A little bored so i am going to ask a QH question :D

- jagr2713

A little bored so i am going to ask a QH question :D

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- jagr2713

The function F satisfies
\[(x-y)f(x+y)-(x+y)f(x-y)+4xy(x ^{2}-y ^{2})=0\]
whenever \[\left| x \right|\neq \left| y \right|\]
Given that f(1)=2. what is
f(x)

- jagr2713

@dan815 @ganeshie8

- jagr2713

Anybody? 12mins have gone by D:

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## More answers

- anonymous

need help

- jagr2713

Of course

- anonymous

i need help

- hartnn

Its an even function.

- jagr2713

?? what

- hartnn

you have any choices??
Its an even function,
because when you plug in x=0, y not =0,
you get f(y) = f(-y)

- jagr2713

i had choices lol
this is off this site dan gave me and i had OB so i just asked it lol

- sparrow2

can f(x)=2x?

- jagr2713

but i have the answer i just need you guys the figure it out :D
sorta like a mini riddle or whatever

- sparrow2

i have no idea beside f(x)=2x :D

- jagr2713

Nope thats not the answer :/

- Empty

I'd definitely substitute the elliptic substitution lol.
a=x+y, b=x-y

- Empty

This will give you:
\[b f(a) - a f(b)+(a^2-b^2)ab=0 \] At least that's nicer to look at don't you think so?

- freckles

\[\text{ If } x+y=1 \text{ then } y=-x+1 \text{ and so } x-y=2x-1 \\ \text{ so we have } \\ (2x-1)f(1)-f(2x-1)+4x(-x+1)(2x-1)=0 \\ (2x-1)(2)-f(2x-1)+4x(-x+1)(2x-1)=0 \\ f(2x-1)=\frac{4x(-x+1)(2x-1)}{2(2x-1)}\]

- freckles

\[u=2x-1 \\ \text{ then } x=\frac{u+1}{2}\]

- freckles

\[f(u)=\frac{4 \cdot \frac{u+1}{2}(-\frac{u+1}{2}+1)(u)}{2u} \text{ assuming } 2x-1=u \neq 0 \\ \text{ we have } \\ f(u)=2 \frac{u+1}{2}(\frac{-u-1}{2}+1)\]

- freckles

\[f(u)=(u+1)(\frac{-u-1+2}{2}) \\ f(u)=(u+1)(\frac{-u+1}{2})\]

- freckles

\[f(u)=\frac{1}{2}(1+u)(1-u) \\ f(u)=\frac{1}{2}(1-u^2)\]

- freckles

I think I made a mistake somewhere since f(1) doesn't equal 2 here

- freckles

|dw:1436382316562:dw|

- hartnn

\((2x-1)(2)-f(2x-1)+4x(-x+1)(2x-1)=0 \)
next step?
\((2x-1)(2)+4x(-x+1)(2x-1)=f(2x-1)\)
right?

- freckles

omg I made a mistake

- freckles

thanks @hartnn

- freckles

u=2x-1
(u+1)/2=x
\[f(u)=u(2)+4(\frac{u+2}{2})(-\frac{u+1}{2}+1)u\]

- freckles

lol

- freckles

\[f(u)=u(2)+4(\frac{u+1}{2})(-\frac{u+1}{2}+1)u\]

- freckles

which should give you after simplifying
\[f(u)=3u-u^3\]
this does have f(1)=3-1=2

- ganeshie8

That x+y=1 is really a very clever substitution!

- freckles

I guess we could have went the other way and chose x-y=1

- freckles

should end up with same thing

- hartnn

why just x+y =1 ??
why not x+y =2 or say 2000?

- freckles

because f(1)=2

- freckles

like in that x+y thingy was in the f( ) thingy

- sparrow2

so you have u=2x-1 so f(u)=f(2x-1)? don;'t get

- freckles

why can't u be 2x-1 @sparrow2

- sparrow2

it can but they want f(x) not f(2x-1) or did i miss smth?

- freckles

I found f(u)=3u-u^3
replace u's with x

- ganeshie8

u or x or some variable, they all are just dummy

- sparrow2

i understand but when you replace you get f(2x-1) as function not f(x)

- freckles

we were looking for f( )

- freckles

I found f( )

- freckles

well we found f( )*

- ganeshie8

I think @sparrow2 's question makes sense if f() is a sequence
then f(2x-1) does not tell any info about f(2x)

- freckles

but f( ) is a function
so If I found f(2x-1)
then I can find f(u) by replacing all the x's with (u+1)/2

- hartnn

nice work! :)

- sparrow2

it will be good if you just write what is f(x)

- freckles

I did

- sparrow2

so f(u) and f(x) is the same?

- freckles

"I found f(u)=3u-u^3
replace u's with x
"

- freckles

\[f(x)=3x-x^3\]

- sparrow2

u=2x-1 (as you wrote) and you just write that u=x

- freckles

u doesn't equal x

- freckles

f( ) is a machine

- freckles

the input changes the output based on the input

- sparrow2

i know what is a fucntion :)

- freckles

like f(goldfish)=3(goldfish)-(goldfish)^3

- sparrow2

but this function has very specific argument

- freckles

I don't think understand what you are saying

- ganeshie8

I think sparrow has the same question that trouble me when i first looked at ur solution but i ignored it
Let say F(n) gives the n'th fibonacci number
then if we have a formula for F(2n-1), then does that mean we also have a formula for F(2n) ?

- ganeshie8

functions work slightly differently as the given function was assumed to be continuous and nice hmm

- sparrow2

by the way i am wondering in f(2x)=x argument is 2x yes?

- ganeshie8

f(2x) = x
let 2x = u
f(u) = u/2
f(x) = x/2
see anythign wrong ? i dont see any..

- sparrow2

i mean in f(2x)=x argument is 2x yes?

- ganeshie8

yes, by argument if it means input

- sparrow2

and when you need to put the argument on x axis, you put 2x and not x yes? like when x=1(this isn't argument) i put 2x=2 so 2 into the x axis yes?

- Empty

\[f(x)=3x-x^3\]

- freckles

@Empty did you do something different than using x+y=1 (or x-y=1)?

- freckles

if so I would like to see your method too

- jagr2713

well @Empty almost has it

- freckles

what f(x)=3x-x^3 is also what I got

- freckles

and it does work @jagr2713

- jagr2713

you to @freckles

- freckles

so if that isn't it what is the answer @jagr2713

- jagr2713

lol
lets see what empty gets

- jagr2713

Ok well its\[-x ^{3}+3x\]
Almost got it @freckles and @Empty

- freckles

do you know addition is commutative ?

- freckles

a+b is the same as b+a

- jagr2713

yea but you forgot the +

- freckles

ummm no

- freckles

\[3x-x^3=+3x-x^3=+3x+(-x^3) \\ \text{ change order } \\ =(-x^3)+3x \\ =-x^3+3x\]

- jagr2713

yea but above i saw you both but -x3-3x not +3x

- Empty

Substitute \(a=x+y\) and \(b=x-y\) to get:
\[b f(a) - a f(b)+(a^2-b^2)ab=0 \]
Now let's rearrange and substitute so that it looks like this:
\[af(b)-bf(a)=(a^2-b^2)ab\]
substitute in b=a+h and rearrange to get:
\[\frac{f(a+h)-f(a)}{h}=\frac{f(a)}{a}-(2a+h)(a+h)\]
Let h go to zero... ;)
\[f'(a)=\frac{f(a)}{a}-2a^2\]
Solve the differential equation to get:
\[f(a)=ca-a^3\]
Plug in f(1)=2 to solve for c.

- freckles

we both wrote 3x-x^3 :p

- jagr2713

but its + not - :D
dont know who to give the medal D: both did a great job and you to ganeshie

- freckles

3x-x^3 is the same thing as -x^3+3x
what are you talking about?

- freckles

verify with some numbers if you are not sure

- jagr2713

i am confused now lol

- freckles

like do you know that
-3+5 is the same as 5-3?

- freckles

like you had -x^3+3x is the same as
3x-x^3

- freckles

it is the same with numbers

- freckles

because those are numbers

- Empty

Woah forget all this, we're right it's ok, what's important is admire how I turned this into a differential equation :O

- freckles

I'm just trying to convince him addition is commutative @empty would like him to understand our answers since he asked for them :p

- jagr2713

Hm thanks guys

- freckles

if you don't trust me or not sure what I'm talking about
maybe you will believe wolfram http://www.wolframalpha.com/input/?i=3x-x%5E3%3D-x%5E3%2B3x

- ganeshie8

Wow! From the differential equation, it seems f(x) = cx - x^3 satisfies the given functional equation, if we forget about initial conditions !

- freckles

The differential equation thing is pretty clever.

- Empty

Yeah I hadn't even realized that this was actually more general and that f(1)=2 is only a particular solution to this functional equation. Fascinating! I was lucky.

- Empty

What's interesting to me though is how f(x) is not an even function.

- freckles

I hope I wasn't come out mean by the way.
None of my argument above was intended to be mean.
well if x=0 and y doesn't then we have
\[-yf(y)=yf(-y)\]
and if y doesn't equal 0 we have
\[-f(y)=f(-y)\]
this actually means f is odd right?

- freckles

and our function was odd

- Empty

Wait nevermind if I look at the relation
\(bf(a)-af(b)+(a^2-b^2)ab=0\) we can plug in \(b=-a\) to get that it's odd. I guess I was confused by hartnn's comment earlier and didn't check it, but if I plug it in now it's odd.

- Empty

Does the condition \(|x| \ne |y|\) really mean anything for this problem?

- freckles

\[\left| x \right|\neq \left| y \right| \implies x=y \text{ or } x=-y \\ \text{ so pluggin this in we get } \\ \]
\[(x-y)f(x+y)-(x+y)f(x-y)+4xy(x ^{2}-y ^{2})=0\]
if x=y we have:
\[(x-x)f(x+x)-(x+x)f(x-x)-4xx(x-x)(x+x)=0 \\ 0f(2x)-2xf(0)-4x^2(0)(2x)=0 \\ -2x f(0)=0 \\ \text{ so } x=0 \text{ or } f(0)=0 \\ \text{ but I don't think we can say } \\ \text{ but I don't think we can say } x=0 \text{ since } x \text{ varies } \\ \\ \text{so } f(0)=0 \\ \text{ but we had found } f(x)=3x-x^3 \\ \text{ and so } f(0)=0 \text{ is actually satisfied so hmmm weird }\]

- freckles

and we should see something similar with the other case

- freckles

I said that little implies statement a little wrong

- freckles

didn't include the not equals part

- freckles

so I don't know why they put that |x| couldn't be |y|

- Empty

Yeah, weird huh.

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