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Computing Natural Logarithm of 2
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[QUOTE="JeromePl, post: 3212631, member: 319218"] I got this question and I know a bit on how to start it, but not sure which direction is best: By considering the power series 1/(1 + x) and 1/(1 - x) show that: so I do the differentiation - ln(1 + x) = [SUP]x[/SUP][SUB]0[/SUB] du/1 + u = x - x[SUP]2[/SUP]/2 + x[SUP]3[/SUP]/3 - x[SUP]4[/SUP]/4 ... which equals - = sigma (upper infinity and lower is k=0) (-1)[SUP]k[/SUP] x[SUP]k+1[/SUP]/k+1 and for ln(1 - x) I do the same... but how do I show this on MATLAB and what else do I need to do? The next question is then: Hence show that ln (1+x/1-x) = 2 (x + x[SUP]3[/SUP]/3 + x[SUP]5[/SUP]/5 + x[SUP]7[/SUP]/7...) = 2 sigma (upper is infinity and lower is k = 0) x[SUP]2k +1[/SUP]/2k +1 then is says determine the range of values of x for which each of the above series will converge. Next question: Define S[SUB]n[/SUB](x) and Si[SUB]n[/SUB](x) to be partial sums for the infinite series above: S[SUB]n[/SUB](x) = Sigma (upper is n and lower k = 0 ) (-1)[SUP]k[/SUP] x[SUP]k+1[/SUP]/K+1 S'[SUB]n[/SUB](x) = 2 Sigma (n is upper and lower k=0) x[SUP]2k+1[/SUP]/2k +1 for the above write two MATLAB functions. Using suitable values for x construct a table of estimates for ln(2) and the errors E[SUB]n[/SUB] = abs(S[SUB]n[/SUB] - ln(2)) and E'[SUB]n[/SUB] = abs(S'[SUB]n[/SUB] - ln(2)). using the table of data, estimate the rate of convergence for the first series S[SUB]n[/SUB]. the estimate how many terms would be needed in each series to ensure an accuracy of 5 decimal places. [/QUOTE]
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