Mathematics-III (BT-301) - Unit 2 Short Notes

How unit 2 is examined

Differentiation and integration from tabulated data, then solving linear systems by direct (Gauss, Crout) and iterative (Gauss-Seidel) methods; Simpson's 3/8 rule and Gauss-Seidel carry the marks, all asked as 7-mark numericals.

Numerical Differentiation

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Definition. Numerical differentiation finds $f'(x)$, $f''(x)$ at a tabulated point by differentiating an interpolation polynomial fitted through the table.

Key points.

1. For equally spaced $x$ near the start of the table use Newton's forward formula, with $u=\dfrac{x-x_0}{h}$.
2. Build the forward difference table first, since the formulas use $\Delta y_0,\Delta^2 y_0,\dots$ from the top row.
3. Differentiating with respect to $x$ uses $\dfrac{du}{dx}=\dfrac1h$, which is why $h$ and $h^2$ appear in the denominators.
4. The point need not be a table entry; $u$ is then a fraction, such as $0.5$.

Formula. $$f'(x)=\frac1h\left[\Delta y_0+\frac{2u-1}{2}\Delta^2y_0+\frac{3u^2-6u+2}{6}\Delta^3y_0+\cdots\right]$$ $$f''(x)=\frac1{h^2}\left[\Delta^2y_0+(u-1)\Delta^3y_0+\frac{6u^2-18u+11}{12}\Delta^4y_0+\cdots\right]$$

Example. $h=0.2$, $u=\frac{1.1-1}{0.2}=0.5$. Differences: $\Delta y_0=0.128$, $\Delta^2y_0=0.288$, $\Delta^3y_0=0.048$, higher differences $0$.

Here $2u-1=0$ and $3u^2-6u+2=-0.25$, so $f'(1.1)=\frac1{0.2}\left[0.128+\frac{-0.25}{6}(0.048)\right]=5(0.128-0.002)=0.63$.

$f''(1.1)=\frac1{0.04}\left[0.288+(0.5-1)(0.048)\right]=25(0.264)=6.6$.

Answer: $f'(1.1)=0.63$, $f''(1.1)=6.6$.

==Differentiate Newton's forward formula term by term, remembering $du/dx=1/h$.==

Asked: [7 marks] (Jun 2023) Find the first and second derivatives at $x=1.1$ from the table $x=1,1.2,1.4,1.6,1.8,2.0$; $F(x)=0,.1280,.5440,1.2960,2.4320,4.00$.

Numerical integration

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Definition. Numerical integration approximates $\int_a^b y\,dx$ by replacing $y$ with an interpolating polynomial over $n$ equal strips of width $h=\dfrac{b-a}{n}$.

Key points.

1. It is used when the integral has no closed form or the function is known only as a table.
2. The ordinates are $y_0,y_1,\dots,y_n$ at $x_i=a+ih$, so $n+1$ values are needed for $n$ strips.
3. Fitting a degree-1 polynomial gives the trapezoidal rule, degree 2 gives Simpson's 1/3, degree 3 gives Simpson's 3/8.
4. More strips give better accuracy; Simpson's rules are more accurate than the trapezoidal rule for the same $h$.

Example. $\int_0^\pi\sin x\,dx$, $n=10$, $h=\pi/10$ (exact value $2$).

Trapezoidal: $\frac h2[0+2(6.3138)]=0.15708\times12.6276=1.9835$. Simpson 1/3: odd sum $=3.2360$, even sum $=3.0778$; $\frac h3[0+4(3.2360)+2(3.0778)]=0.10472\times19.0996=2.0001$.

Answer: Trapezoidal $\approx1.9835$; Simpson's 1/3 $\approx2.0001$.

==Find $h=(b-a)/n$, tabulate $y_i$, then substitute into the rule: ends once, odd ordinates by 4, even by 2 (Simpson).==

Asked: [7 marks] (Nov 2022) Evaluate $\int_0^\pi\sin x\,dx$ by dividing the range into 10 equal parts using (i) Trapezoidal rule (ii) Simpson's 1/3 rule.

Trapezoidal rule

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Definition. The trapezoidal rule treats each strip as a trapezium, replacing the curve by straight chords.

Formula. $$\int_a^b y\,dx\approx\frac h2\left[(y_0+y_n)+2(y_1+y_2+\cdots+y_{n-1})\right]$$

Key points.

1. It works for any number of strips $n$, odd or even.
2. Ends are taken once and every interior ordinate twice, because each is shared by two trapezia.
3. The error is of order $h^2$, so it is the least accurate of the three rules.
4. Area of one strip is $\frac h2(y_i+y_{i+1})$, and the rule is the sum of these.

Simpson's 1/3rd rule

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Definition. Simpson's 1/3 rule replaces the curve over each pair of strips by a parabola through three consecutive points.

Formula. $$\int_a^b y\,dx\approx\frac h3\left[(y_0+y_n)+4(y_1+y_3+\cdots)+2(y_2+y_4+\cdots)\right]$$

Key points.

1. The number of strips $n$ must be even, so the number of ordinates is odd.
2. Odd-position ordinates carry weight 4 and even-position interior ordinates weight 2.
3. Its error is of order $h^4$, far better than the trapezoidal rule.
4. It is exact for polynomials up to degree 3.
5. Use the quoted decimal places only at the end; keep 5 places in $y$.

Example. $\int_2^{10}\frac{dx}{1+x}$, $n=8$, $h=1$.

Ends $=0.42424$; odd $(y_1,y_3,y_5,y_7)=0.64167$; even $(y_2,y_4,y_6)=0.45397$. $I=\frac13[0.42424+4(0.64167)+2(0.45397)]=\frac13(3.89885)=1.29962$.

Answer: $I\approx1.300$ (exact $\ln\frac{11}{3}=1.2993$).

<mark>Simpson's 1/3: $\frac h3[\text{first}+\text{last}+4(\text{odd})+2(\text{even})]$, with $n$ even.</mark>

Asked: [7 marks] (Dec 2025) Calculate by Simpson's 1/3 rule (up to 3 places of decimal) $\int_2^{10}\frac{dx}{1+x}$ by dividing the range into eight equal parts.

Simpson's 3/8th rule

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Definition. Simpson's 3/8 rule fits a cubic through every four consecutive ordinates, i.e. three strips, and integrates it.

Formula. $$\int_a^b y\,dx\approx\frac{3h}{8}\left[(y_0+y_n)+3(y_1+y_2+y_4+y_5+\cdots)+2(y_3+y_6+\cdots)\right]$$

Key points.

1. The number of strips $n$ must be a multiple of 3.
2. Ordinates at positions 3, 6, 9,... are shared between two cubics, so they get weight 2.
3. All other interior ordinates get weight 3, and the two end ordinates weight 1.
4. For a single block of three strips the formula reduces to $\frac{3h}{8}(y_0+3y_1+3y_2+y_3)$.
5. It is exact for cubics and has the same order of error as the 1/3 rule but a slightly larger constant.
6. Choose it when $n$ is a multiple of 3 but not of 2.

Example. $\int_0^6\frac{dx}{1+x^2}$, $n=6$, $h=1$.

Ends $=1.02703$; $3(y_1+y_2+y_4+y_5)=3(0.79728)=2.39184$; $2y_3=0.2$. $I=\frac38(1.02703+2.39184+0.2)=\frac38(3.61887)=1.3571$.

Answer: $I\approx1.3571$.

Other papers' data. Table $h=0.1$, $x=0$ to $0.3$: $\frac{3(0.1)}8(1+3(0.9975)+3(0.99)+0.9776)=0.0375\times7.9401=0.2978$. For $\int_0^6\frac{e^x}{1+x}dx$, $h=1$: $y=1,1.3591,2.4630,5.0214,10.9196,24.7355,57.6327$; ends $58.6327$, $3(1.3591+2.4630+10.9196+24.7355)=118.4304$, $2y_3=10.0428$; $I=\frac38(187.1059)=70.165$.

Answer frame. Open with "Simpson's 3/8 rule: $\int y\,dx=\frac{3h}{8}[\dots]$, valid when $n$ is a multiple of 3"; write $h=(b-a)/n$, then the $x$-$y$ table; state the formula with the weights 1, 3, 3, 2, 3, 3, 1; substitute grouped sums; close with the boxed value.

Pitfall: Giving weight 2 to the wrong ordinates; only $y_3,y_6,\dots$ take 2, all other interior ordinates take 3.

<mark>Simpson's 3/8 rule: $\frac{3h}{8}[(y_0+y_n)+3(\text{rest})+2(y_3+y_6+\cdots)]$, $n$ a multiple of 3.</mark>

Asked: [7 marks] (Nov 2022, Jun 2023, Jun 2025) Using Simpson's 3/8 rule, evaluate $\int_0^6\frac{dx}{1+x^2}$ by dividing the range into 6 equal parts. Asked: [7 marks] (Nov 2022, Jun 2023, Jun 2025) Find the solution using Simpson's 3/8 rule for the table $x=0,0.1,0.2,0.3,0.4$; $y=1,0.9975,0.99,0.9776,0.8604$. Asked: [7 marks] (Nov 2022, Jun 2023, Jun 2025) Find $\int_0^6\frac{e^x}{1+x}dx$ approximately using Simpson's 3/8th rule.

Solution of Simultaneous Linear Algebraic Equations by Gauss's Elimination

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Definition. Gauss elimination reduces the augmented matrix $[A|B]$ to upper triangular form $[U|C]$ by row operations, then finds the unknowns by back substitution.

Key points.

1. Write the augmented matrix $[A|B]$ first.
2. Use the first row's pivot to make all entries below it zero, then the second row's pivot for the next column.
3. Only row operations are allowed: swapping rows, scaling a row, adding a multiple of one row to another.
4. If a pivot is zero, interchange with a lower row that has a non-zero entry.
5. Back substitution starts with the last equation, giving $z$, then $y$, then $x$.

Example. $2x-y-3z=9,\ x-y-2z=6,\ x-y-z=2$.

$$\left[\begin{array}{ccc|c}2&-1&-3&9\\1&-1&-2&6\\1&-1&-1&2\end{array}\right]\xrightarrow{R_2\to 2R_2-R_1,\ R_3\to 2R_3-R_1}\left[\begin{array}{ccc|c}2&-1&-3&9\\0&-1&-1&3\\0&-1&1&-5\end{array}\right]\xrightarrow{R_3\to R_3-R_2}\left[\begin{array}{ccc|c}2&-1&-3&9\\0&-1&-1&3\\0&0&2&-8\end{array}\right]$$

Back substitution: $2z=-8\Rightarrow z=-4$; $-y-z=3\Rightarrow y=1$; $2x=9+y+3z=9+1-12=-2\Rightarrow x=-1$. Check in eq. 3: $-1-1+4=2$. Consistent.

Answer: $x=-1,\ y=1,\ z=-4$.

Answer frame. Open with "Gauss elimination converts $[A|B]$ to upper triangular form and back-substitutes"; write the augmented matrix; show each row operation labelled; back substitute from the last row; verify in one original equation and state the solution.

<mark>Reduce $[A|B]$ to triangular form by row operations, then back-substitute from the last equation.</mark>

Asked: [7 marks] (Jun 2025) Apply Gauss Elimination method to solve $2x-y-3z=9,\ x-y-2z=6,\ x-y-z=2$.

Gauss's Jordan method

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Definition. Gauss-Jordan reduces $[A|B]$ all the way to $[I|X]$, so the solution is read off directly with no back substitution.

Key points.

1. It uses the same row operations as Gauss elimination but clears entries above each pivot as well as below.
2. Each pivot row is divided by its pivot to make the diagonal entry 1.
3. It needs more operations than Gauss elimination, so it is used when the inverse is wanted.
4. Applying it to $[A|I]$ gives $[I|A^{-1}]$, the matrix inverse.

Crout's method

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Definition. Crout's method factorises $A=LU$ with $L$ lower triangular and $U$ upper triangular having unit diagonal, then solves $AX=B$ in two triangular steps.

Key points.

1. Write $A=LU$ with $L=\begin{bmatrix}l_{11}&0&0\\l_{21}&l_{22}&0\\l_{31}&l_{32}&l_{33}\end{bmatrix}$, $U=\begin{bmatrix}1&u_{12}&u_{13}\\0&1&u_{23}\\0&0&1\end{bmatrix}$.
2. Equate $LU$ to $A$: the first column of $L$ equals the first column of $A$, then $u_{1j}=a_{1j}/l_{11}$, and so on column by column.
3. Solve $LY=B$ by forward substitution, then $UX=Y$ by back substitution.
4. The factors can be reused for other right-hand sides.

Example. $2x+3y-z=5,\ 3x+2y+z=10,\ x-5y+3z=0$.

$l_{11}=2,\ l_{21}=3,\ l_{31}=1$; $u_{12}=1.5,\ u_{13}=-0.5$; $l_{22}=2-3(1.5)=-2.5$, $l_{32}=-5-1(1.5)=-6.5$; $u_{23}=\frac{1-3(-0.5)}{-2.5}=-1$; $l_{33}=3-1(-0.5)-(-6.5)(-1)=-3$.

$LY=B$: $y_1=2.5$; $3(2.5)-2.5y_2=10\Rightarrow y_2=-1$; $1(2.5)-6.5(-1)-3y_3=0\Rightarrow y_3=3$. $UX=Y$: $z=3$; $y-z=-1\Rightarrow y=2$; $x+1.5y-0.5z=2.5\Rightarrow x=1$.

Answer: $x=1,\ y=2,\ z=3$.

Answer frame. Open with "Crout's method writes $A=LU$, $U$ with unit diagonal"; write $AX=B$; give the $L$, $U$ entries in order; solve $LY=B$, then $UX=Y$; close with the solution.

==Crout: $A=LU$ with $U$ unit-diagonal; solve $LY=B$ forward, then $UX=Y$ backward.==

Asked: [7 marks] (Jun 2023) Solve $2x+3y-z=5,\ 3x+2y+z=10,\ x-5y+3z=0$ by Crout's method.

Jacobi's method

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Definition. Jacobi's method is an iterative solver that computes every new value from the previous iteration's values only.

Key points.

1. The system must be diagonally dominant for convergence, and each equation is solved for its diagonal unknown.
2. Iteration: $x^{(k+1)}=\frac1{a_{11}}(b_1-a_{12}y^{(k)}-a_{13}z^{(k)})$, and likewise for $y,z$.
3. All three new values are computed together from the old ones.
4. It converges more slowly than Gauss-Seidel.

Gauss-Seidal method

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Definition. Gauss-Seidel is an iterative method that uses each newly computed value immediately in the remaining equations of the same iteration.

Formula. For $a_1x+b_1y+c_1z=d_1$ etc., diagonally dominant: $$x^{(k+1)}=\tfrac1{a_1}(d_1-b_1y^{(k)}-c_1z^{(k)}),\ \ y^{(k+1)}=\tfrac1{b_2}(d_2-a_2x^{(k+1)}-c_2z^{(k)}),\ \ z^{(k+1)}=\tfrac1{c_3}(d_3-a_3x^{(k+1)}-b_3y^{(k+1)})$$

Key points.

1. First check diagonal dominance: each diagonal coefficient must exceed the sum of the others in its row, which guarantees convergence.
2. Rearrange each equation to give its diagonal unknown in terms of the others.
3. Start from $x^{(0)}=y^{(0)}=z^{(0)}=0$ unless told otherwise.
4. Always use the latest available value, so $x^{(k+1)}$ is used at once for $y$ and $z$.
5. Stop when two successive iterations agree to the required decimal places.
6. It converges about twice as fast as Jacobi.

Example. $10x+y+z=12,\ 2x+10y+z=13,\ 2x+2y+10z=14$. Diagonal dominance: $10>2$, $10>3$, $10>4$.

Answer: $x=y=z=1$.

Other papers' systems, same steps from $(0,0,0)$: $20x+y-2z=17,\ 3x+20y-z=-18,\ 2x-3y+20z=25$ gives iteration 1 $(0.85,-1.0275,1.0109)$, iteration 2 $(1.0025,-0.9998,0.9998)$, iteration 3 $(1,-1,1)$, so $x=1,\ y=-1,\ z=1$. $27x+6y-z=85,\ 6x+15y+2z=72,\ x+y+54z=110$ gives $(3.1481,3.5407,1.9132)$, $(2.4322,3.5720,1.9258)$, $(2.4257,3.5729,1.9260)$, $(2.4255,3.5730,1.9260)$, so $x=2.4255,\ y=3.5730,\ z=1.9260$.

Answer frame. Open with "Gauss-Seidel is an iterative method using the latest values"; test diagonal dominance; write the three iteration equations; start at $(0,0,0)$; tabulate iterations; close when values repeat to the required decimals.

Pitfall: Using old $x$ in the $y$ equation; that is Jacobi, not Gauss-Seidel.

<mark>In Gauss-Seidel every newly found value is used immediately in the next equation of the same iteration.</mark>

Asked: [7 marks] (Nov 2022, Jun 2023, Dec 2025) Using Gauss-Seidel iteration, solve $10x+y+z=12,\ 2x+10y+z=13,\ 2x+2y+10z=14$. Asked: [7 marks] (Nov 2022, Jun 2023, Dec 2025) Solve $20x+y-2z=17,\ 3x+20y-z=-18,\ 2x-3y+20z=25$ by Gauss-Seidel method. Asked: [7 marks] (Nov 2022, Jun 2023, Dec 2025) Solve $27x+6y-z=85,\ 6x+15y+2z=72,\ x+y+54z=110$ by Gauss-Seidel iteration method.

Relaxation method

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Definition. Relaxation is an iterative method that repeatedly reduces the largest residual of the system to zero.

Key points.

1. For $Ax=b$ the residual is $R=b-Ax$, computed for the starting guess.
2. Pick the residual of largest magnitude and change the corresponding unknown by $\delta x_i=R_i/a_{ii}$, which makes that residual zero.
3. Update all residuals, since changing $x_i$ alters each $R_j$ by $-a_{ji}\delta x_i$.
4. Stop when all residuals are negligible; over-relaxation uses a factor larger than 1 to speed convergence.

Last-minute revision

• $h=\dfrac{b-a}{n}$; ordinates are $n+1$.
• Trapezoidal: $\frac h2[\text{ends}+2(\text{rest})]$, any $n$.
• Simpson 1/3: $\frac h3[\text{ends}+4(\text{odd})+2(\text{even})]$, $n$ even.
• Simpson 3/8: $\frac{3h}8[\text{ends}+3(\text{rest})+2(y_3,y_6,\dots)]$, $n$ multiple of 3.
• $\int_0^\pi\sin x\,dx$, $n=10$: trapezoidal $1.9835$, Simpson $2.0001$.
• $\int_0^6\frac{dx}{1+x^2}$ by 3/8 rule $=1.3571$; $\int_2^{10}\frac{dx}{1+x}$ by 1/3 rule $=1.300$.
• $f'(1.1)=0.63$, $f''(1.1)=6.6$ with $u=0.5$, $h=0.2$.
• Gauss elimination example: $(-1,1,-4)$; Crout example: $(1,2,3)$.
• Gauss-Seidel: $(1,1,1)$; $(1,-1,1)$; $(2.4255,3.5730,1.9260)$.
• Crout uses $U$ with unit diagonal; Gauss-Jordan ends at $[I|X]$.
• Iterative methods need diagonal dominance.

Memory hooks

• Simpson weights: 1/3 goes "1-4-2-4-1", 3/8 goes "1-3-3-2-3-3-1".
• 3/8 needs $n$ in the three-times table; 1/3 needs $n$ even.
• Gauss-Seidel is "Jacobi with instant updates".
• Gauss stops at a triangle, Jordan goes on to the identity.
• Crout: $L$ carries the pivots, $U$ carries the 1s.

Coverage checklist

• Numerical Differentiation: Jun 2023 derivatives at $x=1.1$.
• Numerical integration: Nov 2022 $\sin x$ by trapezoidal and Simpson's 1/3.
• Trapezoidal rule: formula and key points (not asked recently).
• Simpson's 1/3rd rule: Dec 2025 $\int_2^{10}\frac{dx}{1+x}$.
• Simpson's 3/8th rule: Nov 2022, Jun 2023, Jun 2025, the three integrals.
• Solution of Simultaneous Linear Algebraic Equations by Gauss's Elimination: Jun 2025.
• Gauss's Jordan method: definition and key points (not asked recently).
• Crout's method: Jun 2023.
• Jacobi's method: definition and key points (not asked recently).
• Gauss-Seidal method: Nov 2022, Jun 2023, Dec 2025, the three systems.
• Relaxation method: definition and key points (not asked recently).