Almaz Khusnutdinov•Math enthusiast, love coding and mathematics
Published 8/24/2025
Newton's Method
Problem 1.1
Let f(x)=cosx−x. (a) Derive the Newton update. (b) With x0=0.7,compute x1,x2,x3. (c) Give r to 9 correct decimals.Solution:(a) f′(x)=−sinx−1,xn+1=xn−f′(xn)f(xn)=xn−−sinxn−1cosxn−xn.(b) x0=0.7:x1=0.7−−sin(0.7)−1cos(0.7)−0.7=0.7394364978480582,x2=0.7394364978480582−−sin(x1)−1cos(x1)−x1=0.7390851604651074,x3=0.7390851604651074−−sin(x2)−1cos(x2)−x2=0.7390851332151608.(c) The values stabilize quickly; with ∣x3−x2∣≈2.72×10−8,r≈0.739085133(9 correct decimals).Answer: xn+1=xn−−sinxn−1cosxn−xn,x1≈0.7394364978,x2≈0.7390851605,x3≈0.7390851332,r≈0.739085133.
Problem 1.2
Let f(x)=x3−5. (a) Derive and simplify the Newton iteration. (b) With x0=1.5,compute x1,x2,x3.(c) Estimate 35.Answer: xn+1=xn−3xn2xn3−5=32xn+xn25;x1≈1.7407407407,x2≈1.7105164618,x3≈1.7099761175;35≈1.7099761175.
Problem 1.3
Suppose f has a simple root r (i.e., f(r)=0,f′(r)=0) and f′′ is continuous near r.Show Newton’s method is locally quadratic.Answer (outline): 1.Taylor expand at r:f(x)=f′(r)(x−r)+21f′′(r)(x−r)2+o((x−r)2).2.Write the Newton map N(x)=x−f′(x)f(x)and substitute the expansion with x=r+e.3.Obtain the error recursion e+=N(r+e)−r=2f′(r)f′′(r)e2+o(e2).4.Conclude ∣en+1∣≤Cen2for small en,hence quadratic convergence.
Problem 1.4
Let f(x)=(x−1)2(1+x).(a) Write the modified Newton step for multiplicity m=2.(b) With x0=3,compute x1,x2.Answer: xn+1=xn−f′(xn)2f(xn),f′(x)=2(x−1)(1+x)+(x−1)2.x1=3−2⋅2⋅4+222(22⋅4)=3−2032=1.4,x2=1.4−2⋅0.4⋅2.4+0.422(0.42⋅2.4)≈1.0307692308.
Problem 1.5
To compute a−1/2,apply Newton to f(x)=x−2−a.(a) Derive xn+1=21xn(3−axn2).(b) For a=10,x0=0.3,compute x1,x2,x3.Answer: xn+1=21xn(3−axn2);x1=0.3150000000,x2=0.3162206250,x3≈0.3162277658(≈1/10).
Differentials
Linearization. Affine Functions. Approximations.
Problem 2.1
Use differentials to approximate 99.Take f(x)=x,x0=100,Δx=−1.Provide the estimate and justify with dy=f′(x0)dx.Solution: f′(x)=2x1⇒f′(100)=201.Set dx=Δx=−1⇒dy=f′(100)dx=201⋅(−1)=−0.05.Linear estimate: f(100+Δx)≈f(100)+dy=10−0.05=9.95.99≈9.95.
Problem 2.2
Use differentials to approximate (1.01)12.Let f(x)=(1+x)12,x0=0,Δx=0.01.Compute f(x0)+f′(x0)Δx.Answer: f(0)=1,f′(x)=12(1+x)11⇒f′(0)=12;(1.01)12≈1+12⋅0.01=1.12.
Problem 2.3
A length L is measured as 25.0cm with possible error ∣ΔL∣≤0.2cm.For A(L)=L,use differentials to estimate the maximum absolute error ∣ΔA∣and the relative error A∣ΔA∣.Answer: dA=2L1dL⇒∣ΔA∣≈2251⋅0.2=101⋅0.2=0.02,A∣ΔA∣≈250.02=50.02=0.004(0.4%).
Problem 2.4
Use differentials to estimate e0.05cos(0.05).Let f(x)=excosx,x0=0,Δx=0.05.Compute f(x0)+f′(x0)Δx.Answer: f(0)=1,f′(x)=ex(cosx−sinx)⇒f′(0)=1;e0.05cos(0.05)≈1+1⋅0.05=1.05.
Problem 2.5
For y=x3+x,at x0=2 a change Δy=0.10 occurs. Use differentials to estimate the corresponding Δx.Give the approximate new x value.Answer: dy=(3x2+1)dx⇒dx≈3⋅22+1Δy=130.10≈0.0076923077,x≈2+dx≈2.0076923077.
Linear Predictions. Differentials.
Problem 3.1
For y=ln(1+2x),at x0=1 a change Δy=−0.03 occurs. Use differentials to estimate the corresponding Δx.Give the approximate new x value.Answer: Δx≈−0.045,x≈0.955.
Problem 3.2
For y=x+2x,at x0=9 a change Δy=0.50 occurs. Use differentials to estimate the corresponding Δx.Give the approximate new x value.Answer: Δx≈133≈0.2307692308,x≈9+133≈9.2307692308.
Problem 3.3
For y=1+xx,at x0=4 a change Δy=−0.02 occurs. Use differentials to estimate the corresponding Δx.Give the approximate new x value.Answer: Δx≈−0.5,x≈3.5.
