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Gauss-Seidel Method

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Introduction

The Gauss-Seidel method, named after Carl Friedrich Gauss and Philipp Ludwig von Seidel, is an iterative algorithm for solving systems of linear equations. Like the Jacobi method, it's particularly useful for large, sparse systems where direct methods may be computationally expensive.

The Gauss-Seidel method is an improvement over the Jacobi method, as it uses updated values as soon as they are available in each iteration. This often leads to faster convergence.

Key Concepts

Iterative Approximation

Like the Jacobi method, the Gauss-Seidel method starts with a rough guess and then keeps improving that guess step by step. The big difference is that as soon as it calculates a better value for a variable, it immediately uses it in the same round of calculations.

For Example: Suppose we are solving the system of equations represented by the matrix:

A=(412351113),b=(473)A = \begin{pmatrix} 4 & 1 & 2 \\ 3 & 5 & 1 \\ 1 & 1 & 3 \end{pmatrix}, \quad b = \begin{pmatrix} 4 \\ 7 \\ 3 \end{pmatrix}

Here, we need to find the values of x, y, and z such that:

4x+1y+2z=43x+5y+1z=71x+1y+3z=3\begin{aligned} 4x + 1y + 2z &= 4 \\ 3x + 5y + 1z &= 7 \\ 1x + 1y + 3z &= 3 \end{aligned}

In the Gauss-Seidel method, once we find an updated value for x in the first equation, we use that value immediately in the second equation to find y, and so on. This speeds up the process.

Convergence

The Gauss-Seidel method usually reaches the correct answer faster than the Jacobi method. It works best when dealing with special types of matrices, like those that are symmetric and positive-definite, or ones that are "diagonally dominant."

For Example: If you use the Gauss-Seidel method on the matrix from the example above, it will converge faster because the matrix is close to being diagonally dominant. This means it’s well-suited for this kind of solution.

Diagonal Dominance

Just like with the Jacobi method, if a matrix is "diagonally dominant," it helps ensure that the method will work and find a solution. A matrix is diagonally dominant if, in each row, the number on the diagonal (main number in the row) is bigger than the sum of all the other numbers in that row.

In our matrix:

∣4∣≥∣1∣+∣2∣(4≥3)∣5∣≥∣3∣+∣1∣(5≥4)∣3∣≥∣1∣+∣1∣(3≥2)\begin{aligned} |4| &\geq |1| + |2| \quad (4 \geq 3) \\ |5| &\geq |3| + |1| \quad (5 \geq 4) \\ |3| &\geq |1| + |1| \quad (3 \geq 2) \end{aligned}

Since each diagonal element is larger than or equal to the sum of the other numbers in the row, the matrix is diagonally dominant. This ensures that the Gauss-Seidel method will converge to the correct solution.

∣aii∣≥∑j≠i∣aij∣|a_{ii}| \geq \sum_{j \neq i} |a_{ij}|

The Gauss-Seidel Method Algorithm

Consider a system of linear equations Ax=bAx = b:

a11x1+a12x2+⋯+a1nxn=b1a21x1+a22x2+⋯+a2nxn=b2⋮an1x1+an2x2+⋯+annxn=bn\begin{align} a_{11}x_1 + a_{12}x_2 + \cdots + a_{1n}x_n &= b_1 \\ a_{21}x_1 + a_{22}x_2 + \cdots + a_{2n}x_n &= b_2 \\ \vdots \\ a_{n1}x_1 + a_{n2}x_2 + \cdots + a_{nn}x_n &= b_n \end{align}

Steps of the Gauss-Seidel Method

  1. Start with an initial guess for the solution:x(0)=(x1(0),x2(0),…,xn(0))x^{(0)} = (x_1^{(0)}, x_2^{(0)}, \ldots, x_n^{(0)})
  2. For each ii from 1 to nn, compute the next approximation using:
    xi(k+1)=1aii(bi−∑j<iaijxj(k+1)−∑j>iaijxj(k))x_i^{(k+1)} = \frac{1}{a_{ii}} \left(b_i - \sum_{j < i} a_{ij}x_j^{(k+1)} - \sum_{j > i} a_{ij}x_j^{(k)}\right)
  3. Repeat step 2 until the solution converges or a maximum number of iterations is reached.

The key difference from the Jacobi method is in step 2, where we use the most recent values (xj(k+1)x_j^{(k+1)}) for the variables that have already been updated in the current iteration.

Convergence Criteria and Tolerance

The Gauss-Seidel method continues to run until it meets one of these stopping conditions:

  • The solution stops changing significantly:

    This means that the difference between the current guess and the previous guess for each variable becomes very small. If the largest change for any variable is smaller than a certain small number (called the "tolerance"), we can say the solution has "converged."

    max⁡i∣xi(k+1)−xi(k)∣<tolerance\max_{i} |x_i^{(k+1)} - x_i^{(k)}| < \text{tolerance}

    Example: Suppose after one iteration, the difference between your current guess and previous guess for x is 0.001, for y is 0.002, and for z is 0.0005. If your tolerance is 0.001, the algorithm will stop because the largest difference (0.002 for y) is still within the tolerance range.

  • The residual is small:

    The residual is the difference between the result when you plug your current guess into the system of equations and the actual results you're solving for (in vector b). When this difference becomes smaller than the tolerance, the solution is close enough to the real answer.

    ∥Ax(k)−b∥<tolerance\|Ax^{(k)} - b\| < \text{tolerance}

    Example: In a system of equations, if the matrix A and vector b are given, and your current solution guess gives a small residual (like 0.001), then the system is close enough to the true solution.

  • A maximum number of iterations is reached:

    Sometimes, the method may take too long to reach the solution. To prevent it from running forever, we set a maximum number of iterations. If the method hasn’t converged by then, we stop and either accept the current approximation or try different approaches.

    Example: You might set the maximum number of iterations to 100. If after 100 steps, the solution hasn't converged (meaning it doesn't meet the other conditions), you stop and analyze why it hasn't converged.

Where and When the Gauss-Seidel Method is Used

The Gauss-Seidel method is widely used in various fields of science and engineering:

  • Numerical Weather Prediction: Used in atmospheric models to solve systems of equations describing atmospheric dynamics.
  • Image Processing: Applied in image reconstruction and noise reduction algorithms.
  • Computational Fluid Dynamics: Used to solve discretized versions of the Navier-Stokes equations.
  • Electrical Circuit Analysis: Employed to determine voltages and currents in complex circuits.
  • Structural Engineering: Used in finite element analysis for stress and strain calculations.

The method is particularly useful when:

  • The system is too large for direct methods.
  • The matrix is sparse, as the method only requires matrix-vector products.
  • An approximate solution is acceptable, as the method can be stopped after a fixed number of iterations.
  • The matrix is diagonally dominant or symmetric positive-definite, ensuring convergence.

Example

Let's solve the following system of equations using the Gauss-Seidel method:

10x1−x2+2x3=6−x1+11x2−x3=252x1−x2+10x3=−11\begin{align} 10x_1 - x_2 + 2x_3 &= 6 \\ -x_1 + 11x_2 - x_3 &= 25 \\ 2x_1 - x_2 + 10x_3 &= -11 \end{align}

Step 1: Initial Guess

Let's start with an initial guess of x(0)=(0,0,0)x^{(0)} = (0, 0, 0).

Step 2: First Iteration

In the Gauss-Seidel method, we immediately use the updated value of each variable as soon as it's calculated. Unlike the Jacobi method, where each variable's new value is used only in the next iteration, Gauss-Seidel uses the most recent values within the same iteration.

x1(1)=110(6−(0)−2(0))=0.6x2(1)=111(25−(−0.6)−(0))=2.3273x3(1)=110(−11−2(0.6)−(−2.3273))=−0.9873\begin{align} x_1^{(1)} &= \frac{1}{10}(6 - (0) - 2(0)) = 0.6 \\ x_2^{(1)} &= \frac{1}{11}(25 - (-0.6) - (0)) = 2.3273 \\ x_3^{(1)} &= \frac{1}{10}(-11 - 2(0.6) - (-2.3273)) = -0.9873 \end{align}

Here, the updated value of x1(1)=0.6x_1^{(1)} = 0.6 is immediately used to calculate x2(1)x_2^{(1)}, and both updated values are used to calculate x3(1)x_3^{(1)} within the same iteration.

Step 3: Second Iteration

x1(2)=110(6−(2.3273)−2(−0.9873))=0.9987x2(2)=111(25−(−0.9987)−(−0.9873))=2.0896x3(2)=110(−11−2(0.9987)−(−2.0896))=−0.9909\begin{align} x_1^{(2)} &= \frac{1}{10}(6 - (2.3273) - 2(-0.9873)) = 0.9987 \\ x_2^{(2)} &= \frac{1}{11}(25 - (-0.9987) - (-0.9873)) = 2.0896 \\ x_3^{(2)} &= \frac{1}{10}(-11 - 2(0.9987) - (-2.0896)) = -0.9909 \end{align}

Again, the updated values of x1x_1 and x2x_2 are used immediately in the same iteration to calculate x3x_3.

Step 4: Third Iteration

x1(3)=110(6−(2.0896)−2(−0.9909))=0.9989x2(3)=111(25−(−0.9989)−(−0.9909))=2.089x3(3)=110(−11−2(0.9989)−(−2.089))=−0.991\begin{align} x_1^{(3)} &= \frac{1}{10}(6 - (2.0896) - 2(-0.9909)) = 0.9989 \\ x_2^{(3)} &= \frac{1}{11}(25 - (-0.9989) - (-0.9909)) = 2.089 \\ x_3^{(3)} &= \frac{1}{10}(-11 - 2(0.9989) - (-2.089)) = -0.991 \end{align}

Step 5: Convergence Check

The process continues until the differences between consecutive values of x1x_1, x2x_2, and x3x_3 become smaller than the predefined tolerance or the maximum number of iterations is reached.

Final Result

After several iterations, the solution converges to:

x1≈1.0000,x2≈2.0000,x3≈−1.0000x_1 \approx 1.0000, \quad x_2 \approx 2.0000, \quad x_3 \approx -1.0000

We stop iterating when the solution has stabilized to the desired accuracy.

Advanced Practice Questions

Practice Problem 1

Solve the following system of equations using the Gauss-Seidel method:

4x1−x2+x3=72x1+5x2+x3=8x1+x2+3x3=3\begin{align} 4x_1 - x_2 + x_3 &= 7 \\ 2x_1 + 5x_2 + x_3 &= 8 \\ x_1 + x_2 + 3x_3 &= 3 \end{align}

Solution for Problem 1

Step 1: Initial Guess

We start with the initial guess x(0)=(0,0,0)x^{(0)} = (0, 0, 0).

Step 2: First Iteration

x1(1)=14(7−(0)+(0))=1.75x2(1)=15(8−2(1.75)−(0))=0.7x3(1)=13(3−(1.75)−0.7)=0.1833\begin{align} x_1^{(1)} &= \frac{1}{4}(7 - (0) + (0)) = 1.75 \\ x_2^{(1)} &= \frac{1}{5}(8 - 2(1.75) - (0)) = 0.7 \\ x_3^{(1)} &= \frac{1}{3}(3 - (1.75) - 0.7) = 0.1833 \end{align}

Step 3: Second Iteration

x1(2)=14(7−(0.7)+(0.1833))=1.6208x2(2)=15(8−2(1.6208)−(0.1833))=0.7151x3(2)=13(3−(1.6208)−0.7151)=0.2214\begin{align} x_1^{(2)} &= \frac{1}{4}(7 - (0.7) + (0.1833)) = 1.6208 \\ x_2^{(2)} &= \frac{1}{5}(8 - 2(1.6208) - (0.1833)) = 0.7151 \\ x_3^{(2)} &= \frac{1}{3}(3 - (1.6208) - 0.7151) = 0.2214 \end{align}

Step 4: Third Iteration

x1(3)=14(7−(0.7151)+(0.2214))=1.6266x2(3)=15(8−2(1.6266)−(0.2214))=0.705x3(3)=13(3−(1.6266)−0.705)=0.2228\begin{align} x_1^{(3)} &= \frac{1}{4}(7 - (0.7151) + (0.2214)) = 1.6266 \\ x_2^{(3)} &= \frac{1}{5}(8 - 2(1.6266) - (0.2214)) = 0.705 \\ x_3^{(3)} &= \frac{1}{3}(3 - (1.6266) - 0.705) = 0.2228 \end{align}

Final Result for Problem 1

After several iterations, the solution converges to:

x1≈1.627,x2≈0.705,x3≈0.223x_1 \approx 1.627, \quad x_2 \approx 0.705, \quad x_3 \approx 0.223

Practice Problem 2

Solve the system of equations below, starting with the initial guess x(0)=(1,1,1)x^{(0)} = (1, 1, 1):

8x1−2x2+x3=5−x1+6x2+2x3=7x1+x2+4x3=3\begin{align} 8x_1 - 2x_2 + x_3 &= 5 \\ -x_1 + 6x_2 + 2x_3 &= 7 \\ x_1 + x_2 + 4x_3 &= 3 \end{align}

Solution for Problem 2

Step 1: Initial Guess

Starting with x(0)=(1,1,1)x^{(0)} = (1, 1, 1).

Step 2: First Iteration

x1(1)=18(5+2(1)−(1))=0.75x2(1)=16(7+(0.75)−2(1))=1.2917x3(1)=14(3−(0.75)−(1.2917))=0.2396\begin{align} x_1^{(1)} &= \frac{1}{8}(5 + 2(1) - (1)) = 0.75 \\ x_2^{(1)} &= \frac{1}{6}(7 + (0.75) - 2(1)) = 1.2917 \\ x_3^{(1)} &= \frac{1}{4}(3 - (0.75) - (1.2917)) = 0.2396 \end{align}

Step 3: Second Iteration

x1(2)=18(5+2(1.2917)−(0.2396))=0.9203x2(2)=16(7+(0.9203)−2(0.2396))=1.3734x3(2)=14(3−(0.9203)−(1.3734))=0.1766\begin{align} x_1^{(2)} &= \frac{1}{8}(5 + 2(1.2917) - (0.2396)) = 0.9203 \\ x_2^{(2)} &= \frac{1}{6}(7 + (0.9203) - 2(0.2396)) = 1.3734 \\ x_3^{(2)} &= \frac{1}{4}(3 - (0.9203) - (1.3734)) = 0.1766 \end{align}

Step 4: Third Iteration

x1(3)=18(5+2(1.3734)−(0.1766))=0.9463x2(3)=16(7+(0.9463)−2(0.1766))=1.4039x3(3)=14(3−(0.9463)−(1.4039))=0.1624\begin{align} x_1^{(3)} &= \frac{1}{8}(5 + 2(1.3734) - (0.1766)) = 0.9463 \\ x_2^{(3)} &= \frac{1}{6}(7 + (0.9463) - 2(0.1766)) = 1.4039 \\ x_3^{(3)} &= \frac{1}{4}(3 - (0.9463) - (1.4039)) = 0.1624 \end{align}

Final Result for Problem 2

The solution converges to:

x1≈0.946,x2≈1.404,x3≈0.162x_1 \approx 0.946, \quad x_2 \approx 1.404, \quad x_3 \approx 0.162

Question:

(b) Perform four iterations (rounded to four decimal places) using (i) Jacobi Method and (ii) Gauss-Seidel method, for the following system of equations.

[5−5−11−41−21−6][x1x2x3]=[−8−4−18]\begin{bmatrix} 5 & -5 & -1 \\ 1 & -4 & 1 \\ -2 & 1 & -6 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} -8 \\ -4 \\ -18 \end{bmatrix}

With x(0)=(0,0,0)Tx^{(0)} = (0, 0, 0)^T. The exact solution is (1,2,3)T(1, 2, 3)^T.

Which method gives better approximation to the exact solution?

Solution:

We are given the system of equations:

5x1−5x2−x3=−8x1−4x2+x3=−4−2x1+x2−6x3=−18\begin{align} 5x_1 - 5x_2 - x_3 &= -8 \\ x_1 - 4x_2 + x_3 &= -4 \\ -2x_1 + x_2 - 6x_3 &= -18 \end{align}

We will solve this system using both Jacobi and Gauss-Seidel methods, starting with the initial guess x(0)=(0,0,0)Tx^{(0)} = (0, 0, 0)^T. The exact solution is (1,2,3)T(1, 2, 3)^T.

Jacobi Method

Iteration 1

x1(1)=−8+5(0)+(0)5=−1.6x2(1)=−4−(0)−(0)−4=1x3(1)=−18+2(0)−(0)−6=3\begin{align} x_1^{(1)} &= \frac{-8 + 5(0) + (0)}{5} = -1.6 \\ x_2^{(1)} &= \frac{-4 - (0) - (0)}{-4} = 1 \\ x_3^{(1)} &= \frac{-18 + 2(0) - (0)}{-6} = 3 \end{align}

Iteration 2

x1(2)=−8+5(1)+(3)5=0x2(2)=−4−(0)−(3)−4=2.1x3(2)=−18+2(−1.6)−(1)−6=2.7333\begin{align} x_1^{(2)} &= \frac{-8 + 5(1) + (3)}{5} = 0 \\ x_2^{(2)} &= \frac{-4 - (0) - (3)}{-4} = 2.1 \\ x_3^{(2)} &= \frac{-18 + 2(-1.6) - (1)}{-6} = 2.7333 \end{align}

Iteration 3

x1(3)=−8+5(2.1)+(2.7333)5=1.0133x2(3)=−4−(0)−(2.7333)−4=2.1833x3(3)=−18+2(0)−(1.9133)−6=2.7\begin{align} x_1^{(3)} &= \frac{-8 + 5(2.1) + (2.7333)}{5} = 1.0133 \\ x_2^{(3)} &= \frac{-4 - (0) - (2.7333)}{-4} = 2.1833 \\ x_3^{(3)} &= \frac{-18 + 2(0) - (1.9133)}{-6} = 2.7 \end{align}

Iteration 4

x1(4)=−8+5(2.1833)+(2.7)5=1.0933x2(4)=−4−(1.0133)−(2.7)−4=2.1783x3(4)=−18+2(1.0133)−(2.1783)−6=2.6894\begin{align} x_1^{(4)} &= \frac{-8 + 5(2.1833) + (2.7)}{5} = 1.0933 \\ x_2^{(4)} &= \frac{-4 - (1.0133) - (2.7)}{-4} = 2.1783 \\ x_3^{(4)} &= \frac{-18 + 2(1.0133) - (2.1783)}{-6} = 2.6894 \end{align}

Gauss-Seidel Method

Iteration 1

x1(1)=−8+5(0)+(0)5=−1.6x2(1)=−4−(−1.6)−(0)−4=0.6x3(1)=−18+2(−1.6)−(0.6)−6=2.9333\begin{align} x_1^{(1)} &= \frac{-8 + 5(0) + (0)}{5} = -1.6 \\ x_2^{(1)} &= \frac{-4 - (-1.6) - (0)}{-4} = 0.6 \\ x_3^{(1)} &= \frac{-18 + 2(-1.6) - (0.6)}{-6} = 2.9333 \end{align}

Iteration 2

x1(2)=−8+5(0.6)+(2.9333)5=0x2(2)=−4−(0)−(2.9333)−4=1.7333x3(2)=−18+2(0)−(1.7333)−6=2.6222\begin{align} x_1^{(2)} &= \frac{-8 + 5(0.6) + (2.9333)}{5} = 0 \\ x_2^{(2)} &= \frac{-4 - (0) - (2.9333)}{-4} = 1.7333 \\ x_3^{(2)} &= \frac{-18 + 2(0) - (1.7333)}{-6} = 2.6222 \end{align}

Iteration 3

x1(3)=−8+5(1.7333)+(2.6222)5=1.0311x2(3)=−4−(1.0311)−(2.6222)−4=1.9133x3(3)=−18+2(1.0311)−(1.9133)−6=2.695\begin{align} x_1^{(3)} &= \frac{-8 + 5(1.7333) + (2.6222)}{5} = 1.0311 \\ x_2^{(3)} &= \frac{-4 - (1.0311) - (2.6222)}{-4} = 1.9133 \\ x_3^{(3)} &= \frac{-18 + 2(1.0311) - (1.9133)}{-6} = 2.695 \end{align}

Iteration 4

x1(4)=−8+5(1.9133)+(2.695)5=1.0404x2(4)=−4−(1.0404)−(2.695)−4=1.934x3(4)=−18+2(1.0404)−(1.934)−6=2.6961\begin{align} x_1^{(4)} &= \frac{-8 + 5(1.9133) + (2.695)}{5} = 1.0404 \\ x_2^{(4)} &= \frac{-4 - (1.0404) - (2.695)}{-4} = 1.934 \\ x_3^{(4)} &= \frac{-18 + 2(1.0404) - (1.934)}{-6} = 2.6961 \end{align}

Conclusion

After performing four iterations for both methods, we obtained the following approximations:

Results:

  • Jacobi Method:
    • x1≈1.0933x_1 \approx 1.0933
    • x2≈2.1783x_2 \approx 2.1783
    • x3≈2.6894x_3 \approx 2.6894
  • Gauss-Seidel Method:
    • x1≈1.0404x_1 \approx 1.0404
    • x2≈1.934x_2 \approx 1.934
    • x3≈2.6961x_3 \approx 2.6961

Comparison:

Comparing the approximations to the exact solution (1,2,3)(1, 2, 3), we observe:

  • The Gauss-Seidel method converges faster than the Jacobi method.
  • The approximations from Gauss-Seidel are closer to the exact solution.

Heat Distribution in a Metal Plate

Consider a square metal plate divided into a 3x3 grid. The temperature along the edges is fixed, and we want to find the steady-state temperature distribution inside the plate. The temperatures along the edges are as follows:

  • Top edge: 100°C
  • Bottom edge: 0°C
  • Left edge: 75°C
  • Right edge: 50°C

The temperature at each interior point is the average of its four neighboring points. Use the Gauss-Seidel method to approximate the temperature at the center point of the plate after two iterations. Start with an initial guess of 0°C for the interior points.

Solution

Let's label the center point as TT. The four corner points of the interior will be labeled T1T_1, T2T_2, T3T_3, and T4T_4 (clockwise from top-left).

The equation for the center point T is:

T=14(T1+T2+T3+T4)T = \frac{1}{4}(T_1 + T_2 + T_3 + T_4)

And the equations for the corner points are:

T1=14(100+75+T+T2)T2=14(100+50+T+T1)T3=14(50+0+T+T4)T4=14(75+0+T+T3)\begin{align} T_1 &= \frac{1}{4}(100 + 75 + T + T_2) \\ T_2 &= \frac{1}{4}(100 + 50 + T + T_1) \\ T_3 &= \frac{1}{4}(50 + 0 + T + T_4) \\ T_4 &= \frac{1}{4}(75 + 0 + T + T_3) \end{align}

Initial Guess

We start with T=T1=T2=T3=T4=0T = T_1 = T_2 = T_3 = T_4 = 0.

First Iteration

T1=14(100+75+0+0)=43.75T2=14(100+50+0+43.75)=48.4375T3=14(50+0+0+0)=12.5T4=14(75+0+0+12.5)=21.875T=14(43.75+48.4375+12.5+21.875)=31.640625\begin{align} T_1 &= \frac{1}{4}(100 + 75 + 0 + 0) = 43.75 \\ T_2 &= \frac{1}{4}(100 + 50 + 0 + 43.75) = 48.4375 \\ T_3 &= \frac{1}{4}(50 + 0 + 0 + 0) = 12.5 \\ T_4 &= \frac{1}{4}(75 + 0 + 0 + 12.5) = 21.875 \\ T &= \frac{1}{4}(43.75 + 48.4375 + 12.5 + 21.875) = 31.640625 \end{align}

Second Iteration

T1=14(100+75+31.640625+48.4375)=63.7695T2=14(100+50+31.640625+63.7695)=61.3525T3=14(50+0+31.640625+21.875)=25.8789T4=14(75+0+31.640625+25.8789)=33.1299T=14(63.7695+61.3525+25.8789+33.1299)=46.0327\begin{align} T_1 &= \frac{1}{4}(100 + 75 + 31.640625 + 48.4375) = 63.7695 \\ T_2 &= \frac{1}{4}(100 + 50 + 31.640625 + 63.7695) = 61.3525 \\ T_3 &= \frac{1}{4}(50 + 0 + 31.640625 + 21.875) = 25.8789 \\ T_4 &= \frac{1}{4}(75 + 0 + 31.640625 + 25.8789) = 33.1299 \\ T &= \frac{1}{4}(63.7695 + 61.3525 + 25.8789 + 33.1299) = 46.0327 \end{align}

After two iterations, the approximate temperature at the center point of the plate is 46.0327°C.

This example demonstrates how the Gauss-Seidel method can be applied to solve heat distribution problems, which are common in thermodynamics and materials science. The method quickly converges towards the steady-state solution, providing a good approximation even after just a few iterations.

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