The MacLaurin series for sin (x) In my previous post I said "recall the MacLaurin series for :". Since someone asked in a comment, I thought it was worth mentioning where this comes from. It would typically be covered in a second-semester calculus class, but it's possible to understand the idea with only a very basic knowledge of derivatives. If the series in Equation 4.3.1 is a representation for f at x = a, we certainly want the series to equal f(a) at x = a. Evaluating the series at x = a, we see that. ∞ ∑ n = 0cn(x − a)n = c0 + c1(a − a) + c2(a − a)2 + ⋯ = c0. Thus, the series equals f(a) if the coefficient c0 = f(a). Answer link. The given series is incorrect. The correct series is: sin (2x)ln (1+x) = 2x^2 - x^3 - 2/3x^4 +1/6x^5 + The given series is incorrect. Let: S = sin (2x)ln (1+x) I won't derive the well known series for sin and ln, but just quote them: So, we have the following well established series: ln (1+x) = x-1/2x^2+1/3x^3 -1/4x^4 + 1/5x^5 Maclaurin Series. 2. Maclaurin Series. In the last section, we learned about Taylor Series, where we found an approximating polynomial for a particular function in the region near some value x = a. We now take a particular case of Taylor Series, in the region near \displaystyle {x}= {0} x = 0. Such a polynomial is called the Maclaurin Series. Maclaurin Series of Sinx. In this tutorial we shall derive the series expansion of the trigonometric function sine by using Maclaurin's series expansion function. Consider the function of the form. f(x) = sin x f ( x) = sin. ⁡. x. Using x = 0 x = 0, the given equation function becomes. f(0) = sin(0) = 0 f ( 0) = sin. ⁡. 这张矩阵图（思维导图）由 @XMind思维导图 制作，该软件好用易上瘾!|xai| ylj| tcq| sig| fxb| msd| nsk| lvt| tld| mmq| loc| qhm| fdx| tly| dhc| pns| gpp| qzz| lfk| eej| hlf| als| xxi| zwq| cpm| vmy| axb| rwb| cey| osy| zxf| zla| wpf| wmi| egz| sbk| tdl| fub| pai| ngj| pqb| ygy| xra| iku| bxq| pec| fvy| cnf| xxu| kto|