Introduction: A Polynomial Model for Change
A Taylor polynomial centered at \(a\) approximates a function near \(a\). When the question is about how much the function changes from \(x=a\) to \(x=a+h\), the relevant quantity is not \(f(a+h)\) by itself, but the difference \(f(a+h)-f(a)\).
Subtracting the function’s value at the center removes the constant term of its Taylor polynomial. What remains separates the modeled change into contributions: a linear contribution, a quadratic contribution, a cubic contribution, and so on. The linear term describes the first-order effect of the displacement \(h\); higher-order terms adjust that model.
The Taylor Polynomial and Its Change Form
Suppose \(f\) has derivatives through order \(n\) at \(a\), where \(n\) is a positive integer. The degree-\(n\) Taylor polynomial for \(f\) centered at \(a\) is
Here \(f^{(0)}(a)\) means \(f(a)\), and \(0!=1\). Setting \(x=a+h\) gives \(x-a=h\), so
At the center, every positive power of \(x-a\) is zero. Thus \(T_n(a)=f(a)\). Subtracting this value yields the Taylor-polynomial model for the change:
The first terms, when present, are
The linear contribution is \(f'(a)h\). It is the change predicted by the tangent-line model. The quadratic and higher-order contributions modify that prediction. A term can contribute zero if its derivative coefficient is zero, or if the corresponding power of \(h\) is zero.
Why Subtraction Produces the Change Formula
The change formula follows directly from the definition of the Taylor polynomial. This short derivation also explains why the constant term must disappear.
Substitute \(x=a+h\), so each power \((x-a)^k\) becomes \(h^k\).
Substitute \(x=a\). Every term with \(k\geq 1\) is zero, leaving \(T_n(a)=f(a)\).
The \(k=0\) term in \(T_n(a+h)\) is \(f(a)\), so it cancels with \(T_n(a)\). The terms indexed from \(1\) to \(n\) remain.
In equations, the cancellation is
This is an identity for the Taylor polynomial. It does not, by itself, claim that the polynomial change equals the actual function change. The latter is approximated by the polynomial change; the quality of that approximation depends on the function and on the displacement. No error bound is implied by this formula alone.
Worked Example: An Exact Change for a Cubic
Worked Example: An Exact Change for a Cubic
Let \(f(x)=x^3-2x\), and model the change from \(a=1\) to \(a+h=1.2\), so \(h=0.2\). Use the degree-3 Taylor polynomial.
First calculate the derivatives and their values at the center:
The Taylor polynomial is therefore
At \(x=1.2\), the displacement is \(h=0.2\). The contributions to the modeled change are
Adding these contributions gives
Check by evaluating the polynomial at both inputs. At the center, \(T_3(1)=-1\). At \(1.2\),
Thus \(T_3(1.2)-T_3(1)=-0.672-(-1)=0.328\), matching the sum of the contributions. Since \(f\) is a cubic and we used its degree-3 Taylor polynomial, this change is exact. Direct substitution confirms it: \(f(1.2)=1.728-2.4=-0.672\), while \(f(1)=-1\), so \(f(1.2)-f(1)=0.328\).
How the Terms Contribute
For a small displacement, the powers \(h^2,h^3,\ldots\) are often smaller in magnitude than \(h\). This helps explain why the linear term is a useful first model of change. But “higher order” does not mean “always negligible”: the derivative coefficients matter, and the displacement may not be small. The polynomial records the contributions without deciding in advance that any one of them can be ignored.
The sign of each contribution depends on both its derivative coefficient and the power of \(h\). If \(h\) is negative, an odd power of \(h\) is negative and an even power is positive. A negative derivative coefficient can reverse the contribution’s sign. Adding signed terms—not just their magnitudes—is essential.
Some terms may be absent. If \(f''(a)=0\), the quadratic contribution is zero. If \(f\) has symmetry that forces certain Taylor coefficients to vanish, the corresponding powers do not appear. The change formula simply includes the coefficients that are nonzero.
Worked Example: Exponential Change Through Cubic Order
Worked Example: Exponential Change Through Cubic Order
Let \(f(x)=e^x\), centered at \(a=1\), and take \(h=0.1\). Find the degree-3 Taylor-polynomial model for the change.
Every derivative of \(e^x\) is \(e^x\), so the first three derivatives at \(1\) are all \(e\). The change contributions are
Therefore,
For a check, the polynomial centered at \(1\) is
Thus \(T_3(1)=e\), and substituting \(x=1.1\) gives
Subtracting \(T_3(1)=e\) leaves \(631e/6000\), as before. This is the polynomial’s modeled change, not an assertion that the exponential function’s actual change is exactly that value.
Worked Example: A Missing Linear Contribution
Worked Example: A Missing Linear Contribution
Let \(f(x)=\cos x\), centered at \(a=0\), and consider \(h=0.2\). Use the degree-4 Taylor polynomial.
The values needed at the center are \(f(0)=1\), \(f'(0)=0\), \(f''(0)=-1\), \(f'''(0)=0\), and \(f^{(4)}(0)=1\). Thus
There is no linear contribution because \(f'(0)=0\), and no cubic contribution because \(f'''(0)=0\). The remaining contributions to the change are
Since \(-0.02=-300/15000\), the total is
Check by substitution: \(T_4(0)=1\), while \(T_4(0.2)=1-0.02+1/15000\). Their difference is \(-0.02+1/15000=-299/15000\), which agrees with the term-by-term calculation.
A Common Mistake: Keeping the Constant Term
The constant term \(f(a)\) is part of the Taylor polynomial’s estimate for the function value. It is not part of the change, because it is present at both the starting and ending inputs and cancels when the values are subtracted.
Worked Example: Correcting a Constant-Term Error
Take \(f(x)=x^2\), center \(a=2\), and displacement \(h=0.1\). The degree-2 Taylor polynomial is exact here:
A common incorrect calculation substitutes \(h=0.1\) into \(T_2(a+h)\) and calls the whole result the change:
This is the modeled value \(T_2(2.1)\), not the modeled change. At the starting point, \(T_2(2)=4\), so the correction is to subtract that starting value:
The contribution formula gives the same result: \(f'(2)h=4(0.1)=0.4\), and \(f''(2)h^2/2=2(0.01)/2=0.01\), so \(0.4+0.01=0.41\). Finally, check against the actual function: \(f(2.1)-f(2)=(2.1)^2-2^2=4.41-4=0.41\). The polynomial is exact because \(f\) is a degree-2 polynomial.
Using the Change Model Carefully
To apply the formula reliably, specify the center \(a\), the displacement \(h\), and the degree \(n\). Then evaluate the derivatives at the center, include the factorial in every coefficient, and sum the signed contributions from \(k=1\) through \(n\). A final subtraction check—\(T_n(a+h)-T_n(a)\)—helps catch the common error of reporting a function-value estimate as a change.
The formula is useful because it organizes local change by powers of the displacement. It makes clear which part is linear, which parts correct the linear model, and which coefficients vanish. For a polynomial of degree at most \(n\), the model gives the exact change. For other functions, it is a polynomial approximation; deciding how accurate it is requires additional information about the function and is a separate question.
Check Your Understanding
Answer each question using the Taylor-polynomial change formula.
- Why does the constant term \(f(a)\) disappear when forming \(T_n(a+h)-T_n(a)\)?
- Write the degree-3 Taylor-polynomial model for the change in terms of \(f'(a)\), \(f''(a)\), \(f'''(a)\), and \(h\).
- If \(f'(a)=0\), which contribution is absent, and what terms may still contribute to the change?
- For \(f(x)=x^2\), centered at \(a=3\), and \(h=-0.1\), calculate the linear and quadratic contributions to the change model.
- When does a degree-\(n\) Taylor polynomial give an exact change for a polynomial function?
