Question bank — Polynomial interpolation — Lagrange form, Newton's divided differences

True — Vandermonde system square hai aur uska determinant ${\prod }_{i<j}(x_j-x_i)\ne 0$ hai distinct nodes ke liye, isliye coefficients uniquely pin ho jaate hain.

False — unique interpolant ki degree $\le n$ hoti hai; ye neeche ja sakti hai. Teen collinear points ek degree-1 line dete hain, genuine quadratic nahi.

False — uniqueness theorem ke according sirf ek hi degree-$\le n$ interpolant hota hai; dono forms alag recipes hain jo identical curve mein expand hoti hain.

False — divided differences apne arguments mein symmetric hote hain, isliye nodes ko reorder karne se kisi bhi bracket ki value nahi badlti.

True — har $L_i$ mein har doosre node ka ek factor hai, isliye naya node saare numerators aur denominators badal deta hai; Newton's form sirf ek term append karta hai.

True — constant polynomial $P(x)=y_0$ already har point ko hit karta hai, aur uniqueness se ye the interpolant hai; saare higher divided differences zero hain.

False — high-degree equispaced interpolation ends ke paas violently oscillate karta hai (Runge phenomenon); accuracy better nahi, worse ho sakti hai.

True — ye $\frac{y_1-y_0}{x_1-x_0}$ ke barabar hai, chord ka ordinary rise-over-run.

True — $E(x)=f(x)-P(x)$ mein factor ${\prod }_{i=0}^n(x-x_i)$ hai, jo har node par zero hota hai, isliye wahan interpolant exact hai.

False — leading coefficient dono taraf highest divided difference $f[x_0,\dots ,x_n]$ ke barabar hai, kyunki ye single degree-$n$ contribution hai aur polynomial unique hai.

Denominator missing hai: ${\prod }_{j\ne i}(x_i-x_j)$ se divide kiye bina $L_i(x_i)\ne 1$ hoga, toh $P(x_i)\ne y_i$. Normalization zaroori hai.

Ek repeated node distinctness tod deta hai aur Vandermonde row duplicate ban jaata hai, uniqueness khatam kar deta hai; standard formulas $x_i-x_j$ se divide karte hain, jo $0$ ban jaata hai.

Point $\xi$ unknown hai (sirf exist karne ki guarantee hai), isliye formula $\max |f^{(n+1)}|$ ke through ek bound deta hai, exact number nahi.

Tum top diagonal lete ho — $f[x_0],f[x_0,x_1],f[x_0,x_1,x_2],\dots$ — ye $x_0$ se shuru hone wali forward form ke coefficients hain.

Basis polynomials ke beech cancellations degree ko kam kar sakti hain; sum ki degree $\le n$ hoti hai, exactly $n$ tabhi jab top divided difference nonzero ho.

Isko $f$ ko $(n+1)$-baar differentiable hona chahiye interval par; Rolle-theorem argument utne derivatives use karta hai.

Taaki $k$-th term saare pehle-se-fix nodes $x_0,\dots ,x_{k-1}$ par zero ho jaaye; isse tum usse add kar sako bina pehle matched values ko disturb kiye.

Node product ${\prod }_{i=0}^n(x-x_i)$ distances $|x-x_i|$ ka product hai; interval ke andar ek point $x$ ke liye ye distances average moderate hoti hain, lekin boundary ke paas wale $x$ ke liye door nodes tak distances badi hoti hain, isliye poora product wahan swell karta hai aur error amplify hota hai.

Ye nodes interval ke ends ki taraf cluster karte hain, jo $\left|{\prod }_{i=0}^n(x-x_i)\right|$ ke maximum ko shrink karta hai, error product ko flat karta hai instead of blow up hone dene ke.

Ek naya point sirf ek term append karta hai (next divided difference times ek naya product factor); saare pehle wale coefficients valid rehte hain, toh koi recomputation nahi.

Ek divided difference discrete "slope of slopes" hai; saare nodes ko merge karne par har difference quotient ek derivative ban jaata hai, Taylor series coefficient recover karta hai.

Ye Vandermonde determinant ${\prod }_{i<j}(x_j-x_i)$ ko nonzero banata hai, isliye coefficient system invertible hai — ye ek genuine existence-and-uniqueness statement hai.

Splines kai low-degree pieces use karte hain, ek high-degree polynomial ke wild global oscillations se bachte hain aur joins par smooth rehte hain.

Constant $P(x)=y_0$ — degree-$0$ polynomial, ek single point ka unique interpolant.

Degree $1$ — unse guzarne wali line already fit hai, isliye second divided difference $0$ hai aur quadratic term disappear ho jaata hai.

Ek factor $x_i-x_j=0$ denominator mein aayega, isliye $L_i$ undefined ho jaata hai — method ke liye strictly distinct nodes zaroori hain.

Saare zero, kyunki har secant slope $0$ hai aur slopes-of-slopes $0$ rehte hain; sirf $f[x_i]=c$ bachta hai.

Har jagah exactly zero — tab $f^{(n+1)}\equiv 0$, isliye error formula $0$ deta hai aur $P$ f ko perfectly reproduce karta hai.

Yahan haan, kyunki $e^x$ entire hai bounded derivatives ke saath, lekin ye automatic nahi hai — $\frac{1}{1+25x^2}$ jaisi functions ke liye $[-1,1]$ par equispaced interpolation diverge karta hai (Runge).