Markdown Equation Editor Online - Complex Mathematical Equations
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🔢 Advanced Equation Systems
Create complex mathematical systems with our markdown equation editor online :
Linear Equation Systems
Solve systems of linear equations with clear presentation:
{ 2 x + 3 y − z = 7 x − y + 2 z = − 1 3 x + y + z = 6 (1) \begin{cases}
2x + 3y - z = 7 \\
x - y + 2z = -1 \\
3x + y + z = 6
\end{cases}
\tag{1} ⎩ ⎨ ⎧ 2 x + 3 y − z = 7 x − y + 2 z = − 1 3 x + y + z = 6 ( 1 )
Matrix Form : The system (1) can be written as A x = b A\mathbf{x} = \mathbf{b} A x = b :
[ 2 3 − 1 1 − 1 2 3 1 1 ] [ x y z ] = [ 7 − 1 6 ] (2) \begin{bmatrix}
2 & 3 & -1 \\
1 & -1 & 2 \\
3 & 1 & 1
\end{bmatrix}
\begin{bmatrix}
x \\ y \\ z
\end{bmatrix}
=
\begin{bmatrix}
7 \\ -1 \\ 6
\end{bmatrix}
\tag{2} 2 1 3 3 − 1 1 − 1 2 1 x y z = 7 − 1 6 ( 2 )
Solution : Using Cramer's rule:
x = det ( A x ) det ( A ) , y = det ( A y ) det ( A ) , z = det ( A z ) det ( A ) (3) x = \frac{\det(A_x)}{\det(A)}, \quad y = \frac{\det(A_y)}{\det(A)}, \quad z = \frac{\det(A_z)}{\det(A)}
\tag{3} x = det ( A ) det ( A x ) , y = det ( A ) det ( A y ) , z = det ( A ) det ( A z ) ( 3 )
Differential Equation Systems
First-Order System : Consider the coupled system:
d x d t = a x + b y d y d t = c x + d y \begin{align}
\frac{dx}{dt} &= ax + by \tag{4a} \\
\frac{dy}{dt} &= cx + dy \tag{4b}
\end{align} d t d x d t d y = a x + b y = c x + d y ( 4a ) ( 4b )
Matrix Exponential Solution : The solution is given by:
[ x ( t ) y ( t ) ] = e A t [ x 0 y 0 ] (5) \begin{bmatrix} x(t) \\ y(t) \end{bmatrix} = e^{At} \begin{bmatrix} x_0 \\ y_0 \end{bmatrix}
\tag{5} [ x ( t ) y ( t ) ] = e A t [ x 0 y 0 ] ( 5 )
where A = [ a b c d ] A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} A = [ a c b d ] and e A t = ∑ n = 0 ∞ ( A t ) n n ! e^{At} = \sum_{n=0}^{\infty} \frac{(At)^n}{n!} e A t = ∑ n = 0 ∞ n ! ( A t ) n .
📊 Advanced Mathematical Structures
Piecewise Functions
Define complex functions with multiple conditions:
f ( x ) = { x 2 + 2 x + 1 if x ≤ − 1 2 x − 1 if − 1 < x < 2 x 3 − 4 x + 2 if x ≥ 2 (6) f(x) = \begin{cases}
x^2 + 2x + 1 & \text{if } x \leq -1 \\
2x - 1 & \text{if } -1 < x < 2 \\
x^3 - 4x + 2 & \text{if } x \geq 2
\end{cases}
\tag{6} f ( x ) = ⎩ ⎨ ⎧ x 2 + 2 x + 1 2 x − 1 x 3 − 4 x + 2 if x ≤ − 1 if − 1 < x < 2 if x ≥ 2 ( 6 )
Continuity Analysis : Check continuity at boundary points:
At x = − 1 x = -1 x = − 1 : lim x → − 1 − f ( x ) = 0 \lim_{x \to -1^-} f(x) = 0 lim x → − 1 − f ( x ) = 0 and f ( − 1 ) = 0 f(-1) = 0 f ( − 1 ) = 0 ✓
At x = 2 x = 2 x = 2 : lim x → 2 − f ( x ) = 3 \lim_{x \to 2^-} f(x) = 3 lim x → 2 − f ( x ) = 3 and lim x → 2 + f ( x ) = 2 \lim_{x \to 2^+} f(x) = 2 lim x → 2 + f ( x ) = 2 ✗
Optimization Problems
Lagrange Multipliers : For constrained optimization:
∇ f ( x , y ) = λ ∇ g ( x , y ) g ( x , y ) = 0 \begin{align}
\nabla f(x,y) &= \lambda \nabla g(x,y) \tag{7a} \\
g(x,y) &= 0 \tag{7b}
\end{align} ∇ f ( x , y ) g ( x , y ) = λ ∇ g ( x , y ) = 0 ( 7a ) ( 7b )
Economic Example : Maximize utility U ( x , y ) = x y U(x,y) = xy U ( x , y ) = x y subject to budget constraint p x + q y = I px + qy = I p x + q y = I :
{ ∂ U ∂ x = y = λ p ∂ U ∂ y = x = λ q p x + q y = I ⟹ { x ∗ = I 2 p y ∗ = I 2 q (8) \begin{cases}
\frac{\partial U}{\partial x} = y = \lambda p \\
\frac{\partial U}{\partial y} = x = \lambda q \\
px + qy = I
\end{cases}
\implies
\begin{cases}
x^* = \frac{I}{2p} \\
y^* = \frac{I}{2q}
\end{cases}
\tag{8} ⎩ ⎨ ⎧ ∂ x ∂ U = y = λ p ∂ y ∂ U = x = λ q p x + q y = I ⟹ { x ∗ = 2 p I y ∗ = 2 q I ( 8 )
Advanced Equation Features
🎯 Equation Numbering & References
Automatic Numbering
Our mathematical equation editor provides sophisticated numbering:
Theorem 1 (Fundamental Theorem of Calculus): If f f f is continuous on [ a , b ] [a,b] [ a , b ] , then:
∫ a b f ( x ) d x = F ( b ) − F ( a ) (FTC) \int_a^b f(x) \, dx = F(b) - F(a)
\tag{FTC} ∫ a b f ( x ) d x = F ( b ) − F ( a ) ( FTC )
where F ′ ( x ) = f ( x ) F'(x) = f(x) F ′ ( x ) = f ( x ) .
Corollary 1.1 : From equation (FTC), we derive the evaluation formula:
[ F ( x ) ] a b = F ( b ) − F ( a ) (9) \left[ F(x) \right]_a^b = F(b) - F(a)
\tag{9} [ F ( x ) ] a b = F ( b ) − F ( a ) ( 9 )
Sub-equation Systems
For related equations, use sub-numbering:
E = m c 2 p = γ m v E 2 = ( p c ) 2 + ( m c 2 ) 2 \begin{align}
E &= mc^2 \tag{10a} \\
p &= \gamma mv \tag{10b} \\
E^2 &= (pc)^2 + (mc^2)^2 \tag{10c}
\end{align} E p E 2 = m c 2 = γm v = ( p c ) 2 + ( m c 2 ) 2 ( 10a ) ( 10b ) ( 10c )
These are Einstein's energy-momentum relations (10a-c).
🔬 Scientific Applications
Quantum Mechanics Equations
Time-Dependent Schrödinger Equation :
i ℏ ∂ ∂ t Ψ ( r , t ) = H ^ Ψ ( r , t ) (TDSE) i\hbar \frac{\partial}{\partial t} \Psi(\mathbf{r}, t) = \hat{H} \Psi(\mathbf{r}, t)
\tag{TDSE} i ℏ ∂ t ∂ Ψ ( r , t ) = H ^ Ψ ( r , t ) ( TDSE )
Time-Independent Form : For stationary states Ψ ( r , t ) = ψ ( r ) e − i E t / ℏ \Psi(\mathbf{r}, t) = \psi(\mathbf{r})e^{-iEt/\hbar} Ψ ( r , t ) = ψ ( r ) e − i Et /ℏ :
H ^ ψ ( r ) = E ψ ( r ) (TISE) \hat{H} \psi(\mathbf{r}) = E \psi(\mathbf{r})
\tag{TISE} H ^ ψ ( r ) = E ψ ( r ) ( TISE )
Harmonic Oscillator : The energy eigenvalues are:
E n = ℏ ω ( n + 1 2 ) , n = 0 , 1 , 2 , … (11) E_n = \hbar\omega\left(n + \frac{1}{2}\right), \quad n = 0, 1, 2, \ldots
\tag{11} E n = ℏ ω ( n + 2 1 ) , n = 0 , 1 , 2 , … ( 11 )
Statistical Mechanics
Partition Function : The canonical partition function is:
Z = ∑ i e − β E i = Tr ( e − β H ^ ) (12) Z = \sum_{i} e^{-\beta E_i} = \text{Tr}(e^{-\beta \hat{H}})
\tag{12} Z = i ∑ e − β E i = Tr ( e − β H ^ ) ( 12 )
Thermodynamic Relations : From Z Z Z , we derive:
F = − k B T ln Z U = − ∂ ln Z ∂ β S = k B ( ln Z + β U ) \begin{align}
F &= -k_B T \ln Z \tag{13a} \\
U &= -\frac{\partial \ln Z}{\partial \beta} \tag{13b} \\
S &= k_B(\ln Z + \beta U) \tag{13c}
\end{align} F U S = − k B T ln Z = − ∂ β ∂ ln Z = k B ( ln Z + β U ) ( 13a ) ( 13b ) ( 13c )
📐 Advanced Mathematical Proofs
Convergence Proofs
Theorem 2 : The series ∑ n = 1 ∞ 1 n 2 \sum_{n=1}^{\infty} \frac{1}{n^2} ∑ n = 1 ∞ n 2 1 converges to π 2 6 \frac{\pi^2}{6} 6 π 2 .
Proof Strategy : We use Fourier analysis. Consider the function:
f ( x ) = x 2 , x ∈ [ − π , π ] (14) f(x) = x^2, \quad x \in [-\pi, \pi]
\tag{14} f ( x ) = x 2 , x ∈ [ − π , π ] ( 14 )
The Fourier series expansion gives:
x 2 = π 2 3 + 4 ∑ n = 1 ∞ ( − 1 ) n n 2 cos ( n x ) (15) x^2 = \frac{\pi^2}{3} + 4\sum_{n=1}^{\infty} \frac{(-1)^n}{n^2} \cos(nx)
\tag{15} x 2 = 3 π 2 + 4 n = 1 ∑ ∞ n 2 ( − 1 ) n cos ( n x ) ( 15 )
Setting x = 0 x = 0 x = 0 :
0 = π 2 3 + 4 ∑ n = 1 ∞ ( − 1 ) n n 2 (16) 0 = \frac{\pi^2}{3} + 4\sum_{n=1}^{\infty} \frac{(-1)^n}{n^2}
\tag{16} 0 = 3 π 2 + 4 n = 1 ∑ ∞ n 2 ( − 1 ) n ( 16 )
Setting x = π x = \pi x = π :
π 2 = π 2 3 + 4 ∑ n = 1 ∞ 1 n 2 (17) \pi^2 = \frac{\pi^2}{3} + 4\sum_{n=1}^{\infty} \frac{1}{n^2}
\tag{17} π 2 = 3 π 2 + 4 n = 1 ∑ ∞ n 2 1 ( 17 )
Therefore: ∑ n = 1 ∞ 1 n 2 = π 2 6 \sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6} ∑ n = 1 ∞ n 2 1 = 6 π 2 ∎
Complex Analysis
Residue Theorem : For a simple closed contour C C C and function f f f analytic except at isolated singularities:
∮ C f ( z ) d z = 2 π i ∑ k Res ( f , z k ) (18) \oint_C f(z) \, dz = 2\pi i \sum_{k} \text{Res}(f, z_k)
\tag{18} ∮ C f ( z ) d z = 2 πi k ∑ Res ( f , z k ) ( 18 )
Application : Evaluate ∫ − ∞ ∞ d x 1 + x 2 \int_{-\infty}^{\infty} \frac{dx}{1+x^2} ∫ − ∞ ∞ 1 + x 2 d x
The integrand f ( z ) = 1 1 + z 2 f(z) = \frac{1}{1+z^2} f ( z ) = 1 + z 2 1 has simple poles at z = ± i z = \pm i z = ± i .
Residue Calculation :
Res ( 1 1 + z 2 , i ) = lim z → i ( z − i ) ⋅ 1 ( z − i ) ( z + i ) = 1 2 i (19) \text{Res}\left(\frac{1}{1+z^2}, i\right) = \lim_{z \to i} (z-i) \cdot \frac{1}{(z-i)(z+i)} = \frac{1}{2i}
\tag{19} Res ( 1 + z 2 1 , i ) = z → i lim ( z − i ) ⋅ ( z − i ) ( z + i ) 1 = 2 i 1 ( 19 )
Result : ∫ − ∞ ∞ d x 1 + x 2 = 2 π i ⋅ 1 2 i = π \int_{-\infty}^{\infty} \frac{dx}{1+x^2} = 2\pi i \cdot \frac{1}{2i} = \pi ∫ − ∞ ∞ 1 + x 2 d x = 2 πi ⋅ 2 i 1 = π ✓
Equation Management System
📋 Equation Templates
Common Equation Types
Our complex equations editor includes templates for:
Quadratic Equations : a x 2 + b x + c = 0 ax^2 + bx + c = 0 a x 2 + b x + c = 0
x = − b ± b 2 − 4 a c 2 a (Quadratic) x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\tag{Quadratic} x = 2 a − b ± b 2 − 4 a c ( Quadratic )
Parametric Equations : Circle with radius r r r :
x ( t ) = r cos ( t ) y ( t ) = r sin ( t ) \begin{align}
x(t) &= r\cos(t) \tag{20a} \\
y(t) &= r\sin(t) \tag{20b}
\end{align} x ( t ) y ( t ) = r cos ( t ) = r sin ( t ) ( 20a ) ( 20b )
Polar Equations : Rose curve with n n n petals:
r = a cos ( n θ ) , n ∈ N (21) r = a\cos(n\theta), \quad n \in \mathbb{N}
\tag{21} r = a cos ( n θ ) , n ∈ N ( 21 )
Advanced Templates
Heat Equation : One-dimensional heat diffusion:
∂ u ∂ t = α ∂ 2 u ∂ x 2 (22) \frac{\partial u}{\partial t} = \alpha \frac{\partial^2 u}{\partial x^2}
\tag{22} ∂ t ∂ u = α ∂ x 2 ∂ 2 u ( 22 )
Wave Equation : One-dimensional wave propagation:
∂ 2 u ∂ t 2 = c 2 ∂ 2 u ∂ x 2 (23) \frac{\partial^2 u}{\partial t^2} = c^2 \frac{\partial^2 u}{\partial x^2}
\tag{23} ∂ t 2 ∂ 2 u = c 2 ∂ x 2 ∂ 2 u ( 23 )
Laplace Equation : Two-dimensional potential:
∂ 2 u ∂ x 2 + ∂ 2 u ∂ y 2 = 0 (24) \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0
\tag{24} ∂ x 2 ∂ 2 u + ∂ y 2 ∂ 2 u = 0 ( 24 )
🔗 Cross-Reference System
Equation References
Reference equations naturally in text:
From the fundamental equation (FTC), we can derive...
The system of equations (1) has a unique solution when...
Using the residue theorem (18), we evaluate...
Theorem-Proof Structure
Definition 1 : A function f : R → R f: \mathbb{R} \to \mathbb{R} f : R → R is continuous at a a a if:
lim x → a f ( x ) = f ( a ) (Def-Continuity) \lim_{x \to a} f(x) = f(a)
\tag{Def-Continuity} x → a lim f ( x ) = f ( a ) ( Def-Continuity )
Theorem 3 : If f f f and g g g are continuous at a a a , then f + g f + g f + g is continuous at a a a .
Proof : Using definition (Def-Continuity), we need to show:
lim x → a ( f ( x ) + g ( x ) ) = f ( a ) + g ( a ) (25) \lim_{x \to a} (f(x) + g(x)) = f(a) + g(a)
\tag{25} x → a lim ( f ( x ) + g ( x )) = f ( a ) + g ( a ) ( 25 )
By the limit properties:
lim x → a ( f ( x ) + g ( x ) ) = lim x → a f ( x ) + lim x → a g ( x ) = f ( a ) + g ( a ) (26) \lim_{x \to a} (f(x) + g(x)) = \lim_{x \to a} f(x) + \lim_{x \to a} g(x) = f(a) + g(a)
\tag{26} x → a lim ( f ( x ) + g ( x )) = x → a lim f ( x ) + x → a lim g ( x ) = f ( a ) + g ( a ) ( 26 )
Therefore, f + g f + g f + g is continuous at a a a . ∎
Advanced Formatting Options
🎨 Equation Styling
Highlighted Equations
Important results can be emphasized:
e i π + 1 = 0 (Euler’s Identity) \boxed{e^{i\pi} + 1 = 0}
\tag{Euler's Identity} e iπ + 1 = 0 ( Euler’s Identity )
This is considered one of the most beautiful equations in mathematics.
Multi-line Derivations
Show step-by-step calculations:
∫ 0 π sin 2 ( x ) d x = ∫ 0 π 1 − cos ( 2 x ) 2 d x = 1 2 ∫ 0 π ( 1 − cos ( 2 x ) ) d x = 1 2 [ x − sin ( 2 x ) 2 ] 0 π = 1 2 [ π − 0 ] = π 2 \begin{align}
\int_0^{\pi} \sin^2(x) \, dx &= \int_0^{\pi} \frac{1 - \cos(2x)}{2} \, dx \tag{27} \\
&= \frac{1}{2} \int_0^{\pi} (1 - \cos(2x)) \, dx \tag{28} \\
&= \frac{1}{2} \left[ x - \frac{\sin(2x)}{2} \right]_0^{\pi} \tag{29} \\
&= \frac{1}{2} \left[ \pi - 0 \right] = \frac{\pi}{2} \tag{30}
\end{align} ∫ 0 π sin 2 ( x ) d x = ∫ 0 π 2 1 − cos ( 2 x ) d x = 2 1 ∫ 0 π ( 1 − cos ( 2 x )) d x = 2 1 [ x − 2 sin ( 2 x ) ] 0 π = 2 1 [ π − 0 ] = 2 π ( 27 ) ( 28 ) ( 29 ) ( 30 )
📊 Specialized Equation Types
Economic Models
Cobb-Douglas Production Function :
Y = A K α L 1 − α (31) Y = A K^{\alpha} L^{1-\alpha}
\tag{31} Y = A K α L 1 − α ( 31 )
Marginal Products :
M P K = ∂ Y ∂ K = α A K α − 1 L 1 − α M P L = ∂ Y ∂ L = ( 1 − α ) A K α L − α \begin{align}
MPK &= \frac{\partial Y}{\partial K} = \alpha A K^{\alpha-1} L^{1-\alpha} \tag{32a} \\
MPL &= \frac{\partial Y}{\partial L} = (1-\alpha) A K^{\alpha} L^{-\alpha} \tag{32b}
\end{align} MP K MP L = ∂ K ∂ Y = α A K α − 1 L 1 − α = ∂ L ∂ Y = ( 1 − α ) A K α L − α ( 32a ) ( 32b )
Engineering Applications
Control Theory : State-space representation:
x ˙ = A x + B u y = C x + D u \begin{align}
\dot{\mathbf{x}} &= A\mathbf{x} + B\mathbf{u} \tag{33a} \\
\mathbf{y} &= C\mathbf{x} + D\mathbf{u} \tag{33b}
\end{align} x ˙ y = A x + B u = C x + D u ( 33a ) ( 33b )
Transfer Function : From state-space to frequency domain:
G ( s ) = C ( s I − A ) − 1 B + D (34) G(s) = C(sI - A)^{-1}B + D
\tag{34} G ( s ) = C ( s I − A ) − 1 B + D ( 34 )
Export & Collaboration
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Academic Papers : LaTeX-ready equations with proper numbering
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Version Control : Track equation changes over time
Comments : Annotate complex derivations
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