arkhe

Thanks I will do!

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!

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.

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!

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...