17.2 Taylor polynomial[2D2]
[1FB]Let \(a∈\overlineℝ\) and \(I\) be a neighborhood of \(a\). Let \(f,g:I→ℝ\). We will say that ”\(f(x)=o(g(x))\) for \(x\) tending to \(a\)” if 1
This notation reads like ”f is small o of g”.
If \(g(x)≠ 0\) for \(x≠ a\), then equivalently we can write
We will say that ”\(f(x)=O(g(x))\) for \(x\) tending to \(a\)” if if there is a constant \(c{\gt}0\) and a neighborhood \(J\) of \(a\) for which \(∀ x∈ J, |f(x)|≤ c |g(x)|\).
Again, if \(g(x)≠ 0\) for \(x≠ a\), then equivalently we can write
This notation reads like ”f is big O of g”.
For further information, and more notations, see [ 50 ] .
In the following for simplicity we consider only the case in which \(\lim _{x→ a}g(x)=0\); moreover in Taylor’s expansion we always have that \(g(x)=(x-a)^ n\) with \(n≥ 1\) integer. 2
[1FC]Attention! The symbols ”small o” and ”big O” are used differently from other symbols of mathematics. In fact, they can represent different functions, even in the same context! For example, if we write
the two symbols ”\(o(x)\)” on the right and left represent different functions. Particular care must therefore be taken in showing the properties used in the calculus. When many of such symbols are present, it is advisable to replace them with placeholder function symbols, as in the following examples.
Let’s see two examples. Let \(a=0\) for simplicity.
[1FD]We informally state this property.
To prove it, we convert it into a precise statement. First of all, let’s rewrite it like this.
So let’s prove it. From the hypotheses,
then
[1FF]We informally state this second property
We rewrite it like this.
We note that, for \(x≠ 0\) small, \(x^ n+f(x)\) is not zero, as there is a neighborhood in which \(|f(x)|≤ |x^ n/2|\). As a hypothesis we have that \(\lim _{x→ 0}f(x)x^{-n}=0\) and \(\lim _{x→ 0}g(x)/(x^ n+f(x))=0\) then
but
while
- E393
[1FG] Let \(a=0\) for simplicity. Rewrite the following relations, and prove them.
If \(n≥ m≥ 1\) then
\[ O(x^ n)+O(x^ m) = O(x^ m), \quad o(x^ n)+O(x^ m) = O(x^ m),\quad x^ n+O(x^ m) = O(x^ m)\quad . \]If \(n{\gt} m≥ 1\) then
\[ O(x^ n)+o(x^ m) = o(x^ m),\quad x^ n+o(x^ m) = o(x^ m). \]For \(n,m≥ 1\)
\begin{eqnarray*} x^ n O(x^ m)& =& O(x^{n+m})\\ x^ n o(x^ m)& =& o(x^{n+m})\\ O(x^ n) O(x^ m)& =& O(x^{n+m})\\ o(x^ n) O(x^ m)& =& o(x^{n+m}) \end{eqnarray*}- \[ ∫_ 0^ y O(x^ n)\, {\mathbb {d}}x=O(y^{n+1}) \quad ∫_ 0^ y o(x^ n)\, {\mathbb {d}}x=o(y^{n+1}) \quad . \]
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\(f(x)\) |
= |
\(p(x) + o(x^ 4)\) |
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\((\cos (x))^ 2\) |
= |
\(+o(x^ 4)\) |
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\((\cos (x))^ 3\) |
= |
\(+o(x^ 4)\) |
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\(\cos (x)e^ x\) |
= |
\(+o(x^ 4)\) |
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\(\cos (\sin (x))\) |
= |
\(+o(x^ 4)\) |
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\(\sin (\cos (x))\) |
= |
\(+o(x^ 4)\) |
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\(\log (\log (e+x))\) |
= |
\(+o(x^ 3)\) |
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\((1+x)^{1/x}\) |
= |
\(+o(x^ 3)\) |
(A little imagination is required to address the last two. To reduce the computations, develop the last two only up to \(o(x^ 3)\)).
Hidden solution: [UNACCESSIBLE UUID ’1FK’] [1FM]Find a rational approximation of \(\cos (1)\) with error less than \(1/(10!) ∼ 2.10^{-7}\) Hidden solution: [UNACCESSIBLE UUID ’1FN’] [1FP]Write Taylor’s polynomial of \((1+x)^𝛼\) with \(𝛼 ∈ ℝ ⧵ ℕ\). (Infer a generalization of the binomial symbol \(\binom 𝛼 k \)). The associated Taylor series is called binomial series, it converges for \(|x|{\lt}1\).
Hidden solution: [UNACCESSIBLE UUID ’1FQ’] [1FR] Prerequisites:374.Note:From an idea in Apostol’s book [ 3 ] , Chapter 7.3.Write Taylor’s polynomial (around \(x_ 0 = 0\)) for \(-\log (1 - x)\), integrating
and compare the ”remainder”
thus obtained with with the "integral remainder" of \(f(x) = -\log (1 - x)\) (as presented in Exercise 374).
Proceed similarly for \(\arctan (x)\), integrating
Hidden solution: [UNACCESSIBLE UUID ’1FS’] [1FT]Prerequisites:374,4.Difficulty:.Evaluate for which \(r {\gt} 0\) we have that the Taylor remainder of \(f (x) = - \log (1 - x)\) is infinitesimal in \(n\), uniformly for \(|x| {\lt} r\); this, using the remainder seen in ??, using the integral remainder or using the Lagrange remainder.
Hidden solution: [UNACCESSIBLE UUID ’1FV’]
See also exercise 374.