UNIT 1 • FINDING ROOTS OF EQUATIONS

Newton-Raphson
Tangent Line Method

The Newton-Raphson Method is a second-order root-finding algorithm that uses the tangent slope derivative $f'(x_n)$ at the current point $x_n$ to slide down the curve directly toward the zero crossing with quadratic speed!

HOW NEWTON-RAPHSON WORKSQUADRATIC CONVERGENCE

x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}

1. Current State (x_n)Start with initial guess

x_0

2. Compute Derivative f'(x_n)Evaluate tangent slope derivative

f'(x_n) \neq 0

3. Tangent Intercept StepSubtract ratio

\frac{f(x_n)}{f'(x_n)}

to slide directly along tangent line to

x_{n+1}

Step-by-Step Worked Example

HAND-HOLDING TUTORIAL

OUR PROBLEM TO SOLVE

Find the root of $f(x) = x^2 - 5 = 0$ starting with initial guess $x_0 = 2$ .

STEP 01 • DERIVE SYMBOLIC DERIVATIVE

Find derivative $f'(x)$ and formulate iteration formula:

f(x) = x^2 - 5 \implies f'(x) = 2x

x_{n+1} = x_n - \frac{x_n^2 - 5}{2x_n} = \frac{2x_n^2 - x_n^2 + 5}{2x_n} = \frac{x_n^2 + 5}{2x_n}

STEP 02 • COMPUTE FIRST NEWTON STEP (x₁)

Plug initial guess $x_0 = 2$ into function and derivative:

f(2) = 2^2 - 5 = -1, \quad f'(2) = 2(2) = 4

x_1 = 2 - \frac{-1}{4} = 2 + 0.25 = 2.25

STEP 03 • COMPUTE SECOND NEWTON STEP (x₂)

Evaluate at $x_1 = 2.25$ :

f(2.25) = (2.25)^2 - 5 = 5.0625 - 5 = +0.0625, \quad f'(2.25) = 2(2.25) = 4.5

x_2 = 2.25 - \frac{0.0625}{4.5} = 2.25 - 0.01388889 = 2.236111

STEP 04 • SUBSEQUENT ITERATIONS WALKTHROUGH

Iteration 3:

Function Value:

f(2.236111) = (2.236111)^2 - 5 = +0.000193

Derivative Value:

f'(2.236111) = 4.472222

Next Guess:

x_3 = 2.236111 - \frac{0.000193}{4.472222} = 2.236068

STEP 05 • ITERATION SUMMARY TABLE

Iteration (k)	Current Guess (x_n)	f(x_n)	f'(x_n)	Next Guess (x_n+1)	Error (\|x_n+1 - x_n\|)
1	2.000000	-1.000000	4.000000	2.250000	0.250000
2	2.250000	+0.062500	4.500000	2.236111	0.013889
3	2.236111	+0.000193	4.472222	2.236068	0.000043
4	2.236068	0.000000	4.472136	2.236068	0.000000

FINAL CONCLUSION

Approximated Root: x ≈ 2.236068 (√5)

\text{Exact Root } \sqrt{5} \approx 2.236067977, \quad (2.236068)^2 - 5 = 0

Newton-Raphson exhibits quadratic convergence, effectively doubling the number of correct digits on every single iteration!

LABORATORY

Interactive Live Calculator & Visualizer

TANGENT SOLVER LABORATORY

Newton-Raphson Interactive Engine

PreviousIteration Method

Unit 1•4 of 28

NextGauss-Seidel Method

∑

f(x)∫dxπ

NETZ Engine

Evaluating numerical algorithm

UNIT 1 • FINDING ROOTS OF EQUATIONS

Newton-Raphson
Tangent Line Method

HOW NEWTON-RAPHSON WORKSQUADRATIC CONVERGENCE

x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}

1. Current State (x_n)Start with initial guess

x_0

2. Compute Derivative f'(x_n)Evaluate tangent slope derivative

f'(x_n) \neq 0

3. Tangent Intercept StepSubtract ratio

\frac{f(x_n)}{f'(x_n)}

to slide directly along tangent line to

x_{n+1}

Step-by-Step Worked Example

HAND-HOLDING TUTORIAL

OUR PROBLEM TO SOLVE

Find the root of $f(x) = x^2 - 5 = 0$ starting with initial guess $x_0 = 2$ .

STEP 01 • DERIVE SYMBOLIC DERIVATIVE

Find derivative $f'(x)$ and formulate iteration formula:

f(x) = x^2 - 5 \implies f'(x) = 2x

x_{n+1} = x_n - \frac{x_n^2 - 5}{2x_n} = \frac{2x_n^2 - x_n^2 + 5}{2x_n} = \frac{x_n^2 + 5}{2x_n}

STEP 02 • COMPUTE FIRST NEWTON STEP (x₁)

Plug initial guess $x_0 = 2$ into function and derivative:

f(2) = 2^2 - 5 = -1, \quad f'(2) = 2(2) = 4

x_1 = 2 - \frac{-1}{4} = 2 + 0.25 = 2.25

STEP 03 • COMPUTE SECOND NEWTON STEP (x₂)

Evaluate at $x_1 = 2.25$ :

f(2.25) = (2.25)^2 - 5 = 5.0625 - 5 = +0.0625, \quad f'(2.25) = 2(2.25) = 4.5

x_2 = 2.25 - \frac{0.0625}{4.5} = 2.25 - 0.01388889 = 2.236111

STEP 04 • SUBSEQUENT ITERATIONS WALKTHROUGH

Iteration 3:

Function Value:

f(2.236111) = (2.236111)^2 - 5 = +0.000193

Derivative Value:

f'(2.236111) = 4.472222

Next Guess:

x_3 = 2.236111 - \frac{0.000193}{4.472222} = 2.236068

STEP 05 • ITERATION SUMMARY TABLE

Iteration (k)	Current Guess (x_n)	f(x_n)	f'(x_n)	Next Guess (x_n+1)	Error (\|x_n+1 - x_n\|)
1	2.000000	-1.000000	4.000000	2.250000	0.250000
2	2.250000	+0.062500	4.500000	2.236111	0.013889
3	2.236111	+0.000193	4.472222	2.236068	0.000043
4	2.236068	0.000000	4.472136	2.236068	0.000000

FINAL CONCLUSION

Approximated Root: x ≈ 2.236068 (√5)

\text{Exact Root } \sqrt{5} \approx 2.236067977, \quad (2.236068)^2 - 5 = 0

Newton-Raphson exhibits quadratic convergence, effectively doubling the number of correct digits on every single iteration!

LABORATORY

Interactive Live Calculator & Visualizer

TANGENT SOLVER LABORATORY

Newton-Raphson Interactive Engine

PreviousIteration Method

Unit 1•4 of 28

NextGauss-Seidel Method

Newton-Raphson Tangent Line Method

Step-by-Step Worked Example

Approximated Root: x ≈ 2.236068 (√5)

Interactive Live Calculator & Visualizer

Newton-Raphson Interactive Engine

NETZ Engine

Newton-Raphson Tangent Line Method

Step-by-Step Worked Example

Approximated Root: x ≈ 2.236068 (√5)

Interactive Live Calculator & Visualizer

Newton-Raphson Interactive Engine

Newton-Raphson
Tangent Line Method

Newton-Raphson
Tangent Line Method