MS-A0111 - Differential and Integral Calculus 1, Lecture, 5.9.2022-19.10.2022
Kurssiasetusten perusteella kurssi on päättynyt 19.10.2022 Etsi kursseja: MS-A0111
Differential and Integral Calculus
5. Taylor polynomial
Taylor polynomial
Definition: Taylor polynomial
Let be
times differentiable at the point
. Then the Taylor polynomial
\begin{align}
P_n(x)&=P_n(x;x_0)\\\
&=f(x_0)+f'(x_0)(x-x_0)+\frac{f''(x_0)}{2!}(x-x_0)^2+ \\
& \dots +\frac{f^{(n)}(x_0)}{n!}(x-x_0)^n\\
&=\sum_{k=0}^n\frac{f^{(k)}(x_0)}{k!}(x-x_0)^k\\
\end{align}
is the best polynomial approximation of degree
(with respect to the derivative) for a function
, close to the point
.
Note. The special case is often called the Maclaurin polynomial.
If is
times differentiable at
, then the Taylor polynomial
has the same derivatives at
as the function
, up to the order
(of the derivative).
The reason (case ): Let
so that
\begin{align}
P_n'(x)&=c_1+2c_2x+3c_3x^2+\dots +nc_nx^{n-1}, \\
P_n''(x)&=2c_2+3\cdot 2 c_3x\dots +n(n-1)c_nx^{n-2} \\
P_n'''(x)&=3\cdot 2 c_3\dots +n(n-1)(n-2)c_nx^{n-3} \\
\dots && \\
P^{(k)}(x)&=k!c_k + x\text{ terms} \\
\dots & \\
P^{(n)}(x)&=n!c_n \\
P^{(n+1)}(x)&=0.
\end{align}
From these way we obtain the coefficients one by one:
\begin{align}
c_0= P_n(0)=f(0) &\Rightarrow c_0=f(0) \\
c_1=P_n'(0)=f'(0) &\Rightarrow c_1=f'(0) \\
2c_2=P_n''(0)=f''(0) &\Rightarrow c_2=\frac{1}{2}f''(0) \\
\vdots & \\
k!c_k=P_n^{(k)}(0)=f^{(k)}(0) &\Rightarrow c_k=\frac{1}{k!}f^{(k)}(0). \\
\vdots &\\
n!c_n=P_n^{(n)}(0)=f^{(n)}(0) &\Rightarrow c_k=\frac{1}{n!}f^{(n)}(0).
\end{align}
Starting from index we cannot pose any new conditions, since
.
Taylor's Formula
If the derivative exists and is continuous on
some interval
, then
and the error term
satisfies
at some point
.
If there is a constant
(independent of
) such that
for all
, then
as
.
\neq omitted here (mathematical induction or integral).
Examples of Maclaurin polynomial approximations: \begin{align} \frac{1}{1-x} &\approx 1+x+x^2+\dots +x^n =\sum_{k=0}^{n}x^k\\ e^x&\approx 1+x+\frac{1}{2!}x^2+\frac{1}{3!}x^3+\dots + \frac{1}{n!}x^n =\sum_{k=0}^{n}\frac{x^k}{k!}\\ \ln (1+x)&\approx x-\frac{1}{2}x^2+\frac{1}{3}x^3-\dots + \frac{(-1)^{n-1}}{n}x^n =\sum_{k=1}^{n}\frac{(-1)^{k-1}}{k}x^k\\ \sin x &\approx x-\frac{1}{3!}x^3+\frac{1}{5!}x^5-\dots +\frac{(-1)^n}{(2n+1)!}x^{2n+1} =\sum_{k=0}^{n}\frac{(-1)^k}{(2k+1)!}x^{2k+1}\\ \cos x &\approx 1-\frac{1}{2!}x^2+\frac{1}{4!}x^4-\dots +\frac{(-1)^n}{(2n)!}x^{2n} =\sum_{k=0}^{n}\frac{(-1)^k}{(2k)!}x^{2k} \end{align}
Example
Which polynomial approximates the function
in the interval
so that the absolute value of the error is less than
?
We use Taylor's Formula for at
. Then
independently of
and the point
.
Also, in the interval in question, we have
.
The requirement will be satisfied (at least) if
This inequality must be solved by trying different values of
; it is
true for
.
The required approximation is achieved with , which fo sine
is the same as
.
Check from graphs: is not enough, so the theoretical bound is sharp!
Taylor polynomial and extreme values
If , then also some higher derivatives may be zero:
Then the behaviour of
near
is determined by the
leading term (after the constant term
)
of the Taylor polynomial.
This leads to the following result:
Extreme values
Newton's method
The first Taylor polynomial is the same as
the linearization of
at the point
. This can be used
in some simple approximations and numerical methods.
Newton's method
The equation can be solved approximately by
choosing a starting point
(e.g. by looking at the graph) and defining
for
This leads to a sequence
,
whose terms usually give better and better approximations for a zero of
.
The recursion formula is based on the geometric idea of finding an
approximative zero of by using its linearization (i.e. the tangent line).
Example
Find an approximate value of by using Newton's method.
We use Newton's method for the function and initial value
.
The recursion formula becomes
from which we obtain
,
,
and so on.
By experimenting with these values, we find that the number of correct
decimal places doubles at each step, and gives already 100 correct
decimal places, if intermediate steps are calculated with enough precision.
Taylor series
Taylor series
If the error term in Taylor's Formula goes to zero as
increases,
then the limit of the Taylor polynomial
is the Taylor series of
(= Maclaurin series for
).
The Taylor series of is of the form
This is an example of a power series.
The Taylor series can be formed as soon as has derivatives of all
orders at
and they are substituted into this formula.
There are two problems related to this: Does the Taylor series converge for all values of
?
Answer: Not always; for example, the function
has a Maclaurin series (= geometric series)
converging only for
, although the function is differentiable for all
:
If the series converges for some , then does its sum
equal
? Answer: Not always; for example, the function
satisfies
for all
(elementary but
difficult calculation). Thus its
Maclaurin series is identically zero and converges to
only at
.
Conclusion: Taylor series should be studied carefully using the error terms. In practice, the series are formed by using some well known basic series.
Examples
\begin{align}
\frac{1}{1-x} &= \sum_{k=0}^{\infty} x^k,\ \ |x|< 1 \\
e^x &= \sum_{k=0}^{\infty} \frac{1}{k!}x^k, \ \ x\in \mathbb{R} \\
\sin x &= \sum_{k=0}^{\infty} \frac{(-1)^{k}}{(2k+1)!} x^{2k+1}, \ \ x\in \mathbb{R} \\
\cos x &= \sum_{k=0}^{\infty} \frac{(-1)^{k}}{(2k)!} x^{2k},\ \ x\in \mathbb{R} \\
(1+x)^r &= 1+\sum_{k=1}^{\infty} \frac{r(r-1)(r-2)\dots (r-k+1)}{k!}x^k,
|x|<1
\end{align}
The last is called the Binomial Series and is valid for all . If
, then starting from
, all the coefficients are zero and in the beginning
Compare this to the Binomial Theorem:
for
.
Power series
Definition: Power series
A power series is of the form
The point
is the centre and the
are the coefficients of the
series.
There are only three essentially different cases:
Abel's Theorem.
- The power series converges only for
(and then it consists of the constant
only)
- The power series converges for all
- The power series converges on some interval
(and possibly in one or both of the end points), and diverges for other values of
.
The number is the radius of convergence of the series. In the first two cases we say that
or
respectively.
Example
For which values of the variable does the power series
converge?
We use the ratio test with . Then
as
. By the ratio test, the series converges for
,
and diverges for
. In the border-line cases
the general term of the series
does not tend to zero, so the series diverges.
Result: The series converges for , and diverges otherwise.
Definition: Sum function
In the interval where the series converges, we can
define a function
by setting
\begin{equation}
\label{summafunktio}
f(x) = \sum_{k=0}^{\infty} c_k(x-x_0)^k, \tag{1}
\end{equation}
which is called the sum function of the power series.
The sum function is continuous and differentiable on
.
Moreover, the derivative
can be calculated by differentiating the sum function
term by term:
Note. The constant term
disappears and the series starts with
.
The differentiated series converges in the same interval
; this may sound a bit surprising because of the
extra coefficient
.
Example
Find the sum function of the power series
This series is obtained by differentiating termwise the geometric series
(with ). Therefore,
\begin{align}
1+2x+3x^2+4x^3+\dots &= D(1+x+x^2+x^3+x^4+\dots ) \\
&= \frac{d}{dx}\left( \frac{1}{1-x}\right) = \frac{1}{(1-x)^2}.
\end{align}
Multiplying with
we obtain
which is valid for
.
In the case we can also integrate the sum function
termwise:
Often the definite integral can be extended up to the end points of the
interval of convergence, but this is not always the case.
Example
Calculate the sum of the alternating harmonic series.
Let us first substitute to the geometric series. This yields
By integrating both sides from
to
we obtain
Note. Extending the limit of integration all the way up to
should be justified
more rigorously here. We shall return to integration later on the course.
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