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jagr2713
 one year ago
A little bored so i am going to ask a QH question :D
jagr2713
 one year ago
A little bored so i am going to ask a QH question :D

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jagr2713
 one year ago
Best ResponseYou've already chosen the best response.1The function F satisfies \[(xy)f(x+y)(x+y)f(xy)+4xy(x ^{2}y ^{2})=0\] whenever \[\left x \right\neq \left y \right\] Given that f(1)=2. what is f(x)

jagr2713
 one year ago
Best ResponseYou've already chosen the best response.1Anybody? 12mins have gone by D:

hartnn
 one year ago
Best ResponseYou've already chosen the best response.3you have any choices?? Its an even function, because when you plug in x=0, y not =0, you get f(y) = f(y)

jagr2713
 one year ago
Best ResponseYou've already chosen the best response.1i had choices lol this is off this site dan gave me and i had OB so i just asked it lol

jagr2713
 one year ago
Best ResponseYou've already chosen the best response.1but i have the answer i just need you guys the figure it out :D sorta like a mini riddle or whatever

sparrow2
 one year ago
Best ResponseYou've already chosen the best response.2i have no idea beside f(x)=2x :D

jagr2713
 one year ago
Best ResponseYou've already chosen the best response.1Nope thats not the answer :/

Empty
 one year ago
Best ResponseYou've already chosen the best response.1I'd definitely substitute the elliptic substitution lol. a=x+y, b=xy

Empty
 one year ago
Best ResponseYou've already chosen the best response.1This will give you: \[b f(a)  a f(b)+(a^2b^2)ab=0 \] At least that's nicer to look at don't you think so?

freckles
 one year ago
Best ResponseYou've already chosen the best response.3\[\text{ If } x+y=1 \text{ then } y=x+1 \text{ and so } xy=2x1 \\ \text{ so we have } \\ (2x1)f(1)f(2x1)+4x(x+1)(2x1)=0 \\ (2x1)(2)f(2x1)+4x(x+1)(2x1)=0 \\ f(2x1)=\frac{4x(x+1)(2x1)}{2(2x1)}\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.3\[u=2x1 \\ \text{ then } x=\frac{u+1}{2}\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.3\[f(u)=\frac{4 \cdot \frac{u+1}{2}(\frac{u+1}{2}+1)(u)}{2u} \text{ assuming } 2x1=u \neq 0 \\ \text{ we have } \\ f(u)=2 \frac{u+1}{2}(\frac{u1}{2}+1)\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.3\[f(u)=(u+1)(\frac{u1+2}{2}) \\ f(u)=(u+1)(\frac{u+1}{2})\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.3\[f(u)=\frac{1}{2}(1+u)(1u) \\ f(u)=\frac{1}{2}(1u^2)\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.3I think I made a mistake somewhere since f(1) doesn't equal 2 here

freckles
 one year ago
Best ResponseYou've already chosen the best response.3dw:1436382316562:dw

hartnn
 one year ago
Best ResponseYou've already chosen the best response.3\((2x1)(2)f(2x1)+4x(x+1)(2x1)=0 \) next step? \((2x1)(2)+4x(x+1)(2x1)=f(2x1)\) right?

freckles
 one year ago
Best ResponseYou've already chosen the best response.3u=2x1 (u+1)/2=x \[f(u)=u(2)+4(\frac{u+2}{2})(\frac{u+1}{2}+1)u\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.3\[f(u)=u(2)+4(\frac{u+1}{2})(\frac{u+1}{2}+1)u\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.3which should give you after simplifying \[f(u)=3uu^3\] this does have f(1)=31=2

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1That x+y=1 is really a very clever substitution!

freckles
 one year ago
Best ResponseYou've already chosen the best response.3I guess we could have went the other way and chose xy=1

freckles
 one year ago
Best ResponseYou've already chosen the best response.3should end up with same thing

hartnn
 one year ago
Best ResponseYou've already chosen the best response.3why just x+y =1 ?? why not x+y =2 or say 2000?

freckles
 one year ago
Best ResponseYou've already chosen the best response.3like in that x+y thingy was in the f( ) thingy

sparrow2
 one year ago
Best ResponseYou've already chosen the best response.2so you have u=2x1 so f(u)=f(2x1)? don;'t get

freckles
 one year ago
Best ResponseYou've already chosen the best response.3why can't u be 2x1 @sparrow2

sparrow2
 one year ago
Best ResponseYou've already chosen the best response.2it can but they want f(x) not f(2x1) or did i miss smth?

freckles
 one year ago
Best ResponseYou've already chosen the best response.3I found f(u)=3uu^3 replace u's with x

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1u or x or some variable, they all are just dummy

sparrow2
 one year ago
Best ResponseYou've already chosen the best response.2i understand but when you replace you get f(2x1) as function not f(x)

freckles
 one year ago
Best ResponseYou've already chosen the best response.3we were looking for f( )

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1I think @sparrow2 's question makes sense if f() is a sequence then f(2x1) does not tell any info about f(2x)

freckles
 one year ago
Best ResponseYou've already chosen the best response.3but f( ) is a function so If I found f(2x1) then I can find f(u) by replacing all the x's with (u+1)/2

sparrow2
 one year ago
Best ResponseYou've already chosen the best response.2it will be good if you just write what is f(x)

sparrow2
 one year ago
Best ResponseYou've already chosen the best response.2so f(u) and f(x) is the same?

freckles
 one year ago
Best ResponseYou've already chosen the best response.3"I found f(u)=3uu^3 replace u's with x "

sparrow2
 one year ago
Best ResponseYou've already chosen the best response.2u=2x1 (as you wrote) and you just write that u=x

freckles
 one year ago
Best ResponseYou've already chosen the best response.3the input changes the output based on the input

sparrow2
 one year ago
Best ResponseYou've already chosen the best response.2i know what is a fucntion :)

freckles
 one year ago
Best ResponseYou've already chosen the best response.3like f(goldfish)=3(goldfish)(goldfish)^3

sparrow2
 one year ago
Best ResponseYou've already chosen the best response.2but this function has very specific argument

freckles
 one year ago
Best ResponseYou've already chosen the best response.3I don't think understand what you are saying

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1I 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(2n1), then does that mean we also have a formula for F(2n) ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1functions work slightly differently as the given function was assumed to be continuous and nice hmm

sparrow2
 one year ago
Best ResponseYou've already chosen the best response.2by the way i am wondering in f(2x)=x argument is 2x yes?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1f(2x) = x let 2x = u f(u) = u/2 f(x) = x/2 see anythign wrong ? i dont see any..

sparrow2
 one year ago
Best ResponseYou've already chosen the best response.2i mean in f(2x)=x argument is 2x yes?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1yes, by argument if it means input

sparrow2
 one year ago
Best ResponseYou've already chosen the best response.2and 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?

freckles
 one year ago
Best ResponseYou've already chosen the best response.3@Empty did you do something different than using x+y=1 (or xy=1)?

freckles
 one year ago
Best ResponseYou've already chosen the best response.3if so I would like to see your method too

jagr2713
 one year ago
Best ResponseYou've already chosen the best response.1well @Empty almost has it

freckles
 one year ago
Best ResponseYou've already chosen the best response.3what f(x)=3xx^3 is also what I got

freckles
 one year ago
Best ResponseYou've already chosen the best response.3and it does work @jagr2713

freckles
 one year ago
Best ResponseYou've already chosen the best response.3so if that isn't it what is the answer @jagr2713

jagr2713
 one year ago
Best ResponseYou've already chosen the best response.1lol lets see what empty gets

jagr2713
 one year ago
Best ResponseYou've already chosen the best response.1Ok well its\[x ^{3}+3x\] Almost got it @freckles and @Empty

freckles
 one year ago
Best ResponseYou've already chosen the best response.3do you know addition is commutative ?

freckles
 one year ago
Best ResponseYou've already chosen the best response.3a+b is the same as b+a

jagr2713
 one year ago
Best ResponseYou've already chosen the best response.1yea but you forgot the +

freckles
 one year ago
Best ResponseYou've already chosen the best response.3\[3xx^3=+3xx^3=+3x+(x^3) \\ \text{ change order } \\ =(x^3)+3x \\ =x^3+3x\]

jagr2713
 one year ago
Best ResponseYou've already chosen the best response.1yea but above i saw you both but x33x not +3x

Empty
 one year ago
Best ResponseYou've already chosen the best response.1Substitute \(a=x+y\) and \(b=xy\) to get: \[b f(a)  a f(b)+(a^2b^2)ab=0 \] Now let's rearrange and substitute so that it looks like this: \[af(b)bf(a)=(a^2b^2)ab\] 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: \[f(a)=caa^3\] Plug in f(1)=2 to solve for c.

freckles
 one year ago
Best ResponseYou've already chosen the best response.3we both wrote 3xx^3 :p

jagr2713
 one year ago
Best ResponseYou've already chosen the best response.1but its + not  :D dont know who to give the medal D: both did a great job and you to ganeshie

freckles
 one year ago
Best ResponseYou've already chosen the best response.33xx^3 is the same thing as x^3+3x what are you talking about?

freckles
 one year ago
Best ResponseYou've already chosen the best response.3verify with some numbers if you are not sure

jagr2713
 one year ago
Best ResponseYou've already chosen the best response.1i am confused now lol

freckles
 one year ago
Best ResponseYou've already chosen the best response.3like do you know that 3+5 is the same as 53?

freckles
 one year ago
Best ResponseYou've already chosen the best response.3like you had x^3+3x is the same as 3xx^3

freckles
 one year ago
Best ResponseYou've already chosen the best response.3it is the same with numbers

freckles
 one year ago
Best ResponseYou've already chosen the best response.3because those are numbers

Empty
 one year ago
Best ResponseYou've already chosen the best response.1Woah forget all this, we're right it's ok, what's important is admire how I turned this into a differential equation :O

freckles
 one year ago
Best ResponseYou've already chosen the best response.3I'm just trying to convince him addition is commutative @empty would like him to understand our answers since he asked for them :p

freckles
 one year ago
Best ResponseYou've already chosen the best response.3if you don't trust me or not sure what I'm talking about maybe you will believe wolfram http://www.wolframalpha.com/input/?i=3xx%5E3%3Dx%5E3%2B3x

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1Wow! From the differential equation, it seems f(x) = cx  x^3 satisfies the given functional equation, if we forget about initial conditions !

freckles
 one year ago
Best ResponseYou've already chosen the best response.3The differential equation thing is pretty clever.

Empty
 one year ago
Best ResponseYou've already chosen the best response.1Yeah 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
 one year ago
Best ResponseYou've already chosen the best response.1What's interesting to me though is how f(x) is not an even function.

freckles
 one year ago
Best ResponseYou've already chosen the best response.3I 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
 one year ago
Best ResponseYou've already chosen the best response.3and our function was odd

Empty
 one year ago
Best ResponseYou've already chosen the best response.1Wait nevermind if I look at the relation \(bf(a)af(b)+(a^2b^2)ab=0\) 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
 one year ago
Best ResponseYou've already chosen the best response.1Does the condition \(x \ne y\) really mean anything for this problem?

freckles
 one year ago
Best ResponseYou've already chosen the best response.3\[\left x \right\neq \left y \right \implies x=y \text{ or } x=y \\ \text{ so pluggin this in we get } \\ \] \[(xy)f(x+y)(x+y)f(xy)+4xy(x ^{2}y ^{2})=0\] if x=y we have: \[(xx)f(x+x)(x+x)f(xx)4xx(xx)(x+x)=0 \\ 0f(2x)2xf(0)4x^2(0)(2x)=0 \\ 2x f(0)=0 \\ \text{ so } x=0 \text{ or } f(0)=0 \\ \text{ but I don't think we can say } \\ \text{ but I don't think we can say } x=0 \text{ since } x \text{ varies } \\ \\ \text{so } f(0)=0 \\ \text{ but we had found } f(x)=3xx^3 \\ \text{ and so } f(0)=0 \text{ is actually satisfied so hmmm weird }\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.3and we should see something similar with the other case

freckles
 one year ago
Best ResponseYou've already chosen the best response.3I said that little implies statement a little wrong

freckles
 one year ago
Best ResponseYou've already chosen the best response.3didn't include the not equals part

freckles
 one year ago
Best ResponseYou've already chosen the best response.3so I don't know why they put that x couldn't be y
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