Mathematics-III (BT-401) - Unit 2 Short Notes

How unit 2 is examined

This unit covers numerical differentiation, the trapezoidal and Simpson rules, direct solution of linear systems (Gauss, Gauss-Jordan, Crout) and iterative solution (Jacobi, Gauss-Seidel, relaxation); Simpson's rules, Numerical Differentiation and Gauss-Seidel carry the most marks, and every question is a 7-mark numerical.

Numerical Differentiation

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Definition. Numerical differentiation finds $\frac{dy}{dx}$, $\frac{d^2y}{dx^2}$ at a tabulated point by differentiating the Newton interpolation polynomial through equally spaced data; <mark>near the start of the table use Newton's forward formula, near the end use the backward formula.</mark>

Key points.

1. The data must be equally spaced with step $h$, and the point must lie near the start (forward) or the end (backward) of the table.
2. Build the difference table first: $\Delta y_i = y_{i+1}-y_i$ and $\nabla y_i = y_i - y_{i-1}$, then higher orders.
3. The derivative formulas come from differentiating Newton's interpolation formula with $u=(x-x_0)/h$, giving the series below at $u=0$.
4. Each higher derivative divides by one more power of $h$, so $f'$ has $1/h$, $f''$ has $1/h^2$, $f'''$ has $1/h^3$.
5. Use the last non-zero difference and stop there; a difference column that is zero ends the series.
6. The answer is an approximation whose accuracy falls as the derivative order rises.

Formula (forward, at $x_0$).

$$f'(x_0)=\frac1h\Big[\Delta y_0-\frac{\Delta^2y_0}{2}+\frac{\Delta^3y_0}{3}-\frac{\Delta^4y_0}{4}+\cdots\Big]$$

$$f''(x_0)=\frac1{h^2}\Big[\Delta^2y_0-\Delta^3y_0+\frac{11}{12}\Delta^4y_0-\cdots\Big],\quad f'''(x_0)=\frac1{h^3}\Big[\Delta^3y_0-\frac32\Delta^4y_0+\cdots\Big]$$

Formula (backward, at $x_n$).

$$f'(x_n)=\frac1h\Big[\nabla y_n+\frac{\nabla^2y_n}{2}+\frac{\nabla^3y_n}{3}+\cdots\Big],\quad f''(x_n)=\frac1{h^2}\Big[\nabla^2y_n+\nabla^3y_n+\frac{11}{12}\nabla^4y_n+\cdots\Big]$$

Example (Dec 2020). $h=0.5$, $x_0=1.5$:

$f'(1.5)=\frac{1}{0.5}\left[3.625-1.5+0.25\right]=4.75$; $f''(1.5)=\frac{1}{0.25}[3.000-0.750]=9$; $f'''(1.5)=\frac{0.75}{0.125}=6$.

Answer: $f'=4.75,\ f''=9,\ f'''=6$.

Example (Jun 2020, backward, $h=0.1$). $\nabla y=0.14196$, $\nabla^2y=0.01350$, $\nabla^3y=0.00127$ at $x=0.4$. $\frac{dy}{dx}=10[0.14196+0.00675+0.00042]=1.4913$; $\frac{d^2y}{dx^2}=100[0.01350+0.00127]=1.477$.

Answer frame. Open with "the table is equally spaced, so Newton's forward (or backward) formula is used"; draw the difference table; write the derivative formula for each derivative asked; substitute $h$ and the differences; close with the boxed values.

Pitfall: Using the forward formula at the last point (or backward at the first) gives a wrong answer; also forgetting $h^2$ or $h^3$ in the denominator.

Asked: [7 marks] (May 2019, Jun 2020) Find the first and second order derivative of $f(x)$ at $x=1.5$ from the table $x$: 1.5 to 4.0 step 0.5, $f(x)$: 3.375, 7.000, 13.625, 24.000, 38.875, 59.000. Asked: [7 marks] (Dec 2020) Find the first, second and third derivative of $f(x)$ at $x=1.5$ from the same data. Asked: [7 marks] (Jun 2020) Given $x$: 0.1, 0.2, 0.3, 0.4 and $y$: 1.10517, 1.22140, 1.34986, 1.49182, find $\frac{dy}{dx}$ and $\frac{d^2y}{dx^2}$ at $x=0.4$.

Numerical integration: Trapezoidal rule

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Definition. The trapezoidal rule replaces the curve between consecutive ordinates by straight lines and adds the areas of the trapezoids: ==$\int_a^b f\,dx\approx\frac h2[y_0+y_n+2(y_1+\dots+y_{n-1})]$, with $h=\frac{b-a}{n}$.==

Key points.

1. It works for any number $n$ of strips, odd or even.
2. Ends carry weight 1 and every interior ordinate weight 2, all multiplied by $h/2$.
3. The error is of order $h^2$, so more strips give a better value.
4. Compute ordinates in radians for $\sin x$.

Example (Jun 2023). $\int_{0.2}^{1.4}(\sin x-\log_e x+e^x)dx$, $n=6$, $h=0.2$. Ordinates: 3.0295, 2.7975, 2.8976, 3.1660, 3.5598, 4.0698, 4.7042. Sum $=\frac{0.2}{2}[3.0295+4.7042+2(16.4907)]=4.0715$. Answer: $\approx 4.0715$.

Asked: [7 marks] (Jun 2023) Compute $\int_{0.2}^{1.4}(\sin x-\log_e x+e^x)dx$ by the trapezoidal rule.

Simpson's 1/3rd and 3/8 rules

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Definition. Simpson's rules fit parabolas (1/3 rule, two strips at a time) or cubics (3/8 rule, three strips at a time) through the ordinates; <mark>Simpson's 1/3 rule is $\int_a^b f\,dx\approx\frac h3[y_0+y_n+4(y_1+y_3+\cdots)+2(y_2+y_4+\cdots)]$, and it needs $n$ even.</mark>

Key points.

1. For the 1/3 rule the number of strips $n$ must be even; the odd-numbered ordinates get weight 4 and the even interior ones weight 2.
2. For the 3/8 rule $n$ must be a multiple of 3; every third interior ordinate gets weight 2 and all others weight 3.
3. The step is $h=\frac{b-a}{n}$ for both rules, and the 3/8 rule uses the factor $\frac{3h}{8}$.
4. The 1/3 rule is exact for polynomials up to degree 3 and is generally more accurate than the trapezoidal rule for the same $h$.
5. When the paper gives a table of values, use those ordinates directly and take $n$ from the number of points.
6. When no $n$ is given, choose it yourself: $n=6$ suits both rules at once.
7. To compare with the actual value, integrate exactly and quote the error.

Formula (3/8 rule).

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

Example (Dec 2020 and Jun 2023, $\pi$). $\int_0^1\frac{dx}{1+x^2}=\frac\pi4$, $n=6$, $h=\frac16$. Ordinates: 1, 0.97297, 0.90000, 0.80000, 0.69231, 0.59016, 0.50000.

Answer: $\pi\approx3.1416$ by both rules.

Results for the other asked integrals (each computed with the stated $n$):

For $\int_0^1\frac{x^2}{1+x^3}dx$ the ordinates are $0,\ 0.06154,\ 0.22222,\ 0.39560,\ 0.5$ and the exact integral is $\frac13\ln2$.

Example (Nov 2022, given data). $\int_0^4e^xdx$, $h=1$: $\frac13[1+54.60+4(2.72+20.09)+2(7.39)]=\frac{161.62}{3}=53.873$. Exact value $e^4-1=53.598$, so the error is $0.275$. Answer: 53.873 against 53.598.

Example (Jun 2022, trapezoidal and 3/8). $\int_0^1\log x\cos x\,dx$ is singular at 0 because $\log 0$ is undefined; take $f(0)=0$ as the convention, and use $n=3$, $h=\frac13$ with ordinates $0,\ -1.0381,\ -0.3186,\ 0$ (at $x=1$, $\log1=0$). Trapezoidal: $\frac{1}{6}[0+0+2(-1.3567)]=-0.4523$. Simpson 3/8: $\frac18[0+3(-1.3567)]=-0.5088$. Say plainly that the true value is about $-0.946$ and both are rough because of the singularity.

Answer frame. Open with the rule's formula and its condition on $n$ (even for 1/3, multiple of 3 for 3/8); write $h$ and the table of $x$, $y$; substitute into the formula and show the weighted sums (odd, even, ends) separately; close with the boxed value and, if asked, the comparison with the exact value or $\pi=4\times$ integral.

Pitfall: Using the 1/3 rule with an odd $n$, or giving weight 3 to $y_3$ in the 3/8 rule (it gets weight 2).

Asked: [7 marks] (May 2019, Jun 2023, Jun 2026) Apply Simpson's 1/3 rule to evaluate $\int_0^1\frac{dx}{1+x}$; use Simpson's rule for $\int_1^2\frac{dx}{x}$; calculate $\log_e7$ by Simpson's 1/3 rule correct to four decimal places. Asked: [7 marks] (Dec 2020, Jun 2023) Use Simpson's 1/3 and 3/8 rule to evaluate $\int_0^1\frac{dx}{1+x^2}$ and hence obtain the approximate value of $\pi$ in each case. Asked: [7 marks] (Dec 2020) Using Simpson 1/3 and 3/8 rules find $\int_0^1\frac{dx}{1+x^4}$. Asked: [7 marks] (Jun 2020) Find $\log2$ from $\int_0^1\frac{x^2}{1+x^3}dx$ using Simpson's 1/3 rule with four equal parts. Asked: [7 marks] (Jun 2022) Evaluate $\int_0^1\log x\cos x\,dx$ by (i) Trapezoidal rule (ii) Simpson 3/8 rule. Asked: [7 marks] (Nov 2022) Given $e^0=1,e^1=2.72,e^2=7.39,e^3=20.09,e^4=54.60$, find $\int_0^4e^xdx$ by Simpson's 1/3 rule and compare with the actual value. Asked: [7 marks] (Nov 2022) Using Simpson's 3/8 rule solve $\int_4^{5.2}\log_ex\,dx$. Asked: [7 marks] (Nov 2022) Evaluate by Simpson's rule $\int_0^1e^{-x^2}dx$.

Solution of Simultaneous Linear Algebraic Equations by Gauss's Elimination

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Definition. Gauss elimination solves $AX=B$ by reducing the augmented matrix $[A|B]$ to upper triangular form using row operations (forward elimination), then finding the unknowns from the last equation upward (back substitution).

Key points.

1. Eliminate $x$ from rows 2 and 3 using row 1, then $y$ from row 3 using row 2.
2. Choose the largest available pivot (partial pivoting) to reduce round-off error.
3. Back substitution gives $z$ first, then $y$, then $x$.
4. Always verify by substituting in all three equations.

Example (Dec 2020). $[A|B]$: $R_1=[2,-6,8\,|\,24]$; $R_2-2.5R_1=[0,19,-23\,|\,-58]$; $R_3-1.5R_1=[0,10,-10\,|\,-20]$; $R_3-\frac{10}{19}R_2=[0,0,\frac{40}{19}\,|\,\frac{200}{19}]$. Then $z=5$, $19y=-58+115\Rightarrow y=3$, $2x=24+18-40\Rightarrow x=1$. Answer: $x=1,\ y=3,\ z=5$.

Asked: [7 marks] (Dec 2020) Using Gauss elimination solve $2x-6y+8z=24$, $5x+4y-3z=2$, $3x+y+2z=16$.

Gauss's Jordan

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Definition. Gauss-Jordan method continues Gauss elimination until $A$ becomes the identity matrix, so the right-hand column is the solution directly and no back substitution is needed.

Key points.

1. Form $[A|B]$, make each pivot 1, and clear the entries above and below it.
2. It needs more row operations than Gauss elimination but gives the answer by reading the last column.
3. It is also the method for finding $A^{-1}$ by reducing $[A|I]$.
4. Check the answer in the original equations.

Example (May 2019). $[A|B]$ for $10x+y+z=12,\ 2x+10y+z=13,\ x+y+5z=7$: $R_1\to R_1/10$ gives $[1,0.1,0.1|1.2]$; $R_2-2R_1=[0,9.8,0.8|10.6]$, $R_3-R_1=[0,0.9,4.9|5.8]$; $R_2/9.8=[0,1,0.0816|1.0816]$; $R_1-0.1R_2$ and $R_3-0.9R_2$ leave $[0,0,4.8265|4.8265]$, so $z=1$; clearing the last column gives $x=1,\ y=1$. Answer: $x=y=z=1$.

Asked: [7 marks] (May 2019) Solve $10x+y+z=12;\ 2x+10y+z=13;\ x+y+5z=7$ by Gauss-Jordan method.

Crout's methods

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Definition. Crout's method factorises $A=LU$ with $L$ lower triangular and $U$ upper triangular with unit diagonal, then solves $LY=B$ by forward substitution and $UX=Y$ by back substitution.

Key points.

1. Write $u_{ii}=1$; the first column of $L$ is the first column of $A$, and the first row of $U$ is $u_{1j}=a_{1j}/l_{11}$.
2. Then $l_{ij}=a_{ij}-\sum_{k<j}l_{ik}u_{kj}$ for $i\ge j$ and $u_{ij}=\frac{1}{l_{ii}}\big(a_{ij}-\sum_{k<i}l_{ik}u_{kj}\big)$ for $i<j$.
3. Solve $LY=B$ from the top, then $UX=Y$ from the bottom.

Example (Jun 2022). $A=\begin{pmatrix}1&1&1\\3&1&-3\\1&-2&-5\end{pmatrix}$, $B=(1,5,10)^T$. $L$: $l_{11}=1,\ l_{21}=3,\ l_{31}=1$; $U$: $u_{12}=1,\ u_{13}=1$; $l_{22}=1-3=-2$, $l_{32}=-2-1=-3$, $u_{23}=\frac{-3-3}{-2}=3$, $l_{33}=-5-1+9=3$. $LY=B$: $y_1=1,\ y_2=\frac{5-3}{-2}=-1,\ y_3=\frac{10-1-3}{3}=2$. $UX=Y$: $x_3=2,\ x_2=-1-6=-7,\ x_1=1+7-2=6$. Answer: $x_1=6,\ x_2=-7,\ x_3=2$.

Asked: [7 marks] (Jun 2022) Solve $x_1+x_2+x_3=1$, $3x_1+x_2-3x_3=5$, $x_1-2x_2-5x_3=10$ using Crout's method.

Jacobi's iteration

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Definition. Jacobi's method solves a diagonally dominant system by solving each equation for its diagonal unknown and computing every new value from the previous iteration only; ==$x^{(k+1)}=\frac{1}{a_{11}}\big(b_1-a_{12}y^{(k)}-a_{13}z^{(k)}\big)$, and similarly for $y$ and $z$.==

Key points.

1. The method converges when the coefficient matrix is diagonally dominant: $|a_{ii}|>\sum_{j\ne i}|a_{ij}|$ in each row; rearrange the equations to achieve this before starting.
2. Solve equation $i$ for the $i$th unknown to get the iteration formulas.
3. Start from $(0,0,0)$ unless a guess is given.
4. In each iteration all three values are computed together from the old values, so it is simultaneous replacement.
5. Stop when two successive iterations agree to the required decimals.
6. It is slower than Gauss-Seidel, which uses the newest values at once.

Example (Nov 2022). $x=\frac{17-y+2z}{20},\ y=\frac{-18-3x+z}{20},\ z=\frac{25-2x+3y}{20}$; diagonal 20 dominates 3.

Answer: $x=1,\ y=-1,\ z=1$. For $4x+y+z=7,\ x+5y+z=-8,\ x+y+6z=6$ (Jun 2026) use $x=\frac{7-y-z}{4}$, $y=\frac{-8-x-z}{5}$, $z=\frac{6-x-y}{6}$; iteration 1 is $(1.75,-1.6,1)$, iteration 2 $(1.9,-2.15,0.975)$, and the values settle near $(2.05,-2.21,1.03)$.

Answer frame. Open with "the system is diagonally dominant, so Jacobi's iteration converges"; write the three iteration formulas; tabulate iterations from $(0,0,0)$; close with the values agreeing to the required decimals.

Asked: [7 marks] (Nov 2022, Jun 2026) Solve by Jacobi's iteration method $20x+y-2z=17$, $3x+20y-z=-18$, $2x-3y+20z=25$; and $4x+y+z=7$, $x+5y+z=-8$, $x+y+6z=6$.

Gauss-Seidal iteration

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Definition. The Gauss-Seidel method is Jacobi's method with immediate updating: each newly computed unknown is used at once in the remaining equations of the same iteration; ==$x^{(k+1)}=\frac{b_1-a_{12}y^{(k)}-a_{13}z^{(k)}}{a_{11}},\ y^{(k+1)}=\frac{b_2-a_{21}x^{(k+1)}-a_{23}z^{(k)}}{a_{22}},\ z^{(k+1)}=\frac{b_3-a_{31}x^{(k+1)}-a_{32}y^{(k+1)}}{a_{33}}.==$

Key points.

1. Check diagonal dominance first, and rearrange so the largest coefficients lie on the diagonal.
2. Solve equation 1 for $x$, equation 2 for $y$, and equation 3 for $z$.
3. Start with $(0,0,0)$, or with $y=z=0$ in the first equation.
4. Always substitute the latest available value of each variable.
5. It converges roughly twice as fast as Jacobi's method for the same system.
6. Stop when successive iterations agree to the required accuracy and state the values.

Example (May 2019, Jun 2020, Jun 2023). $10x+y+z=12,\ x+10y+z=12,\ x+y+10z=12$: $x=\frac{12-y-z}{10}$, etc.

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

Other asked systems. $10x+y+z=12,\ 2x+10y+z=13,\ 3x+2y+11z=16$: iteration 1 gives $(1.2,\ 1.06,\ 0.9345)$, iteration 2 $(1.0005,\ 1.0064,\ 0.9987)$, converging to $x=y=z=1$. $27x+6y-z=85,\ 6x+15y+2z=72,\ x+y+54z=110$: iteration 1 gives $(3.1481,\ 3.5407,\ 1.9132)$, iteration 4 gives $x=2.4255,\ y=3.573,\ z=1.926$.

Answer frame. Open with the diagonal dominance check and the iteration formulas; tabulate iterations, showing that each row uses the newest values; close with the converged values to the accuracy where two iterations agree.

Pitfall: Using the old $x$ in the $y$ equation turns Gauss-Seidel into Jacobi and gives a different iteration table.

Asked: [7 marks] (May 2019, Jun 2020, Jun 2023) Solve $10x+y+z=12;\ x+10y+z=12;\ x+y+10z=12$ by Gauss-Seidel iteration; solve $27x+6y-z=85$, $6x+15y+2z=72$, $x+y+54z=110$. Asked: [7 marks] (Jun 2020) Solve $10x+y+z=12$; $2x+10y+z=13$; $3x+2y+11z=16$ by Gauss-Seidel method.

Relaxation method

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Definition. The relaxation method solves $AX=B$ by starting from a guess, computing the residuals $R=B-AX$, and repeatedly reducing the numerically largest residual to zero.

Key points.

1. Arrange the equations with the largest diagonal terms and compute all residuals for the first guess.
2. To zero the largest residual $R_i$, change $x_i$ by $\delta x_i=R_i/a_{ii}$.
3. Every other residual changes by $R_j\leftarrow R_j-a_{ji}\,\delta x_i$; repeat until all residuals are negligibly small.
4. Over-relaxation uses a slightly larger change to speed convergence.

Last-minute revision

• Forward derivative: $f'=\frac1h[\Delta-\frac{\Delta^2}{2}+\frac{\Delta^3}{3}]$, $f''=\frac1{h^2}[\Delta^2-\Delta^3]$; backward uses $+$ signs with $\nabla$.
• 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: $\frac{3h}{8}[\text{ends}+3\,\text{others}+2\,y_3,y_6]$, $n$ multiple of 3.
• $\int_0^1\frac{dx}{1+x^2}=\frac\pi4$, so $\pi=4\times$ integral; $n=6$ gives 3.1416.
• Gauss elimination: triangular form then back substitution; Gauss-Jordan: identity on the left.
• Crout: $A=LU$ with $u_{ii}=1$, then $LY=B$, $UX=Y$.
• Jacobi: old values only; Gauss-Seidel: newest values at once; both need diagonal dominance.
• Answers: Dec 2020 Gauss elimination $(1,3,5)$; Crout $(6,-7,2)$; Jacobi $(1,-1,1)$.
• Relaxation: zero the largest residual by $\delta x=R/a_{ii}$.

Memory hooks

• "Simpson 1/3 = even, 3/8 = three's multiple."
• "Jacobi is Jealous of nobody, he waits for the whole round; Seidel uses the news at once."
• "Forward for the front, backward for the back."
• "Crout keeps the ones in U, not L."

Coverage checklist

• Numerical Differentiation: May 2019/Jun 2020 (first and second derivative), Dec 2020 (three derivatives), Jun 2020 (backward).
• Numerical integration: Trapezoidal rule: Jun 2023.
• Simpson’s 1/3rd and 3/8 rules: $\int\frac{dx}{1+x}$, $\int\frac{dx}{x}$ and $\ln7$, $\pi$ and $\frac{1}{1+x^4}$, $\ln2$, $\log x\cos x$, $e^x$ comparison, $\ln x$ (3/8), $e^{-x^2}$.
• Solution of Simultaneous Linear Algebraic Equations by Gauss’s Elimination: Dec 2020.
• Gauss’s Jordan: May 2019.
• Crout’s methods: Jun 2022.
• Jacobi’s: Nov 2022, Jun 2026.
• Gauss-Seidal: May 2019, Jun 2020, Jun 2023.
• Relaxation method: not asked recently.