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Math Help - How to Solve for Function of Two Variables

  1. #1
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    How to Solve for Function of Two Variables

    if f((x+y)/3) = (2+f(x) +f(y))/3 for all real x, y and f (2) =2 and f if differentiable function
    Then can we find f (x) .in question nothing is said about f(x)(no further information is given)
    but they are finding the answer as 2x+2
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  2. #2
    Behold, the power of SARDINES!
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    Quote Originally Posted by ayushdadhwal View Post
    if f((x+y)/3) = (2+f(x) +f(y))/3 for all real x, y and f’ (2) =2 and f if differentiable function
    Then can we find f (x) .in question nothing is said about f(x)(no further information is given)
    but they are finding the answer as 2x+2
    Take the partial derivative with respect to y to get

    \frac{1}{3}\frac{\partial f}{\partial y}\left(\frac{x+y}{3} \right)=\frac{1}{3}\frac{\partial f}{\partial y}(y)

    Now if we evaluate this at the point x=3x y=0)we get (I will now use primes to represent the derivative with respect to y)

    f'(x)=f'(0)

    Since the derivative is constant the function must me linear! If we plug in x=2 we get

    f'(2)=f'(0) =2

    So the equation must have the form

    f(x)=2x+b

    Can you finish from here?
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  3. #3
    Senior Member abhishekkgp's Avatar
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    Quote Originally Posted by ayushdadhwal View Post
    if f((x+y)/3) = (2+f(x) +f(y))/3 for all real x, y and f’ (2) =2 and f if differentiable function
    Then can we find f (x) .in question nothing is said about f(x)(no further information is given)
    but they are finding the answer as 2x+2
    this is not a very rigorous solution but is good to see.
    put x=0,y=0 to get f(0)=2.

    the given condition becomes
    f(\frac{0+x+y}{3})=\frac{f(0)+f(x)+f(y)}{3}.....(1)
    note that the centroid of the triangle OAB with O=(0,f(0)),\,A=(x,f(x)),\,B=(y,f(y)) is (\frac{0+x+y}{3},\frac{f(0)+f(x)+f(y)}{3}).
    i am assuming that the curve does not change its concavity(this is where it becomes non-rigorous but this may be established by putting y=x and then using second derivative test).
    The centroid of the triangle is inside the triangle but from (1) the centroid is on the curve Y=f(X) and hence outside ot on the triangle .... to see this draw a curve which is concave downwards and mark the three points OAB(this part is also non-rigorous)

    so the curve of 'f' has to be a straight line.
    put f(X)=aX+b and use f'(2)=2.
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  4. #4
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    Quote Originally Posted by abhishekkgp View Post
    this is not a very rigorous solution but is good to see.
    put x=0,y=0 to get f(0)=2.

    the given condition becomes
    f(\frac{0+x+y}{3})=\frac{f(0)+f(x)+f(y)}{3}.....(1)
    note that the centroid of the triangle OAB with O=(0,f(0)),\,A=(x,f(x)),\,B=(y,f(y)) is (\frac{0+x+y}{3},\frac{f(0)+f(x)+f(y)}{3}).
    i am assuming that the curve does not change its concavity(this is where it becomes non-rigorous but this may be established by putting y=x and then using second derivative test).
    The centroid of the triangle is inside the triangle but from (1) the centroid is on the curve Y=f(X) and hence outside ot on the triangle .... to see this draw a curve which is concave downwards and mark the three points OAB(this part is also non-rigorous)

    so the curve of 'f' has to be a straight line.
    put f(X)=aX+b and use f'(2)=2.
    Sir can you draw the figure ?
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  5. #5
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    Quote Originally Posted by ayushdadhwal View Post
    Sir can you draw the figure ?
    i am assuming that the curve does not change its concavity(this is where it becomes non-rigorous but this may be established by putting y=x and then using second derivative test).
    The centroid of the triangle is inside the triangle but from (1) the centroid is on the curve Y=f(X) and hence outside ot on the triangle .... to see this draw a curve which is concave downwards and mark the three points OAB(this part is also non-rigorous
    bhaiya i cant understand even a single line .but it looks beautiful. can you explain in a very simple way whole of the above part
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