QCSunny Lab

Equation, Calculus & Limit Solver方程求解、微积分与极限计算器

Solve polynomial equations (Quadratic ax²+bx+c=0, Cubic ax³+bx²+cx+d=0), 2x2 linear systems, numerical calculus (derivatives & integrals), and function limits via Richardson extrapolation.代数方程求解、微积分与极限工具:支持自然输入方程或方程组智能识别、一元二次/三次方程求根(含复数虚根)、二元一次方程组、数值微积分以及函数极限求解。

Solve essential algebraic equations, numerical calculus, and mathematical limits in your browser. Handles quadratic equations ax² + bx + c = 0 (with real and complex conjugate roots using the numerically stable Citardauq formula), cubic equations via Cardano formula, and 2×2 linear equation systems.

Also features a numerical calculus engine: evaluates first-order derivatives f'(x₀) via a 5-point symmetric stencil, definite integrals ∫[a,b] f(x) dx via Adaptive Simpson Quadrature, and limits (including indeterminate forms like sin(x)/x as x→0) via Richardson extrapolation across geometric perturbation sequences.

Frequently asked questions

Can it solve quadratic equations with negative discriminant (Δ < 0)?

Yes. When Δ < 0, it computes both complex conjugate roots in the standard format u ± vi.

How does it solve indeterminate limits like sin(x)/x as x → 0?

Direct substitution gives 0/0 (NaN). The solver samples decreasing perturbations h = 10⁻¹ ... 10⁻⁶ around x₀ and applies Richardson extrapolation (10·f(h/10) - f(h))/9 to eliminate first-order truncation error and cleanly converge to 1.

How is the definite integral computed?

It uses Adaptive Simpson Quadrature with recursive sub-interval refinement (tolerance 1e-9), adjusting sample density where curvature is high while computing smooth intervals in fractions of a millisecond.

中文说明

在浏览器本地实时求解经典代数方程、数值微积分与数学极限:支持一元二次方程(采用高精度 Citardauq 公式消除巨额灾难性抵消,自动输出实根与共轭复数虚根)、基于卡尔丹公式的一元三次方程解析求解,以及二元一次线性方程组。

同时集成高精度数值微积分与极限求解器:利用五点对称差商计算导数切线斜率 f'(x₀),利用自适应辛普森积分法(Adaptive Simpson Quadrature)以极少函数采样高精度求解定积分,并通过几何步长理查森外推算法(Richardson Extrapolation)稳健化解 0/0 不定型并精确计算单侧与双侧极限(如 sin(x)/x 在 x 趋向 0 时的极限为 1)。

常见问题

判别式 Δ < 0 时能求解虚根吗?

可以。当 Δ < 0 时,计算器会自动输出共轭复数解 u ± vi。

如何求解像 sin(x)/x 在 x 趋向 0 这样的 0/0 不定式极限?

直接代入会产生 0/0 无法求值。求解器在趋近点附近按几何递减步长 h = 10⁻¹ ... 10⁻⁶ 采样,并通过理查森外推公式 (10·f(h/10) - f(h))/9 消除主导截断误差,稳定消除浮点抖动并精确收敛至极限值 1。

定积分是如何计算的?

基于自适应辛普森积分法(Adaptive Simpson Quadrature),递归评估局部误差(容差 1e-9),在平滑区域只需十余次计算即可完成,在剧烈振荡区域自动加密网格,兼顾极高精度与毫秒级速度。