doi:10.1016/0022-314X(73)90050-4In 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 ...
Find the Taylor series expansion of tan(x): taylor(tan(x), x) ans = (2*x^5)/15 + x^3/3 + x Rewrite the tangent function in terms of the sine and cosine functions: rewrite(tan(x), 'sincos') ans = sin(x)/cos(x) Rewrite the tangent function in terms of the exponential fun...
You're talking a Taylor expansion of either $\arctan(x)$ around $x=1$, or of $\arctan(1+z)$ around $z=0$. But I'm afraid that in both cases the Taylor expansion requires you to evaluate $\arctan^{(n)}(1)$. Jun 5, 2013 #10 SweatingBear 119 0 I like Serena said: ...
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() ...
∫ tan x dx / (1 + x) Indefinite Integrals: There are many types of indefinite integral or definite integrals, that can be solved with the help of the series expansion of the function. We expand the integrand using the Taylor series or the Maclaurin series expansion formula. ...
https://math.stackexchange.com/questions/1922648/antiderivative-of-e2-arctanx If you want a way to proceed, we can make use of the Taylor expansion of the arctangent function: arctan(x)=∑k=0+∞2k+1(−1)k x2k+1 Hence you get: ∫e2(∑k=0+∞2k+1(−1)k x2k+1) dx ......
Find the first 6 non-zero terms of the Taylor series expanded at the given value of x=a. \displaystyle a. \ f(x) = \sin(x), \ \ a = \pi/4 \\ \displaystyle b. \ f(x) = \frac{1}{x}, \ \ a=2 Find the first three non zero terms of...
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...
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)...
x−arctan(x) 評估 −arctan(x)+x 對x 微分 x2+1x2 圖表