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

True — the Vandermonde system is square and its determinant ${\prod }_{i<j}(x_j-x_i)\ne 0$ for distinct nodes, so the coefficients are uniquely pinned.

False — the unique interpolant has degree $\le n$; it can drop lower. Three collinear points give a degree-1 line, not a genuine quadratic.

False — by the uniqueness theorem there is only one degree-$\le n$ interpolant; the two forms are different recipes that expand to the identical curve.

False — divided differences are symmetric in their arguments, so reordering nodes never changes the value of any bracket.

True — every $L_i$ contains a factor for each other node, so a new node changes all numerators and denominators; Newton's form only appends one term.

True — the constant polynomial $P(x)=y_0$ already hits every point, and by uniqueness it is the interpolant; all higher divided differences are zero.

False — high-degree equispaced interpolation oscillates violently near the ends (the Runge phenomenon); accuracy can get worse, not better.

True — it equals $\frac{y_1-y_0}{x_1-x_0}$, the ordinary rise-over-run of the chord.

True — $E(x)=f(x)-P(x)$ contains the factor ${\prod }_{i=0}^n(x-x_i)$, which vanishes at each node, so the interpolant is exact there.

False — the leading coefficient equals the highest divided difference $f[x_0,\dots ,x_n]$ either way, since it is the single degree-$n$ contribution and the polynomial is unique.

Missing the denominator: without dividing by ${\prod }_{j\ne i}(x_i-x_j)$ we get $L_i(x_i)\ne 1$, so $P(x_i)\ne y_i$. The normalization is essential.

A repeated node breaks distinctness and makes a Vandermonde row duplicate, killing uniqueness; the standard formulas divide by $x_i-x_j$, which becomes $0$.

The point $\xi$ is unknown (only guaranteed to exist), so the formula gives a bound via $\max |f^{(n+1)}|$, not an exact number.

You take the top diagonal — $f[x_0],f[x_0,x_1],f[x_0,x_1,x_2],\dots$ — those are the coefficients of the forward form starting at $x_0$.

Cancellations between the basis polynomials can lower the degree; the sum has degree $\le n$, exactly $n$ only when the top divided difference is nonzero.

It needs $f$ to be $(n+1)$-times differentiable on the interval; the Rolle-theorem argument uses that many derivatives.

So the $k$-th term vanishes at all previously-fixed nodes $x_0,\dots ,x_{k-1}$; this lets you add it without disturbing values already matched.

The node product ${\prod }_{i=0}^n(x-x_i)$ is a product of the distances $|x-x_i|$; for a point $x$ inside the interval those distances are on average moderate, but for $x$ near a boundary the distances to the far nodes are large, so the whole product swells there and amplifies the error.

They cluster nodes toward the interval ends, which shrinks the maximum of $\left|{\prod }_{i=0}^n(x-x_i)\right|$, flattening the error product instead of letting it blow up.

A new point only appends one term (the next divided difference times a new product factor); all earlier coefficients stay valid, so no recomputation.

A divided difference is a discrete "slope of slopes"; letting all nodes merge turns each difference quotient into a derivative, recovering the Taylor series coefficient.

It makes the Vandermonde determinant ${\prod }_{i<j}(x_j-x_i)$ nonzero, so the coefficient system is invertible — a genuine existence-and-uniqueness statement.

Splines use many low-degree pieces, avoiding the wild global oscillations of one high-degree polynomial while staying smooth at the joins.

The constant $P(x)=y_0$ — a degree-$0$ polynomial, the unique interpolant of a single point.

Degree $1$ — the line through them already fits, so the second divided difference is $0$ and the quadratic term disappears.

A factor $x_i-x_j=0$ appears in a denominator, so $L_i$ is undefined — the method requires strictly distinct nodes.

All zero, because every secant slope is $0$ and slopes-of-slopes stay $0$; only $f[x_i]=c$ survives.

Exactly zero everywhere — then $f^{(n+1)}\equiv 0$, so the error formula gives $0$ and $P$ reproduces $f$ perfectly.

Here yes, because $e^x$ is entire with bounded derivatives, but this is not automatic — for functions like $\frac{1}{1+25x^2}$ on $[-1,1]$ equispaced interpolation diverges (Runge).