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The error of a Taylor polynomial is the next term with the derivative at an unknown point in between, so it is bounded rather than computed.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to state Taylor's theorem with its differentiability hypothesis and recognise it as the mean value theorem one order up, write the Lagrange remainder with the correct derivative and factorial, bound it by taking a maximum of that derivative over the whole interval rather than evaluating it at the centre, prove that a series represents its function by showing the remainder tends to zero, explain why the factorial wins for the exponential and trigonometric functions and not for the geometric series, and solve the bound for the order a given accuracy needs.
The last lesson built Taylor series and ended with a warning: a function need not equal its series. $e^{-1/x^2}$ has a Maclaurin series that converges everywhere, to the zero function, and the function is not zero.
So something more is needed. That something is the remainder $R_n = f - T_n$, and the statement "$f$ equals its Taylor series on an interval" is exactly the statement "$R_n \to 0$ there". Taylor's theorem gives a formula for $R_n$ that can be bounded, and that is the whole business of this lesson — closing the gap, and getting an error bar for free while doing it.
The remainder after the $n$th-order Taylor polynomial is $R_n(x) = f(x) - T_n(x)$: the exact error, defined without any formula.
The Lagrange form of it is $\dfrac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}$ for some unspecified $c$ between $a$ and $x$. It looks like the next term of the series, with the derivative evaluated at a mystery point instead of at the centre.
The order $n$ is the degree of the polynomial used. A tolerance is the accuracy asked for; solving the bound for $n$ says what order meets it.
Taylor's theorem. If $f$ is $(n+1)$ times differentiable on an interval containing $a$ and $x$, then $$f(x) = T_n(x) + R_n(x), \qquad R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!}(x - a)^{n+1}$$ for some $c$ strictly between $a$ and $x$.
It generalises the mean value theorem, which is the case $n = 0$: $f(x) = f(a) + f'(c)(x-a)$. The same existence-without-location, one order up.
The bound. Since $c$ is unknown, take $M \ge |f^{(n+1)}|$ over the whole interval between $a$ and $x$. Then $$|R_n(x)| \le \frac{M\,|x-a|^{n+1}}{(n+1)!}.$$ Two things make this small: a small step, raised to a high power, and a factorial in the denominator that outgrows everything.
Proving a series represents its function. $f$ equals its Taylor series on a set exactly when $R_n \to 0$ there.
Reading it backwards. Setting the bound below a tolerance and solving for $n$ says how many terms an accuracy needs. That is how $\sin$, $\cos$ and $e^x$ are computed in practice.
Another way: picture
$\sin x$ with $T_3$ drawn beside it. Near the origin the two are indistinguishable; by $x = 2$ a visible gap has opened; by $x = 4$ the polynomial has dived away and the sine is still oscillating. The gap at each point is $R_3(x)$, and the bound $\frac{|x|^4}{24}$ is the envelope that contains it — tight near zero and generous far out, which is exactly the shape of the picture.
Another way: steps
The bound $\dfrac{M|x-a|^{n+1}}{(n+1)!}$ has a power on top and a factorial underneath. For fixed $x$, $\dfrac{R^{n}}{n!} \to 0$ however large $R$ is — which is the growth-rate fact from lesson 17, and the reason $e^x$ is represented by its series on the whole line and not merely near $0$.
It is worth seeing how the race goes. For $R = 10$ the terms $\frac{10^n}{n!}$ increase until $n = 10$, peaking around $2755$, and only then begin to fall — and they fall fast, reaching $10^{-7}$ by $n = 40$. So a Taylor estimate of $e^{10}$ from the origin is useless with ten terms and excellent with forty.
That pattern is general: for a step of size $h$, the bound worsens until $n \approx h$ and improves sharply afterwards. It is why series are used near the centre and why calculators reduce an argument into a small interval before expanding — computing $\sin(1000)$ by series about $0$ would be absurd, and computing $\sin$ of the equivalent angle in $[0, \pi/4]$ takes four terms.
And where the factorial cannot win, the representation fails: $\frac{1}{1-x}$ has $f^{(n)}(0) = n!$, exactly cancelling the factorial, which leaves $|x|^{n+1}$ — tending to zero only for $|x| < 1$.
Evaluating the derivative at the centre. The theorem says $c$ is somewhere between, unknown. Only a bound over the whole interval is usable.
Using the wrong derivative. After $T_n$ the remainder involves $f^{(n+1)}$ — one more than the order, matching the next term.
Dividing by $n!$ instead of $(n+1)!$. Same off-by-one, in the other factor.
Treating the bound as the error. It is an upper bound, often a generous one. The true error is usually smaller, and for an alternating series the sharper Leibniz bound may be available instead.
Concluding a representation from convergence. The series converging is necessary and not sufficient; $R_n \to 0$ is the actual condition, and $e^{-1/x^2}$ separates them.
Bounding over the wrong interval. It runs from the centre to the point, both included — not just near the point.
Forgetting the hypothesis. $f$ must be $(n+1)$ times differentiable on the interval. A corner anywhere in between and the theorem does not apply at all.
The Lagrange form looks like a formula, so it invites being evaluated: substitute a number for $c$ and read off the error. There is no number to substitute. The theorem asserts that some $c$ works and gives no way to find it, and $c$ moves as $x$ and $n$ move.
The habit that follows is to use $c = a$, because that is the only point in sight with a known derivative. The resulting numbers are not bounds — they are the value the remainder would have if $c$ happened to be the centre, and there is no reason to think it is. For $e^x$ at $x = 1$ with $n = 3$, using $c = 0$ gives $\frac{1}{24} \approx 0.042$; the true error is $0.0516$. The "bound" is smaller than the error, which makes it worse than no bound at all.
The correct move is to give up on $c$ entirely and bound $|f^{(n+1)}|$ over the whole interval where $c$ could be. That is often crude — for $e^x$ on $[0,1]$ you use $e$, the value at the far end — and crude is fine. An upper bound that is generous is still a guarantee; a sharp number that might be wrong is not.
This is the same discipline the mean value theorem required in Calculus I, and Taylor's theorem is that theorem one order up. Existence theorems tell you something exists; they are used by bounding, never by evaluating.
Estimate $\sqrt{e}$ using $T_3$ for $e^x$ about $0$ at $x = \tfrac12$: $T_3\left(\tfrac12\right) = 1 + \tfrac12 + \tfrac18 + \tfrac{1}{48} = \dfrac{79}{48} \approx 1.6458$.
The polynomial part.
Remainder: $f^{(4)}(x) = e^{x}$, and on $[0, \tfrac12]$ that is at most $e^{1/2} < 1.65$. So $|R_3| \le \dfrac{1.65 \cdot (1/2)^{4}}{4!} = \dfrac{1.65}{384} < 0.0043$.
Bounded over the interval, using a crude bound for $e^{1/2}$ itself.
So $\sqrt e$ is between $1.6415$ and $1.6501$. The true value is $1.64872$, comfortably inside — and the estimate came with a certificate rather than a hope.
Claim: $\sin x = \sum_{n\ge0}\dfrac{(-1)^{n}x^{2n+1}}{(2n+1)!}$ for every real $x$, not merely near $0$.
A statement about the remainder, not about convergence.
Every derivative of $\sin$ is $\pm\sin$ or $\pm\cos$, so $|f^{(n+1)}| \le 1$ on any interval whatever. Hence $|R_n(x)| \le \dfrac{|x|^{n+1}}{(n+1)!}$.
One bound, valid everywhere.
For fixed $x$, $\dfrac{|x|^{n+1}}{(n+1)!} \to 0$, because the factorial beats any fixed power. So $R_n \to 0$ and the series equals the function at every $x$.
The factorial does the work.
$T_2(x) = 1 - \dfrac{x^{2}}{2}$, so $T_2(0.3) = 1 - 0.045 = 0.955$.
The polynomial first.
The remainder uses $f^{(3)}(x) = \sin x$, which is at most $\sin(0.3) < 0.3$ on $[0, 0.3]$ — a bound over the interval, not a value at a point.
One more derivative than the order.
$|R_2| \le \dfrac{0.3 \times 0.3^{3}}{3!} = \dfrac{0.0081}{6} = 0.00135$. The true value is $0.955336\ldots$, so the actual error is about $0.00034$ — well inside the bound, as an upper bound should be.
A second-order Taylor polynomial at $a$ is used at $a + \dfrac{1}{3}$, and $|f'''| \le 2$ throughout. Bound the error.
Answer:
Taylor's theorem says $R_n(x) = \dfrac{f^{(n+1)}(c)}{(n+1)!}(x - a)^{n+1}$ for some $c$. What is known about $c$?
Put the steps of bounding the error of a Taylor polynomial into the order you do them.
Number the steps in order (write the number in the box):
With every derivative bounded by $1$ and a step of $h = \dfrac{1}{4}$, give the Lagrange bound $\dfrac{h^{n+1}}{(n+1)!}$ for each order $n$.
| Bound | |
|---|---|
| $n = 1$ | |
| $n = 2$ | |
| $n = 3$ |
Match each function to the set where its Maclaurin series converges to it.
| Every real number | Every real number, by the same bounded-derivative argument | Only $(-1, 1)$ | Only the single point $0$ | |
|---|---|---|---|---|
| $e^{x}$ | ||||
| $\sin x$ | ||||
| $\dfrac{1}{1 - x}$ | ||||
| $e^{-1/x^{2}}$ for $x \ne 0$, and $0$ at $x = 0$ |
For $\sin x$ near $0$ every derivative is bounded by $1$, so the Lagrange bound at $x = \dfrac{1}{4}$ is $\dfrac{1}{(n+1)!\,4^{\,n+1}}$. Using $n = 2$, what is the reciprocal of that bound?
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
A second-order Taylor polynomial at $a$ is used at $a + \dfrac{1}{2}$, and $|f'''| \le 5$ throughout. Bound the error.
Answer:
You can attach a guaranteed error bar to a Taylor estimate and prove a series equals its function. Say in your own words why $c$ cannot be evaluated and what is done instead. Next: curves that are not graphs of functions at all.
10. Your turn: bound the error in using $T_2$ for $\cos x$ about $0$ at $x = 0.3$, step 3