Mathematics-III (BT-401) - Unit 1 Short Notes

How unit 1 is examined

This unit covers root finding (Newton-Raphson, Regula-Falsi, bisection), finite differences and operators, and interpolation; interpolation, Newton-Raphson, Regula-Falsi and operator proofs carry most marks.

Solution of polynomial and transcendental equations

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Definition. A polynomial equation is $a_0x^n+\dots+a_n=0$; a transcendental equation contains trigonometric, exponential or logarithmic terms, such as $xe^x-3=0$.

Key points.

1. Such equations rarely have a closed-form root, so numerical methods give an approximate root.
2. By the intermediate value theorem, if $f(a)f(b)<0$ and $f$ is continuous, a root lies in $(a,b)$.
3. Bisection and Regula-Falsi are bracketing methods; Newton-Raphson is an open method needing $f'$.
4. <mark>A root is located by a sign change of $f(x)$ and then refined by iteration.</mark>

Bisection method

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Definition. ==Bisection repeatedly halves an interval $[a,b]$ with $f(a)f(b)<0$, taking the midpoint $c=\frac{a+b}{2}$ and keeping the half where the sign changes.==

Key points.

1. Start with $f(2)=-3<0$ and $f(3)=13>0$ for $f(x)=x^3-3x-5$, so a root lies in $(2,3)$.
2. If $f(a)f(c)<0$ set $b=c$, otherwise set $a=c$; the interval halves each step.
3. Successive midpoints for the paper's equation are 2.5, 2.25, 2.375, 2.3125, 2.28125, 2.2656, 2.2734, 2.2773, 2.2793, 2.2783, 2.2788, so the root is $\mathbf{2.279}$ to three decimals.
4. Convergence is slow but sure, gaining about one binary digit per step.

Asked: [7 marks] (May 2019) Find a root of $f(x)=x^3-3x-5$ by the bisection method, correct to three decimal places.

Newton-Raphson method

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Definition. <mark>Newton-Raphson improves a guess $x_n$ by replacing the curve with its tangent at $x_n$ and taking the tangent's x-intercept as $x_{n+1}$.</mark>

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

Key points.

1. The tangent at $(x_n,f(x_n))$ is $y-f(x_n)=f'(x_n)(x-x_n)$; setting $y=0$ gives the formula.
2. Choose $x_0$ by a sign change of $f$, preferably at the end where $|f|$ is smaller.
3. Convergence is quadratic, so correct digits roughly double each step.
4. It fails if $f'(x_n)=0$ or $x_0$ is far from the root; a sufficient condition is $|f\,f''|<f'^2$ near the root.
5. Stop when two successive iterates agree to the required decimals.
6. For $\sqrt N$ use $f=x^2-N$, giving $x_{n+1}=\frac12\left(x_n+\frac{N}{x_n}\right)$.

Example. $f=x^4-x-10$, $f'=4x^3-1$, $f(2)=4>0$, $f(1)<0$, take $x_0=2$.

Root = 1.8556 (1.856 to three decimals).

Results for the other papers (same method, $x_0$ in brackets):

• $x^3+2x^2+10x-20=0$, $f'=3x^2+4x+10$ ($x_0=1$): root 1.3688.
• $e^{2x}-e^x-2=0$, $f'=2e^{2x}-e^x$ ($x_0=1$): root 0.6931 ($=\ln 2$).
• $x^3-3x-5=0$, $f'=3x^2-3$ ($x_0=2$): 2.3333, 2.2806, 2.27902, root 2.2790.
• $\sqrt{12}$, $f=x^2-12$ ($x_0=3.5$): 3.464286, 3.464102, 3.4641.

Answer frame. Open with the formula and $f$, $f'$; find $x_0$ from a sign change; tabulate $n,x_n,f,f',x_{n+1}$; close with the root correct to the asked decimals.

Pitfall: Stopping while iterates still differ in the last asked decimal, or using degrees instead of radians for trig terms.

Asked: [7 marks] (Jun 2022, Nov 2022) Find the real root of $x^4-x-10=0$ (near $x=2$, three decimals) by Newton-Raphson. Asked: [7 marks] (May 2019) Solve $x^3+2x^2+10x-20=0$ by Newton-Raphson. Asked: [7 marks] (Dec 2020) Find the real root of $e^{2x}-e^x-2=0$ by Newton-Raphson. Asked: [7 marks] (Jun 2023) Find the real root of $x^3-3x-5=0$ to four decimals by Newton-Raphson. Asked: [7 marks] (Jun 2026) Apply Newton-Raphson to evaluate $\sqrt{12}$ approximately.

Regula-Falsi method

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Definition. <mark>Regula-Falsi (false position) draws the chord through $(a,f(a))$ and $(b,f(b))$ with $f(a)f(b)<0$ and takes its x-intercept as the next approximation.</mark>

Formula. $$x_2=\frac{a f(b)-b f(a)}{f(b)-f(a)}$$

Key points.

1. Bracket the root by a sign change, then compute $x_2$ from the formula.
2. Replace $b$ by $x_2$ if $f(a)f(x_2)<0$, otherwise replace $a$ by $x_2$, so the root stays bracketed.
3. Convergence is linear, faster than bisection but slower than Newton-Raphson.
4. It needs no derivative and always converges for a continuous $f$.

Example. $f=x\log_{10}x-1.2$, $f(2)=-0.59794$, $f(3)=0.23136$.

Root = 2.74065.

Other papers. $x^3-9x+1=0$: $f(2)=-9$, $f(3)=1$, iterates 2.9, 2.94156, 2.94278, 2.94282, root 2.9428. $xe^x-3=0$: $f(1)=-0.28$, $f(2)=11.78$, root 1.050 (1.0499).

Answer frame. Open with $f$ and the bracket; write the formula; tabulate $a,b,x_2,f(x_2)$; close with the root to the asked decimals.

Asked: [7 marks] (Jun 2020, Nov 2022, Jun 2023) Find the real root of $x\log_{10}x-1.2=0$ to five decimals by Regula-Falsi; also $xe^x-3=0$ to three decimals. Asked: [7 marks] (Jun 2020) Find a real root of $x^3-9x+1=0$ by the method of false position.

Finite differences

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Definition. ==For step $h$, $\Delta f(x)=f(x+h)-f(x)$ and $\Delta^n f=\Delta(\Delta^{n-1}f)$.==

Key points.

1. A polynomial of degree $n$ has constant $n$th difference $\Delta^nx^n=n!\,h^n$ and zero $(n+1)$th difference.
2. Build the forward difference table and stop when a column becomes constant; that column gives the degree.
3. Table $x=0..4$, $f=3,6,11,18,27$: $\Delta f=3,5,7,9$; $\Delta^2f=2,2,2$; $\Delta^3f=0$. Newton's formula with $u=x$ gives $f=3+3x+\frac{x(x-1)}{2}\cdot2$, so $\mathbf{f(x)=x^2+2x+3}$.
4. $\Delta^6(ax-1)(bx^2-1)(cx^3-1)$: the product has degree 6 with leading term $abc\,x^6$, so $\Delta^6=6!\,abc=\mathbf{720abc}$.
5. Proof of $\Delta^n\sin(ax+b)$: $\Delta\sin(ax+b)=2\sin\frac{ah}{2}\sin\left[ax+b+\frac{ah+\pi}{2}\right]$, using $\sin A-\sin B$. Applying $\Delta$ again multiplies by $2\sin\frac{ah}{2}$ and adds $\frac{ah+\pi}{2}$ to the phase, so by induction the result holds for $n$.

Answer frame. Open with the definition of $\Delta$; for the table draw the difference table and stop at the constant column; for the proof do $n=1$ then induction; close with the result.

Asked: [7 marks] (Dec 2020) From the table $x=0..4$, $f=3,6,11,18,27$ find $f(x)$ by finite differences. Asked: [7 marks] (Jun 2022) Prove $\Delta^n\sin(ax+b)=\left(2\sin\frac{ah}{2}\right)^n\sin\left[ax+b+n\frac{ah+\pi}{2}\right]$. Asked: [7 marks] (Jun 2023) Evaluate $\Delta^6(ax-1)(bx^2-1)(cx^3-1)$, $h=1$.

Relation between operators

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Definition. ==$Ef(x)=f(x+h)$, $\Delta=E-1$, $\nabla=1-E^{-1}$, $\delta=E^{1/2}-E^{-1/2}$, $\mu=\frac12\left(E^{1/2}+E^{-1/2}\right)$, and $I$ is the identity.==

Key points.

1. $E^{-1}f(x)=f(x-h)$, so $\nabla=1-E^{-1}$ and $E=1+\Delta=(1-\nabla)^{-1}$.
2. $\delta^2=E-2+E^{-1}$, hence $\mu^2=\frac14(E+2+E^{-1})=1+\frac{\delta^2}{4}$.
3. Since $\Delta-\nabla=E-2+E^{-1}$ and $\Delta\nabla=(E-1)(1-E^{-1})=E-2+E^{-1}$, we get $\Delta\nabla=\nabla\Delta=\delta^2=\Delta-\nabla$.
4. Derive $\Delta$ and $\nabla$ in terms of $\delta$: $\delta\mu=\frac12(E-E^{-1})$ and $\mu=\sqrt{1+\delta^2/4}$, so $$\frac{\delta^2}{2}+\delta\sqrt{1+\frac{\delta^2}{4}}=\frac{E-2+E^{-1}}{2}+\frac{E-E^{-1}}{2}=E-1=\Delta,$$ $$-\frac{\delta^2}{2}+\delta\sqrt{1+\frac{\delta^2}{4}}=\frac{-E+2-E^{-1}+E-E^{-1}}{2}=1-E^{-1}=\nabla.$$
5. Prove $(I+\Delta)(I-\nabla)$: $=E\cdot E^{-1}=I$, and expanding, $I+\Delta-\nabla-\Delta\nabla=I$ because $\Delta-\nabla=\Delta\nabla$. The printed right side $\Delta\nabla$ equals $\delta^2$, not $I$, so state the identity as $=I$ and show $\Delta-\nabla-\Delta\nabla=0$.
6. Prove $\frac{\Delta^2}{E}x^3=(E-2+E^{-1})x^3=(x+h)^3-2x^3+(x-h)^3=6xh^2$.
7. Prove $\Delta^n0^{n+1}=\frac{n(n+1)}{2}\Delta^n0^n$ ($h=1$, $O$ read as $0$): write $x^{n+1}=x^{(n+1)}+\frac{n(n+1)}{2}x^{(n)}+\dots$ in factorial powers $x^{(k)}=x(x-1)\cdots(x-k+1)$. Applying $\Delta^n$ at $x=0$ kills every term except $x^{(n)}$, and $\Delta^nx^{(n)}=n!=\Delta^n0^n$.

Answer frame. Open by defining $E,\Delta,\nabla,\delta,\mu$; write each side in $E$; simplify line by line; close with "hence proved".

Asked: [7 marks] (Dec 2020) Derive $\Delta=\frac{\delta^2}{2}+\delta\sqrt{1+\frac{\delta^2}{4}}$ and $\nabla=-\frac{\delta^2}{2}+\delta\sqrt{1+\frac{\delta^2}{4}}$. Asked: [7 marks] (Jun 2022) Prove $\Delta^nO^{n+1}=\frac{n(n+1)}{2}\Delta^nO^n$. Asked: [7 marks] (Nov 2022) Prove $(I+\Delta)(I-\nabla)\equiv\Delta\nabla$. Asked: [7 marks] (Jun 2026) Prove (i) $\frac{\Delta^2}{E}x^3=6xh^2$; (ii) $\mu^2=1+\frac{\delta^2}{4}$.

Interpolation using Newton's forward and backward difference formulae

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Definition. ==For equally spaced $x_0,x_0+h,\dots$ Newton's forward formula is $y_x=y_0+u\Delta y_0+\frac{u(u-1)}{2!}\Delta^2y_0+\frac{u(u-1)(u-2)}{3!}\Delta^3y_0+\cdots$ with $u=\frac{x-x_0}{h}$.==

Key points.

1. Derivation: $y_x=E^uy_0=(1+\Delta)^uy_0$, expanded by the binomial series.
2. Backward formula: $y_x=y_n+v\nabla y_n+\frac{v(v+1)}{2!}\nabla^2y_n+\cdots$, $v=\frac{x-x_n}{h}$.
3. Use forward near the start of the table and backward near the end.
4. Build the difference table first, then substitute $u$; the top row is used for forward.
5. Stop where differences become constant or negligible; the result is a polynomial fitted through all points.
6. Prove $u_0+\frac{u_1x}{1!}+\frac{u_2x^2}{2!}+\cdots=e^x\left[u_0+x\Delta u_0+\frac{x^2}{2!}\Delta^2u_0+\cdots\right]$: since $u_n=E^nu_0$, LHS $=e^{xE}u_0=e^{x(1+\Delta)}u_0=e^xe^{x\Delta}u_0$, and expanding $e^{x\Delta}$ gives the bracket.

Example. $\sin52^\circ$, $h=5$, $u=\frac{52-45}{5}=1.4$.

$y=0.7071+1.4(0.0589)+\frac{1.4(0.4)}{2}(-0.0057)+\frac{1.4(0.4)(-0.6)}{6}(-0.0007)$, so $\mathbf{\sin52^\circ=0.7880}$.

Other papers. Net premium (age 25): $u=1.25$, $\Delta=154,191,224$, $\Delta^2=37,33$, $\Delta^3=-4$ (units $10^{-5}$), giving 0.01625. $\log3375$: use $\log337.5$ with $x_0=310$, $u=2.75$ (the printed 3.5440680 is a misprint for 2.5440680), giving 2.52827, so 3.52827. Data $x=1,3,\dots,11$, $y=3,14,19,21,23,28$: $\Delta^3=3$ (constant), so with $u=\frac{x-1}{2}$ the cubic is $\frac{x^3-21x^2+159x-91}{16}$ and $\mathbf{f(2)=\frac{151}{16}=9.4375}$. Cubic through $(0,1),(1,0),(2,1),(3,10)$: $y=x^3-2x^2+1$, so $\mathbf{y(4)=33}$.

Answer frame. Open with the formula and $h$; draw the difference table; give $u$; substitute term by term; close with the value and a reasonableness check.

Asked: [7 marks] (Dec 2020, Jun 2020, Jun 2022) Use Newton's formula to find the net premium at age 25 from Age 20, 24, 28, 32 and premium 0.01427, 0.01581, 0.01772, 0.01996; also find $\log3375$ from $\log x$ at $x=310,\dots,360$. Asked: [7 marks] (May 2019, Jun 2026) Find a polynomial through $x=1,3,\dots,11$, $F=3,14,19,21,23,28$ and compute $f(2)$; fit a cubic through $(0,1),(1,0),(2,1),(3,10)$ and find $y(4)$. Asked: [7 marks] (Dec 2020) Prove $u_0+\frac{u_1x}{1!}+\frac{u_2x^2}{2!}+\cdots=e^x\left[u_0+x\Delta u_0+\frac{x^2}{2!}\Delta^2u_0+\cdots\right]$. Asked: [7 marks] (Dec 2020) Given $\sin45^\circ=0.7071$, $\sin50^\circ=0.7660$, $\sin55^\circ=0.8192$, $\sin60^\circ=0.8660$, find $\sin52^\circ$ by Newton's interpolation. Asked: [7 marks] (Jun 2020) Find $\sin52^\circ$ from the same data by any method of interpolation.

Interpolation with unequal intervals: Newton's divided difference

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Definition. ==For unequally spaced $x_i$, the divided difference is $f[x_0,x_1]=\frac{f(x_1)-f(x_0)}{x_1-x_0}$, and Newton's formula is $f(x)=f_0+(x-x_0)f[x_0,x_1]+(x-x_0)(x-x_1)f[x_0,x_1,x_2]+\cdots$.==

Key points.

1. Higher divided differences are $f[x_0,\dots,x_k]=\frac{f[x_1,\dots,x_k]-f[x_0,\dots,x_{k-1}]}{x_k-x_0}$.
2. They are symmetric in their arguments, and the $n$th divided difference of a degree-$n$ polynomial is constant.
3. The top diagonal of the table supplies the coefficients of the formula.
4. Example, $\log_{10}656$ from $x=654,658,659,661$, $y=2.8156,2.8182,2.8189,2.8202$: first differences $0.00065,0.00070,0.00065$; second $0.00001,-0.0000167$; third $-0.0000038$. Then $\log_{10}656=2.8156+2(0.00065)+(2)(-2)(0.00001)+(2)(-2)(-3)(-0.0000038)=\mathbf{2.8168}$.
5. For $x=4,5,7,10,11,13$, $F=48,100,294,900,1210,2028$: first $52,97,202,310,409$; second $15,21,27,33$; third $1,1,1$, so the cubic gives $\mathbf{f(8)=448}$, $\mathbf{f(15)=3150}$.
6. For $x=1,2,3,4,7$, $f=2,4,8,16,128$: coefficients $2,\,2,\,1,\,\frac13,\,\frac{11}{90}$ give $\mathbf{f(5)=\frac{494}{15}=32.93}$.

Answer frame. Open by noting the intervals are unequal; draw the divided difference table; write the formula with the top-row entries; substitute $x$; close with the value.

Asked: [7 marks] (May 2019, Nov 2022, Jun 2023) Using Newton's divided difference formula find $\log_{10}656$ from $\log_{10}654,658,659,661$; find $f(5)$ from the table $x=1,2,3,4,7$; find $f(8)$ and $f(15)$ from the table $x=4,5,7,10,11,13$.

Lagrange's formulae

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Definition. ==For any $n+1$ points, $f(x)=\sum_{i=0}^{n}f_i\prod_{j\ne i}\frac{x-x_j}{x_i-x_j}$.==

Key points.

1. It needs no difference table and works for equal or unequal spacing.
2. Each basis term equals 1 at its own $x_i$ and 0 at the others.
3. For $f(5)$ from $x=1,2,3,4,7$, $f=2,4,8,16,128$ the weights are $-\frac13,\frac85,-3,\frac83,\frac1{15}$, so $f(5)=-\frac23+\frac{32}5-24+\frac{128}3+\frac{128}{15}=\mathbf{\frac{494}{15}=32.93}$.
4. The whole sum must be recomputed if a point is added, unlike Newton's divided difference.

Asked: [7 marks] (Jun 2020) Apply Lagrange's formula to find $f(5)$, given $f(1)=2$, $f(2)=4$, $f(3)=8$, $f(4)=16$, $f(7)=128$.

Last-minute revision

• Root by sign change: $f(a)f(b)<0$.
• Bisection: $c=\frac{a+b}{2}$; root of $x^3-3x-5$ is 2.279.
• Newton-Raphson: $x_{n+1}=x_n-\frac{f}{f'}$, quadratic convergence; $x^4-x-10\to1.8556$, $\sqrt{12}\to3.4641$.
• Regula-Falsi: $x_2=\frac{af(b)-bf(a)}{f(b)-f(a)}$; $x\log_{10}x=1.2\to2.74065$.
• $\Delta=E-1$, $\nabla=1-E^{-1}$, $\delta=E^{1/2}-E^{-1/2}$, $\mu^2=1+\frac{\delta^2}{4}$.
• $\Delta^nx^n=n!h^n$; $\Delta^6=720abc$.
• Forward: $u=\frac{x-x_0}{h}$; backward: $v=\frac{x-x_n}{h}$.
• $\sin52^\circ=0.7880$; premium at 25 is 0.01625.
• Divided difference: $\log_{10}656=2.8168$.
• Lagrange and divided difference both give $f(5)=\frac{494}{15}$.

Memory hooks

• Newton-Raphson = tangent, Regula-Falsi = chord, bisection = halve.
• "Forward from the front, backward from the back."
• $E$ shifts, $\Delta=E-1$; $\nabla$ looks back, $\delta$ is central.
• Unequal gaps mean divided difference or Lagrange.
• Constant $n$th difference means a degree-$n$ polynomial.

Coverage checklist

• Solution of polynomial and transcendental equations: definition, bracketing.
• Bisection method: May 2019 ($x^3-3x-5$).
• Newton-Raphson method: $x^4-x-10$, cubic, $e^{2x}-e^x-2$, $x^3-3x-5$, $\sqrt{12}$.
• Regula-Falsi method: $x\log_{10}x-1.2$, $xe^x-3$, $x^3-9x+1$.
• Finite differences: table $f(x)$, $\Delta^n\sin(ax+b)$, $\Delta^6$ product.
• Relation between operators: $\Delta,\nabla$ in $\delta$, $\Delta^nO^{n+1}$, $(I+\Delta)(I-\nabla)$, $\frac{\Delta^2}{E}x^3$ and $\mu^2$.
• Interpolation using Newton's forward and backward difference formulae: premium, $\log3375$, polynomial and cubic, $e^x$ proof, $\sin52^\circ$ (two).
• Interpolation with unequal intervals: Newton's divided difference: $\log_{10}656$, $f(5)$, $f(8)$ and $f(15)$.
• Lagrange's formulae: $f(5)$.