Exercises — SymPy — symbolic algebra, calculus, ODE solving

KYA: Brackets open karna = sab kuch multiply out karna. WHY woh tool: expand "multiply out" wala pencil move hai; factor iska reverse hai, simplify ek heuristic hai jo yeh kare ya na kare.

x = symbols('x')
expand((x + 3)*(x - 3))     # -> x**2 - 9

Result: x**2 - 9.

KYA: roots dhundhna. WHY solve: solve(eq, x) eq ko eq = 0 maanta hai aur symbolic roots ki list return karta hai — ek akela number nahi.

x = symbols('x')
solve(x**2 - 7*x + 12, x)    # -> [3, 4]

Result: [3, 4]. Dhyan raho yeh ek list hai, isliye 3 lene ke liye sol[0] index karna hoga.

KYA: symbol swap karo, exactness rakho. WHY .subs phir simplify: .subs x ko sqrt(3) se swap karta hai lekin expression un-collapsed chhod deta hai; simplify isko combine karta hai. Kaisa dikhta hai: $(\sqrt{3}-2{)}^2=3-4\sqrt{3}+4=7-4\sqrt{3}$.

x = symbols('x')
e = (x - 2)**2
simplify(e.subs(x, sqrt(3)))   # -> 7 - 4*sqrt(3)

Result: 7 - 4*sqrt(3). Agar decimal chahiye toh .evalf() add karo → 0.0718....

KYA: ek product differentiate karna. WHY diff: yeh known rules (product + power + trig) automatically apply karta hai.

x = symbols('x')
diff(x**3 * sin(x), x)     # -> x**3*cos(x) + 3*x**2*sin(x)

Result: x**3*cos(x) + 3*x**2*sin(x). Do rules: product rule $(uv{)}^{\prime }=u^{\prime }v+u{v}^{\prime }$ jisme $u=x^3$ (power rule deta hai $u^{\prime }=3x^2$) aur $v=\sin x$ (deta hai $v^{\prime }=\cos x$).

KYA: $[0,2]$ par curve ke neeche ka area. WHY tuple form (x,a,b): yeh Fundamental Theorem ${\int }_a^{b}f=F(b)-F(a)$ apply karta hai. Haath se: $F=x^3+x$, toh $F(2)-F(0)=(8+2)-0=10$.

x = symbols('x')
integrate(3*x**2 + 1, (x, 0, 2))   # -> 10

Result: 10 (exact integer, 10.0 nahi).

KYA: $\cos x$ ko $0$ ke around order 7 tak (lekin 7 include nahi) expand karo.

x = symbols('x')
cos(x).series(x, 0, 7)
# -> 1 - x**2/2 + x**4/24 - x**6/720 + O(x**7)

Result: 1 - x**2/2 + x**4/24 - x**6/720 + O(x**7). KYUN sirf even powers: $x^n$ ka Taylor coefficient $\frac{f^{(n)}(0)}{n!}$ hota hai (dekho Taylor and Maclaurin Series). $\cos$ ke liye, $0$ par har odd derivative $\pm \sin (0)=0$ hota hai, isliye woh coefficients zero ho jaate hain. Geometrically, $\cos$ ek even function hai ($y$-axis ke across mirror-symmetric), aur even functions mein sirf even-power terms hote hain. Figure s01 dekho: mirror symmetry hi visual reason hai.

KYA: ek quadratic ke roots jiske koi real solutions nahi hain.

x = symbols('x')
solve(x**2 + x + 1, x)
# -> [-1/2 - sqrt(3)*I/2, -1/2 + sqrt(3)*I/2]

Result: [-1/2 - sqrt(3)*I/2, -1/2 + sqrt(3)*I/2]. KYUN yeh form: quadratic formula deta hai $x=\frac{-1\pm \sqrt{1-4}}{2}=\frac{-1\pm \sqrt{-3}}{2}$. Kyunki $\sqrt{-3}=\sqrt{3}i$ (jahan $i^2=-1$), SymPy isko exact aur complex rakhta hai crash karne ki bajay. Yahi complex-root logic hai jo characteristic equation method mein $r=\pm i$ ko oscillations mein badle ke liye use hota hai.

KYA: ${\lim }_{h\to 0}\frac{(x+h{)}^3-x^3}{h}$ compute karo. WHY limit aur diff nahi: hum derive karna chahte hain, assume nahi — yeh Limits and the definition of the derivative ki definition hai.

x, h = symbols('x h')
limit(((x + h)**3 - x**3)/h, h, 0)   # -> 3*x**2
diff(x**3, x)                        # -> 3*x**2  (matches!)

Result: dono 3*x**2 dete hain. Kaisa dikhta hai: expand karne par, $(x+h{)}^3-x^3=3x^{2}h+3x{h}^2+h^3$; $h$ se divide karo taaki $3x^2+3xh+h^2$ mile; $h\to 0$ hone par sirf $3x^2$ bachta hai.

KYA: teen tools chain kiye — series, .removeO(), aur lambdify. KYUN chain karo: series ek symbolic polynomial deta hai; .removeO() bookkeeping wala $O(x^6)$ strip karta hai taaki woh real polynomial ban jaye; lambdify isko fast numeric function mein compile karta hai evaluation ke liye.

x = symbols('x')
p = sin(x).series(x, 0, 6).removeO()   # x - x**3/6 + x**5/120
f = lambdify(x, p, 'math')
f(0.5)          # -> 0.4794270833333333
math.sin(0.5)   # -> 0.479425538604203

Polynomial 0.47942708... deta hai, true value 0.47942553... hai — yeh 5 decimals tak agree karte hain kyunki humne $x^5$ tak ke terms rakhe. Kaisa dikhta hai: figure s02 mein polynomial (orange) true $\sin$ (blue) par overlay hai; 0 ke paas yeh chipke hain, door jaane par alag ho jaate hain.

FORECAST: $y^{\prime }=3y$ matlab "growth rate, size ke proportional hai" → exponential $e^{3x}$. General solution $C_1{e}^{3x}$; $y(0)=5$ se $C_1=5$ pin hota hai. Toh predict hai $y=5e^{3x}$.

x = symbols('x')
y = Function('y')
sol = dsolve(Eq(y(x).diff(x) - 3*y(x), 0), y(x), ics={y(0): 5})
sol            # -> Eq(y(x), 5*exp(3*x))

Result: Eq(y(x), 5*exp(3*x)) — forecast se match karta hai. Verify: $y^{\prime }=15e^{3x}$ aur $3y=15e^{3x}$, toh $y^{\prime }-3y=0$ ✓ aur $y(0)=5e^0=5$ ✓.

KYA: structural identity aur mathematical equality mein farq samjho.

x = symbols('x')
(x + 2)**2 == x**2 + 4*x + 4        # -> False  (different trees!)
simplify((x + 2)**2 - (x**2 + 4*x + 4)) == 0   # -> True
((x + 2)**2).equals(x**2 + 4*x + 4)            # -> True

KYUN False: == expression tree ki shape compare karta hai. (x+2)**2 ek power node ke roop mein store hai; x**2+4*x+4 ek sum node hai — different structures, isliye == False hai chahe woh functions ki tarah equal hon. Fix: subtract karo aur simplify karke 0 dekho, ya .equals use karo.

KYA: improper integrals (upper limit oo hai, SymPy ki infinity).

x = symbols('x')
integrate(1/x**2, (x, 1, oo))   # -> 1
integrate(1/x,   (x, 1, oo))    # -> oo

Results: 1 aur oo. KYUN yeh farq: $1/x^2$ ka antiderivative $-1/x$ hai, jo $x\to \infty$ par $0$ approach karta hai, isliye area bounded hai: $[-1/x{]}_1^{\infty }=0-(-1)=1$. Lekin $1/x$ ka antiderivative $\ln x$ hai, jo bina bound ke badhta hai, isliye area infinite hai. Edge lesson: do curves jo dono zero ki taraf shrink hoti hain, wildly different total areas enclose kar sakti hain — $1/x^2$ itni tezi se shrink hoti hai ki ho jaata hai, $1/x$ nahi hoti.

KYA: simplification is baat par depend karta hai ki SymPy ko assume karne ki permission kya hai.

x = symbols('x')
simplify(sqrt(x**2))          # -> sqrt(x**2)  (stays, or Abs(x))
p = symbols('p', positive=True)
simplify(sqrt(p**2))          # -> p

KYUN: general real $x$ ke liye, $\sqrt{x^2}=|x|$ (socho $x=-3$: $\sqrt{9}=3=|-3|$, nahi $-3$). SymPy absolute value drop karne se mana kar deta hai kyunki woh prove nahi kar sakta ki $x\ge 0$ hai. Jab tum positive=True declare karte ho, assumption isko sirf p tak simplify karne deta hai. Edge lesson: wahi expression ke do correct answers ho sakte hain sign domain par depend karke — SymPy ko hamesha apni assumptions batao.

KYA: solve, list indexing, aur .evalf combine karo.

x = symbols('x')
roots = solve(x**2 - 2, x)     # -> [-sqrt(2), sqrt(2)]
pos = roots[1]                 # sqrt(2)
pos                            # exact: sqrt(2)
pos.evalf(6)                   # -> 1.41421

Results: exact root sqrt(2), decimal 1.41421. KYUN pehle index phir evalf: solve exact symbolic roots ki list return karta hai; hum positive wala pick karte hain, algebra ke liye exact rakhte hain, aur sirf ekdum end mein float mein collapse karte hain.

(x+2)**2 == x**2+4*x+4 False kyun return karta hai? == expression-tree structure compare karta hai, mathematical value nahi. simplify(a-b)==0 ya a.equals(b) use karo.

series ko lambdify karne se pehle kya call karna zaroori hai? .removeO(), taaki O(x**n) order term strip ho jaye jo numerically evaluable nahi hai.

${\int }_1^{\infty }\frac{1}{x}dx$ infinite kyun hai lekin ${\int }_1^{\infty }\frac{1}{x^2}dx$ finite kyun hai? Antiderivative $\ln x$ bina bound ke badhta hai; $-1/x$ ek finite limit ($1$) approach karta hai.

sqrt(x**2) ko x mein simplify kyun nahi karta? Kyunki general real $x$ ke liye yeh $|x|$ ke barabar hai. Reduction allow karne ke liye positive=True declare karo.