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as far as I remember this series has a very poor accuracy to calculate the Natural Logarithm, is that why the exercise of the letter c asked the relative error? lol. Now let's try using the function I created, sum with progression of 20 ( n=20 ), no matter the exercise of the letter a was legal and let me use any value of n and any value of x between 1 e -1 : equacao(0.3,20)īingoooo ln(1-x) é igual a Σ(-(x^n)/n) huahuahua ONLY NOOOOO. When typing help log this help appeared on the log function, this is exactly what we need right? let's compare using a x that is within the range that the exercise asks for, for example x=0.3 log(1-0.3) īut let's see if the Taylor series is correct? the return of this function is the same as ln(1-x) ? does it have some ready-made matlab function that computes this? What's up? help logĬompute the natural logarithm, 'ln (X)', for each element of X. this is what asks for the letter a of the question, you may have to create within this function a if that ensures that your x is within the range that the exercise asks for. To write a function using the equation of the statement is very simple, the sigma ( Σ ) of the formula tells us that you must sum the equation (-(x^n)/n), the iteration goes from 1 to infinito, does not mean that we will be calculating this to infinity, in fact the iteration will end when the value of n reaches the limit defined by the user, in the statement it leaves some things explicit, use x between -1 e 1, it also says ln(1-x) is equal to Σ(-(x^n)/n), I will not do everything for you, this is the function that does exactly the calculation of the formula Σ(-(x^n)/n) : function F = equacao(x,n) I was not going to respond, I understand that people need help, I will do my part, the question formula looks a lot like the Taylor series to find natural logarithm (as far as I remember the university lol).