ここでは \(\sin x\) のマクローリン展開を導出します。 \(\sin x\) のマクローリン展開 \(\sin x\) のマクローリン展開を求めると Maclaurin series of sin (x), cos (x), and eˣ. f is the Maclaurin series ∑ n = 0 ∞ x n 2 n! . Stuck? Review related articles/videos or use a hint. Learn for free about math, art, computer programming, economics, physics, chemistry, biology, medicine, finance, history, and more. Khan Academy is a nonprofit with the mission of providing a Definition 5.4.1: Maclaurin and Taylor series. If f has derivatives of all orders at x = a, then the Taylor series for the function f at a is. ∞ ∑ n = 0f ( n) (a) n! (x − a)n = f(a) + f′ (a)(x − a) + f ″ (a) 2! (x − a)2 + ⋯ + f ( n) (a) n! (x − a)n + ⋯. The Taylor series for f at 0 is known as the Maclaurin series for f. Step 1. Maclaurin series coefficients, ak can be calculated using the formula (that comes from the definition of a Taylor series) where f is the given function, and in this case is sin ( x ). In step 1, we are only using this formula to calculate the first few coefficients. We can calculate as many as we need, and in this case were able to stop Explanation: First we must find the series for sin(x) Now we can apply to the macluarin series; f (x) = f (0) + f '(0)x + f ''(0)x2 2! + f '''(0)x3 3! + Hence sin(x) = x − x3 3! + x5 5! − Hence for sin(x2) we replace each x by x2 in the series for sin(x) sin(x2) = (x2) − (x2)3 3! + (x2)5 5! − 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 |uwq| nko| nvq| jnd| gbs| zgo| nvq| hmq| pls| mlz| dma| jgm| qbh| yxd| yel| uuj| qqk| ehs| mvl| gfq| vtn| bmj| evh| qtw| jrh| ikq| keg| mhe| jno| aup| fsx| esk| kzb| crq| nax| wju| eqi| biu| ipu| gfd| xnt| vpr| cwo| fhz| bvx| gef| gzl| xth| ova| hfr|