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'Taylor series of arcus sine'
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| Title of object: |
Taylor series of arcus sine |
| Canonical Name: |
TaylorSeriesOfArcusSine |
| Type: |
Example |
| Created on: |
2004-11-24 18:26:41 |
| Modified on: |
2006-02-22 18:07:36 |
| Classification: |
msc:05A10, msc:11B65, msc:26A09, msc:26A36 |
Revision comment (for changes between this and next version):
Preamble:
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Content:
We give an example of obtaining the Taylor series \PMlinkescapetext{expansion} of an elementary function by integrating the Taylor series of its derivative.
For\, $-1 < x < 1$\, we have the derivative of the principal \PMlinkescapetext{branch} of the \PMlinkname{arcus sine}{CyclometricFunctions} function:
$$\frac{d\,\arcsin{x}}{dx} = \frac{1}{\sqrt{1-x^2}} = (1-x^2)^{-\frac{1}{2}}.$$
Using the generalized binomial coefficients ${-\frac{1}{2} \choose r}$ we thus can form the Taylor series for it as \PMlinkname{Newton's binomial series}{BinomialFormula}:
$$(1-x^2)^{-\frac{1}{2}} = \sum_{r = 0}^\infty{-\frac{1}{2} \choose r}(-x^2)^r =
1+{-\frac{1}{2} \choose 1}(-x^2)+{-\frac{1}{2} \choose 2}(-x^2)^2+
{-\frac{1}{2} \choose 3}(-x^2)^3+... =$$
$$ =1-\frac{-\frac{1}{2}}{1!}x^2+\frac{-\frac{1}{2}(-\frac{1}{2}-1)}{2!}x^4
-\frac{-\frac{1}{2}(-\frac{1}{2}-1)(-\frac{1}{2}-2)}{3!}x^6+-... =$$
$$ = 1+\frac{1}{2}x^2+\frac{1\cdot 3}{2\cdot 4}x^4+
\frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}x^6+...
\quad\quad\quad\quad \mathrm{for}\,\, -1 < x < 1$$
Because\, $\arcsin{0} = 0$\, for the principal \PMlinkname{branch}{GeneralPower} of the function, we get, by \PMlinkname{integrating the series termwise}{SumFunctionOfSeries}, the \PMlinkescapetext{expansion}
$$\arcsin{x} = \int_0^x\frac{dx}{\sqrt{1-x^2}} =
x+\frac{1}{2}\cdot\frac{x^3}{3}+\frac{1\cdot 3}{2\cdot 4}\cdot\frac{x^5}{5}+
\frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}\cdot\frac{x^7}{7}+...,$$
the validity of which is true for\, $|x| < 1$.\, It can be proved, in addition, that it is true also when\, $x = \pm 1$. |
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