arctanx的泰勒展开式为:arctanx = x - 1/3x^3 + 1/5x^5 - 1/7x^7 + 1/9x^9 + ... + (-1)^(n+1) / (2n-1) * x^(2n-1),在x≤ 1的范围内有效。 arctanx的泰勒展开 泰勒展开式的基本概念 泰勒展开式(Taylor Expansion)是一种用函数在某点的信息描述...
In a paper of Hirzebruch ( Ann. of Math. 60 (1954), 213–235) the following number-theoretical question occurs in a problem (Problem 5, p. 217): Determine those coefficients in the Taylor expansion of tan x whose reciprocals are in Z. We give here the complete solution. In 1960, ...
ans = tan(x)^2 + 1 ans = 2*tan(x)*(tan(x)^2 + 1) Find the indefinite integral of the tangent function: int(tan(x), x) ans = -log(cos(x)) Find the Taylor series expansion of tan(x): taylor(tan(x), x) ans = (2*x^5)/15 + x^3/3 + x Rewrite the tangent functi...
return(TaylorExpansion); } double snowcos(double x) { x = (PI / 2) - x; return sin(x); } double snowtan(double x) { return (snowsin(x) / snowcos(x)); } double snowcot(double x) { return (1 / snowtan(x)); } int main() ...
代数输入 三角输入 微积分输入 矩阵输入 x−arctan(x) 求值 −arctan(x)+x 关于x 的微分 x2+1x2 图表 测验 Algebra x−arctan(x)
Explain with the definition, Taylor polynomial of degree n with center x = a. What is the use of hyperbolic functions? Formulate a precise definition of \lim_{x\to-\infty}f(x) = -\infty Then use your definition to prove that \lim_{x\to-\infty}(1+x^3)=-\infty ...
Find the limit using series expansion: limx→0x−16x3−tan−1xx5, wheretan−1x=∑n=0∞(−1)nx2n+12n+1. Taylor Series: The Taylor series can be very useful when it comes to handling limits of complica...
In 1671, James Gregory, and independently, Leibniz in 1673, discovered the Taylor series expansion for arctangent: displaystylearctanz=z-fracz³3+fracz⁵5-fracz⁷7+cdots This series, sometimes called the Gregory–Leibniz series, equals textstylefracpi4 when evaluated with displaystylez=1....
这种情况稍微复杂一些,因为我们需要利用洛必达法则(L'Hôpital's rule)或者泰勒展开(Taylor expansion)来求解。使用洛必达法则: 洛必达法则允许我们在求极限时,对分子和分母同时求导。在这里, [ \lim_{{x \to 0}} \frac{\arctan x}{x} = \lim_{{x \to 0}} \frac{\frac{d}{dx}(\arctan x)...
TaylorExpansion += sum; i += 2; } while (myabs(sum) > 1e-15); return(TaylorExpansion); } double snowcos(double x) { x = (PI / 2) - x; return sin(x); } double snowtan(double x) { return (snowsin(x) / snowcos(x)); } double snowcot(double x) { return (1 / snow...