UNIT 1 • FINDING ROOTS OF EQUATIONS

Fixed Point Iteration
Recurrence Method

The Fixed Point Iteration Method rewrites an equation $f(x) = 0$ into the self-feeding form $x = \phi(x)$ . Starting from an initial guess $x_0$ , each output becomes the next input, drawing a staircase or cobweb diagram straight into the fixed point!

RECURRENCE RELATION & CONVERGENCECOBWEB STABILITY

x_{k} = \phi(x_{k-1})

|\phi'(x)| < 1 \quad (\text{Strict Convergence Condition})

1. Rewrite f(x) = 0Isolate

x

to create iteration form

x = \phi(x)

2. Test Slope DerivativeEnsure

|\phi'(x)| < 1

near the root for rapid convergence.

3. Feed Next GuessPlug

x_{k}

back into

\phi

until differences vanish.

Step-by-Step Worked Example

HAND-HOLDING TUTORIAL

OUR PROBLEM TO SOLVE

Find the root of $f(x) = x^3 - x - 1 = 0$ starting at initial guess $x_0 = 1.5$ .

STEP 01 • REWRITE & VERIFY CONVERGENCE

Isolate $x$ to define $\phi(x)$ :

\phi(x) = (x + 1)^{1/3}

\phi'(x) = \frac{1}{3(x+1)^{2/3}} \implies |\phi'(1.5)| = \frac{1}{3(2.5)^{2/3}} \approx 0.1809 < 1 \quad (\text{Strictly Converges!})

STEP 02 • COMPUTE FIRST ITERATION (x₁)

Plug initial guess $x_0 = 1.5$ into $\phi(x)$ :

x_1 = \phi(1.5) = (1.5 + 1)^{1/3} = (2.5)^{1/3} \approx 1.357209

STEP 03 • COMPUTE SECOND ITERATION (x₂)

Feed $x_1 = 1.357209$ back into $\phi(x)$ :

x_2 = \phi(1.357209) = (1.357209 + 1)^{1/3} = (2.357209)^{1/3} \approx 1.330861

STEP 04 • SUBSEQUENT ITERATIONS WALKTHROUGH

Iteration 3:

Calculation:

x_3 = (1.330861 + 1)^{1/3} = (2.330861)^{1/3} \approx 1.325884

Iteration 4:

Calculation:

x_4 = (1.325884 + 1)^{1/3} = (2.325884)^{1/3} \approx 1.324942

STEP 05 • ITERATION SUMMARY TABLE

Iteration (k)	Input Guess (x_{k-1})	Output Next Guess (x_k)	Difference (\|x_k - x_{k-1}\|)
1	1.500000	1.357209	0.142791
2	1.357209	1.330861	0.026348
3	1.330861	1.325884	0.004977
4	1.325884	1.324942	0.000942
5	1.324942	1.324764	0.000178
6	1.324764	1.324730	0.000034

FINAL CONCLUSION

Approximated Fixed Point: x ≈ 1.3247

\text{Exact Root } \alpha \approx 1.324718, \quad (1.324718)^3 - (1.324718) - 1 = 0

Continuous self-feeding evaluation quickly stabilizes at the exact root 1.32472 because the slope derivative condition $|\phi'(x)| < 1$ holds!

LABORATORY

Interactive Live Calculator & Visualizer

FIXED-POINT SOLVER LABORATORY

Fixed-Point Iteration Engine

PreviousFalse Position Method

Unit 1•3 of 28

NextNewton-Raphson Method

∑

f(x)∫dxπ

NETZ Engine

Evaluating numerical algorithm

UNIT 1 • FINDING ROOTS OF EQUATIONS

Fixed Point Iteration
Recurrence Method

RECURRENCE RELATION & CONVERGENCECOBWEB STABILITY

x_{k} = \phi(x_{k-1})

|\phi'(x)| < 1 \quad (\text{Strict Convergence Condition})

1. Rewrite f(x) = 0Isolate

x

to create iteration form

x = \phi(x)

2. Test Slope DerivativeEnsure

|\phi'(x)| < 1

near the root for rapid convergence.

3. Feed Next GuessPlug

x_{k}

back into

\phi

until differences vanish.

Step-by-Step Worked Example

HAND-HOLDING TUTORIAL

OUR PROBLEM TO SOLVE

Find the root of $f(x) = x^3 - x - 1 = 0$ starting at initial guess $x_0 = 1.5$ .

STEP 01 • REWRITE & VERIFY CONVERGENCE

Isolate $x$ to define $\phi(x)$ :

\phi(x) = (x + 1)^{1/3}

\phi'(x) = \frac{1}{3(x+1)^{2/3}} \implies |\phi'(1.5)| = \frac{1}{3(2.5)^{2/3}} \approx 0.1809 < 1 \quad (\text{Strictly Converges!})

STEP 02 • COMPUTE FIRST ITERATION (x₁)

Plug initial guess $x_0 = 1.5$ into $\phi(x)$ :

x_1 = \phi(1.5) = (1.5 + 1)^{1/3} = (2.5)^{1/3} \approx 1.357209

STEP 03 • COMPUTE SECOND ITERATION (x₂)

Feed $x_1 = 1.357209$ back into $\phi(x)$ :

x_2 = \phi(1.357209) = (1.357209 + 1)^{1/3} = (2.357209)^{1/3} \approx 1.330861

STEP 04 • SUBSEQUENT ITERATIONS WALKTHROUGH

Iteration 3:

Calculation:

x_3 = (1.330861 + 1)^{1/3} = (2.330861)^{1/3} \approx 1.325884

Iteration 4:

Calculation:

x_4 = (1.325884 + 1)^{1/3} = (2.325884)^{1/3} \approx 1.324942

STEP 05 • ITERATION SUMMARY TABLE

Iteration (k)	Input Guess (x_{k-1})	Output Next Guess (x_k)	Difference (\|x_k - x_{k-1}\|)
1	1.500000	1.357209	0.142791
2	1.357209	1.330861	0.026348
3	1.330861	1.325884	0.004977
4	1.325884	1.324942	0.000942
5	1.324942	1.324764	0.000178
6	1.324764	1.324730	0.000034

FINAL CONCLUSION

Approximated Fixed Point: x ≈ 1.3247

\text{Exact Root } \alpha \approx 1.324718, \quad (1.324718)^3 - (1.324718) - 1 = 0

Continuous self-feeding evaluation quickly stabilizes at the exact root 1.32472 because the slope derivative condition $|\phi'(x)| < 1$ holds!

LABORATORY

Interactive Live Calculator & Visualizer

FIXED-POINT SOLVER LABORATORY

Fixed-Point Iteration Engine

PreviousFalse Position Method

Unit 1•3 of 28

NextNewton-Raphson Method

Fixed Point Iteration Recurrence Method

Step-by-Step Worked Example

Approximated Fixed Point: x ≈ 1.3247

Interactive Live Calculator & Visualizer

Fixed-Point Iteration Engine

NETZ Engine

Fixed Point Iteration Recurrence Method

Step-by-Step Worked Example

Approximated Fixed Point: x ≈ 1.3247

Interactive Live Calculator & Visualizer

Fixed-Point Iteration Engine

Fixed Point Iteration
Recurrence Method

Fixed Point Iteration
Recurrence Method