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Thread: Finding a Non-linear Map That Satisfies af(x) = f(ax)

  1. #1
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    Finding a Non-linear Map That Satisfies af(x) = f(ax)

    Linear maps satisfy both conditions L1 and L2 below. I have to find a map R^2 -> R that satisfies ONLY L2 and not L1. I have been thinking about this for quite a while, but I haven't made that much progress. Do I have to use absolute values in some way?

    It would be nice to see the actual solution or at least get some clear hints.

    x and y are vectors.

    L1: f(x+y) = f(x) + f(y)
    L2: af(x) = f (ax)
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  2. #2
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    Quote Originally Posted by Koaske View Post
    Linear maps satisfy both conditions L1 and L2 below. I have to find a map R^2 -> R that satisfies ONLY L2 and not L1. I have been thinking about this for quite a while, but I haven't made that much progress. Do I have to use absolute values in some way?

    It would be nice to see the actual solution or at least get some clear hints.

    x and y are vectors.

    L1: f(x+y) = f(x) + f(y)
    L2: af(x) = f (ax)
    Think in terms of polar coordinates. What about the map [r,\theta]\mapsto rf(\theta), for some suitable function f(\theta) ?
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    Quote Originally Posted by Opalg View Post
    Think in terms of polar coordinates. What about the map [r,\theta]\mapsto rf(\theta), for some suitable function f(\theta) ?
    Your polar coordinate hint led me to try trigonometric functions, but I couldn't find a solution there. I have been trying pretty much every possible polar coordinate function I can think of, but the problem is that I just can't seem to find this "suitable" function. Whenever I find a map that satisfies L2, it always satisfies L1 too. More help would be apprectiated.
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  4. #4
    Senior Member roninpro's Avatar
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    Have you tried the norm function?

    Send (x,y)\to |(x,y)|=\sqrt{x^2+y^2}.
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by roninpro View Post
    Have you tried the norm function?

    Send (x,y)\to |(x,y)|=\sqrt{x^2+y^2}.
    This doesn't work since \|\alpha (x,y)\|=\sqrt{\alpha^2x^2+\alpha^2y^2}=|\alpha|\|(  x,y)\|.
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  6. #6
    Senior Member roninpro's Avatar
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    It seems to be a good place to look though - perhaps it can be modified to include a sign.

    How about (x,y)\mapsto \sqrt[3]{x^3+y^3}?
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  7. #7
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    Quote Originally Posted by Koaske View Post
    Your polar coordinate hint led me to try trigonometric functions, but I couldn't find a solution there. I have been trying pretty much every possible polar coordinate function I can think of, but the problem is that I just can't seem to find this "suitable" function. Whenever I find a map that satisfies L2, it always satisfies L1 too. More help would be appreciated.
    You need a function f(\theta) that satisfies f(\theta+\pi) = -f(\theta), so as to take care of multiplication by a negative scalar. But you can't use f(\theta) = \cos\theta or f(\theta) = \sin\theta, because they lead to linear functions. The function I had in mind was f(\theta) = \cos^3\theta.
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