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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 realvalued, differentiable function F(x) that satisfies the functional equation F(x+y) = (F(x) + F(y))/ (1F(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 realvalued. 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! $$\begin{align*} &\text{If $f(0)$ exists, then let } & x = y &= 0 \\ &\text{so that} & f(0) &= {2f(0) \over 1  f^2(0)} \\ && \implies f(0) \left(1 + f^2(0)\right) &= 0 \\ &&\implies 1 + f^2(0) &= 0 \qquad \text{or} \qquad f(0) = 0 \\[8pt] &\text{if $f$ is a real valued function} & f^2(0) &\ne 1 \implies f(0) = 0 \\[12pt] &\text{holding $x$ fixed, we have} & f'(x+y) &= {f'(y)\big(1f(x)f(y)\big)+f'(y)f(x)\big(f(x) + f(y)\big) \over \big(1  f(x)f(y)\big)^2} \qquad \text{using the quotient rule} \\ &\text{now setting $y=0$} & f'(x) &= f'(0) + f'(0)f^2(x) \\ &\text{writing $a = f'(0)$, we get} & f'(x) &= a\left(1+f^2(x)\right) \end{align*} $$ 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  
Math Team Joined: Dec 2013 From: Colombia Posts: 7,671 Thanks: 2651 Math Focus: Mainly analysis and algebra  Quote:
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 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,671 Thanks: 2651 Math Focus: Mainly analysis and algebra  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 
Global Moderator Joined: Dec 2006 Posts: 20,814 Thanks: 2155  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 
Global Moderator Joined: Dec 2006 Posts: 20,814 Thanks: 2155  

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