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 January 10th, 2015, 06:04 PM #1 Newbie   Joined: Jan 2015 From: Davis Posts: 1 Thanks: 0 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
 January 19th, 2015, 10:00 AM #2 Global Moderator   Joined: Dec 2006 Posts: 20,814 Thanks: 2155 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?
 January 19th, 2015, 11:12 AM #3 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,671 Thanks: 2651 Math Focus: Mainly analysis and algebra 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.
 January 21st, 2015, 03:03 PM #4 Global Moderator   Joined: Dec 2006 Posts: 20,814 Thanks: 2155 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.
 January 23rd, 2015, 04:21 AM #5 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 I think if you are asked to "show that F(0)= 0" you can reasonably assume that F(0) exists!
January 23rd, 2015, 07:19 AM   #6
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 Originally Posted by skipjack 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.

January 23rd, 2015, 07:48 AM   #7
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 Originally Posted by Country Boy 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.

January 23rd, 2015, 03:07 PM   #8
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Quote:
 Originally Posted by Country Boy . . . 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.

 January 23rd, 2015, 04:22 PM #9 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,671 Thanks: 2651 Math Focus: Mainly analysis and algebra 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.
January 23rd, 2015, 06:30 PM   #10
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Quote:
 Originally Posted by v8archie 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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