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JoelFishel Frequent poster
Post Number: 96
 Posted on Friday, 01 February, 2013  06:14 pm:  
I noticed that Neill Cooper said this in a thread I made earlier but I've just learnt about Maclaurin/Taylor series and find this post really interesting now: An example. This week with my Upper Sixth A2 Further Mathematicians we were looking at Maclaurin expansions and when they were valid. The discussion turned to ln(1+x) when x=1. The formula book and text book says it is a valid expansion (1  1/2 + 1/3 etc). I started explaining conditional and absolute convergence. But since only half the class of 21 students were interested I could not spend too long discussing it. If we has smaller classes I could do more such digressions. I was wondering if (not necessarily Neill) could expand on this point about ln(1+x) centred around x=1 because I actually want to use ln(2) in a SMS question and I assume i'm not allowed to use calculators for it so I was wondering how far you have to go with this expansion for it to become useful? 
lebesgue Veteran poster
Post Number: 3060
 Posted on Saturday, 02 February, 2013  09:46 am:  
As this series for ln(1+x) is rather slowly convergent, this is not so convenient to use for computation of ln(2), say. The reason is that the coefficients decrease to 0 with linear speed only, so after n steps we still have an error of order constant/n. So, even to achieve precision to the three decimal digits, we need to take thousand of terms, or so. To compare, the series sin(x), cos(x), e^{x} converge much faster, as their nth term decrease so fast as constant/n!, which already for n=10 is extremely small. 
JoelFishel Frequent poster
Post Number: 98
 Posted on Saturday, 02 February, 2013  10:40 am:  
Ok, I see what you mean. Is there any reasonable way to approximate log(2) without a calculator? The question gives no reference to whether or not calculators can be used and it doesn't give any log values either. It wants me to work out the most significant digit of 2^400. I've ended up just using a calculator tbh. 
lebesgue Veteran poster
Post Number: 3061
 Posted on Saturday, 02 February, 2013  10:59 am:  
Main difficulty may be that this definition uses logarithm to the base e, which you may need to be computed first. But as the series for e^{x} is simple and much faster converging, you may look for x such that e^{x}=2 (by trial and error, say, or using Newton method), this 'root' x will be ln(2) then. 
JoelFishel Frequent poster
Post Number: 100
 Posted on Saturday, 02 February, 2013  11:07 am:  
nice idea.. The problem i'm doing is actually log base 10 though which makes things even trickier? 
lebesgue Veteran poster
Post Number: 3062
 Posted on Saturday, 02 February, 2013  11:19 am:  
No, log base any integer is usually simpler. For example, looking for log(A) base 10 would need to look for x, or rather its rational approximation p/q, such that 10^{x} is approximately A, or 10^{p/q}=A, or 10^{p}=A^{q}. So study just integral powers of A, A^{q} for few smaller values of q, and when it is close to a power of 10, 10^{p}, then p/q is an approximation of log(A) base 10. Of course, one can make some estimates how precise this would be, but this is not so simple and efficient either. 
lebesgue Veteran poster
Post Number: 3063
 Posted on Saturday, 02 February, 2013  11:55 am:  
For example, if A=2 and you want to know log(2) base 10. Here 2^{10}=1024 which is quite close to 10^{3}, so 3/10 would be our first approximation of that logarithm. But as the powers grow extremely fast, we can't achieve high precision very easily, for that purpose the power series methods are better, but 'nice' series is available for base e only; other logarithms may need some prefactor to compute first. 
JoelFishel Frequent poster
Post Number: 101
 Posted on Sunday, 03 February, 2013  04:12 pm:  
Ye i get you, thanks for the help! 
