How do u work out non-integer powers without a calculator?

[ivan77]

I have been wondering how this is worked out without just plugging it into the calculator, is it possible to do it on paper or does it take a long time to work out? I feel i should already know this but i have never learned it in school or in my maths course.

Its obviously not a ratio as: 9^1.5=27 but 9^2= 81.

[elfstone]

Well, if the decimal number is easy i know how to do it. A power of 0.5 equals square root, so 9^1.5 = 9^1.0 x 9^0.5 = 9x3 = 27.

A power of 0.33 i suppose is the cubic root. I don't know how to generalize this though.

[Dave]

The way to approximate them is to use Taylor expansions, but I daresay you probably haven't done those yet. So, in a word, probably not

[ivan77]

Are taylor expansions similiar to those used to find natural logs ie: e^2 for example.

[Dave]

You might have seen the formula:

 

[math]e^x = 1 + x + \frac{1}{2!}x^2 + \frac{1}{3!}x^3 + \cdots[/math]

 

This is the Taylor expansion for e^x.

 

Maybe my answer isn't quite correct; you can also use the binomial theorem to expand something like [math](1+x)^{1/2}[/math] and then get a pretty good approximation from that.

[ydoaPs]

how bout [math]a^\frac{b}{c}=(\sqrt[c]{a})^b[/math]? doesn't that work?

[Zeo]

I was going to say try using logarithms. We just learned about it last week in Pre-Cal . . . so maybe I'm not entirely sure what I'm talking about but . . . what's your problem anyway?

[Dave]

Not if b/c is, say, 1/2 and a is something nasty like 23. You can guesstimate, but it's a lot nicer if you use something like the binomial formula because you end up with putting decimals into polynomials, which is a lot easier.

[ydoaPs]

good point. i guess ur right.

[Dave]

It's the way most calculators work things like this out (or so I'm told). It's pretty accurate as well.

[Ducky Havok]

Not if b/c is, say, 1/2 and a is something nasty like 23. You can guesstimate, but it's a lot nicer if you use something like the binomial formula because you end up with putting decimals into polynomials, which is a lot easier.

 

you could still use differentials to approximate it pretty close though.

Like [math]\sqrt{23}[/math],

you could say [math]x=25, dx=-2[/math], then

[math]\frac{1}{2\sqrt{x}}dx+ \sqrt{x}[/math], which would give you

[math]-1/5+5 [/math], which is 4.8.

 

That's pretty close to the square root of 23. I'm only saying this because I don't understand how to use the binomial formula though. Can you please explain it some? (I might know it, just not know that that's what it is. Most likely I don't though)

[Dave]

Ack! For my perspective on seperating dx's and all of that, have a look at the "dt ramblings" thread in the Calculus forum

 

But yeah, that's just guesstimation really. Well, the binomial formula is this:

 

[math](1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!} x^3 \cdots[/math]

 

I'm pretty sure you know that formula