M3 Numerical Methods — Formula Sheet kuch kuch important topics

 

M3 Numerical Methods — Formula Sheet

1. Newton-Raphson Method

xn+1=xnf(xn)f(xn)\boxed{x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}}

Yaad rakho:

New x=Old xFunctionDerivative\boxed{\text{New }x=\text{Old }x-\frac{\text{Function}}{\text{Derivative}}}

2. Regula-Falsi Method

Pehle interval (a,b)(a,b) choose karo such that:

f(a)f(b)<0\boxed{f(a)f(b)<0}

Formula:

x=af(b)bf(a)f(b)f(a)\boxed{x=\frac{a f(b)-b f(a)}{f(b)-f(a)}}

Equivalent form:

x=af(a)(ba)f(b)f(a)\boxed{x=a-\frac{f(a)(b-a)}{f(b)-f(a)}}

Phir f(x)f(x) ka sign check karke aa ya bb replace karte hain.


3. Newton Forward Interpolation

Kab? Jab required xx, table ke beginning ke paas ho.

y=y0+uΞ”y0+u(u1)2!Ξ”2y0+u(u1)(u2)3!Ξ”3y0+\boxed{ y= 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 }

Where:

u=xx0h\boxed{u=\frac{x-x_0}{h}}

and

h=x1x0h=x_1-x_0

Pattern:

u,u(u1),u(u1)(u2)u,\quad u(u-1),\quad u(u-1)(u-2)

Matlab Forward = minus pattern.


4. Newton Backward Interpolation

Kab? Jab required xx, table ke end ke paas ho.

y=yn+uyn+u(u+1)2!2yn+u(u+1)(u+2)3!3yn+\boxed{ y= y_n +u\nabla y_n +\frac{u(u+1)}{2!}\nabla^2y_n +\frac{u(u+1)(u+2)}{3!}\nabla^3y_n+\cdots }

Where:

u=xxnh\boxed{u=\frac{x-x_n}{h}}

Pattern:

u,u(u+1),u(u+1)(u+2)u,\quad u(u+1),\quad u(u+1)(u+2)

Matlab Backward = plus pattern.

πŸ”₯ Forward vs Backward trick

Forward → beginning → u1,u2\boxed{\text{Forward → beginning → }u-1,u-2} Backward → end → u+1,u+2\boxed{\text{Backward → end → }u+1,u+2}

5. Trapezoidal Rule

Given:

abf(x)dx\int_a^b f(x)\,dx

Step size:

h=ban\boxed{h=\frac{b-a}{n}}

Formula:

I=h2[y0+yn+2(y1+y2++yn1)]\boxed{ I=\frac{h}{2} [ y_0+y_n+ 2(y_1+y_2+\cdots+y_{n-1}) ] }

Coefficient pattern

1,  2,  2,  2,,  2,  1\boxed{1,\;2,\;2,\;2,\ldots,\;2,\;1}

6. Simpson's 1/3 Rule

Condition:

n=even\boxed{n=\text{even}}

Step size:

h=ban\boxed{h=\frac{b-a}{n}}

Formula:

I=h3[y0+yn+4(y1+y3+y5+)+2(y2+y4+y6+)]\boxed{ I= \frac{h}{3} [ y_0+y_n +4(y_1+y_3+y_5+\cdots) +2(y_2+y_4+y_6+\cdots) ] }

⭐ Coefficient pattern

1,  4,  2,  4,  2,  4,,  1\boxed{1,\;4,\;2,\;4,\;2,\;4,\ldots,\;1}

Easy trick:

Odd positions → 4
Even positions → 2


7. Simpson's 3/8 Rule

Condition:

n=multiple of 3\boxed{n=\text{multiple of 3}}

Like:

n=3,6,9,12,n=3,6,9,12,\ldots

Step size:

h=ban\boxed{h=\frac{b-a}{n}}

Formula:

I=3h8[y0+yn+3(y1+y2+y4+y5+)+2(y3+y6+y9+)]\boxed{ I= \frac{3h}{8} [ y_0+y_n+ 3(y_1+y_2+y_4+y_5+\cdots) + 2(y_3+y_6+y_9+\cdots) ] }

⭐ Coefficient pattern

1,  3,  3,  2,  3,  3,  2,,  1\boxed{1,\;3,\;3,\;2,\;3,\;3,\;2,\ldots,\;1}

Easy trick:

Every third interior point → 2
Remaining interior points → 3


8. Runge-Kutta Fourth Order — First Order

Given differential equation:

dydx=f(x,y)\boxed{\frac{dy}{dx}=f(x,y)}

We calculate four values:

k1=hf(xn,yn)\boxed{k_1=hf(x_n,y_n)} k2=hf(xn+h2,yn+k12)\boxed{ k_2= hf\left( x_n+\frac h2, y_n+\frac{k_1}{2} \right) } k3=hf(xn+h2,yn+k22)\boxed{ k_3= hf\left( x_n+\frac h2, y_n+\frac{k_2}{2} \right) } k4=hf(xn+h,yn+k3)\boxed{ k_4= hf(x_n+h,y_n+k_3) }

Finally:

yn+1=yn+16(k1+2k2+2k3+k4)\boxed{ y_{n+1} = y_n+ \frac{1}{6} (k_1+2k_2+2k_3+k_4) }

πŸ”₯ RK4 memory trick

Remember:

1,  2,  2,  1\boxed{1,\;2,\;2,\;1}

because final formula:

k1+2k2+2k3+k46\frac{k_1+2k_2+2k_3+k_4}{6}

9. Runge-Kutta Fourth Order — Second Order

Suppose:

d2ydx2=f(x,y,dydx)\boxed{\frac{d^2y}{dx^2}=f\left(x,y,\frac{dy}{dx}\right)}

Pehle put:

z=dydx\boxed{z=\frac{dy}{dx}}

Then second-order equation ko two first-order equations mein convert karo:

dydx=z\boxed{\frac{dy}{dx}=z}

and

dzdx=f(x,y,z)\boxed{\frac{dz}{dx}=f(x,y,z)}

Ab RK4 simultaneously dono par lagta hai.

For yy:

k1=hznk_1=hz_n

and for zz:

l1=hf(xn,yn,zn)l_1=hf(x_n,y_n,z_n)

Then:

k2=h(zn+l12)k_2=h\left(z_n+\frac{l_1}{2}\right) l2=hf(xn+h2,yn+k12,zn+l12)l_2= hf\left( x_n+\frac h2, y_n+\frac{k_1}{2}, z_n+\frac{l_1}{2} \right)

Then:

k3=h(zn+l22)k_3=h\left(z_n+\frac{l_2}{2}\right) l3=hf(xn+h2,yn+k22,zn+l22)l_3= hf\left( x_n+\frac h2, y_n+\frac{k_2}{2}, z_n+\frac{l_2}{2} \right)

Then:

k4=h(zn+l3)k_4=h(z_n+l_3) l4=hf(xn+h,yn+k3,zn+l3)l_4= hf(x_n+h,y_n+k_3,z_n+l_3)

Finally:

yn+1=yn+k1+2k2+2k3+k46\boxed{ y_{n+1} = y_n+ \frac{k_1+2k_2+2k_3+k_4}{6} }

and

zn+1=zn+l1+2l2+2l3+l46\boxed{ z_{n+1} = z_n+ \frac{l_1+2l_2+2l_3+l_4}{6} }

🧠 One-Page Memory Table

MethodFormula / Main Trick
Newton-Raphsonxn+1=xnf(xn)f(xn)\displaystyle x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}
Regula-Falsix=af(b)bf(a)f(b)f(a)\displaystyle x=\frac{af(b)-bf(a)}{f(b)-f(a)}
Newton Forwardu=(xx0)/hu=(x-x_0)/h, minus pattern
Newton Backwardu=(xxn)/hu=(x-x_n)/h, plus pattern
Trapezoidalh2[1,2,2,,2,1]\frac h2[1,2,2,\ldots,2,1]
Simpson 1/3h3[1,4,2,4,2,,1]\frac h3[1,4,2,4,2,\ldots,1], nn even
Simpson 3/83h8[1,3,3,2,3,3,2,,1]\frac{3h}{8}[1,3,3,2,3,3,2,\ldots,1], nn multiple of 3
RK4 First Orderk1,k2,k3,k4k_1,k_2,k_3,k_4, final weights 1,2,2,1
RK4 Second OrderPut z=yz=y', then use kk's + ll's

⭐ Sabse important formulas

Exam ke liye agar priority banaoge, to especially ye patterns pakke rakho:

Newton-Raphson: Function/Derivative\boxed{\text{Newton-Raphson: Function/Derivative}} Forward: 1,2,3\boxed{\text{Forward: }-1,-2,-3} Backward: +1,+2,+3\boxed{\text{Backward: }+1,+2,+3} Trapezoidal: 1,2,2,2,1\boxed{\text{Trapezoidal: }1,2,2,2,1} Simpson 1/3: 1,4,2,4,2,1\boxed{\text{Simpson 1/3: }1,4,2,4,2,1} Simpson 3/8: 1,3,3,2,3,3,1\boxed{\text{Simpson 3/8: }1,3,3,2,3,3,1} RK4: 1,2,2,1\boxed{\text{RK4: }1,2,2,1}

Next best step is har method ka ek RGPV-level numerical solve karna, kyunki formula dekhna easy hai but exam mein question se h,n,u,k1,k2h,n,u,k_1,k_2 etc. kaise nikalte hain wahi main part hai.

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