riaology
Unit 3 · Lesson 7 of 7

A tangent can guide a root search.

A numerical solver predicts where a local tangent hits zero, then repeats the prediction from that new location.

Change the input. Follow the mathematics.

What you'll be able to do
  • Set up the model and calculate its iterate after the chosen number of steps.
  • Connect the equation to the changing visual.
  • Explain the assumptions and identify a misleading interpretation.
Predict before exploring

What will change if you increase initial guess? Predict the direction, then explain the response.

A model you can changeMove one control at a time.
Read the modelxn+1=xn−f(xn)f′(xn)=12(xn+2xn)x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}=\frac12\left(x_n+\frac2{x_n}\right)

Newton iteration for the positive square root of 2. Initial guess is positive so zero-derivative failure is avoided in this example.

Input x versus x²−2. Function, Last tangent. Selected iterate after the chosen number of steps: 1.5 . The tangent intercept supplies the next guess. Newton can fail for other functions or starting values; convergence here is a property of this positive-root example.-305.52144x²−2Input x
FunctionLast tangent · dashed

iterate after the chosen number of stepsx1=1.5,∣x1−2∣=0.0857864x_1=1.5,\quad|x_1-\sqrt2|=0.0857864

The tangent intercept supplies the next guess. Newton can fail for other functions or starting values; convergence here is a property of this positive-root example.

Assumptions, units, and limits

Newton iteration for the positive square root of 2. Initial guess is positive so zero-derivative failure is avoided in this example.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

01

From the situation to an equation

Newton's iteration subtracts residual divided by derivative. The derivative must be nonzero and the next iterate must remain in an appropriate domain. A small residual and stable iterate are useful checks, but convergence is not guaranteed for every starting point.

xn+1=xn−f(xn)f′(xn)=12(xn+2xn)x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}=\frac12\left(x_n+\frac2{x_n}\right)
02

Keep the reasoning visible

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

The tangent intercept supplies the next guess. Newton can fail for other functions or starting values; convergence here is a property of this positive-root example.

03

What makes this model useful?

Newton iteration for the positive square root of 2. Initial guess is positive so zero-derivative failure is avoided in this example.

Newton can diverge, cycle, or leave the domain. The positive square-root example is deliberately well behaved.

A worked example

Calculate the starting scenario

  1. Use initial guess = 1 ; iterations = 1 .
  2. x1=1.5,∣x1−2∣=0.0857864x_1=1.5,\quad|x_1-\sqrt2|=0.0857864
  3. The tangent intercept supplies the next guess. Newton can fail for other functions or starting values; convergence here is a property of this positive-root example. The checked quantity is iterate after the chosen number of steps.
Make the idea your own

Calculate. Compare. Explain.

Numeric checks allow 1% rounding error. Values near zero use a small absolute tolerance. Work through the setup before opening a solution.

Calculate and explain

Rebuild the starting calculation

Calculate the iterate after the chosen number of steps using initial guess 1 , iterations 1 . State the meaning of its sign and units.

Checked quantity: iterate after the chosen number of steps · dimensionless. Follow the definition for finite-cutoff and numerical-model questions.

Hint

Newton's iteration subtracts residual divided by derivative. The derivative must be nonzero and the next iterate must remain in an appropriate domain. A small residual and stable iterate are useful checks, but convergence is not guaranteed for every starting point.

Explore the visual solution
Read the modelxn+1=xn−f(xn)f′(xn)=12(xn+2xn)x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}=\frac12\left(x_n+\frac2{x_n}\right)

Newton iteration for the positive square root of 2. Initial guess is positive so zero-derivative failure is avoided in this example.

Input x versus x²−2. Function, Last tangent. Selected iterate after the chosen number of steps: 1.5 . The tangent intercept supplies the next guess. Newton can fail for other functions or starting values; convergence here is a property of this positive-root example.-305.52144x²−2Input x
FunctionLast tangent · dashed

iterate after the chosen number of stepsx1=1.5,∣x1−2∣=0.0857864x_1=1.5,\quad|x_1-\sqrt2|=0.0857864

The tangent intercept supplies the next guess. Newton can fail for other functions or starting values; convergence here is a property of this positive-root example.

Assumptions, units, and limits

Newton iteration for the positive square root of 2. Initial guess is positive so zero-derivative failure is avoided in this example.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

The controls explore variations. These written steps retain the original question's inputs.

Newton's iteration subtracts residual divided by derivative. The derivative must be nonzero and the next iterate must remain in an appropriate domain. A small residual and stable iterate are useful checks, but convergence is not guaranteed for every starting point.

xn+1=xn−f(xn)f′(xn)=12(xn+2xn)x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}=\frac12\left(x_n+\frac2{x_n}\right)
Calculate and explain

Apply the idea to a new scenario

Starting at x₀=2, perform two Newton steps for x2−2x^2-2 and report x₂.

Checked quantity: iterate after the chosen number of steps · dimensionless. Follow the definition for finite-cutoff and numerical-model questions.

Hint

Newton's iteration subtracts residual divided by derivative. The derivative must be nonzero and the next iterate must remain in an appropriate domain. A small residual and stable iterate are useful checks, but convergence is not guaranteed for every starting point.

Explore the visual solution
Read the modelxn+1=xn−f(xn)f′(xn)=12(xn+2xn)x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}=\frac12\left(x_n+\frac2{x_n}\right)

Newton iteration for the positive square root of 2. Initial guess is positive so zero-derivative failure is avoided in this example.

Input x versus x²−2. Function, Last tangent. Selected iterate after the chosen number of steps: 1.41667 . The tangent intercept supplies the next guess. Newton can fail for other functions or starting values; convergence here is a property of this positive-root example.-4.2504.8752144x²−2Input x
FunctionLast tangent · dashed

iterate after the chosen number of stepsx2=1.41667,∣x2−2∣=0.0024531x_2=1.41667,\quad|x_2-\sqrt2|=0.0024531

The tangent intercept supplies the next guess. Newton can fail for other functions or starting values; convergence here is a property of this positive-root example.

Assumptions, units, and limits

Newton iteration for the positive square root of 2. Initial guess is positive so zero-derivative failure is avoided in this example.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

The controls explore variations. These written steps retain the original question's inputs.

Newton's iteration subtracts residual divided by derivative. The derivative must be nonzero and the next iterate must remain in an appropriate domain. A small residual and stable iterate are useful checks, but convergence is not guaranteed for every starting point.

xn+1=xn−f(xn)f′(xn)=12(xn+2xn)x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}=\frac12\left(x_n+\frac2{x_n}\right)
Challenge the interpretation

Which explanation survives the model?

Choose the statement you can justify, then use the visual to explain why the others fail.

Test it with the visual
Read the modelxn+1=xn−f(xn)f′(xn)=12(xn+2xn)x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}=\frac12\left(x_n+\frac2{x_n}\right)

Newton iteration for the positive square root of 2. Initial guess is positive so zero-derivative failure is avoided in this example.

Input x versus x²−2. Function, Last tangent. Selected iterate after the chosen number of steps: 1.5 . The tangent intercept supplies the next guess. Newton can fail for other functions or starting values; convergence here is a property of this positive-root example.-305.52144x²−2Input x
FunctionLast tangent · dashed

iterate after the chosen number of stepsx1=1.5,∣x1−2∣=0.0857864x_1=1.5,\quad|x_1-\sqrt2|=0.0857864

The tangent intercept supplies the next guess. Newton can fail for other functions or starting values; convergence here is a property of this positive-root example.

Assumptions, units, and limits

Newton iteration for the positive square root of 2. Initial guess is positive so zero-derivative failure is avoided in this example.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

The reasoning. Newton's iteration subtracts residual divided by derivative. The derivative must be nonzero and the next iterate must remain in an appropriate domain. A small residual and stable iterate are useful checks, but convergence is not guaranteed for every starting point.

Your learning path

Can you explain the connection?

Mark complete when you can set up the calculation and explain what the visual is showing. Opening a page never completes it automatically.

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