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March 21st, 2018, 07:40 AM   #1
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Is there a way to get a fractional exponent from whole number exponents?

Is there a function or operator that I can apply to get the equivalent of the right side of this equation with whole Numbers on the left?

f (a^(1/wholeNumberX) , a^(1/wholeNumberY) ) = a^(1.3)

Last edited by Sixophrenia; March 21st, 2018 at 08:12 AM.
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March 21st, 2018, 07:56 AM   #2
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$\dfrac{1}{1.3} = \dfrac{1}{1} \div 1.3 = \dfrac{1}{1} \div \dfrac{13}{10} = \dfrac{1}{1} * \dfrac{10}{13} = \dfrac{10}{13}.$

$f(a,\ j,\ k) = a^{(j/k)} \implies f(a,\ 10,\ 13) = a^{(10/13)} = a^{(1/1.3)}.$
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March 21st, 2018, 08:13 AM   #3
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Quote:
Originally Posted by JeffM1 View Post
$\dfrac{1}{1.3} = \dfrac{1}{1} \div 1.3 = \dfrac{1}{1} \div \dfrac{13}{10} = \dfrac{1}{1} * \dfrac{10}{13} = \dfrac{10}{13}.$

$f(a,\ j,\ k) = a^{(j/k)} \implies f(a,\ 10,\ 13) = a^{(10/13)} = a^{(1/1.3)}.$
I think I made a mistake, but it looks like there's enough room in the algebra to accomplish what I was attempting; representations for fractional exponents.
thanks

oh wait.
actually. I can't do that 10/13.
I know that sucks. but it's a limitation of the programming language I'm in. Solidity.

Maybe a Taylor series would work? I don't know. I am barely stringing things together.
I may have to try to find a different way of figuring this out.

Last edited by Sixophrenia; March 21st, 2018 at 08:35 AM.
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March 21st, 2018, 08:38 AM   #4
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Quote:
Originally Posted by Sixophrenia View Post
I think I made a mistake, but it looks like there's enough room in the algebra to accomplish what I was attempting; representations for fractional exponents.
thanks

oh wait.
actually. I can't do that 10/13.
I know that sucks. but it's a limitation of the programming language I'm in. Solidity.

Maybe a Taylor series would work? I don't know. I am barely stringing things together.
I may have to try to find a different way of figuring this out.
If you can do division in this computer, do you think you can use Taylor series. Good luck.
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March 21st, 2018, 09:04 AM   #5
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Originally Posted by JeffM1 View Post
If you can do division in this computer, do you think you can use Taylor series. Good luck.

It's just that division requires a number type that can't be directly supported.
No floating point integers in Solidity.

I think I basically have to abstract that with real integers
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