Zveza med vrsto in funkcijo

Iz E-študij, proste zakladnice študentskega znanja

Skoči na: navigacija, iskanje

Zveza med potenčno vrsto in funkcijo:

$\frac{1}{1-x} = 1 + x + x^2 + x^3 + ... = \sum_{n=0}^\infty x_n$

$1 + x + x 2 + ... = f (x)$

vrsta konvergira na intervalu (-1,1)

$1 + x (1 + x + x 2 + ...) = f (x)$

$1+ x \cdot f(x) = f(x)$

$f(x) = \frac{1}{1-x}$

$\frac{1}{1+y^2} = \left. \frac{1}{1-x} \right|_{x=-y^2}$

$\frac{1}{1+y^2} = \frac{1}{1-(-y^2} = 1 - y^2 + y^4 - y^6 + ... = \sum_{n=0}^\infty (-1)^n y^{2n}$

$\frac{1}{1+x^2} \sum_{n=0}^\infty (-1)^n x^{2n}$

$\frac{x^3}{x+2} = \frac{\frac{x^3}{2}}{1 + \frac{x}{2}} = \frac{x^3}{2} \frac{1}{1+\frac{x}{2}} = \frac{x^3}{2} ( 1 - \frac{x}{2} +\frac{x^2}{2^2} -\frac{x^3}{2^3} +\frac{x^4}{2^4} - ...$

$\frac{1}{x^2-x-2} = \frac{1}{(x-2)(x+1)} = \frac{A}{x-2} + \frac{B}{x+1}$

$1 = A \cdot (x+1) + B \cdot (x-2)$

x = - 1

x = 2

$1 = -3B \Rightarrow B = -\frac13$

$1 = 3A \Rightarrow A = \frac13$

$\frac{A}{x-2} + \frac{B}{x+1} = \frac{\frac13}{x-2} + \frac{-\frac13}{x+1} =$

$\frac{1}{(x-1)^2} = (x-1)^{-2}$

$\int (x-1)^{-2} dx = - \frac{1}{x-1} = \frac{1}{1-x}$

$\left( \frac{1}{1-x} \right) ' = \frac{1}{(x-1)^2}$

$\frac{1}{(x-1)^2} = (1 + x + x^2 + ...)' = 1 + 2x + 3x^2 = \sum_{n=0}^\infty (n+1)x^n$

$(\log (x-1))' = \frac{1}{x-1} = \frac{-1}{1-x}$

torej je

$\log (x-1) = - \int \frac{1}{1-x} dx = - \int \sum_{n=0}^n x^n dx = -\sum_{n=0}^\infty ( \int x^n dx ) = \sum_{n=0}^\infty \frac{x^{n+1}}{n+1} + C$

$(\log (x-1)) = C - x - \frac{x^2}{2} - \frac{x^3}{3} - \frac{x^4}{4} - ...$

$(arctg x)' = \frac{1}{1+x^2}$

$arctg x = \int (1 - y^2 + y^4 - y^6 + ...) dx = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} +...$

$f(x) = \sum a_n x^n$

$x \rightarrow x-a = y$

Razvij funkcijo:

$f(x-a) = \sum a_n (x a)^n$

$f(y) = \sum a_n y^n$

$f(x) = \sum a_n( x a )^n$

x - a = y

$f(a+y) = \sum a_n x^n$

$f(x) = \sum_{n=0}^\infty a_n (x-a)^n = a_0 + a_1 (x-a) + a_2(x-a)^2 +$

$f (a) = a 0$

$f'(x) = \sum_{n=1}^\infty n a_n (x-a)^{n-1} = a_1 + 2 a_2 (x-a) + 3 a_3 (x-a)^2 + ...$

 $f'(a) = a 1$

$f''(x) = \sum_{n=2}^\infty n(n-1)(x-a)^{n-2} = 2 a_2 + 6 a_3 (x-a) + 12 a_4 (x-a) ^2 + ... <math>\,\! f''(a) = 2 a_2$

$\,\! a_n = \frac{f^{(n)}(a)}{n!}$

Taylorjeva vrsta

Taylorjeva vrsta funkcije f okoli točke a:

Če imamo funkcijo f, ki se da razviti v potenčno vrsto okoli točke a je razvoj takšen:

Vzpostavljeno iz »http://www.e-studij.si/Zveza_med_vrsto_in_funkcijo«

Kategoriji: UL/FRI/VSP-RI/ANA2 | Matematika

Osebna orodja

Tiskanje/izvoz

orodja