the functional equation for which $f(0)$ does not exist. It is proper to state that you are not considering those unless you prove that $f(0)$ exists.
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January 23rd, 2015, 03:07 PM   #8
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Quote:
Originally Posted by Country Boy View Post
. . . if you are asked to "show that F(0)= 0" you can reasonably assume that F(0) exists!
The problem was to show that "F necessarily satisfies F(0) = 0". The word "necessarily" would be redundant unless it's there to emphasize that no assumption is needed. The opening paragraph begins "Suppose we wish to find" - it doesn't tell you that any function F exists with the properties given in the paragraph.
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January 23rd, 2015, 04:22 PM   #9
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I realise that I said above that I should state that we are assuming that $f$ is differentiable, but that is a condition that we were given - so I didn't need to state it.
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January 23rd, 2015, 06:30 PM   #10
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Quote:
Originally Posted by v8archie View Post
There are an infinite number of functions that are solutions to the functional equation for which $\frac\pi2$ is not special at all.
What did you mean here by "are solutions to the functional equation"?
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