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jagr2713

  • one year ago

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

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  1. jagr2713
    • one year ago
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    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
    • one year ago
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    @dan815 @ganeshie8

  3. jagr2713
    • one year ago
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    Anybody? 12mins have gone by D:

  4. anonymous
    • one year ago
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    need help

  5. jagr2713
    • one year ago
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    Of course

  6. anonymous
    • one year ago
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    i need help

  7. hartnn
    • one year ago
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    Its an even function.

  8. jagr2713
    • one year ago
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    ?? what

  9. hartnn
    • one year ago
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    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
    • one year ago
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    i had choices lol this is off this site dan gave me and i had OB so i just asked it lol

  11. sparrow2
    • one year ago
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    can f(x)=2x?

  12. jagr2713
    • one year ago
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    but i have the answer i just need you guys the figure it out :D sorta like a mini riddle or whatever

  13. sparrow2
    • one year ago
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    i have no idea beside f(x)=2x :D

  14. jagr2713
    • one year ago
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    Nope thats not the answer :/

  15. Empty
    • one year ago
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    I'd definitely substitute the elliptic substitution lol. a=x+y, b=x-y

  16. Empty
    • one year ago
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    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
    • one year ago
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    \[\text{ If } x+y=1 \text{ then } y=-x+1 \text{ and so } x-y=2x-1 \\ \text{ so we have } \\ (2x-1)f(1)-f(2x-1)+4x(-x+1)(2x-1)=0 \\ (2x-1)(2)-f(2x-1)+4x(-x+1)(2x-1)=0 \\ f(2x-1)=\frac{4x(-x+1)(2x-1)}{2(2x-1)}\]

  18. freckles
    • one year ago
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    \[u=2x-1 \\ \text{ then } x=\frac{u+1}{2}\]

  19. freckles
    • one year ago
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    \[f(u)=\frac{4 \cdot \frac{u+1}{2}(-\frac{u+1}{2}+1)(u)}{2u} \text{ assuming } 2x-1=u \neq 0 \\ \text{ we have } \\ f(u)=2 \frac{u+1}{2}(\frac{-u-1}{2}+1)\]

  20. freckles
    • one year ago
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    \[f(u)=(u+1)(\frac{-u-1+2}{2}) \\ f(u)=(u+1)(\frac{-u+1}{2})\]

  21. freckles
    • one year ago
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    \[f(u)=\frac{1}{2}(1+u)(1-u) \\ f(u)=\frac{1}{2}(1-u^2)\]

  22. freckles
    • one year ago
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    I think I made a mistake somewhere since f(1) doesn't equal 2 here

  23. freckles
    • one year ago
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    |dw:1436382316562:dw|

  24. hartnn
    • one year ago
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    \((2x-1)(2)-f(2x-1)+4x(-x+1)(2x-1)=0 \) next step? \((2x-1)(2)+4x(-x+1)(2x-1)=f(2x-1)\) right?

  25. freckles
    • one year ago
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    omg I made a mistake

  26. freckles
    • one year ago
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    thanks @hartnn

  27. freckles
    • one year ago
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    u=2x-1 (u+1)/2=x \[f(u)=u(2)+4(\frac{u+2}{2})(-\frac{u+1}{2}+1)u\]

  28. freckles
    • one year ago
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    lol

  29. freckles
    • one year ago
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    \[f(u)=u(2)+4(\frac{u+1}{2})(-\frac{u+1}{2}+1)u\]

  30. freckles
    • one year ago
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    which should give you after simplifying \[f(u)=3u-u^3\] this does have f(1)=3-1=2

  31. ganeshie8
    • one year ago
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    That x+y=1 is really a very clever substitution!

  32. freckles
    • one year ago
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    I guess we could have went the other way and chose x-y=1

  33. freckles
    • one year ago
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    should end up with same thing

  34. hartnn
    • one year ago
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    why just x+y =1 ?? why not x+y =2 or say 2000?

  35. freckles
    • one year ago
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    because f(1)=2

  36. freckles
    • one year ago
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    like in that x+y thingy was in the f( ) thingy

  37. sparrow2
    • one year ago
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    so you have u=2x-1 so f(u)=f(2x-1)? don;'t get

  38. freckles
    • one year ago
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    why can't u be 2x-1 @sparrow2

  39. sparrow2
    • one year ago
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    it can but they want f(x) not f(2x-1) or did i miss smth?

  40. freckles
    • one year ago
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    I found f(u)=3u-u^3 replace u's with x

  41. ganeshie8
    • one year ago
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    u or x or some variable, they all are just dummy

  42. sparrow2
    • one year ago
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    i understand but when you replace you get f(2x-1) as function not f(x)

  43. freckles
    • one year ago
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    we were looking for f( )

  44. freckles
    • one year ago
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    I found f( )

  45. freckles
    • one year ago
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    well we found f( )*

  46. ganeshie8
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    nice work! :)

  49. sparrow2
    • one year ago
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    it will be good if you just write what is f(x)

  50. freckles
    • one year ago
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    I did

  51. sparrow2
    • one year ago
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    so f(u) and f(x) is the same?

  52. freckles
    • one year ago
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    "I found f(u)=3u-u^3 replace u's with x "

  53. freckles
    • one year ago
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    \[f(x)=3x-x^3\]

  54. sparrow2
    • one year ago
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    u=2x-1 (as you wrote) and you just write that u=x

  55. freckles
    • one year ago
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    u doesn't equal x

  56. freckles
    • one year ago
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    f( ) is a machine

  57. freckles
    • one year ago
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    the input changes the output based on the input

  58. sparrow2
    • one year ago
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    i know what is a fucntion :)

  59. freckles
    • one year ago
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    like f(goldfish)=3(goldfish)-(goldfish)^3

  60. sparrow2
    • one year ago
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    but this function has very specific argument

  61. freckles
    • one year ago
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    I don't think understand what you are saying

  62. ganeshie8
    • one year ago
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    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
    • one year ago
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    functions work slightly differently as the given function was assumed to be continuous and nice hmm

  64. sparrow2
    • one year ago
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    by the way i am wondering in f(2x)=x argument is 2x yes?

  65. ganeshie8
    • one year ago
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    f(2x) = x let 2x = u f(u) = u/2 f(x) = x/2 see anythign wrong ? i dont see any..

  66. sparrow2
    • one year ago
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    i mean in f(2x)=x argument is 2x yes?

  67. ganeshie8
    • one year ago
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    yes, by argument if it means input

  68. sparrow2
    • one year ago
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    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
    • one year ago
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    \[f(x)=3x-x^3\]

  70. freckles
    • one year ago
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    @Empty did you do something different than using x+y=1 (or x-y=1)?

  71. freckles
    • one year ago
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    if so I would like to see your method too

  72. jagr2713
    • one year ago
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    well @Empty almost has it

  73. freckles
    • one year ago
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    what f(x)=3x-x^3 is also what I got

  74. freckles
    • one year ago
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    and it does work @jagr2713

  75. jagr2713
    • one year ago
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    you to @freckles

  76. freckles
    • one year ago
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    so if that isn't it what is the answer @jagr2713

  77. jagr2713
    • one year ago
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    lol lets see what empty gets

  78. jagr2713
    • one year ago
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    Ok well its\[-x ^{3}+3x\] Almost got it @freckles and @Empty

  79. freckles
    • one year ago
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    do you know addition is commutative ?

  80. freckles
    • one year ago
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    a+b is the same as b+a

  81. jagr2713
    • one year ago
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    yea but you forgot the +

  82. freckles
    • one year ago
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    ummm no

  83. freckles
    • one year ago
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    \[3x-x^3=+3x-x^3=+3x+(-x^3) \\ \text{ change order } \\ =(-x^3)+3x \\ =-x^3+3x\]

  84. jagr2713
    • one year ago
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    yea but above i saw you both but -x3-3x not +3x

  85. Empty
    • one year ago
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    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
    • one year ago
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    we both wrote 3x-x^3 :p

  87. jagr2713
    • one year ago
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    but its + not - :D dont know who to give the medal D: both did a great job and you to ganeshie

  88. freckles
    • one year ago
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    3x-x^3 is the same thing as -x^3+3x what are you talking about?

  89. freckles
    • one year ago
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    verify with some numbers if you are not sure

  90. jagr2713
    • one year ago
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    i am confused now lol

  91. freckles
    • one year ago
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    like do you know that -3+5 is the same as 5-3?

  92. freckles
    • one year ago
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    like you had -x^3+3x is the same as 3x-x^3

  93. freckles
    • one year ago
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    it is the same with numbers

  94. freckles
    • one year ago
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    because those are numbers

  95. Empty
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    Hm thanks guys

  98. freckles
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    The differential equation thing is pretty clever.

  101. Empty
    • one year ago
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    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
    • one year ago
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    What's interesting to me though is how f(x) is not an even function.

  103. freckles
    • one year ago
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    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
    • one year ago
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    and our function was odd

  105. Empty
    • one year ago
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    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
    • one year ago
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    Does the condition \(|x| \ne |y|\) really mean anything for this problem?

  107. freckles
    • one year ago
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    \[\left| x \right|\neq \left| y \right| \implies x=y \text{ or } x=-y \\ \text{ so pluggin this in we get } \\ \] \[(x-y)f(x+y)-(x+y)f(x-y)+4xy(x ^{2}-y ^{2})=0\] 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
    • one year ago
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    and we should see something similar with the other case

  109. freckles
    • one year ago
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    I said that little implies statement a little wrong

  110. freckles
    • one year ago
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    didn't include the not equals part

  111. freckles
    • one year ago
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    so I don't know why they put that |x| couldn't be |y|

  112. Empty
    • one year ago
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    Yeah, weird huh.

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