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The polynomial matching a function's derivatives at a point, the Lagrange and Peano remainders, and why smooth is not analytic.
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 build a Taylor polynomial from the derivatives at a centre, write the remainder in Lagrange form and turn it into a numerical error bound by bounding the next derivative, and choose between the Lagrange, integral and Peano forms according to the question. You will also be able to say why a Taylor series can converge to something other than the function it came from.
The mean value theorem, and the reading of the derivative as the coefficient of the best linear approximation. Taylor's theorem is that reading carried to higher order, and its proof is the mean value theorem applied repeatedly.
The Taylor polynomial of order $n$ about a centre matches the function's derivatives there up to that order, and the remainder is the difference between the function and it. A function is analytic near a point when its remainders there tend to zero, which is strictly stronger than being smooth.
The Taylor polynomial of order $n$ about $c$ is $$P_n(x) = \sum_{k=0}^{n} \frac{f^{(k)}(c)}{k!}(x - c)^k,$$ the unique polynomial of degree at most $n$ whose derivatives at $c$ up to order $n$ match those of $f$.
Taylor's theorem (Lagrange form). If $f$ is $n+1$ times differentiable on an interval containing $c$ and $x$, then $$f(x) = P_n(x) + \frac{f^{(n+1)}(p)}{(n+1)!}(x - c)^{n+1}$$ for some $p$ strictly between $c$ and $x$.
With $n = 0$ this is exactly the mean value theorem, and the general case is proved the same way: build an auxiliary function vanishing at both ends and apply Rolle.
The remainder is the theorem. A polynomial that approximates with no statement about the error is a guess. The Lagrange form converts a bound on the next derivative into a numerical bound on the error, and the unknown point $p$ disappears in the process.
Smooth is not analytic. $f$ is analytic near $c$ when its Taylor series converges to $f$ there — that is, when the remainders tend to zero. The function $e^{-1/x^2}$, extended by $0$, is smooth with every derivative zero at $0$, so its Taylor series is identically zero and the function is not. Over $\mathbb{R}$, having every derivative is strictly weaker than being the sum of one's own series.
Another way: picture
The tangent line matches the height and the slope at the centre. The next polynomial also matches the bending, the next matches the rate at which the bending changes, and so on. Each extra term buys one more order of agreement at the centre and says nothing in itself about how far that agreement reaches.
Another way: steps
To approximate with a guaranteed error:
Step 4 is where the unknown point is eliminated, and skipping it leaves an approximation with no control.
| Form | Statement | Use |
|---|---|---|
| Lagrange | $\dfrac{f^{(n+1)}(p)}{(n+1)!}(x-c)^{n+1}$ | numerical bounds from a derivative bound |
| integral | $\displaystyle\int_c^x \dfrac{f^{(n+1)}(t)}{n!}(x-t)^n\,dt$ | exact; estimate the integral directly |
| Peano | $o\!\left((x-c)^n\right)$ as $x \to c$ | limits, where no size is needed |
The Peano form asks least of $f$ — only that $f^{(n)}(c)$ exists — and says least: the error divided by $(x-c)^n$ tends to zero. It settles limits such as $\lim_{x\to0}\frac{x - \sin x}{x^3}$ in one line and gives no numerical bound at all.
The integral form needs $f^{(n+1)}$ to be integrable and gives the remainder exactly, which matters when the next derivative is large somewhere and small on most of the interval. It also has the advantage of naming no unknown point.
For $e^x$ about $0$ the remainder after $n$ terms is $\frac{e^p}{(n+1)!}x^{n+1}$ with $p$ between $0$ and $x$. On any fixed interval $e^p$ is bounded and $\frac{|x|^{n+1}}{(n+1)!} \to 0$, so the remainders vanish and the series converges to the function everywhere.
For $\ln(1+x)$ about $0$ the series converges only for $-1 < x \le 1$, and the reason is visible in the coefficients rather than in any failure of smoothness on the real line.
For $\frac{1}{1+x^2}$ about $0$ the function is smooth on all of $\mathbb{R}$ and the series converges only for $|x| < 1$. Nothing on the real line explains the radius $1$; the explanation is a singularity off the real line, at $\pm i$, and that observation is one of the reasons complex analysis is a separate course.
It does not say the series converges. The theorem is about a polynomial of finite order and a remainder. Convergence of the infinite series is a separate question and often has a bounded answer.
Convergence is not convergence to the function. The gap between the two is the remainder, and it must be shown to vanish. The flat function at $0$ is the standing counterexample.
More terms is not automatically better far away. Each extra term buys one more order of agreement at the centre. Outside the radius of convergence the partial sums get worse, not better.
The unknown point cannot be computed. Like the mean value theorem's point, it exists and is not identified. The estimate works by bounding the derivative over all possible positions of it.
Approximate $\sin x$ near $0$ by $x - x^3/6$, the Taylor polynomial of order $3$ (and of order $4$, since the next coefficient vanishes).
Choose the centre and the order.
The remainder is $\dfrac{f^{(5)}(p)}{120}x^5$, and every derivative of sine is bounded by $1$ in size.
Write the remainder, then bound the derivative.
So the error is at most $|x|^5/120$. At $x = 0.5$ that is below $0.0003$, and no knowledge of $p$ was ever needed.
A numerical bound, with the point eliminated.
Compute $\lim_{x \to 0} \dfrac{x - \sin x}{x^3}$.
A limit, not a numerical estimate.
Expand: $\sin x = x - \dfrac{x^3}{6} + o(x^3)$, so the numerator is $\dfrac{x^3}{6} + o(x^3)$.
The Peano form is enough here.
Dividing by $x^3$ gives $\dfrac{1}{6} + o(1) \to \dfrac{1}{6}$. L'Hopital would need three applications and three checks of the form.
The right form for the question asked.
The remainder after $n$ terms at $x = 1$ is $\dfrac{e^p}{(n+1)!}$ with $p$ between $0$ and $1$, so it is at most $\dfrac{3}{(n+1)!}$.
Bound the derivative crudely but honestly.
Solve $\dfrac{3}{(n+1)!} < \dfrac{1}{1000}$: since $7! = 5040$, taking $n = 6$ suffices.
Solve the bound for the order.
So seven terms, from the constant through $x^6$, are enough. The crude bound $e^p \le 3$ cost almost nothing — using $e$ itself would not have changed the answer, which is the usual situation and the reason crude bounds are worth taking.
Give the Taylor coefficients of $f(x) = (1 + x)^{9}$ about $0$, for the powers $x^0$, $x^1$ and $x^2$.
| coefficient | |
|---|---|
| coefficient of the constant term | |
| coefficient of the first power | |
| coefficient of the second power |
A function has $|f''| \le 7$ everywhere. Its first Taylor polynomial about $c$ is used at distance $2$ from $c$. What bound does the Lagrange remainder give for the error?
Answer:
Put the steps of estimating a function by a Taylor polynomial, with a guaranteed error, into order.
Number the steps in order (write the number in the box):
Match each form of the remainder after $8$ terms to the question it answers.
| gives a numerical error bound from a bound on a derivative | gives an exact expression to estimate directly | gives the right statement for computing a limit | is not a remainder, but what fixes the coefficients | |
|---|---|---|---|---|
| the next derivative at an unknown point, times a power over a factorial | ||||
| an integral of the next derivative against a weight | ||||
| an error that divided by the last power kept still tends to zero | ||||
| the polynomial's derivatives at the centre matching the function's |
Here is an argument that a smooth function equals its Taylor series near $0$. Mark the one step that does not follow.
This task has no paper form; do it on a device.
Two smooth functions on the real line have the same derivatives of every order at $0$. What follows?
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Give the Taylor coefficients of $f(x) = (1 + x)^{3}$ about $0$, for the powers $x^0$, $x^1$ and $x^2$.
| coefficient | |
|---|---|
| coefficient of the constant term | |
| coefficient of the first power | |
| coefficient of the second power |
You can build a Taylor polynomial, bound its error with the Lagrange remainder, and choose the remainder form a question needs. Say in your own words why a smooth function need not equal its own Taylor series. Next: the integral, built from the ground up.
10. Your turn: how many terms of the series for $e^x$ are needed to get $e$ to within one thousandth?, step 3