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 March 21st, 2018, 07:40 AM #1 Newbie   Joined: Oct 2017 From: Georgia Posts: 4 Thanks: 0 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.
 March 21st, 2018, 07:56 AM #2 Senior Member   Joined: May 2016 From: USA Posts: 1,310 Thanks: 550 $\dfrac{1}{1.3} = \dfrac{1}{1} \div 1.3 = \dfrac{1}{1} \div \dfrac{13}{10} = \dfrac{1}{1} * \dfrac{10}{13} = \dfrac{10}{13}.$ $f(a,\ j,\ k) = a^{(j/k)} \implies f(a,\ 10,\ 13) = a^{(10/13)} = a^{(1/1.3)}.$
March 21st, 2018, 08:13 AM   #3
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 Originally Posted by JeffM1 $\dfrac{1}{1.3} = \dfrac{1}{1} \div 1.3 = \dfrac{1}{1} \div \dfrac{13}{10} = \dfrac{1}{1} * \dfrac{10}{13} = \dfrac{10}{13}.$ $f(a,\ j,\ k) = a^{(j/k)} \implies f(a,\ 10,\ 13) = a^{(10/13)} = a^{(1/1.3)}.$
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.

March 21st, 2018, 08:38 AM   #4
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Quote:
 Originally Posted by Sixophrenia 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.

March 21st, 2018, 09:04 AM   #5
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 Originally Posted by JeffM1 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

 Tags exponent, exponents, fractional, number

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