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

1. 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)

2. jagr2713

@dan815 @ganeshie8

3. jagr2713

Anybody? 12mins have gone by D:

4. anonymous

need help

5. jagr2713

Of course

6. anonymous

i need help

7. hartnn

Its an even function.

8. jagr2713

?? what

9. 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)

10. jagr2713

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

11. sparrow2

can f(x)=2x?

12. jagr2713

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

13. sparrow2

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

14. jagr2713

Nope thats not the answer :/

15. Empty

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

16. 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?

17. freckles

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

18. freckles

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

19. freckles

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

20. freckles

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

21. freckles

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

22. freckles

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

23. freckles

|dw:1436382316562:dw|

24. 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?

25. freckles

omg I made a mistake

26. freckles

thanks @hartnn

27. freckles

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

28. freckles

lol

29. freckles

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

30. freckles

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

31. ganeshie8

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

32. freckles

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

33. freckles

should end up with same thing

34. hartnn

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

35. freckles

because f(1)=2

36. freckles

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

37. sparrow2

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

38. freckles

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

39. sparrow2

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

40. freckles

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

41. ganeshie8

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

42. sparrow2

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

43. freckles

we were looking for f( )

44. freckles

I found f( )

45. freckles

well we found f( )*

46. 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)

47. 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

48. hartnn

nice work! :)

49. sparrow2

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

50. freckles

I did

51. sparrow2

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

52. freckles

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

53. freckles

$f(x)=3x-x^3$

54. sparrow2

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

55. freckles

u doesn't equal x

56. freckles

f( ) is a machine

57. freckles

the input changes the output based on the input

58. sparrow2

i know what is a fucntion :)

59. freckles

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

60. sparrow2

but this function has very specific argument

61. freckles

I don't think understand what you are saying

62. 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) ?

63. ganeshie8

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

64. sparrow2

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

65. ganeshie8

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

66. sparrow2

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

67. ganeshie8

yes, by argument if it means input

68. 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?

69. Empty

$f(x)=3x-x^3$

70. freckles

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

71. freckles

if so I would like to see your method too

72. jagr2713

well @Empty almost has it

73. freckles

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

74. freckles

and it does work @jagr2713

75. jagr2713

you to @freckles

76. freckles

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

77. jagr2713

lol lets see what empty gets

78. jagr2713

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

79. freckles

do you know addition is commutative ?

80. freckles

a+b is the same as b+a

81. jagr2713

yea but you forgot the +

82. freckles

ummm no

83. freckles

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

84. jagr2713

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

85. 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.

86. freckles

we both wrote 3x-x^3 :p

87. jagr2713

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

88. freckles

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

89. freckles

verify with some numbers if you are not sure

90. jagr2713

i am confused now lol

91. freckles

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

92. freckles

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

93. freckles

it is the same with numbers

94. freckles

because those are numbers

95. 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

96. 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

97. jagr2713

Hm thanks guys

98. 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

99. ganeshie8

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

100. freckles

The differential equation thing is pretty clever.

101. 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.

102. Empty

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

103. 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?

104. freckles

and our function was odd

105. 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.

106. Empty

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

107. 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: $(x-x)f(x+x)-(x+x)f(x-x)-4xx(x-x)(x+x)=0 \\ 0f(2x)-2xf(0)-4x^2(0)(2x)=0 \\ -2x f(0)=0 \\ \text{ so } x=0 \text{ or } f(0)=0 \\ \text{ but I don't think we can say } \\ \text{ but I don't think we can say } x=0 \text{ since } x \text{ varies } \\ \\ \text{so } f(0)=0 \\ \text{ but we had found } f(x)=3x-x^3 \\ \text{ and so } f(0)=0 \text{ is actually satisfied so hmmm weird }$

108. freckles

and we should see something similar with the other case

109. freckles

I said that little implies statement a little wrong

110. freckles

didn't include the not equals part

111. freckles

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

112. Empty

Yeah, weird huh.