Second derivative test — Hessian determinant

4.4.13 · Maths › Multivariable Calculus

WHY does this work? (Derivation from scratch)

Hum jaanna chahte hain ki $f$ ki shape critical point $(a, b)$ ke paas kaisi hai. Taylor expand karo second order tak. Kyunki wahan $f_{x} = f_{y} = 0$ hai, linear terms gayab ho jaate hain aur quadratic part sab kuch control karta hai. Maano $u = x - a$ , $v = y - b$ :

$f (a + u, b + v) \approx f (a, b) + \frac{1}{2} (f_{xx} u^{2} + 2 f_{x y} uv + f_{y y} v^{2})$

Shape poori tarah quadratic form $Q (u, v) = f_{xx} u^{2} + 2 f_{x y} uv + f_{y y} v^{2}$ par depend karti hai. Hum poochhte hain: kya $Q$ hamesha positive hai (bowl), hamesha negative (dome), ya dono signs (saddle)?

Step — complete the square. Maano $f_{xx} \neq = 0$ :

$Q = f_{xx} (u^{2} + \frac{2 f _{x y}}{f _{xx}} uv) + f_{y y} v^{2}$

Yeh step kyun? $f_{xx}$ ko bahar nikaalne se hum $u$ mein square complete kar sakte hain, aur ek messy cross-term ko kuch aisi cheez mein badal sakte hain jiska sign hum analyse kar sakein.

$Q = f_{xx} (u + \frac{f _{x y}}{f _{xx}} v)^{2} + (f_{y y} - \frac{f _{x y}^{2}}{f _{xx}}) v^{2}$

Yeh step kyun? Squared bracket hamesha $\geq 0$ hota hai. Toh sign behaviour ab do coefficients par hinge karti hai.

Doosre coefficient ko multiply karo: $f_{y y} - \frac{f _{x y}^{2}}{f _{xx}} = \frac{f _{xx} f _{y y} - f _{x y}^{2}}{f _{xx}} = \frac{D}{f _{xx}}$

Toh: $Q = f_{xx} (u + \frac{f _{x y}}{f _{xx}} v)^{2} + \frac{D}{f _{xx}} v^{2}$

Ab cases padho:

Agar $D > 0$ aur $f_{xx} > 0$ : dono coefficients $f_{xx}$ aur $D / f_{xx}$ positive hain → $Q > 0$ hamesha → minimum.
Agar $D > 0$ aur $f_{xx} < 0$ : dono coefficients negative → $Q < 0$ hamesha → maximum. (Note: $D > 0$ force karta hai ki $f_{xx}$ aur $f_{y y}$ ka same sign ho, isliye sirf $f_{xx}$ check karna kaafi hai.)
Agar $D < 0$ : $f_{xx}$ aur $D / f_{xx}$ ke opposite signs hain → $Q$ ko positive bhi banaya ja sakta hai ( $v = 0$ lo) ya negative bhi (bracket ko zero karo) → saddle.
Agar $D = 0$ : ek coefficient vanish ho jaata hai, quadratic info ek direction mein flat hai → inconclusive.

Figure — Second derivative test — Hessian determinant

Worked examples

Common mistake Common errors ko steel-man karna

Galti 1: $(f_{x y})^{2}$ term bhool jaana. $D = f_{xx} f_{y y}$ likhna. Kyun sahi lagta hai: "do curvatures ka product" natural lagta hai. Fix: cross-term $f_{x y}$ encode karta hai ki curvature axes kitne tilted hain — isko ignore karna waisa hai jaise Pringle ka tilt ignore karo. Minus sign completing the square se aata hai.

Galti 2: $D > 0$ hone par $f_{y y}$ use karna aur alag answer paana. Kyun sahi lagta hai: "koi bhi second derivative lo." Fix: jab $D > 0$ hota hai, $f_{xx}$ aur $f_{y y}$ ka same sign forced hota hai, toh dono agree karte hain — lekin agar algebra se different signs aayein, toh arithmetic mein galti hai. Standard ke taur par $f_{xx}$ use karo.

Galti 3: $D = 0$ ko "koi extremum nahi" samajhna. Kyun sahi lagta hai: "fail = kuch nahi" jaisa lagta hai. Fix: $D = 0$ matlab inconclusive hai — phir bhi min, max, ya saddle ho sakta hai (e.g. $f = x^{4} + y^{4}$ ka origin par min hai lekin $D = 0$ hai). Seedha investigate karo.

Galti 4: Test ko non-critical point par apply karna. Taylor expansion linear terms tabhi drop karta hai kyunki gradient zero hai. Pehle hamesha $\nabla f = 0$ solve karo.

Recall Feynman: ek 12-saal ke bachche ko samjhao

Socho tum ek pahari par khade ho jahan tumhare paon ke neeche zameen bilkul flat hai (kisi bhi direction mein koi slope nahi). Kya tum ek ghati ke neeche ho, ek pahari ki choti par ho, ya ek mountain pass par ho? Dekho ki zameen kaise curve karti hai. Agar har direction mein upar curve kare, tum ghati mein ho (minimum). Agar har jagah neeche curve kare, tum choti par ho (maximum). Agar ek taraf upar aur doosri taraf neeche curve kare — jaise ghode ki saddle par baithna — tum ek pass par ho. "Hessian determinant" sirf ek number hai jo hum curving se calculate karte hain jo humein batata hai ki yeh teeno mein se kaun sa hai, jaise ek magic detector.

Active recall

f (x, y)

ke liye Hessian determinant

D

kya hai?

D = f_{xx} f_{y y} - (f_{x y})^{2}

Second derivative test apply karne se pehle kya true hona chahiye?

Point critical hona chahiye:

f_{x} = f_{y} = 0

D > 0

aur

f_{xx} > 0

se kya imply hota hai?

Local minimum.

D > 0

aur

f_{xx} < 0

se kya imply hota hai?

Local maximum.

D < 0

se kya imply hota hai?

Saddle point.

D = 0

se kya imply hota hai?

Test inconclusive hai — seedha investigate karo.

Test sirf second-order Taylor terms kyun use karta hai?

Critical point par gradient (linear terms) zero hota hai, isliye quadratic form local shape control karta hai.

Hessian ke eigenvalues ke terms mein

D

kya hai?

D = λ_{1} λ_{2}

(curvatures ka product).

(f_{x y})^{2}

subtract kyun karna chahiye, ignore kyun nahi?

Yeh completing the square se aata hai;

f_{x y}

dono curvature directions ka tilt/coupling encode karta hai.

Origin par

f = x^{2} - y^{2}

ke liye

D

kya hai aur conclusion kya hai?

D = - 4 < 0

, saddle point.

Connections

Critical points and gradient — $\nabla f = 0$ dhundhna.
Taylor series multivariable — is derivation ki foundation.
Hessian matrix — eigenvalues, definiteness, $n$ -dimensional generalisation.
Quadratic forms and definiteness — positive/negative definite ↔ min/max.
Clairaut's theorem — $f_{x y} = f_{y x}$ kyun.
Lagrange multipliers — constrained optimisation ka cousin.

Second derivative test — Hessian determinant

WHAT is the Hessian?

HOW the test works

WHY does this work? (Derivation from scratch)

Worked examples

Active recall

Connections

Concept Map

Deep Dive