Almost every error bound in this course comes from one fact. A smooth function agrees with its Taylor polynomial almost exactly. And "almost" has an honest, computable size.
A Taylor polynomial is a stand-in for a harder function. It's built from derivatives at one point. The remainder term is the exact leftover between the real function and that stand-in. It is not a vague "it's probably close." It's a formula with the same shape as the next term in the series, just evaluated at some in-between point nobody ever has to actually find. Every error bound in this course works the same way: pin down the worst case that unknown point could produce, and that worst case becomes the guarantee.
Suppose f has n+1 continuous derivatives on an interval containing a and x. Then Taylor's theorem with the Lagrange remainder says f(x) = \underbrace{\sum_{k=0}^{n}\frac{f^{(k)}(a)}{k!}(x-a)^k}_{\text{degree-}n\text{ Taylor polynomial }P_n(x)} + \underbrace{\frac{f^{(n+1)}(\xi)}{(n+1)!}(x-a)^{n+1}}_{\text{remainder }R_n(x)} Here \xi is some point strictly between a and x. It exists, but its exact location stays unknown. The remainder is not an approximation of the error. It is an exact formula for it. Its one unknown, \xi, gets bounded rather than pinned down: replace f^{(n+1)}(\xi) with the largest value \left|f^{(n+1)}\right| takes on the whole interval. The equality then becomes a usable inequality.
Watch the remainder shrink
Increase the degree n and watch P_n(x) hug f(x) more tightly. Drag x farther from the center a and notice the same degree buys less accuracy.
Loading interactive…
Worked example
Bounding the error of a linear approximation
-
1
Bound the error of using P_1(x)=1+x to approximate e^x on [0,0.1].
-
2
R_1(x) = \frac{f''(\xi)}{2!}x^2 = \frac{e^\xi}{2}x^2 for some \xi between 0 and x. On [0,0.1], e^\xi \le e^{0.1}\approx1.105.
-
3|R_1(0.1)| \le \frac{1.105}{2}(0.1)^2 \approx 0.0055
-
4
The true value e^{0.1}\approx1.1052 and P_1(0.1)=1.1 differ by about 0.0052 — inside the bound, as guaranteed.
Loading interactive…
Worked example
Why the bound gets worse farther from the center
-
1
Compare the guaranteed error bound for P_2(x) approximating \sin(x) (centered at 0) at x=0.5 versus x=2.
-
2
The remainder bound scales with |x-a|^{3} for a degree-2 polynomial. So moving from x=0.5 to x=2 — four times farther from the center — inflates the worst-case bound by a factor of 4^3=64.
-
3\boxed{\text{same polynomial, 64$\times$ worse guaranteed accuracy}}
This is the entire reason step size matters throughout this course: the same approximation degree buys dramatically less accuracy farther from where it was built.
Loading interactive…
Where this goes next
This course's own error bounds
Every truncation-error formula this course derives makes the same move. Write down two or three terms of a Taylor expansion. Then bound what's left over, using exactly this theorem. That's how this course gets a finite-difference derivative approximation, a quadrature rule, and a Taylor method for an ODE. Nothing new is introduced here. This section exists so that move never has to be re-derived from scratch later.
Quick review
Practice from earlier coursework, pulled in for this topic.
From Calculus II — Integral Calculus & Infinite Series
Could a function have a Taylor series that converges everywhere, yet fails to equal the function anywhere except the center?
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
The series converging (as a series of numbers) is a separate fact from it converging specifically TO f(x) — the remainder must be shown to vanish for that stronger conclusion.
This situation is a known, real phenomenon in analysis, not a logical impossibility.
Polynomials have R_n=0 exactly, so they're the LEAST likely case for this to happen, not the only case.
Why
A Taylor series can converge as a numerical series at every x while its sum disagrees with f(x) away from the center — proving equality to f specifically requires showing the remainder itself vanishes, a strictly stronger fact than mere convergence.
A Taylor series converging as a NUMBER at some x is a different fact from that number actually equaling f(x) -- what extra step bridges the two?
From Calculus II — Integral Calculus & Infinite Series
Once R_n(x)\to0 has been shown for every real x for a particular function, what can be concluded?
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Showing R_n(x)\to0 is a strictly stronger conclusion than mere numerical convergence — it proves the sum EQUALS f(x).
Plenty of non-polynomial functions (like e^x, \sin x, \cos x) have Taylor series that converge to them everywhere.
Converging for every real x is exactly the R=\infty case, not a finite radius.
Why
R_n(x)\to0 for every x means T_n(x)\to f(x) for every x, i.e. the Taylor series converges to f on its entire domain of validity, here all reals.
If the remainder vanishes at every real x, what does that force T_n(x) to do at every one of those x-values?
From Calculus II — Integral Calculus & Infinite Series
What is the logical relationship between f(x)=T_n(x)+R_n(x) and the convergence proof?
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
The identity is exact at every single finite n — it isn't a limiting statement itself.
There's no approximation error in the identity itself; R_n(x) is defined as exactly f(x)-T_n(x).
The identity is what's used TO PROVE f equals its series — assuming that conclusion first would be circular.
Why
R_n(x):=f(x)-T_n(x) is an exact identity for every n, so taking the limit of both sides directly shows T_n(x)\to f(x) whenever R_n(x)\to0.
R_n(x)=f(x)-T_n(x) holds as an exact identity for every n -- what does taking a limit of both sides reveal?