jagr2713
  • jagr2713
A little bored so i am going to ask a QH question :D
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
jagr2713
  • 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
  • jagr2713
@dan815 @ganeshie8
jagr2713
  • jagr2713
Anybody? 12mins have gone by D:

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More answers

anonymous
  • anonymous
need help
jagr2713
  • jagr2713
Of course
anonymous
  • anonymous
i need help
hartnn
  • hartnn
Its an even function.
jagr2713
  • jagr2713
?? what
hartnn
  • 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
  • jagr2713
i had choices lol this is off this site dan gave me and i had OB so i just asked it lol
sparrow2
  • sparrow2
can f(x)=2x?
jagr2713
  • jagr2713
but i have the answer i just need you guys the figure it out :D sorta like a mini riddle or whatever
sparrow2
  • sparrow2
i have no idea beside f(x)=2x :D
jagr2713
  • jagr2713
Nope thats not the answer :/
Empty
  • Empty
I'd definitely substitute the elliptic substitution lol. a=x+y, b=x-y
Empty
  • 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
  • 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
  • freckles
\[u=2x-1 \\ \text{ then } x=\frac{u+1}{2}\]
freckles
  • 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
  • freckles
\[f(u)=(u+1)(\frac{-u-1+2}{2}) \\ f(u)=(u+1)(\frac{-u+1}{2})\]
freckles
  • freckles
\[f(u)=\frac{1}{2}(1+u)(1-u) \\ f(u)=\frac{1}{2}(1-u^2)\]
freckles
  • freckles
I think I made a mistake somewhere since f(1) doesn't equal 2 here
freckles
  • freckles
|dw:1436382316562:dw|
hartnn
  • 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
  • freckles
omg I made a mistake
freckles
  • freckles
thanks @hartnn
freckles
  • freckles
u=2x-1 (u+1)/2=x \[f(u)=u(2)+4(\frac{u+2}{2})(-\frac{u+1}{2}+1)u\]
freckles
  • freckles
lol
freckles
  • freckles
\[f(u)=u(2)+4(\frac{u+1}{2})(-\frac{u+1}{2}+1)u\]
freckles
  • freckles
which should give you after simplifying \[f(u)=3u-u^3\] this does have f(1)=3-1=2
ganeshie8
  • ganeshie8
That x+y=1 is really a very clever substitution!
freckles
  • freckles
I guess we could have went the other way and chose x-y=1
freckles
  • freckles
should end up with same thing
hartnn
  • hartnn
why just x+y =1 ?? why not x+y =2 or say 2000?
freckles
  • freckles
because f(1)=2
freckles
  • freckles
like in that x+y thingy was in the f( ) thingy
sparrow2
  • sparrow2
so you have u=2x-1 so f(u)=f(2x-1)? don;'t get
freckles
  • freckles
why can't u be 2x-1 @sparrow2
sparrow2
  • sparrow2
it can but they want f(x) not f(2x-1) or did i miss smth?
freckles
  • freckles
I found f(u)=3u-u^3 replace u's with x
ganeshie8
  • ganeshie8
u or x or some variable, they all are just dummy
sparrow2
  • sparrow2
i understand but when you replace you get f(2x-1) as function not f(x)
freckles
  • freckles
we were looking for f( )
freckles
  • freckles
I found f( )
freckles
  • freckles
well we found f( )*
ganeshie8
  • 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
  • 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
  • hartnn
nice work! :)
sparrow2
  • sparrow2
it will be good if you just write what is f(x)
freckles
  • freckles
I did
sparrow2
  • sparrow2
so f(u) and f(x) is the same?
freckles
  • freckles
"I found f(u)=3u-u^3 replace u's with x "
freckles
  • freckles
\[f(x)=3x-x^3\]
sparrow2
  • sparrow2
u=2x-1 (as you wrote) and you just write that u=x
freckles
  • freckles
u doesn't equal x
freckles
  • freckles
f( ) is a machine
freckles
  • freckles
the input changes the output based on the input
sparrow2
  • sparrow2
i know what is a fucntion :)
freckles
  • freckles
like f(goldfish)=3(goldfish)-(goldfish)^3
sparrow2
  • sparrow2
but this function has very specific argument
freckles
  • freckles
I don't think understand what you are saying
ganeshie8
  • 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
  • ganeshie8
functions work slightly differently as the given function was assumed to be continuous and nice hmm
sparrow2
  • sparrow2
by the way i am wondering in f(2x)=x argument is 2x yes?
ganeshie8
  • ganeshie8
f(2x) = x let 2x = u f(u) = u/2 f(x) = x/2 see anythign wrong ? i dont see any..
sparrow2
  • sparrow2
i mean in f(2x)=x argument is 2x yes?
ganeshie8
  • ganeshie8
yes, by argument if it means input
sparrow2
  • 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
  • Empty
\[f(x)=3x-x^3\]
freckles
  • freckles
@Empty did you do something different than using x+y=1 (or x-y=1)?
freckles
  • freckles
if so I would like to see your method too
jagr2713
  • jagr2713
well @Empty almost has it
freckles
  • freckles
what f(x)=3x-x^3 is also what I got
freckles
  • freckles
and it does work @jagr2713
jagr2713
  • jagr2713
you to @freckles
freckles
  • freckles
so if that isn't it what is the answer @jagr2713
jagr2713
  • jagr2713
lol lets see what empty gets
jagr2713
  • jagr2713
Ok well its\[-x ^{3}+3x\] Almost got it @freckles and @Empty
freckles
  • freckles
do you know addition is commutative ?
freckles
  • freckles
a+b is the same as b+a
jagr2713
  • jagr2713
yea but you forgot the +
freckles
  • freckles
ummm no
freckles
  • freckles
\[3x-x^3=+3x-x^3=+3x+(-x^3) \\ \text{ change order } \\ =(-x^3)+3x \\ =-x^3+3x\]
jagr2713
  • jagr2713
yea but above i saw you both but -x3-3x not +3x
Empty
  • 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
  • freckles
we both wrote 3x-x^3 :p
jagr2713
  • jagr2713
but its + not - :D dont know who to give the medal D: both did a great job and you to ganeshie
freckles
  • freckles
3x-x^3 is the same thing as -x^3+3x what are you talking about?
freckles
  • freckles
verify with some numbers if you are not sure
jagr2713
  • jagr2713
i am confused now lol
freckles
  • freckles
like do you know that -3+5 is the same as 5-3?
freckles
  • freckles
like you had -x^3+3x is the same as 3x-x^3
freckles
  • freckles
it is the same with numbers
freckles
  • freckles
because those are numbers
Empty
  • 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
  • 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
  • jagr2713
Hm thanks guys
freckles
  • 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
  • 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
  • freckles
The differential equation thing is pretty clever.
Empty
  • 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
  • Empty
What's interesting to me though is how f(x) is not an even function.
freckles
  • 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
  • freckles
and our function was odd
Empty
  • 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
  • Empty
Does the condition \(|x| \ne |y|\) really mean anything for this problem?
freckles
  • 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
  • freckles
and we should see something similar with the other case
freckles
  • freckles
I said that little implies statement a little wrong
freckles
  • freckles
didn't include the not equals part
freckles
  • freckles
so I don't know why they put that |x| couldn't be |y|
Empty
  • Empty
Yeah, weird huh.

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