Choosing a Base Point for Accurate Linearization

Why the Base Point Matters

A linearization approximates a function near a point where its value and first derivatives are known. In practice, the target input may be awkward: its coordinates may make the function value or derivatives difficult to calculate. A nearby point with simpler coordinates can make the calculation manageable.

That convenience has a cost. The farther the chosen point is from the target, the less local the approximation may be. Curvature also matters: a function that bends substantially between the two points can make a linear estimate inaccurate, even when the points seem close. Choosing a base point therefore means balancing easy calculations against closeness and the function’s behavior nearby.

Key idea: Choose a point where the function value and derivatives are easy to compute, but check that it is sufficiently close to the target and that the function does not vary too sharply between them.

Linearization at a Chosen Point

For a differentiable function $f(x,y)$, let $(a,b)$ be the base point. Its linearization there is the affine function formed from the value and first partial derivatives at that same point:

L_{(a,b)}(x,y) = f(a,b)+f_x(a,b)(x-a)+f_y(a,b)(y-b).

To estimate $f$ at a target $(x_0,y_0)$, substitute the target coordinates into $L_{(a,b)}$. The values $f(a,b)$, $f_x(a,b)$, and $f_y(a,b)$ must all be evaluated at the base point. Using a convenient base point does not change the target; it changes the linear function used to estimate the value there.

In three variables, the same construction uses all three first partial derivatives:

L_{(a,b,c)}(x,y,z) = f(a,b,c)+f_x(a,b,c)(x-a)+f_y(a,b,c)(y-b)+f_z(a,b,c)(z-c).

The base point should be in the domain of $f$, and the derivatives used in the formula must exist there. The linearization is most useful for a target near that point; differentiability guarantees a good approximation as the target approaches the base point, but does not by itself promise a specified error at a particular, fixed target.

A Practical Way to Choose

When the target coordinates make direct evaluation inconvenient, look for nearby coordinates that simplify the function and its derivatives. Integers, zeroes, or other inputs that reduce powers and products are common candidates. Then evaluate the complete linearization at the target and assess the distance and curvature between the two points.

Identify candidate points.
Choose points in the domain near the target, favoring coordinates that simplify the function and its derivatives.

Compute each linearization consistently.
At each candidate, calculate the function value and every first partial derivative at that candidate.

Evaluate at the target.
Substitute the target into each linearization to obtain the competing estimates.

Assess likely error.
Consider how far the target is from the base point and, when available, use a second-derivative bound on the segment between them.

Distance alone is not a complete accuracy test. A closer base point is often helpful, but a function’s curvature can vary from place to place. A remainder bound gives a more informative assessment when the required second derivatives can be bounded throughout the relevant region.

Remainder bound: Suppose $f$ has continuous second partial derivatives on a region containing the line segment from $p$ to $q$. If the operator norm of its Hessian is at most $M$ along that segment, then $$ |f(q)-L_p(q)|\leq \frac{M}{2}\|q-p\|^2. $$ Here $L_p$ is the linearization at $p$, and $\|\cdot\|$ is the Euclidean norm.

Why the Remainder Bound Works

Restrict the function to the straight segment from the base point $p$ to the target $q$, by defining $g(t)=f(p+t(q-p))$ for $0\leq t\leq 1$. The chain rule gives $g'(0)=\nabla f(p)\cdot(q-p)$, which is exactly the linear change used by $L_p(q)$. Applying the one-variable Taylor formula with integral remainder gives:

f(q)-L_p(q) = g(1)-g(0)-g'(0) = \int_0^1 (1-t)g''(t)\,dt.

Along the segment, $g''(t)=(q-p)^\top H_f(p+t(q-p))(q-p)$, where $H_f$ is the Hessian matrix of second partial derivatives. The operator norm bound implies $|g''(t)|\leq M\|q-p\|^2$. Therefore:

|f(q)-L_p(q)| \leq \int_0^1(1-t)M\|q-p\|^2\,dt =\frac{M}{2}\|q-p\|^2.

This argument explains both parts of the estimate: the error decreases quadratically with distance, while larger curvature, represented by $M$, allows a larger error. The bound can be conservative; it guarantees an upper limit, not the actual error.

Worked Example: A Simple Nearby Integer Point

Estimate $f(1.04,0.97)$ for $f(x,y)=x^2+xy+y^2$, using $(1,1)$ as the base point. Its coordinates simplify the function and its derivatives.

First compute the function value and partial derivatives:

f(1,1)=1^2+(1)(1)+1^2=3, \qquad f_x=2x+y, \qquad f_y=x+2y.

At $(1,1)$, $f_x(1,1)=2(1)+1=3$ and $f_y(1,1)=1+2(1)=3$. The target increments are $1.04-1=0.04$ and $0.97-1=-0.03$. Thus:

L_{(1,1)}(1.04,0.97) =3+3(0.04)+3(-0.03) =3+0.12-0.09 =3.03.

Check against direct evaluation: $1.04^2=1.0816$, $(1.04)(0.97)=1.0088$, and $0.97^2=0.9409$. Therefore $f(1.04,0.97)=1.0816+1.0088+0.9409=3.0313$, so the estimate’s error, estimate minus exact value, is $3.03-3.0313=-0.0013$.

For this quadratic, the Taylor remainder is exactly $(\Delta x)^2+\Delta x\Delta y+(\Delta y)^2$. Here it is $0.04^2+(0.04)(-0.03)+(-0.03)^2=0.0016-0.0012+0.0009=0.0013$. The estimate is lower by precisely that amount.

Comparing Convenience and Closeness

Two plausible base points can produce different errors. Comparing them makes the trade-off concrete: a point with especially simple coordinates may be easier to use, while a nearby alternative may give a closer estimate.

Worked Example: Comparing Two Base Points

Use $f(x,y)=x^2+xy+y^2$ to estimate its value at $(1.08,0.93)$. Compare the convenient base point $A=(1,1)$ with the closer point $B=(1.1,0.9)$. The exact target value, for checking, is:

$$ f(1.08,0.93)=1.08^2+(1.08)(0.93)+0.93^2 =1.1664+1.0044+0.8649 =3.0357. $$

At $A=(1,1)$, the function value is $3$, and both partial derivatives equal $3$. The target increments are $0.08$ and $-0.07$, so:

$$ L_A(1.08,0.93)=3+3(0.08)+3(-0.07)=3+0.24-0.21=3.03. $$

The error is $3.03-3.0357=-0.0057$. The exact remainder is $0.08^2+(0.08)(-0.07)+(-0.07)^2=0.0064-0.0056+0.0049=0.0057$, confirming the error.

At $B=(1.1,0.9)$, $f(B)=1.1^2+(1.1)(0.9)+0.9^2=1.21+0.99+0.81=3.01$. The derivatives are $f_x(B)=2(1.1)+0.9=3.1$ and $f_y(B)=1.1+2(0.9)=2.9$. The increments from $B$ to the target are $-0.02$ and $0.03$. Therefore:

$$ L_B(1.08,0.93) =3.01+3.1(-0.02)+2.9(0.03) =3.01-0.062+0.087 =3.035. $$

The error is $3.035-3.0357=-0.0007$. The exact remainder is $(-0.02)^2+(-0.02)(0.03)+(0.03)^2=0.0004-0.0006+0.0009=0.0007$, again confirming the error. Here $B$ is closer to the target and gives the more accurate estimate, while $A$ uses simpler coordinates. Both estimates come from correctly computed linearizations.

Checking a Base Point in Three Variables

The same selection principle applies when the input has three coordinates. A nearby point with simple coordinates can make the function and all three first partial derivatives straightforward to calculate.

Worked Example: A Three-Variable Estimate

Estimate $f(1.02,0.99,1.03)$ for $f(x,y,z)=x^2+y^2+z^2+xy$, using $(1,1,1)$ as the base point. The function value there is $1+1+1+1=4$. Its partial derivatives are $f_x=2x+y$, $f_y=2y+x$, and $f_z=2z$, which at $(1,1,1)$ equal $3$, $3$, and $2$, respectively.

The coordinate changes are $0.02$, $-0.01$, and $0.03$. Substitution into the linearization gives:

L_{(1,1,1)}(1.02,0.99,1.03) =4+3(0.02)+3(-0.01)+2(0.03) =4+0.06-0.03+0.06 =4.09.

For a direct check, $1.02^2=1.0404$, $0.99^2=0.9801$, $1.03^2=1.0609$, and $(1.02)(0.99)=1.0098$. Thus $f(1.02,0.99,1.03)=1.0404+0.9801+1.0609+1.0098=4.0912$. The estimate error is $4.09-4.0912=-0.0012$.

The exact quadratic remainder is $(0.02)^2+(-0.01)^2+(0.03)^2+(0.02)(-0.01)=0.0004+0.0001+0.0009-0.0002=0.0012$, which confirms the estimate is lower by $0.0012$.

A Common Mistake: Mixing Base Points

A linearization uses one point consistently. It is not valid to take the function value at a convenient point and combine it with derivatives evaluated at the target. The resulting expression is not the linearization at either point.

Worked Example: Correcting an Inconsistent Calculation

Return to $f(x,y)=x^2+xy+y^2$, with target $(1.08,0.93)$ and base point $(1,1)$. A mistaken calculation keeps $f(1,1)=3$ but uses derivatives at the target: $f_x(1.08,0.93)=2(1.08)+0.93=3.09$ and $f_y(1.08,0.93)=1.08+2(0.93)=2.94$. It then computes:

$$ 3+3.09(0.08)+2.94(-0.07) =3+0.2472-0.2058 =3.0414. $$

This is not a valid linearization estimate because the value and derivatives come from different points. At the chosen base point $(1,1)$, the derivatives are both $3$; the correct estimate is $3+3(0.08)+3(-0.07)=3.03$. Comparing with the exact value $3.0357$, its error is $-0.0057$. The correction is simple: evaluate every ingredient of the linearization at the same base point.

How to Judge Whether the Estimate Is Useful

After calculating an estimate, separate what is known from what is only suggested. The distance from the base point to the target is known, and a remainder bound may be available if the second derivatives can be controlled along the segment. Without such a bound, proximity and smooth-looking behavior are reasons to expect an estimate to work, not a guarantee of a particular error.

In practice, a useful choice often has three features:

The base point lies in the domain and is near the target.
The function value and first derivatives there are easy to evaluate accurately.
The function’s curvature on the intervening segment is not so large that the remainder overwhelms the linear change.

When candidate points are similarly convenient, prefer one that is closer to the target, provided the function behaves regularly between them. When one point is much easier to calculate, compare the convenience gained with the larger distance and, where possible, use a remainder bound to quantify the trade-off. A linearization is an estimate, not an exact value; its usefulness depends on both the selected point and the function near that point.

Check Your Understanding

Answer these questions using the definitions and examples above.

In a linearization at $(a,b)$, where must the function value and both first partial derivatives be evaluated?
Why might a nearby integer-coordinate point be convenient, and what accuracy concern comes with choosing it?
For a twice continuously differentiable function, what quantities appear in the stated remainder bound?
Why does a smaller distance between base point and target not, on its own, guarantee a specified error?
What is wrong with combining a function value at one point with derivatives evaluated at another?