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Math Help - Working out the level of homogeneity

  1. #1
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    Working out the level of homogeneity

    I know that q= 120L1 1/2 K - (L3/√K) has homogenous degree of 2 1/2 but I'm unsure how that has been worked out. I think its the square root thats throwing me, could someone show me how its 2 1/2?

    Thanks in advance.
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  2. #2
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    Re: Working out the level of homogeneity

    In this case, homogeneous means that the sum of the exponents of L and K for the first term is the same as the sum of the exponents of L and K for the second term.

    For the first term, L has exponent 1\,\frac{1}{2} and K has exponent 1, so the sum is 2\,\frac{1}{2}. For the second term, L has exponent 3 and K has exponent -\frac{1}{2} (since K^{-\frac{1}{2}}=\frac{1}{\sqrt{K}}), so the sum is again 2\,\frac{1}{2}.

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  3. #3
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    Re: Working out the level of homogeneity

    Thank you, for some reason I was adding those terms together.
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