IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

reddevil_2009

Find all tame solutions of:::::

`1) f^2(x)=f(x+y)f(x-y)`

`2) f(x+y)+f(x-y)=2f(x)`

`3) f(x+y)=[f(x)f(y)]/[f(x)+f(y)]`

rates assured for complete solution

reddevil_2009

Helpppppppppp

ankurgupta91

2) (c1)x + c

hsbhatt

Ok, the 1st one:

If f(0) = 0, then f(x) = 0 for all x.

Else:

It is easy to see that the function is either positive for all x or negative for all x. Also, if f(x) is a solution, so is -f(x).

So, without loss of generality, assume

$f^2(x) = f(x+y) f(x-y) \implies 2 \log_a f(x) = \log f(x+y) + \log f(x-y) \\ \\<br/>\implies \log f \left( \frac{(x+y) + (x-y)}{2} \right) = \frac{ \log f(x+y) + \log f(x-y)}{2}$

(I have dropped the base a for now, I will bring it back only in the end as typing out the base is a pain!)

Call g(x) = log f(x)

Then $g\left(\frac{x_1+x_2}{2} \right) = \frac{g(x_1)+g(x_2)}{2} \ \text{for any} x_1, x_2 \in \mathbb{R}$

This means the function is neither convex or concave on any interval. This means the function is linear

i.e. g(x) = px+q where p and q are constants.

$\therefore f(x) = a^{px+q} = c \ a^{px}$

hsbhatt

2nd one:

The solution is partly contained in the above post.

Again, we have,

$f(x+y) + f(x-y) = 2f(x) \Rightarrow f \left( \frac{(x+y)+(x-y)}{2} \right) = \frac{f(x+y) + f(x-y)}{2} \\ \\<br/>\Rightarrow f(x) = ax+b$

hsbhatt

I suspect the last one has not been answered till now only because of the choice of colour.

$f(x+y) = \frac{f(x) f(y)}{f(x) + f(y)} \Rightarrow \frac{1}{f(x+y)} = \frac{1}{f(x)} + \frac{1}{f(y)}$

Now, it's straightforward as $g(x) = \frac{1}{f(x)} \implies g(x+y) = g(x)+g(y)$

and that is one of the Cauchy Functional Equations (see http://www.goiit.com/posts/list/algebra-sir-please-reply-me-71982.htm#354815) with the general solution g(x) = kx

Hence the solution is $f(x) = \frac{c}{x}$

hsbhatt

edit: In the first solution I wrote:

"It is easy to see that the function is either positive for all x or negative for all x. Also, if f(x) is a solution, so is -f(x).

So, without loss of generality, assume ..."

The assumption was meant to be: "f(x) > 0 for all x"

hsbhatt

Ok to the list of standard functional equations, add the Jensen's equation:

$f \left(\frac{x+y}{2} \right ) = \frac{f(x)+f(y)}{2}$

with the general solution f(x) = ax+b as indicated above.

The proof, if you dont want to invoke the arguments of convexity and concavity is as follows:

Setting f(0) = a and then y = 0, we get

$f \left( \frac{x}{2} \right) = \frac{f(x)+a}{2}$

So, now

$\frac{f(x)+f(y)}{2} = f \left( \frac{x+y}{2} \right) = \frac{f(x+y)+a}{2} \\ \\<br/>\Rightarrow f(x) + f(y) = f(x+y)+a \ \text{or} \ [f(x)-a]+[f(y)-a] = [f(x+y) - a]$

Setting g(x) = f(x) - a,

we get g(x)+g(y) = g(x+y)

Now we have reduced it to the standard Cauchy functional equation with solution g(x) = mx

Hence f(x) = mx+a.

feynmann

U don't need at all that mumbo-jumbo ( concavity -convexity or the above proof ) to solve the functional eqn

$f \left(\frac{x+y}{2} \right ) = \frac{f(x)+f(y)}{2}$

It is very simple to solve .

first differentiate partially both sides wrt 'y' , yielding

$\frac{f$ ....................(1)

Now put y=0 in both sides , yielding f'(x/2)=f'(0) = constant .......(2)

from (2) it follows that f(x) =mx +c ( proved)

feynmann

Now somebody may object that it( my proof) requires the differentiability of f(x) .

My answer is : Yes , it does require that .

hsbhatt

Generally, while solving functional equations, unless specifically stated, it is not a done thing to assume differentiability.

That is why, the definition of convexity is given in the form of the following inequality

$f \left(\frac{x+y}{2} \right) \le \frac{f(x)+f(y)}{2}$ instead of the more compact f"(x)>0, so that even functions that are differentiable not everywhere (but countably many points) also qualify.

feynmann

I don't require differentiability at every point . If it is differentiable at a single point ( let that be zero point )then as my proof shows it is differentiable at all other points .

hsbhatt

It only proves that where it is differentiable, the derivative is a constant

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Find all tame solutions of:::::

1) f^2(x)=f(x+y)f(x-y)

2) f(x+y)+f(x-y)=2f(x)

3) f(x+y)=[f(x)f(y)]/[f(x)+f(y)]

`1) f^2(x)=f(x+y)f(x-y)`

`2) f(x+y)+f(x-y)=2f(x)`

`3) f(x+y)=[f(x)f(y)]/[f(x)+f(y)]`