How Do You Calculate Noninteger Fractional Exponents?

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I have a function f(x) such that

f(x) = x^{\frac{a}{b}}

where {\frac{a}{b}} is noninteger. Is there an equation to solve this? A series expansion or something? I've looked around and couldn't find anything.
 
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Solve for what? You don't really have an equation that you can solve for anything. The equation just provides a formula for f(x).

Now if you had an equation such as x^{a/b} = 7, then you could solve for x by taking the b/a power of both sides.
 
The classical method to do this involves logarithms and exponentials:

x^{a/b} = \exp((a/b)\ln(x))
 
Mark44 said:
Solve for what? You don't really have an equation that you can solve for anything. The equation just provides a formula for f(x).

Now if you had an equation such as x^{a/b} = 7, then you could solve for x by taking the b/a power of both sides.

To clarify a bit, I'm trying to solve for f(x). By solve, I mean express the solution symbolically and in such a way that the only operations are addition, subtraction, multiplication, and division. In reality, I might consider the use of factorials, sinusoidal functions, special functions, operator functions, etc. to be acceptable. In other words, the question is: how does the calculator solve it? Thanks to g_edgar for answering this question.
 
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dimensionless said:
To clarify a bit, I'm trying to solve for f(x). By solve, I mean express the solution symbolically and in such a way that the only operations are addition, subtraction, multiplication, and division.
That's not "solving" for f(x). As I already said, the equation for f(x) is merely a definition of its formula. What you want to do is write the formula in a different form.
dimensionless said:
In reality, I might consider the use of factorials, sinusoidal functions, special functions, operator functions, etc. to be acceptable. In other words, the question is: how does the calculator solve it? Thanks to g_edgar for answering this question.
Edit: Fixed typo: e2 --> ex
What you're asking about is answered in the part of calculus that deals with power series, such as Taylor and Maclaurin series, and Fourier series, to name a few. A function such as ex has a Maclaurin series 1 + x + x2/2! + x3/3! + ... + xn/n! + ... As you can see, the series representation consists only of addition and multiplication (plus factorials).

As I understand things, calculators use a technique similar to this but not exactly the same, combined with lookup tables, to calculate the various functions that are on a scientific calculator. It's been a long time since I thought about it, but the acronym CORDIC fits in here somehow.
 
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Mark44 said:
That's not "solving" for f(x). As I already said, the equation for f(x) is merely a definition of its formula. What you want to do is write the formula in a different form.

What you're asking about is answered in the part of calculus that deals with power series, such as Taylor and Maclaurin series, and Fourier series, to name a few. A function such as e2
Typo: Mark44 meant ex here

has a Maclaurin series 1 + x + x2/2! + x3/3! + ... + xn/n! + ... As you can see, the series representation consists only of addition and multiplication (plus factorials).

As I understand things, calculators use a technique similar to this but not exactly the same, combined with lookup tables, to calculate the various functions that are on a scientific calculator. It's been a long time since I thought about it, but the acronym CORDIC fits in here somehow.
 
Thanks. ex is what I meant. I fixed it in my post.
 
xa/b=(1+x-1)a/b=1+(a/b)(x-1)+(a/b)(a/b-1)/2!(x-1)2
+(a/b)(a/b-1)(a/b-2)/3!(x-1)3...
(Sorry, can't do better latex..any good guides?)
 
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