thinking about exponents

[arkhe]

Hi there,

So we know that some number [math]b^n[/math] is [math]b[/math] multiplied by itself [math]n[/math] times.

What about roots? The logic would imply that we take [math]b^\frac{1}{n}[/math] as being [math]b[/math] multiplied by itself [math]\frac{1}{n}[/math] times. I'm a little vague on the underlying mechanisms of taking roots and fractional exponents. I am aware that fractional exponents and roots are computed using logarithms but how?

As an example:

[math]10^\frac{5}{2} = 10^{2 + \frac{1}{2}} = 10^2 10^\frac{1}{2} \approx 100(3.162) \approx 316.2[/math]

I can see that breaking the exponent apart gives a sum of a whole power [math]2[/math] and a square root. The whole power is straightforward as a result of taking [math]10[/math] as a factor twice but what about the square root? I'd like to know in more detail how this is calculated without using a calculator and without resorting entirely to a root finding algorithm such as Newton's method. What is the underlying mechanism?

Thanks!

EDIT:

On a side note, as a result of playing around with the idea of raising [math]2[/math] to the following powers in the sequence [math]0,\frac{1}{10},\frac{1}{5},\frac{3}{10},\frac{2}{5},\frac{1}{2},\frac{3}{5},\frac{7}{10},\frac{4}{5},\frac{9}{10},1[/math] I get various values and taking the ratio of the current value and the previous value (starting from the second expression) gives me a constant of [math]\approx 1.072[/math]. ie.

[math]A: 2^0 = 1[/math]

[math]B: 2^{\frac{1}{10}} \approx 1.072 1[/math]

[math]\frac{B}{A} \approx 1.072[/math]

[math]C: 2^{\frac{1}{5}} \approx 1.149 1[/math]

[math]\frac{C}{B} \approx 1.072[/math]

[math]D: 2^{\frac{3}{10}} \approx 1.231 1[/math]

[math]\frac{D}{C} \approx 1.072[/math]

[math]...[/math]

I can see that there must be some kind of relationship here, but not seeing it...

Edited October 7, 2013 by arkhe

[studiot]

I'm afraid your post was a bit rambling and I was unable to properly determine your question or point. You seem to have extracted a number of fractional powers by calculator, but asked how to do this without one.

 

In the days before calculators this would have been done by logarithms. This could be done from tables or directly by using a slide rule which has log scales.

Edited October 7, 2013 by studiot

[Bignose]

[math]\log b^a = a \log b[/math]

 

So, say you need to calculate sqrt(25) = 25^(1/3)

 

Using 'log' to be mean base 10 logarithm:

 

[math] 25^{(1/3)} = 10^{ \log 25^{(1/3)} } = 10^{ (1/3) \log 25 } [/math]

 

log 25 could be found in a table = 1.39794

1/3 that = 0.46598

Then 10^0.46598 could be found in a table = 2.924 and that's your answer.

[arkhe]

Hi there, thanks for the replies and sorry for the lack of clarity in my first post!

 

What I meant was to ask the question of how to calculate the exponentiation without the use of a calculator or log table. I assume then that the log tables were compiled exclusively using root finding algorithms? Could you perhaps shed a little more light on this?

 

In other words, when the log tables were compiled were they created using only algorithms like Newtons method etc. or are there other ways of doing this?

 

Thanks!

[studiot]

Folks created tables because direct calculation of many functions is difficult to impractical.

 

Newton not only found Newton's method which refers to a numerical method of solving equations, that is not much use in creating tables (How do you think it might be used for this purpose?), Newton also did a great deal of work on Finite Differences. These are used to interpolate tables to obtain values in between already tabulated ones. Original tabulated ones were calculated using series methods.

[arkhe]

Thanks for the reply studiot. Could you perhaps give a little introduction on series starting with the simplest example you can think of? I am aware of various series such as [math]e = 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} ...[/math]. Obviously for me this makes no sense at all, and hopefully you can give a really simple (although simple might be impossible example of a series that is relevant.

EDIT: OK so I found the following series for computing the natural logarithm for [math]-1 < x < 1[/math].

[math]ln(1 + x) = \sum_{n=0}^\infty (-1)^{n+1} \frac{x^n}{n}[/math]

But I have no idea how it works. I know this is probably not that simple, but is there perhaps a way to get started on series by means of a simple "introductory" example?

Thanks!

EDIT 2: OK so I have done a bit of reading up on series and I realize that the best way is for me to actually learn all of this stuff first before I can really question the above. I've started with geometric series, so it's going from there onwards.

Edited October 8, 2013 by arkhe

[studiot]

That's a tall order but since you ask I will show how to calculate the natural logs from 1 to 10 using a modification of your formula.

 

Since you appear unsure of the sigma notation I have written out the series it represents each time. We can discuss the sigma notation separately if you like, but you really should look that up first.

 

I have split your formula into two since it is not suitable for numerical calcualtion as it converges so slowly. Further it is only correct for x <1 which has its limitations.

 

[math]\log (1 + x) = x - \frac{{{x^2}}}{2} + \frac{{{x^3}}}{3} - \frac{{{x^4}}}{4} + \frac{{{x^5}}}{5} - \frac{{{x^6}}}{6}...................\bmod (x) < 1[/math]

 

[math]\log (1 - x) = [/math][math] - x - \frac{{{x^2}}}{2} - \frac{{{x^3}}}{3} - \frac{{{x^4}}}{4} - \frac{{{x^5}}}{5} - \frac{{{x^6}}}{6}..................\bmod (x) < 1[/math]

Subtracting the second series from the first

 

[math]f(x) = \log (1 + x) - \log (1 - x)[/math]

 

[math] = \log \frac{{1 + x}}{{1 - x}} = 2\left[ {x + \frac{{{x^2}}}{2} + \frac{{{x^3}}}{3} + \frac{{{x^4}}}{4} + \frac{{{x^5}}}{5} + \frac{{{x^6}}}{6}.........} \right]\bmod (x) < 1[/math]



[math]putx = \frac{1}{5};\frac{1}{6};\frac{1}{7};\frac{1}{8}[/math]

We obtain four simultaneous equations

 

[math]f(1/5) = \log (3/2) = \log 3 - \log 2[/math]

[math]f(1/6) = \log (7/5) = \log 7 - \log 5[/math]

[math]f(1/7) = \log (4/3) = 2\log 2 - \log 3[/math]

[math]f(1/8) = \log (9/7) = 2\log 3 - \log 7[/math]

From which we can calculate log2, log3, log5 and log7

 

Using these we can calculate the rest

[math]\log 4 = 2\log 2[/math]

[math]\log 6 = \log 3 + \log 2[/math]

[math]\log 8 = 3\log 2[/math]

[math]\log 9 = 2\log 3[/math]

[math]\log 10 = \log 5 + \log 2[/math]

 

and of course log1 = 0

Edited October 8, 2013 by studiot

[HalfWit]

Hi there, thanks for the replies and sorry for the lack of clarity in my first post!

 

What I meant was to ask the question of how to calculate the exponentiation without the use of a calculator or log table. I assume then that the log tables were compiled exclusively using root finding algorithms? Could you perhaps shed a little more light on this?

 

In other words, when the log tables were compiled were they created using only algorithms like Newtons method etc. or are there other ways of doing this?

 

Thanks!

If it's the history you're curious about, look up Napier. He's the guy who discovered logarithms ... and having done so, he spent the next 20 years of his life calculating log tables. Talk about dedication. No mechanical calculators were in existence at that time.

 

There are a lot of links about the history. Just google "Napier" and you'll find lots of info.

[arkhe]

Hi there, thanks for the replies. I did do some reading up on the history of Napier and Briggs and the old methods of computing using trig tables. Fascinating really.

 

studiot, thanks I've started with the geometric series, so I'm assuming it will all work out from there as I progress.

 

Thanks!

[studiot]

Successful reading and ask again if you come across anything you are unsure of.

 

Edited October 9, 2013 by studiot

[arkhe]

Thanks I will do!