Second derivative test — Hessian determinant: Question bank

4.4.13 · D5 · Maths › Multivariable Calculus › Second derivative test — Hessian determinant

Quick reminder un tools ka jo test kiye ja rahe hain (sab parent note mein scratch se banaye gaye hain):

Critical point woh hota hai jahan dono first partials zero hon: $f_{x} = f_{y} = 0$ .
Hessian determinant (ek single number jo us point par compute hota hai) hai $D = f_{xx} f_{y y} - (f_{x y})^{2}$ .
$D > 0$ aur $f_{xx} > 0$ → minimum; $D > 0$ aur $f_{xx} < 0$ → maximum; $D < 0$ → saddle; $D = 0$ → test kuch nahi kehta.

True ya false — justify karo

Har statement ya to subtly sahi hai ya subtly galat. Reveal mein reason milega, sirf verdict nahi.

Agar kisi critical point par

f_{xx} > 0

ho, to woh point local minimum hai.

False.

f_{xx} > 0

sirf itna kehta hai ki surface

x

-direction mein upar curve karta hai; agar

D < 0

ho to woh doosri direction mein neeche curve karta hai aur tumhare paas saddle hai. Pehle

D

check karna zaroori hai, aur tabhi

f_{xx}

ka sign dekho.

Agar

D > 0

ho to

f_{xx}

aur

f_{y y}

ka sign same hona chahiye.

True. Kyunki

D = f_{xx} f_{y y} - (f_{x y})^{2} > 0

se milta hai

f_{xx} f_{y y} > (f_{x y})^{2} \geq 0

, to product

f_{xx} f_{y y}

strictly positive hai — matlab dono factors ka sign same hai. Isliye sirf

f_{xx}

check karna kaafi hai.

Agar

D < 0

ho to point phir bhi local minimum ho sakta hai.

False.

D < 0

force karta hai ki dono curvature eigenvalues ke signs opposite hon, isliye

Q

positive aur negative dono values leta hai — yahi saddle ki definition hai. Jab

D < 0

ho tab minimum impossible hai.

D > 0

guarantee karta hai ki point genuine extremum (min ya max) hai, saddle nahi.

True.

D > 0

matlab dono eigenvalues (principal axes ke saath curvatures) ka sign same hai, to surface har direction mein same taraf curve karta hai — ek bowl ya dome. Phir

f_{xx}

ka sign decide karta hai kaun sa.

Ek critical point par jahan

f_{xx} = 0

aur

f_{y y} = 0

ho, point saddle hona chahiye.

True (agar cross-term nonzero ho). Tab

D = 0 \cdot 0 - (f_{x y})^{2} = - (f_{x y})^{2}

. Agar

f_{x y} \neq = 0

to yeh negative hai → saddle. Sirf tab jab

f_{x y}

bhi zero ho,

D = 0

hota hai aur test khamosh ho jaata hai.

Agar

D = 0

ho, to us point par koi extremum nahi hai.

False.

D = 0

matlab inconclusive hai, "kuch nahi" nahi.

f = x^{4} + y^{4}

ke liye origin clearly minimum hai phir bhi wahan

D = 0

hai — second-order Taylor terms flat hain aur tumhe haath se higher order dekhna padta hai.

Hessian hamesha symmetric hota hai, isliye

f_{x y} = f_{y x}

automatically hota hai.

True jab bhi mixed partials continuous hon — yeh Clairaut's theorem hai. Is course ke smooth functions ke liye yeh hamesha hold karta hai, isliye

D

mein sirf ek cross-term

(f_{x y})^{2}

chahiye.

Jab tak Hessian sahi se compute karo, test kisi bhi point par apply ho sakta hai.

False. Poori derivation linear Taylor terms drop karti hai isliye kyunki wahan gradient zero hai. Non-critical point par woh linear terms dominate karte hain aur

D

shape ke baare mein kuch nahi batata.

D = f_{xx} f_{y y}

ek theek shortcut hai jab cross-term chhota ho.

False.

(f_{x y})^{2}

drop karna curvature axes ka tilt chhod deta hai. Chhota

f_{x y}

bhi

D

ko positive se negative kar sakta hai — minus sign ka pura point yahi hai ki directions ke beech coupling matter karta hai.

Error dhundho

Har item mein ek (galat) student ki reasoning di gayi hai. Reveal mein flaw ka naam hai.

(0, 0)

par,

f_{xx} = 2

aur

f_{y y} = - 2

hai, aur

D > 0

hai, to yeh min hai kyunki

f_{xx} > 0

." Flaw dhundho.

Dono premises ek doosre se contradict karte hain:

f_{xx} = 2

f_{y y} = - 2

deta hai

f_{xx} f_{y y} = - 4 < 0

, isliye

D

positive ho hi nahi sakta. Jab

D > 0

ho to dono second derivatives same sign share karte hain — alag signs ek arithmetic slip indicate karta hai.

f_{x} = f_{y} = 0

ka koi solution nahi, isliye koi critical point nahi aur function ka koi minimum nahi." Flaw dhundho.

Interior par minimum ke liye critical point zaroori hai, lekin extrema boundary par ya infinity par bhi ho sakte hain (ya function simply unbounded ho sakta hai). "Koi interior critical point nahi" ka matlab "kahi bhi minimum nahi" nahi hota.

D < 0

hai, to point inconclusive hai — main haath se investigate karoonga." Flaw dhundho.

D < 0

poori tarah decisive hai: yeh saddle hai. Sirf

D = 0

inconclusive hai. Student ne dono failure-jaisi cases ko confuse kiya.

"Maine

D > 0

find kiya, to point maximum hai." Flaw dhundho.

D > 0

akele max aur min mein kabhi distinguish nahi karta — sirf yeh kehta hai "same-sign curvatures." Finish karne ke liye phir bhi

f_{xx}

ka sign padhna padta hai (

> 0

min,

< 0

max).

"Taylor expansion ne diya

f (a, b) + f_{x} u + f_{y} v + \frac{1}{2} Q

, aur maine

Q

mein linear terms rakhe." Flaw dhundho.

Critical point par

f_{x} = f_{y} = 0

hota hai, isliye linear terms identically zero hain aur

Q

purely quadratic hai. Nonzero linear terms rakhna matlab tumne actually critical point par land hi nahi kiya.

D

eigenvalues ka sum hai, isliye

D > 0

matlab positive curvature." Flaw dhundho.

D = det H = λ_{1} λ_{2}

product hai, sum nahi. Product positive hona matlab same sign; trace

f_{xx} + f_{y y} = λ_{1} + λ_{2}

sum hai, aur sirf trace ka sign batata hai min vs max.

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

ke liye, kyunki

f_{x y} = 0

hai, surface mein koi coupling nahi aur

D

zero hona chahiye." Flaw dhundho.

f_{x y} = 0

karta hai

D = f_{xx} f_{y y} - 0 = f_{xx} f_{y y}

, jo is bowl ke liye

2 \cdot 2 = 4 > 0

hai — bilkul nonzero. Zero cross-term ka matlab hai axes tilted nahi hain, nahi ki

D

vanish karta hai.

Why questions

Apne aap ko reason dene par majboor karo, sirf rule nahi.

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

Critical point par linear terms vanish ho jaate hain, isliye quadratic form leading behaviour hai — chhote displacements ke liye yeh dominate karta hai. Higher-order terms negligible hain jab tak quadratic form degenerate na ho (

D = 0

), aur theek usi waqt hum unpar fall back karte hain.

Completing the square

Q

ka sign analyse karne ka sahi tool kyun hai?

Raw quadratic

f_{xx} u^{2} + 2 f_{x y} uv + f_{y y} v^{2}

mein

u

aur

v

mix hote hain, isliye iska sign obvious nahi. Completing the square ise (positive square) plus (coefficient)

\times v^{2}

ke roop mein rewrite karta hai, jahan har squared piece

\geq 0

hai — ab sign purely do coefficients

f_{xx}

aur

D / f_{xx}

par depend karta hai.

Coefficient

D / f_{xx}

completed square mein naturally kyun aata hai?

u

mein completing the square karne par bacha hua

v^{2}

-coefficient

f_{y y} - \frac{f _{x y}^{2}}{f _{xx}}

hota hai, jo common denominator par

\frac{f _{xx} f _{y y} - f _{x y}^{2}}{f _{xx}} = \frac{D}{f _{xx}}

hai. To

D

invented nahi hai — yeh algebra se naturally nikalta hai second curvature ke roop mein.

D < 0

saddle kyun produce karta hai, na ki bahut flat extremum?

D < 0

ke saath, dono coefficients

f_{xx}

aur

D / f_{xx}

ke opposite signs hote hain.

v = 0

lo to

Q

ek sign ka ho jaata hai, phir bracket ko zero karo aur

Q

doosre sign ka — surface genuinely ek direction mein utha aur doosre mein gira. Yeh flat spot nahi, pass-through hai.

Jab

D > 0

ho to tiebreaker ke roop mein

f_{y y}

ki jagah

f_{xx}

conventional kyun hai?

Jab

D > 0

ho dono sign mein agree karte hain, isliye dono kaam karenge;

f_{xx}

sirf standard choice hai. Agar woh disagree karein, tumne galti ki hai — ek built-in consistency check.

D

ko eigenvalues se connect karna ek saath sab kuch kyun explain karta hai?

Symmetric Hessian ke eigenvalues principal axes ke saath curvatures hain. Unka product

D

hai: same sign (

D > 0

) bowl/dome hai, opposite sign (

D < 0

) saddle hai. Pura test sirf "do curvatures ke signs padho" hai.

Hessian ka symmetric hona

- (f_{x y})^{2}

form ke liye zaroori kyun hai?

Single cross-term

(f_{x y})^{2}

assume karta hai

f_{x y} = f_{y x}

; symmetry ke bina do alag off-diagonal entries honge aur

det H = f_{xx} f_{y y} - f_{x y} f_{y x}

hoga. Clairaut's theorem guarantee karta hai ki continuous mixed partials ke liye woh equal hain, jo ise

- (f_{x y})^{2}

tak collapse karta hai.

Edge cases

Woh scenarios jinhe clean rules quietly assume kar lete hain.

Origin par

f = x^{4} + y^{4}

ke liye test kya kehta hai?

Wahan sab second partials zero hain, isliye

D = 0

— inconclusive. Phir bhi origin clearly minimum hai (har displacement

f

badhata hai). Yeh "

D = 0

needs a by-hand argument" ka poster child hai.

Origin par

f = x^{3}

ke liye (do variables ka function,

y

absent) test kya kehta hai?

f_{xx} = 6 x = 0

aur

f_{y y} = f_{x y} = 0

origin par, isliye

D = 0

— inconclusive. Directly,

x^{3}

x < 0

ke liye negative aur

x > 0

ke liye positive jaata hai, isliye yeh na max hai na min: ek degenerate inflection jo test classify nahi kar sakta.

Agar Hessian ke dono eigenvalues zero hon, to

D

kya hai aur kya conclude kar sakte ho?

D = λ_{1} λ_{2} = 0

, isliye test khamosh hai. Quadratic form identically zero hai aur saari shape information higher-order terms mein chhup jaati hai — aage expand karna padega.

Agar

f_{xx} = 0

lekin

f_{y y} \neq = 0

ho to "complete the square" derivation ka kya hoga?

f_{xx}

se divide nahi kar sakte, isliye uski jagah

v

mein completing the square karo (

f_{y y}

use karke). Conclusion identical hai —

u

aur

v

ke roles symmetric hain — kyunki

D

khud dono variables mein symmetric hai.

Kya kisi function ka

D > 0

aur

f_{xx} > 0

ek aise point par ho sakta hai jo sirf local minimum ho, global nahi?

Haan. Test purely local hai — yeh ek infinitesimal neighbourhood mein curvature padhta hai. Function ke kai alag basins ho sakte hain, har ek local min, ek doosre se deeper; Hessian sirf immediate bowl dekhta hai.

Ek critical point par jahan

f

ek direction mein constant ho (plateau), kya hota hai?

Ek curvature zero hai, isliye woh eigenvalue

0

hai aur

D = λ_{1} \cdot 0 = 0

— inconclusive. Surface us direction mein second order tak flat hai, isliye second-order data classify nahi kar sakta.

Kya

D > 0

possible hai jab

f_{xx}

aur

f_{y y}

dono zero hon?

Nahi. Agar dono zero hain to

D = 0 \cdot 0 - (f_{x y})^{2} = - (f_{x y})^{2} \leq 0

, jo kabhi positive nahi ho sakta.

D > 0

ke liye zaroori hai ki dono diagonal entries nonzero hon aur matching sign ki hon.

Recall Self-check: tum kis trap mein sabse zyada fase?

Sabse common trap ::: $D$ check karne se pehle $f_{xx}$ ke sign se min/max conclude karna — hamesha pehle $D$ compute karo. Sabse common "false failure" ::: $D < 0$ ko inconclusive padhna; yeh decisively saddle hai. Sirf $D = 0$ inconclusive hai.

Connections

Second derivative test — Hessian determinant — parent note jis par ye traps targeted hain.
Critical points and gradient — har item assume karta hai ki tumne pehle $\nabla f = 0$ solve kar liya.
Taylor series multivariable — kyun sirf second-order terms aate hain.
Hessian matrix — symmetry, eigenvalues, determinant.
Quadratic forms and definiteness — min/max/saddle ke peeche sign-analysis.
Clairaut's theorem — kyun $f_{x y} = f_{y x}$ .
Lagrange multipliers — is unconstrained test ka constrained cousin.