Mathematics-III (BT-301) Unit 1 Notes - RGPV

How unit 1 is examined

This unit covers root-finding (bisection, Newton-Raphson, Regula-Falsi) and interpolation; Newton-Raphson, finite differences and the interpolation formulae carry the marks, and every past question is a 7-mark numerical.

Solution of polynomial and transcendental equations

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Definition. A polynomial equation has only powers of $x$; a transcendental equation contains $\sin x$, $\cos x$, $e^x$, $\log x$ and similar functions. Such equations have no closed-form root, so we find the root by successive approximation.

Key points.

The iteration method rewrites $f(x)=0$ as $x=\phi(x)$ and repeats $x_{n+1}=\phi(x_n)$.
It converges only if $|\phi'(x)|<1$ near the root.
Start from an $x_0$ where $f$ changes sign, and stop when two successive values agree to the required decimals.
Keep the calculator in radians for $\sin$ and $\cos$.

Example. $3x=\cos x+1$: take $\phi(x)=\tfrac13(\cos x+1)$, so $|\phi'|=\tfrac{|\sin x|}{3}<1$. With $x_0=0.5$: $0.6259,\ 0.6035,\ 0.6078,\ 0.6070,\ 0.6071$. Root $\approx 0.6071$.

Asked: [7 marks] (Nov 2022) Find a positive root of the equation by iteration method $3x=\cos x+1$.

Bisection method

Definition. If $f$ is continuous and $f(a)f(b)<0$, a root lies in $(a,b)$; bisection halves the interval repeatedly, keeping the half where the sign changes.

Key points.

Take the midpoint $c=\dfrac{a+b}{2}$ and evaluate $f(c)$.
If $f(a)f(c)<0$ the root is in $(a,c)$, so set $b=c$; otherwise set $a=c$.
The interval halves each step, so the method always converges, but only linearly and slowly.
Stop when $|b-a|$ is below the required tolerance.

Example. $f(x)=x^3-9x+1$, $f(2)=-9<0$, $f(4)=29>0$.

n	a	b	c	f(c)
1	2	4	3	+1
2	2	3	2.5	-5.875
3	2.5	3	2.75	-2.953
4	2.75	3	2.875	-1.111
5	2.875	3	2.9375	-0.090
6	2.9375	3	2.96875	+0.446

The root lies in (2.9375, 2.96875), about 2.94 after these steps.

Asked: [7 marks] (Jun 2023) Find the root of $x^3-9x+1=0$ between $x=2$ and $x=4$ by the method of bisection.

Newton-Raphson method

Definition. ==Newton-Raphson replaces the curve by its tangent at $x_n$ and takes the point where the tangent cuts the $x$-axis as the next approximation: $x_{n+1}=x_n-\dfrac{f(x_n)}{f'(x_n)}$.==

Key points.

Derivation: the tangent at $(x_n,f(x_n))$ is $y-f(x_n)=f'(x_n)(x-x_n)$; putting $y=0$ gives $x_{n+1}=x_n-f(x_n)/f'(x_n)$.
Choose $x_0$ near the root, where $f$ changes sign, and ensure $f'(x_0)\neq 0$.
Iterate until two successive values agree to the required decimals.
The method has quadratic convergence: $|e_{n+1}|\le c\,|e_n|^2$, so the number of correct digits roughly doubles each step.
It needs $f'(x)$ at every step, and it may diverge if $x_0$ is far from the root or $f'$ is small.
It is faster than bisection and Regula-Falsi, which converge only linearly.

Example. $x^3-2x-5=0$, $x_0=2$: $f(2)=-1$, $f'(x)=3x^2-2$, $f'(2)=10$.

n	$x_n$	$f(x_n)$	$f'(x_n)$	$x_{n+1}$
0	2	-1	10	2.1
1	2.1	0.061	11.23	2.0946
2	2.0946	0.00019	11.1616	2.0946

Next approximation $x_1=2.10$; root $=2.095$ (3 decimals).

Second example. $3x=\cos x+1$: $f=3x-\cos x-1$, $f'=3+\sin x$, $f(0)<0<f(1)$, take $x_0=0.5$: $x_1=0.6085$, $x_2=0.6071$, $x_3=0.6071$. Root $=0.6071$.

Answer frame. Numerical: write $f$ and $f'$, locate the sign change for $x_0$, state the formula, tabulate 2-3 rows, and close with the root to the asked decimals. Convergence: open with "order $p=2$"; write $|e_{n+1}|\le c|e_n|^2$; close by saying the correct digits double each step.

Pitfall: Forgetting radians for $\cos x$ gives a wrong root.

Asked: [7 marks] (Jun 2025, Dec 2025) If $f(x)=x^3-2x-5=0$ and the initial approximation is $x=2$, find the next approximation correct to 2 decimals; also find the root nearer to 2 correct to three decimals.

Asked: [7 marks] (Jun 2023) Using Newton Raphson method, find the real root of $3x=\cos x+1$.

Asked: [7 marks] (Jun 2025) What is the rate of convergence of Newton Raphson method?

Regula-Falsi method

Definition. Regula-Falsi (false position) joins $(a,f(a))$ and $(b,f(b))$ by a chord and takes where it cuts the $x$-axis as the new estimate: $x=\dfrac{a f(b)-b f(a)}{f(b)-f(a)}$.

Key points.

Start with $f(a)f(b)<0$.
Replace the end whose function value has the same sign as $f(x)$, so the root stays bracketed.
Convergence is linear, faster than bisection, and always assured.
Stop when successive values agree to the required decimals.

Example. $f=x\log_{10}x-1.2$, $f(2)=-0.5979<0$, $f(3)=0.2314>0$. Successive values: $2.7210,\ 2.7402,\ 2.7406,\ 2.7406$. Root $=2.7406$.

Asked: [7 marks] (Jun 2025) Using Regula-Falsi method, find the real root of $x\log_{10}x=1.2$, correct to four decimal places.

Finite differences

Definition. For equally spaced $x$ with step $h$: forward difference $\Delta y_r=y_{r+1}-y_r$, backward $\nabla y_r=y_r-y_{r-1}$, and shift operator $Ey_r=y_{r+1}$, so $\Delta=E-1$.

Key points.

Higher differences are repeated: $\Delta^2y_0=y_2-2y_1+y_0$.
$\Delta^n y_0=(E-1)^n y_0$ expands with binomial coefficients: $\Delta^4y_0=y_4-4y_3+6y_2-4y_1+y_0$.
The $n$th difference of a polynomial of degree $n$ is constant, and the $(n+1)$th is zero.
So with $k$ known values, the polynomial is taken of degree $k-1$ and $\Delta^k$ is set to zero.

Example (functions). $\Delta\tan x=\dfrac{\sin h}{\cos(x+h)\cos x}$; $\Delta e^{2x}=e^{2x}(e^{2h}-1)$, so $\Delta^2e^{2x}=e^{2x}(e^{2h}-1)^2$; $\Delta\log x=\log\!\left(1+\dfrac hx\right)$.

Example (missing values). Known $y_0=6,y_1=10,y_3=17,y_5=31$ (4 values, so $\Delta^4y=0$). $\Delta^4y_0=0$: $y_4+6y_2=94$. $\Delta^4y_1=0$: $y_4+y_2=35.75$. Solving: $y_2=11.65$, $y_4=24.1$.

Example (first term). Terms $8,3,0,-1,0$: $\Delta$: $-5,-3,-1,1$; $\Delta^2$: $2,2,2$. Going back, $\Delta^2$ before is 2, $\Delta$ before is $-7$, first term $=8+7=$ 15.

Answer frame. Open with the definition of $\Delta$ and $E$; for missing values, state degree and set $\Delta^4=0$, expand, solve; for the series, extend the difference table backward; close with the answer.

Asked: [7 marks] (Jun 2025) Find (i) $\Delta\tan x$ (ii) $\Delta^2e^{2x}$ (iii) $\Delta\log x$.

Asked: [7 marks] (Jun 2025) Find the missing values ($x=0,5,10,15,20,25$; $y=6,10,-,17,-,31$).

Asked: [7 marks] (Dec 2025) Find the first term of the series whose second and subsequent terms are 8, 3, 0, -1, 0.

Relation between operators

Definition. $E f(x)=f(x+h)$, $\Delta=E-1$, $\nabla=1-E^{-1}$, $\delta=E^{1/2}-E^{-1/2}$, $\mu=\tfrac12(E^{1/2}+E^{-1/2})$.

Key points.

$E=1+\Delta$, so $f(x+nh)=(1+\Delta)^nf(x)$.
$\nabla=\Delta E^{-1}$ and $\Delta\nabla=\nabla\Delta=\delta^2$.
Proofs work by expanding each difference in function values.
Then the terms are added and the values cancel.

Proof. With $h=1$: $\Delta f(2)=f(3)-f(2)$; $\Delta^2f(1)=f(3)-2f(2)+f(1)$; $\Delta^3f(1)=f(4)-3f(3)+3f(2)-f(1)$. Adding $f(3)$ to these gives $f(4)+(1+1+1-3)f(3)+(-1-2+3)f(2)+(1-1)f(1)=f(4)$. Hence proved.

Asked: [7 marks] (Jun 2025) Prove that $f(4)=f(3)+\Delta f(2)+\Delta^2f(1)+\Delta^3f(1)$, taking 1 as the interval of differencing.

Interpolation using Newton's forward difference formulae

Definition. ==For equal spacing $h$, with $u=\dfrac{x-x_0}{h}$: $y(x)=y_0+u\Delta y_0+\dfrac{u(u-1)}{2!}\Delta^2y_0+\dfrac{u(u-1)(u-2)}{3!}\Delta^3y_0+\cdots$==

Key points.

Use it when $x$ lies near the beginning of the table.
Build the forward difference table first; the top diagonal gives $\Delta y_0,\Delta^2y_0,\dots$.
The formula needs equally spaced $x$.
Interpolating a frequency: form cumulative frequencies, interpolate at the upper class limit, then subtract.

Example. $x=2,3,4,5$; $f=2.625,3.454,4.784,6.986$. $\Delta$: $0.829,1.330,2.202$; $\Delta^2$: $0.501,0.872$; $\Delta^3$: $0.371$. $h=1$, $u=1.5$: $f(3.5)=2.625+1.5(0.829)+\dfrac{1.5(0.5)}{2}(0.501)+\dfrac{1.5(0.5)(-0.5)}{6}(0.371)$ $=2.625+1.2435+0.1879-0.0232=$ 4.0332.

Marks table. Cumulative $<40,<50,<60,<70,<80$ = $31,73,124,159,190$; at $x=45$, $u=0.5$ gives $47.87$, so between 40 and 45: $47.87-31\approx$ 17 students.

Answer frame. Open by stating the formula and $u$; draw the difference table; substitute term by term; close with the value.

Asked: [7 marks] (Jun 2025, Dec 2025) Find $f(3.5)$ from the table using Newton's forward interpolation formula; estimate the number of students with marks between 40 and 45 (classes 30-40 to 70-80: 31, 42, 51, 35, 31).

Interpolation using Newton's backward difference formulae

Definition. For $x$ near the end of the table, with $u=\dfrac{x-x_n}{h}$: $y(x)=y_n+u\nabla y_n+\dfrac{u(u+1)}{2!}\nabla^2y_n+\dfrac{u(u+1)(u+2)}{3!}\nabla^3y_n+\cdots$

Key points.

It uses the bottom diagonal of the difference table.
$u$ is negative because $x<x_n$.
Values must be equally spaced.

Example. $x=45,50,55,60$; $\sin x=0.7071,0.7660,0.8192,0.8660$. $\nabla y_n=0.0468$, $\nabla^2y_n=-0.0064$, $\nabla^3y_n=-0.0007$; $h=5$, $u=-0.4$: $\sin58=0.8660-0.4(0.0468)+\dfrac{(-0.4)(0.6)}{2}(-0.0064)+\dfrac{(-0.4)(0.6)(1.6)}{6}(-0.0007)=$ 0.8481.

Asked: [7 marks] (Nov 2022) Given $\sin45=0.7071$, $\sin50=0.7660$, $\sin55=0.8192$, $\sin60=0.8660$, find $\sin58$ by Newton's backward interpolation formula.

Newton's divided difference formulae

Definition. ==$f[x_0,x_1]=\dfrac{f(x_1)-f(x_0)}{x_1-x_0}$, and higher divided differences follow the same rule; $f(x)=f(x_0)+(x-x_0)f[x_0,x_1]+(x-x_0)(x-x_1)f[x_0,x_1,x_2]+\cdots$==

Key points.

It works for unequally spaced $x$.
Divided differences are symmetric in their arguments.
The $n$th divided difference of a polynomial of degree $n$ is constant.
Adding a new point only adds one term.

Example 1. $x=300,304,305,307$: $f[x_0,x_1]=0.00145$, $f[x_1,x_2]=0.0014$, $f[x_2,x_3]=0.0014$; second: $-0.00001,\ 0$; third: $1.43\times10^{-6}$. $f(x)=2.4771+0.00145(x-300)-0.00001(x-300)(x-304)+1.43\times10^{-6}(x-300)(x-304)(x-305)$. At $x=301$: 2.4786.

Example 2. $x=4,5,7,10,11,13$, $f=48,100,294,900,1210,2028$: first: $52,97,202,310,409$; second: $15,21,27,33$; third: $1,1,1$; fourth $0$. $f(x)=48+52(x-4)+15(x-4)(x-5)+(x-4)(x-5)(x-7)=x^3-x^2$. $f(8)=448$, $f(15)=3150$.

Answer frame. Open with the formula; draw the divided difference table; substitute the top diagonal; close with the value.

Asked: [7 marks] (Jun 2023, Dec 2025) Find the solution using Newton's divided difference formula for $x=300,304,305,307$, $F=2.4771,2.4829,2.4843,2.4871$; also find $f(8)$ and $f(15)$ from the table $x=4,5,7,10,11,13$, $f=48,100,294,900,1210,2028$.

Lagrange's formulae

Definition. ==For any $x_i$, equal or unequal, $y(x)=\sum_{i=0}^{n}y_i\prod_{j\neq i}\dfrac{x-x_j}{x_i-x_j}$.==

Key points.

It is valid for unequally spaced data.
Each term is a basis polynomial $l_i(x)$ times $y_i$; $l_i(x_i)=1$ and $l_i(x_j)=0$.
There is no difference table, but adding a point means recomputing everything.
$n+1$ points give a polynomial of degree at most $n$.

Example 1. $x=3,5,7,9,11$; $y=6,24,58,108,74$; at $x=6$ $l_0=-\tfrac5{128}$, $l_1=\tfrac{15}{32}$, $l_2=\tfrac{45}{64}$, $l_3=-\tfrac5{32}$, $l_4=\tfrac3{128}$, so the terms $y_il_i$ are $-0.234,\ 11.25,\ 40.781,\ -16.875,\ 1.734$: $y(6)=36.66$.

Example 2. $x=-1,0,3,6,7$: $f(x)=x^4-3x^3+5x^2-6$; for $x=-1,0,2,3$ with $f=-8,3,1,12$: $f(x)=2x^3-6x^2+3x+3$.

Answer frame. Open with the formula for the given number of points; write each $l_i$ with numerator and denominator; multiply by $y_i$; sum, simplify and state the result.

Asked: [7 marks] (Nov 2022) Using Lagrange's interpolation formula, find $y(6)$ from $x=3,5,7,9,11$, $y=6,24,58,108,74$.

Asked: [7 marks] (Jun 2025) By Lagrange's method find $f(x)$ as a polynomial for $x=-1,0,3,6,7$, $f=3,-6,39,822,1611$; also for $x=-1,0,2,3$, $f=-8,3,1,12$.

Last-minute revision

Bisection: $c=\frac{a+b}{2}$; always converges, linear.
Regula-Falsi: $x=\frac{af(b)-bf(a)}{f(b)-f(a)}$.
Newton-Raphson: $x_{n+1}=x_n-f/f'$; order 2.
$x^3-2x-5$ from 2: $x_1=2.10$, root $2.0946$.
$3x=\cos x+1$: root $0.6071$.
$\Delta=E-1$, $\nabla=1-E^{-1}$, $\Delta^n$ of degree-$n$ polynomial is constant.
Forward: $u=\frac{x-x_0}{h}$; backward: $u=\frac{x-x_n}{h}$.
$f(3.5)=4.0332$; $\sin58=0.8481$.
Divided difference and Lagrange work for unequal spacing.
Missing values: $y_2=11.65$, $y_4=24.1$; first term is $15$.

Memory hooks

Bisection is Bold but slow: the interval just halves.
Newton uses the tangent, Regula-Falsi uses the chord.
Forward is for the front of the table, backward for the back.
Lagrange needs no table; divided differences need one.

Coverage checklist

Solution of polynomial and transcendental equations: Nov 2022 iteration $3x=\cos x+1$.
Bisection method: Jun 2023 $x^3-9x+1$.
Newton-Raphson method: Jun 2025/Dec 2025 $x^3-2x-5$; Jun 2023 $3x=\cos x+1$; Jun 2025 convergence.
Regula-Falsi method: Jun 2025 $x\log_{10}x=1.2$.
Finite differences: Jun 2025 $\Delta$ of functions; Jun 2025 missing values; Dec 2025 first term.
Relation between operators: Jun 2025 proof.
Interpolation using Newton's forward difference formulae: Jun 2025/Dec 2025 $f(3.5)$ and marks table.
Interpolation using Newton's backward difference formulae: Nov 2022 $\sin58$.
Newton's divided difference formulae: Jun 2023/Dec 2025.
Lagrange's formulae: Nov 2022 $y(6)$; Jun 2025 polynomials.

Account

Mathematics-III (BT-301) - Unit 1 Short Notes

How unit 1 is examined

Solution of polynomial and transcendental equations

Bisection method

Newton-Raphson method

Regula-Falsi method

Finite differences

Relation between operators

Interpolation using Newton's forward difference formulae

Interpolation using Newton's backward difference formulae

Newton's divided difference formulae

Lagrange's formulae

Last-minute revision

Memory hooks

Coverage checklist