Second derivative test — Hessian determinant: Visual walkthrough

4.4.13 · D2 · Maths › Multivariable Calculus › Second derivative test — Hessian determinant

Neeche sab kuch sirf do cheezein assume karta hai: tum jaante ho ki function $f (x, y)$ kya hota hai (har point $(x, y)$ pe floor ke upar ek height), aur tum jaante ho ki partial derivative kya hoti hai (us height ki slope agar tum sirf ek direction mein chalo, doosre ko freeze karke). Baaki sab hum yahan build karte hain.

Step 1 — Woh flat spot jahan hum khade hain

KYA. Hum floor pe ek point $(a, b)$ choose karte hain jahan surface har direction mein flat ho. "Flat" ka matlab hai: agar tum east–west chalo toh ground ki slope zero hai, aur agar tum north–south chalo toh ground ki slope zero hai. Symbols mein,

$f_{x} (a, b) = 0 and f_{y} (a, b) = 0.$

Yahan $f_{x}$ east–west slope hai aur $f_{y}$ north–south slope. Jis point pe dono zero hoon use critical point kehte hain (dekho Critical points and gradient).

Yahan se kyun shuru karein. Flat spot hi ek aisi jagah hai jahan max, min, ya saddle chhup sakta hai. Kisi slope pe tum hamesha downhill chal sakte ho, toh tum unme se kisi pe bhi nahi ho. Flatness hi humari permission slip hai ki aage curvature dekhen.

PICTURE. Teen flat spots — ek bowl bottom, ek dome top, ek saddle pass — sab ek hi horizontal tangent plane (amber sheet) share karte hain. Slopes zero hain; sirf bending alag hai.

Figure — Second derivative test — Hessian determinant

Step 2 — Zoom in: linear part mar jaati hai, curvature hi bachti hai

KYA. Hum surface ki shape $(a, b)$ ke paas chahte hain, toh hum $f$ ko ek flat part, ek slope part, aur ek bending part ke sum ke roop mein likhte hain. Yeh second order tak ka Taylor expansion hai. Maano $u = x - a$ aur $v = y - b$ measure karte hain ki hum flat spot se kitna door gaye:

$f (a + u, b + v) \approx centre pe height f (a, b) + slope part f_{x} u + f_{y} v + bending part \frac{1}{2} (f_{xx} u^{2} + 2 f_{x y} uv + f_{y y} v^{2}) .$

$u, v$ — east aur north mein tiny steps.
$f_{xx}$ — east–west slope khud kitna change hoti hai jab tum east jaate ho ( $u$ ke along curvature).
$f_{y y}$ — $v$ ke along curvature.
$f_{x y}$ — coupling: east–west slope kitna change hoti hai jab tum north chalte ho. Yeh ek diagonal twist measure karta hai.

KYU. Kyunki hum ek critical point pe hain, $f_{x} = f_{y} = 0$ , isliye slope part poora vanish ho jaata hai. Constant $f (a, b)$ sirf poori surface ko upar ya neeche shift karta hai aur shape kabhi nahi badalta. Toh bowl vs dome vs saddle ka faisla karne wali ek hi cheez bending part hai.

PICTURE. Wahi surface, magnifying glass se blow up ki gayi. Door se curved dikhti hai; zoom in karne pe flat sheet aur parabolic bending hi bachti hai — slope arrows kuch nahi reh jaate.

Step 3 — Bachne wale ko naam do: quadratic form $Q$

KYA. Constant aur dead linear terms hata do. Jo bachta hai (times $\frac{1}{2}$ ) woh $u$ aur $v$ mein ek single expression hai:

$Q (u, v) = f_{xx} u^{2} + 2 f_{x y} uv + f_{y y} v^{2} .$

Yeh ek quadratic form hai (dekho Quadratic forms and definiteness): $u, v$ mein har term degree two ki hai. Isse left to right padhte hain:

$f_{xx} u^{2}$ — pure east–west bending, hamesha $f_{xx}$ jaisi sign.
$2 f_{x y} uv$ — twist; yeh $(u, v)$ plane ke kuch corners mein positive hai aur kuch mein negative, kyunki $uv$ axes ke across sign flip karta hai.
$f_{y y} v^{2}$ — pure north–south bending.

KYU. Flat spot ke paas height hai $f (a, b) + \frac{1}{2} Q$ . Toh surface jahan $Q > 0$ hoga wahan upar curve karti hai aur jahan $Q < 0$ hoga wahan neeche. Hamara poora problem ek sawaal tak chhota ho jaata hai:

Kya $Q$ hamesha positive hai, hamesha negative, ya dono signs leta hai?

Hamesha positive → valley (minimum). Hamesha negative → peak (maximum). Dono signs → saddle.

PICTURE. $(u, v)$ floor $Q$ ki sign se coloured: cyan jahan $Q > 0$ , amber jahan $Q < 0$ . Bowl poora cyan hai; saddle plane ko cyan aur amber wedges mein split karta hai.

Step 4 — Obstacle: woh middle term $2 f_{x y} uv$

KYA. Agar twist term na hota, life easy hoti: $Q = f_{xx} u^{2} + f_{y y} v^{2}$ , do clean squares jinki signs hum bas dekh lete. Problem $2 f_{x y} uv$ hai jo $u$ aur $v$ ko mix karta hai, toh ek nazar mein sign nahi bata sakte.

KYU tool chahiye. Hum completing the square use karte hain — classic move jo $a u^{2} + b u + c$ ko $a (u + \frac{b}{2 a})^{2} + leftover$ likhta hai. Yeh tool kyun aur factoring kyun nahi? Kyunki perfect square $(\dots)^{2}$ guaranteed $\geq 0$ hota hai; ek baar messy cross-term ko ek square ke andar bottle kar diya, toh jo bache woh ek clean, isolated coefficient ban jaata hai jis pe hum sign-test kar sakte hain. Factoring ke liye pehle roots pata hone chahiye; completing the square kabhi yeh nahi maangta.

Hume assume karna hoga ki $f_{xx} \neq = 0$ taaki ise common factor ke roop mein bahar nikal sakein. (Step 8 mein $f_{xx} = 0$ alag se handle kiya gaya hai.)

PICTURE. $Q$ ki tilted trough: ridge/valley line angle pe chalti hai $f_{x y}$ ki wajah se. Completing the square dono coordinates ko us tilt ke saath line up karne ke liye rotate karna hai.

Step 5 — Algebra karo: square appear hota hai

KYA. $u$ wale do terms se $f_{xx}$ bahar nikalo:

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

Ab bracket ke andar square complete karo. Pattern $u^{2} + 2 k uv = (u + k v)^{2} - k^{2} v^{2}$ with $k = \frac{f _{x y}}{f _{xx}}$ deta hai:

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

Term by term:

$(u + \frac{f _{x y}}{f _{xx}} v)^{2}$ — ek perfect square, kabhi negative nahi. Iska coefficient $f_{xx}$ hai.
$(f_{y y} - \frac{f _{x y}^{2}}{f _{xx}}) v^{2}$ — ek doosra square $v^{2}$ (kabhi negative nahi) times ek leftover coefficient. Woh subtracted $\frac{f _{x y}^{2}}{f _{xx}}$ exactly woh price hai jo twist ne charge ki.

KYU. Ab humne $Q$ ko (positive coefficient?) times (square) + (leftover coefficient) times (square) ke roop mein likha hai. Kyunki dono squares $\geq 0$ hain, $Q$ ki sign purely do saamne wale coefficients se decide hoti hai.

PICTURE. Do coefficient "dials". Ek $f_{xx}$ padhta hai; doosra leftover padhta hai. Unki signs green (positive) ya red (negative) light up hoti hain.

Step 6 — Number $D$ paida hota hai

KYA. Leftover coefficient ko common denominator pe tidy karo:

$f_{y y} - \frac{f _{x y}^{2}}{f _{xx}} = \frac{f _{xx} f _{y y} - f _{x y}^{2}}{f _{xx}} = \frac{D}{f _{xx}}, D := f_{xx} f_{y y} - f_{x y}^{2} .$

Woh numerator $f_{xx} f_{y y} - f_{x y}^{2}$ exactly Hessian matrix $H = (f_{xx} f_{x y} f_{x y} f_{y y})$ ka determinant hai. (Do off-diagonal entries Clairaut's theorem ki wajah se equal hain.) Toh $Q$ ban jaata hai:

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

Ab har symbol earned hai. Do coefficients hain $f_{xx}$ aur $\frac{D}{f _{xx}}$ .

Subtraction kyun, product kyun nahi. Notice karo $D$ hai $f_{xx} f_{y y}$ minus $f_{x y}^{2}$ . Woh $- f_{x y}^{2}$ us twist ki memory hai jo humne absorb ki. Isse ignore karo (parent mein Mistake 1) aur tum poori tilt throw away kar dete ho.

PICTURE. Second derivatives ka two-by-two grid jis pe "diagonal product minus off-diagonal square" arrows draw hain — determinant visible ho gaya.

Step 7 — Teen main cases padhna

KYA. $Q = f_{xx} (square_{1}) + \frac{D}{f _{xx}} (square_{2})$ . Dono squares $\geq 0$ . Toh:

KYU yeh watertight hai. $D < 0$ ke liye humne haath nahi hilaaya — humne construct kiya ek direction jahan $Q > 0$ hai aur ek aur jahan $Q < 0$ . Yahi saddle ki honest definition hai.

PICTURE. Teen side-by-side height plots coefficient signs ke hisaab se keyed: all-up bowl, all-down dome, aur split saddle jiske do coloured escape directions arrowed hain.

Step 8 — Degenerate edges: $D = 0$ aur $f_{xx} = 0$

KYA. Do loose ends jo tidy algebra ne skip kiye.

Case $f_{xx} = 0$ (factor nahi kar saka). Tab $Q = 2 f_{x y} uv + f_{y y} v^{2} = v (2 f_{x y} u + f_{y y} v)$ . Agar $f_{x y} \neq = 0$ toh yeh do independent linear factors ka product hai, toh dono lines cross karte waqt sign change karta hai → saddle ( $D = f_{xx} f_{y y} - f_{x y}^{2} = - f_{x y}^{2} < 0$ ke consistent). Agar $f_{x y} = 0$ bhi ho, tab $Q = f_{y y} v^{2}$ — poore $u$ -axis ke along flat → degenerate, aur actually $D = 0$ .

Case $D = 0$ . Step 6 mein ek coefficient vanish ho jaati hai, toh $Q$ ek single square (ya zero) hai — yeh $\geq 0$ (ya $\leq 0$ ) rehta hai lekin ek poori line ke along zero ko touch karta hai, sirf ek point pe nahi. Us valley-floor line ke along quadratic term koi information nahi deta; actual shape third- ya fourth-order terms se decide hoti hai jo humne throw away kiye. Isliye inconclusive. Example: $f = x^{4} + y^{4}$ ka origin pe genuine minimum hai phir bhi $D = 0$ .

KYU inhe apna step chahiye. Contract: reader ko kabhi ek aisa scenario nahi milna chahiye jo humne nahi dikhaya. $f_{xx} = 0$ factoring break karta hai; $D = 0$ strict inequality break karta hai. Dono real inputs hain, toh dono handle hote hain.

PICTURE. Left: $x^{4} + y^{4}$ , ek flat-bottomed bowl jiska base ek point hai lekin plate jaisa curve karta hai — $D = 0$ , phir bhi min. Right: ek "monkey saddle" gutter jahan $Q$ ek line ke along vanish hota hai. Dono second-order test ko defeat karte hain.

Ek picture mein summary

KYA. Poora logic ek single map pe: flat spot → linear terms drop karo → quadratic form $Q$ → square complete karo → coefficients $f_{xx}$ aur $D / f_{xx}$ → sign combinations → verdict.

Recall Feynman retelling — poori walk plain words mein

Tum wahan khade ho jahan ground tumhare pair ke neeche bilkul level hai — jis taraf bhi mudo slope nahi. Yeh jaanne ke liye ki tum valley mein ho, hill pe ho, ya mountain pass pe, tum flatness ignore karte ho (woh kuch nahi batata) aur study karte ho ki ground kaise curve karta hai. Curving ko ek expression $Q$ ke roop mein likho. Iske do parts hain: ek friendly part (pure east–west aur north–south bending) aur ek annoying twisty part ( $f_{x y} uv$ term) jo sab kuch tilt karta hai. "Completing the square" trick tumhara view rotate karti hai taaki tilt ek neat squared bracket ke andar chhup jaaye, aur peeche do plain dials bachte hain: ek $f_{xx}$ padhta hai, doosra $D / f_{xx}$ padhta hai, jahan $D = f_{xx} f_{y y} - f_{x y}^{2}$ . Agar dono dials positive point karein, ground har jagah upar curve karta hai — valley (min). Dono negative — peak (max). Ek upar, ek neeche — tum saddle pe ho, kyunki tumne ek direction upar jaata mila aur doosra neeche. Aur agar koi dial exactly zero padhe, ground ek poori line ke along flat hai aur curving akela decide nahi kar sakta — tumhe aur close dekhna padega. Woh single number $D$ detector hai; $f_{xx}$ ki sign bas "smile or frown" hai.

Active recall

Yahan Taylor expansion mein linear terms kyun drop hote hain?

Critical point pe

f_{x} = f_{y} = 0

, toh poora slope part zero ho jaata hai.

Square complete karne ke baad

Q

ke do coefficients kya hain?

f_{xx}

aur

D / f_{xx}

, dono ek non-negative square multiply karte hain.

D

mein

- f_{x y}^{2}

kahan se aata hai?

Square complete karte waqt

- k^{2} v^{2}

leftover se, jahan

k = f_{x y} / f_{xx}

D < 0

definitely saddle kyun hai?

Do coefficients ki opposite signs hain, toh

Q

ek direction mein positive aur doosre mein negative hai.

f_{xx} = 0

ho toh derivation ka kya hota hai?

Ise factor out nahi kar sakte;

Q = v (2 f_{x y} u + f_{y y} v)

likho — sign-changing product (saddle) jab tak

f_{x y} = 0

bhi na ho (degenerate,

D = 0

D = 0

inconclusive kyun hai?

Q

ek single square ban jaata hai jo ek poori line ke along vanish hota hai; second-order info wahan flat hai, toh higher terms decide karte hain.

Connections

Second derivative test — Hessian determinant — woh parent rule jise yeh page derive karta hai.
Critical points and gradient — Step 1, jahan $\nabla f = 0$ .
Taylor series multivariable — Step 2, expansion.
Hessian matrix — Step 6, $D = det H$ .
Quadratic forms and definiteness — Steps 3 & 7, $Q$ ka sign analysis.
Clairaut's theorem — $f_{x y} = f_{y x}$ kyun, jo $H$ ko symmetric banata hai.
Lagrange multipliers — is optimisation ka constrained cousin.

Visual walkthrough — Second derivative test — Hessian determinant