Skip to content
Oscar Song Tutoring
9.3

Solve equations using the Newton-Raphson method and other recurrence relations of the form xₙ₊₁ = g(xₙ); understand how such methods can fail.

Draft — not yet indexed

Newton-Raphson method

Worked answers and methods for 9.3 on Edexcel A-level Maths 9MA0.

Explanation

  • Newton-Raphson replaces the curve locally by its tangent, giving xn+1=xn−f(xn)f′(xn)x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)} for a root of f(x)=0f(x)=0.
  • Differentiate first, substitute the current approximation into both ff and f′f', and retain extra calculator digits until the requested final accuracy.
  • For instance, applying Newton-Raphson to f(x)=x2−af(x)=x^2-a gives xn+1=12(xn+axn)x_{n+1}=\frac12\left(x_n+\frac{a}{x_n}\right).
  • The method can fail when f′(xn)=0f'(x_n)=0, when a tangent sends the next value away from the desired root, or when the values enter a cycle; a common error is to continue without checking the iterates.
  • Set f(x)=x3+x−4f(x)=x^3+x-4, so f′(x)=3x2+1f'(x)=3x^2+1.
Newton–Raphson uses the tangent at xnx_n; its intercept with the xx-axis is the next iterate xn+1x_{n+1}.

Worked example

Derive a Newton-Raphson recurrence for x3+x−4=0x^3+x-4=0. Starting with x0=1.3x_0=1.3, find x1x_1 and x2x_2 to 55 decimal places.

  1. 1.Set f(x)=x3+x−4f(x)=x^3+x-4, so f′(x)=3x2+1f'(x)=3x^2+1.
  2. 2.Substitution in the Newton-Raphson formula gives xn+1=xn−xn3+xn−43xn2+1x_{n+1}=x_n-\frac{x_n^3+x_n-4}{3x_n^2+1}.
  3. 3.Using x0=1.3x_0=1.3 gives x1=1.382866…x_1=1.382866\ldots, and substituting this unrounded value gives x2=1.378806…x_2=1.378806\ldots.

Answer: xn+1=xn−xn3+xn−43xn2+1x_{n+1}=x_n-\frac{x_n^3+x_n-4}{3x_n^2+1}.; x1=1.38287x_1=1.38287 and x2=1.37881x_2=1.37881.

Common mistakes

  • Don't choose a starting value where f′(x0)=0f'(x_0)=0, making the first Newton step undefined.
  • Don't use the Newton formula with the derivative evaluated at the new iterate rather than the current one.

Exam tip

Write the Newton recurrence explicitly, retain guard digits between iterations and round only the reported values.

Worked practice

Q1
Tier 1 · Easy

1.

Use one Newton-Raphson step on f(x)=x2−7f(x)=x^2-7 starting from x0=2.5x_0=2.5. Find x1x_1.

(2)

(Total for Question 1 is 2 marks)

Mark scheme

Mark scheme for question 1
QuestionSchemeMarks
1
  • x1=2.65x_1=2.65
2
Notes
Here f′(x)=2xf'(x)=2x. Therefore x1=2.5−2.52−72(2.5)=2.5−−0.755=2.65x_1=2.5-\frac{2.5^2-7}{2(2.5)}=2.5-\frac{-0.75}{5}=2.65.

(2 marks)

Q2
Tier 2 · Standard

2.

Newton-Raphson is applied to f(x)=x3−2x+2f(x)=x^3-2x+2 starting from x0=0x_0=0. Calculate x1x_1, x2x_2 and x3x_3, and explain why the method fails to converge to a root from this starting value.

(4)

(Total for Question 2 is 4 marks)

Mark scheme

Mark scheme for question 2
QuestionSchemeMarks
2
  • x1=1x_1=1, x2=0x_2=0, x3=1x_3=1
  • The iteration enters the cycle 0,1,0,1,…0,1,0,1,\ldots.
4
Notes
Here f′(x)=3x2−2f'(x)=3x^2-2. From x0=0x_0=0, x1=0−f(0)/f′(0)=0−2/(−2)=1x_1=0-f(0)/f'(0)=0-2/(-2)=1. Then x2=1−f(1)/f′(1)=1−1/1=0x_2=1-f(1)/f'(1)=1-1/1=0, so x3=1x_3=1 again. The values repeat in a two-cycle and neither 00 nor 11 is a root, so this starting value does not produce convergence.

(4 marks)

Q3
Tier 3 · Hard

3.

Newton-Raphson is applied to f(x)=x2−af(x)=x^2-a, where a>0a>0, starting from x0=4x_0=4. The first iterate is x1=134x_1=\dfrac{13}{4}. Find aa. Then calculate x2x_2 exactly and x3x_3 to six decimal places, and hence give the positive root of f(x)=0f(x)=0 correct to four decimal places.

(5)

(Total for Question 3 is 5 marks)

Mark scheme

Mark scheme for question 3
QuestionSchemeMarks
3
  • a=10a=10.
  • x2=329104x_2=\dfrac{329}{104} and x3=3.162278x_3=3.162278 to six decimal places.
  • The positive root is 3.16233.1623 correct to four decimal places.
5
Notes
For f(x)=x2−af(x)=x^2-a, Newton-Raphson gives xn+1=12(xn+a/xn)x_{n+1}=\frac12(x_n+a/x_n). Since x0=4x_0=4 and x1=13/4x_1=13/4, 13/4=12(4+a/4)13/4=\frac12(4+a/4), so 13/2=4+a/413/2=4+a/4 and a=10a=10. Then x2=12(13/4+40/13)=329/104x_2=\frac12(13/4+40/13)=329/104. One more iteration gives x3=12(329/104+1040/329)=216401/68432=3.162277881…x_3=\frac12(329/104+1040/329)=216401/68432=3.162277881\ldots. Hence the positive root is 10=3.162277660…=3.1623\sqrt{10}=3.162277660\ldots=3.1623 correct to four decimal places.

(5 marks)

Q4
Tier 1 · Easy

4.

Use one Newton-Raphson step on f(x)=ex−3f(x)=e^x-3, starting with x0=1x_0=1. Give x1x_1 to 44 decimal places.

(2)

(Total for Question 4 is 2 marks)

Mark scheme

Mark scheme for question 4
QuestionSchemeMarks
4
  • x1=1.1036x_1=1.1036
2
Notes
Since f′(x)=exf'(x)=e^x, Newton-Raphson gives x1=1−(e−3)/e=3/e=1.103638…x_1=1-(e-3)/e=3/e=1.103638\ldots. Hence x1=1.1036x_1=1.1036 to 44 decimal places.

(2 marks)

Q5
Tier 2 · Standard

5.

Derive a Newton-Raphson recurrence for x3+x2−6=0x^3+x^2-6=0. Starting with x0=1.5x_0=1.5, find x1x_1 and x2x_2, giving each answer to 44 decimal places.

(4)

(Total for Question 5 is 4 marks)

Mark scheme

Mark scheme for question 5
QuestionSchemeMarks
5
  • xn+1=xn−xn3+xn2−63xn2+2xnx_{n+1}=x_n-\dfrac{x_n^3+x_n^2-6}{3x_n^2+2x_n}; x1=1.5385x_1=1.5385 and x2=1.5377x_2=1.5377.
4
Notes
Take f(x)=x3+x2−6f(x)=x^3+x^2-6, so f′(x)=3x2+2xf'(x)=3x^2+2x. Substitution in xn+1=xn−f(xn)/f′(xn)x_{n+1}=x_n-f(x_n)/f'(x_n) gives the stated recurrence. With x0=1.5x_0=1.5, x1=1.538461…x_1=1.538461\ldots; substituting this unrounded value gives x2=1.537656…x_2=1.537656\ldots. Thus the requested values are 1.53851.5385 and 1.53771.5377.

(4 marks)

Q6
Tier 3 · Hard

6.

Newton-Raphson is to be used to solve x3−3x+3=0x^3-3x+3=0. Explain why x0=1x_0=1 cannot be used. Then start with x0=−2x_0=-2 and calculate x1x_1 and x2x_2, giving each value to 55 decimal places.

(5)

(Total for Question 6 is 5 marks)

Mark scheme

Mark scheme for question 6
QuestionSchemeMarks
6
  • x0=1x_0=1 fails because f′(1)=0f'(1)=0, so the Newton step is undefined.
  • Starting from −2-2, x1=−2.11111x_1=-2.11111 and x2=−2.10384x_2=-2.10384.
5
Notes
Here f′(x)=3x2−3f'(x)=3x^2-3 and xn+1=xn−(xn3−3xn+3)/(3xn2−3)x_{n+1}=x_n-(x_n^3-3x_n+3)/(3x_n^2-3). At x0=1x_0=1 the denominator is zero, so no Newton iterate exists. From x0=−2x_0=-2, f(−2)=1f(-2)=1 and f′(−2)=9f'(-2)=9, giving x1=−2−1/9=−2.111111…x_1=-2-1/9=-2.111111\ldots. Substitution of this unrounded value gives x2=−2.103835…x_2=-2.103835\ldots, hence the stated rounded values.

(5 marks)

Q7
Tier 2 · Standard

7.

Newton-Raphson is used to solve ln⁡x+x−3=0\ln x+x-3=0. Derive the recurrence and, starting with x0=2x_0=2, find x1x_1 and x2x_2, giving each value to 55 decimal places. Use unrounded values in your working.

(4)

(Total for Question 7 is 4 marks)

Mark scheme

Mark scheme for question 7
QuestionSchemeMarks
7
  • xn+1=xn−ln⁡xn+xn−31+1/xnx_{n+1}=x_n-\dfrac{\ln x_n+x_n-3}{1+1/x_n}; x1=2.20457x_1=2.20457 and x2=2.20794x_2=2.20794.
4
Notes
For f(x)=ln⁡x+x−3f(x)=\ln x+x-3, f′(x)=1/x+1f'(x)=1/x+1. Substitution in the Newton-Raphson formula gives xn+1=xn−(ln⁡xn+xn−3)/(1+1/xn)x_{n+1}=x_n-(\ln x_n+x_n-3)/(1+1/x_n). Starting from x0=2x_0=2 gives x1=2.204568546…x_1=2.204568546\ldots. Using this unrounded value gives x2=2.207939227…x_2=2.207939227\ldots, so the requested iterates are 2.204572.20457 and 2.207942.20794.

(4 marks)

Q8
Tier 3 · Hard

8.

Consider applying Newton-Raphson to f(x)=x3f(x)=\sqrt[3]{x}, whose only real root is 00. For xn≠0x_n\ne0, derive the recurrence in its simplest form. Starting with x0=18x_0=\dfrac18, find x1x_1, x2x_2 and x3x_3, and explain why the method does not converge to the root.

(4)

(Total for Question 8 is 4 marks)

Mark scheme

Mark scheme for question 8
QuestionSchemeMarks
8
  • xn+1=−2xnx_{n+1}=-2x_n; x1=−14x_1=-\dfrac14, x2=12x_2=\dfrac12 and x3=−1x_3=-1.
  • The iterates alternate in sign while their distance from 00 doubles, so they diverge.
4
Notes
For non-zero xx, f′(x)=13x−2/3f'(x)=\frac13x^{-2/3}. Newton-Raphson therefore gives xn+1=xn−xn1/3/(13xn−2/3)=xn−3xn=−2xnx_{n+1}=x_n-x_n^{1/3}/(\frac13x_n^{-2/3})=x_n-3x_n=-2x_n. From x0=1/8x_0=1/8, the next values are −1/4-1/4, 1/21/2 and −1-1. Their signs alternate and their magnitudes double, so the sequence moves away from the root 00 rather than converging to it.

(4 marks)

Q9
Tier 3 · Hard

9.

Apply Newton-Raphson to cos⁡x=x\cos x=x by first deriving a recurrence. From x0=0.7x_0=0.7, obtain the first three iterates and report each to 44 decimal places, retaining full calculator precision throughout. Verify by a change of sign between 0.739050.73905 and 0.739150.73915 that the root is 0.73910.7391 to 44 decimal places.

(6)

(Total for Question 9 is 6 marks)

Mark scheme

Mark scheme for question 9
QuestionSchemeMarks
9
  • xn+1=xn+cos⁡xn−xnsin⁡xn+1x_{n+1}=x_n+\dfrac{\cos x_n-x_n}{\sin x_n+1}.
  • x1=0.7394x_1=0.7394, x2=0.7391x_2=0.7391 and x3=0.7391x_3=0.7391.
  • cos⁡(0.73905)−0.73905>0\cos(0.73905)-0.73905>0 and cos⁡(0.73915)−0.73915<0\cos(0.73915)-0.73915<0, so the root is 0.73910.7391 to 44 decimal places.
6
Notes
Take f(x)=cos⁡x−xf(x)=\cos x-x, so f′(x)=−sin⁡x−1f'(x)=-\sin x-1. Newton-Raphson gives xn+1=xn−f(xn)/f′(xn)=xn+(cos⁡xn−xn)/(sin⁡xn+1)x_{n+1}=x_n-f(x_n)/f'(x_n)=x_n+(\cos x_n-x_n)/(\sin x_n+1). From x0=0.7x_0=0.7, unrounded calculation gives x1=0.739436497…x_1=0.739436497\ldots, x2=0.739085160…x_2=0.739085160\ldots and x3=0.739085133…x_3=0.739085133\ldots. Also f(0.73905)=0.0000587989…>0f(0.73905)=0.0000587989\ldots>0 and f(0.73915)=−0.0001085633…<0f(0.73915)=-0.0001085633\ldots<0. Since f′(x)<0f'(x)<0 on this interval, the root is unique there and the bracket pins it as 0.73910.7391 to 44 decimal places.

(6 marks)

Q10
Tier 3 · Hard

10.

Newton-Raphson is to be applied to 1x−15=0\dfrac1x-\dfrac15=0. Derive the recurrence in polynomial form. Starting with x0=1x_0=1, find x1x_1 and x2x_2 exactly. Then show that the starting value x0=10x_0=10 makes the method fail after one step, and explain the failure.

(5)

(Total for Question 10 is 5 marks)

Mark scheme

Mark scheme for question 10
QuestionSchemeMarks
10
  • xn+1=2xn−xn25x_{n+1}=2x_n-\dfrac{x_n^2}{5}.
  • For x0=1x_0=1, x1=95x_1=\dfrac95 and x2=369125x_2=\dfrac{369}{125}.
  • For x0=10x_0=10, x1=0x_1=0, where the original function and the next Newton-Raphson step are undefined.
5
Notes
With f(x)=1/x−1/5f(x)=1/x-1/5, f′(x)=−1/x2f'(x)=-1/x^2. Therefore xn+1=xn−(1/xn−1/5)/(−1/xn2)=2xn−xn2/5x_{n+1}=x_n-(1/x_n-1/5)/(-1/x_n^2)=2x_n-x_n^2/5. From x0=1x_0=1, this gives x1=2−1/5=9/5x_1=2-1/5=9/5 and x2=18/5−81/125=369/125x_2=18/5-81/125=369/125. If x0=10x_0=10, then x1=20−100/5=0x_1=20-100/5=0. Since f(0)f(0) is undefined, the method cannot form another tangent step even though the equation itself has the root x=5x=5.

(5 marks)

Verified exam appearances

We have not yet indexed a verified real-paper appearance for 9.3. Browse the Edexcel A-level Maths 9MA0 past papers directly.

Other points in 9 Numerical methods

Want help turning this into marks?

Bring 9.3 or any tricky specification point, and we can work through the method and exam wording together.

Book a free intro call