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Mariusz M

Rejestracja: 11 Sep 2010
Offline Ostatnio: Feb 16 2018 19:21
*****

Moje posty

W temacie: Procedura Kropki - algorytm

03.12.2017 - 15:40

Pewnie chciałby rozwiązać swoje równanie rekurencyjne 
gdyby po użyciu funkcji tworzącej  z iloczynu szeregów czy tzw splotu ciągów można by  skorzystać to nawet całkiem nieźle by się liczyło

Tyle że iloczyn szeregów to coś takiego \sum_{n=0}^{\infty}\sum_{k=0}^{n}a_{k}b_{n-k}x^{n}

i na pierwszy rzut oka nie widzę aby indeksy się zgadzały 


W temacie: Całka 11

26.11.2017 - 10:01

\int{\frac{3x^2+2x+12}{x^3+4x}\mbox{d}x}=\int{\frac{3(x^2+4)+2x}{x^3+4x}\mbox{d}x}\\</p>\\<p>=\int{\frac{3(x^2+4)}{x(x^2+4)}\mbox{d}x}+\int{\frac{2x}{x(x^2+4)}\mbox{d}x}\\</p>\\<p>=3\int{\frac{1}{x}\mbox{d}x}+2\int{\frac{1}{x^2+4}\mbox{d}x}\\</p>\\<p>=3\int{\frac{1}{x}\mbox{d}x}+\frac{1}{2}\int{\frac{1}{1+\left(\frac{x}{2}\right)^2}\mbox{d}x}\\</p>\\<p>=3\ln{|x|}+\arctan{\left(\frac{x}{2}\right)}+C\\</p>\\<p>


W temacie: Całka niewymierna 7

26.11.2017 - 04:44

\int{\frac{\sqrt{x^2-4}}{x}\mbox{d}x}\\</p>\\<p>\sqrt{x^2-4}=t-x\\</p>\\<p>x^2-4=t^2-2tx+x^2\\</p>\\<p>-4=t^2-2tx\\</p>\\<p>2tx=t^2+4\\</p>\\<p>x=\frac{t^2+4}{2t}\\</p>\\<p>t-x=\frac{2t^2-t^2-4}{2t}=\frac{t^2-4}{2t}\\</p>\\<p>\mbox{d}x=\frac{2t\cdot 2t-2(t^2+4)}{4t^2}\mbox{d}t\\</p>\\<p>\mbox{d}x=\frac{t^2-4}{2t^2}\mbox{d}t\\</p>\\<p>\int{\frac{t^2-4}{2t}\cdot\frac{2t}{t^2+4}\cdot\frac{t^2-4}{2t^2}\mbox{d}t}\\</p>\\<p>\frac{1}{2}\int{\frac{(t^2-4)^2}{t^2(t^2+4)}\mbox{d}t}\\</p>\\<p>\frac{1}{2}\int{\frac{(t^2+4)^2-16t^2}{t^2(t^2+4)}\mbox{d}t}\\</p>\\<p>\frac{1}{2}\int{\frac{t^2+4}{t^2}\mbox{d}t}-8\int{\frac{1}{t^2+4}\mbox{d}t}\\</p>\\<p>\frac{t}{2}-\frac{2}{t}-2\int{\frac{1}{1+\left(\frac{t}{2}\right)^2}\mbox{d}t}\\</p>\\<p>\frac{t}{2}-\frac{2}{t}-4\int{\frac{\frac{1}{2}}{1+\left(\frac{t}{2}\right)^2}\mbox{d}t}\\</p>\\<p>\frac{t}{2}-\frac{2}{t}-4\arctan{\left(\frac{t}{2}\right)}+C\\</p>\\<p>\frac{t^2-4}{2t}-4\arctan{\left(\frac{t}{2}\right)}+C\\</p>\\<p>=\sqrt{x^2-4}-4\arctan{\left(\frac{x+\sqrt{x^2-4}}{2}\right)}+C</p>\\<p>


W temacie: cAŁAK 7

26.11.2017 - 03:43

Metoda Ostrogradskiego też będzie wymagała ośmiu współczynników , (po zastosowaniu wystarczy pobawić się licznikiem aby uzyskać dalszy rozkład)

Tutaj można pobawić się częściami aby uprościć sobie całkę

 

8(4x^2-2x-3)=(8x-2)(4x-1)-26\\</p>\\<p>(8x-2)(4x-1)-8(4x^2-2x-3)=26\\</p>\\<p>\int{\frac{1}{(4x^2-2x-3)^4}\mbox{d}x}=\frac{1}{26}\int{\frac{(8x-2)(4x-1)-8(4x^2-2x-3)}{(4x^2-2x-3)^4}\mbox{d}x}\\</p>\\<p>\int{\frac{1}{(4x^2-2x-3)^4}\mbox{d}x}=\frac{1}{26}\left(\int{\frac{(8x-2)(4x-1)}{(4x^2-2x-3)^4}}-8\int{\frac{(4x^2-2x-3)}{(4x^2-2x-3)^4}\mbox{d}x}\right)\\</p>\\<p>\int{\frac{1}{(4x^2-2x-3)^4}\mbox{d}x}=\frac{1}{26}\left(\int{(4x-1)\frac{(8x-2)}{(4x^2-2x-3)^4}\mbox{d}x}-8\int{\frac{1}{(4x^2-2x-3)^3}\mbox{d}x}\right)\\</p>\\<p>\int{\frac{1}{(4x^2-2x-3)^4}\mbox{d}x}=\frac{1}{26}\left(-\frac{1}{3}\frac{4x-1}{(4x^2-2x-3)^3}+\frac{4}{3}\int{\frac{1}{(4x^2-2x-3)^3}\mbox{d}x}-8\int{\frac{1}{(4x^2-2x-3)^3}\mbox{d}x}\right)\\</p>\\<p></p>\\<p>\int{\frac{1}{(4x^2-2x-3)^4}\mbox{d}x}=\frac{1}{26}\left(-\frac{1}{3}\frac{4x-1}{(4x^2-2x-3)^3}-\frac{20}{3}\int{\frac{1}{(4x^2-2x-3)^3}\mbox{d}x}\right)\\</p>\\<p>\int{\frac{1}{(4x^2-2x-3)^4}\mbox{d}x}=-\frac{1}{78}\frac{4x-1}{(4x^2-2x-3)^3}-\frac{10}{39}\int{\frac{1}{(4x^2-2x-3)^3}\mbox{d}x}\\</p>\\<p>\int{\frac{1}{(4x^2-2x-3)^3}\mbox{d}x}=\frac{1}{26}\int{\frac{(8x-2)(4x-1)-8(4x^2-2x-3)}{(4x^2-2x-3)^3}\mbox{d}x}\\</p>\\<p></p>\\<p>\int{\frac{1}{(4x^2-2x-3)^3}\mbox{d}x}=\frac{1}{26}\left(\int{\frac{(8x-2)(4x-1)}{(4x^2-2x-3)^3}}-8\int{\frac{(4x^2-2x-3)}{(4x^2-2x-3)^3}\mbox{d}x}\right)\\</p>\\<p>\int{\frac{1}{(4x^2-2x-3)^3}\mbox{d}x}=\frac{1}{26}\left(\int{(4x-1)\frac{(8x-2)}{(4x^2-2x-3)^3}\mbox{d}x}-8\int{\frac{1}{(4x^2-2x-3)^2}\mbox{d}x}\right)\\</p>\\<p>\int{\frac{1}{(4x^2-2x-3)^3}\mbox{d}x}=\frac{1}{26}\left(-\frac{1}{2}\frac{4x-1}{(4x^2-2x-3)^2}+2\int{\frac{1}{(4x^2-2x-3)^2}\mbox{d}x}-8\int{\frac{1}{(4x^2-2x-3)^2}\mbox{d}x}\right)\\</p>\\<p></p>\\<p>\int{\frac{1}{(4x^2-2x-3)^3}\mbox{d}x}=\frac{1}{26}\left(-\frac{1}{2}\frac{4x-1}{(4x^2-2x-3)^2}-6\int{\frac{1}{(4x^2-2x-3)^2}\mbox{d}x}\right)\\<br>\\\int{\frac{1}{(4x^2-2x-3)^3}\mbox{d}x}=-\frac{1}{52}\frac{4x-1}{(4x^2-2x-3)^2}-\frac{3}{13}\int{\frac{1}{(4x^2-2x-3)^2}\mbox{d}x}\\</p>\\<p>\int{\frac{1}{(4x^2-2x-3)^2}\mbox{d}x}=\frac{1}{26}\int{\frac{(8x-2)(4x-1)-8(4x^2-2x-3)}{(4x^2-2x-3)^2}\mbox{d}x}\\</p>\\<p>\int{\frac{1}{(4x^2-2x-3)^2}\mbox{d}x}=\frac{1}{26}\left(\int{\frac{(8x-2)(4x-1)}{(4x^2-2x-3)^2}\mbox{d}x}-8\int{\frac{(4x^2-2x-3)}{(4x^2-2x-3)^2}\mbox{d}x}\right)\\</p>\\<p></p>\\<p>\int{\frac{1}{(4x^2-2x-3)^2}\mbox{d}x}=\frac{1}{26}\left((4x-1)\int{\frac{(8x-2)}{(4x^2-2x-3)^2}\mbox{d}x}-8\int{\frac{1}{4x^2-2x-3}\mbox{d}x}\right)\\</p>\\<p>\int{\frac{1}{(4x^2-2x-3)^2}\mbox{d}x}=\frac{1}{26}\left(-\frac{4x-1}{4x^2-2x-3}+4\int{\frac{1}{4x^2-2x-3}\mbox{d}x}-8\int{\frac{1}{4x^2-2x-3}\mbox{d}x}\right)\\</p>\\<p>\int{\frac{1}{(4x^2-2x-3)^2}\mbox{d}x}=\frac{1}{26}\left(-\frac{4x-1}{4x^2-2x-3}-4\int{\frac{1}{4x^2-2x-3}\mbox{d}x}\right)\\</p>\\<p>\int{\frac{1}{(4x^2-2x-3)^2}\mbox{d}x}=-\frac{1}{26}\frac{4x-1}{4x^2-2x-3}-\frac{\sqrt{13}}{169}\int{\frac{4(4x-1+\sqrt{13})-4(4x-1-\sqrt{13})}{16x^2-8x-12}\mbox{d}x}\\<br>\\\int{\frac{1}{(4x^2-2x-3)^2}\mbox{d}x}=-\frac{1}{26}\frac{4x-1}{4x^2-2x-3}-\frac{\sqrt{13}}{169}\left(\int{\frac{4}{(4x-1-\sqrt{13})}\mbox{d}x}-\int{\frac{4}{(4x-1+\sqrt{13})}}\right)\\<br>\\\int{\frac{1}{(4x^2-2x-3)^2}\mbox{d}x}=-\frac{1}{26}\frac{4x-1}{4x^2-2x-3}-\frac{\sqrt{13}}{169}\ln{\left|\frac{4x-1-\sqrt{13}}{4x-1+\sqrt{13}}\right|}\\</p>\\<p>\int{\frac{1}{(4x^2-2x-3)^3}\mbox{d}x}=-\frac{1}{52}\frac{4x-1}{(4x^2-2x-3)^2}-\frac{3}{13}\int{\frac{1}{(4x^2-2x-3)^2}\mbox{d}x}\\</p>\\<p>\int{\frac{1}{(4x^2-2x-3)^3}\mbox{d}x}=-\frac{1}{52}\frac{4x-1}{(4x^2-2x-3)^2}-\frac{3}{13}\left(-\frac{1}{26}\frac{4x-1}{4x^2-2x-3}-\frac{\sqrt{13}}{169}\ln{\left|\frac{4x-1-\sqrt{13}}{4x-1+\sqrt{13}}\right|}\right)\\</p>\\<p>\int{\frac{1}{(4x^2-2x-3)^3}\mbox{d}x}=-\frac{1}{52}\frac{4x-1}{(4x^2-2x-3)^2}+\frac{3}{338}\frac{4x-1}{4x^2-2x-3}+\frac{3\sqrt{13}}{2197}\ln{\left|\frac{4x-1-\sqrt{13}}{4x-1+\sqrt{13}}\right|}\\</p>\\<p>\int{\frac{1}{(4x^2-2x-3)^4}\mbox{d}x}=-\frac{1}{78}\frac{4x-1}{(4x^2-2x-3)^3}-\frac{10}{39}\int{\frac{1}{(4x^2-2x-3)^3}\mbox{d}x}\\</p>\\<p>\int{\frac{1}{(4x^2-2x-3)^4}\mbox{d}x}=-\frac{1}{78}\frac{4x-1}{(4x^2-2x-3)^3}-\frac{10}{39}\left(-\frac{1}{52}\frac{4x-1}{(4x^2-2x-3)^2}+\frac{3}{338}\frac{4x-1}{4x^2-2x-3}+\frac{3\sqrt{13}}{2197}\ln{\left|\frac{4x-1-\sqrt{13}}{4x-1+\sqrt{13}}\right|}\right)\\</p>\\<p>\int{\frac{1}{(4x^2-2x-3)^4}\mbox{d}x}=-\frac{1}{78}\frac{4x-1}{(4x^2-2x-3)^3}+\frac{5}{1014}\frac{4x-1}{(4x^2-2x-3)^2}-\frac{5}{2197}\frac{4x-1}{4x^2-2x-3}-\frac{10\sqrt{13}}{28561}\ln{\left|\frac{4x-1-\sqrt{13}}{4x-1+\sqrt{13}}\right|}+C\\<br>\\


W temacie: Całka 30

26.11.2017 - 01:15

\int{\frac{x}{\left(x^2+2x+2\right)^2}\mbox{d}x}=\frac{a_{1}x+a_{0}}{x^2+2x+2}+\int{\frac{b_{1}x+b_{0}}{x^2+2x+2}\mbox{d}x}\\</p>\\<p>\frac{x}{\left(x^2+2x+2\right)^2}=\frac{a_{1}(x^2+2x+2)-(a_{1}x+a_{0})(2x+2)}{(x^2+2x+2)^2}+\frac{b_{1}x+b_{0}}{x^2+2x+2}\\</p>\\<p>\frac{x}{\left(x^2+2x+2\right)^2}=\frac{a_{1}x^2+2a_{1}x+2a_{1}-(2a_{1}x^2+2a_{1}x+2a_{0}x+2a_{0})}{(x^2+2x+2)^2}+\frac{b_{1}x+b_{0}}{x^2+2x+2}\\</p>\\<p></p>\\<p>\frac{x}{\left(x^2+2x+2\right)^2}=\frac{a_{1}x^2+2a_{1}x+2a_{1}-(2a_{1}x^2+2a_{1}x+2a_{0}x+2a_{0})+(b_{1}x+b_{0})(x^2+2x+2)}{(x^2+2x+2)^2}\\</p>\\<p>x=a_{1}x^2+2a_{1}x+2a_{1}-(2a_{1}x^2+2a_{1}x+2a_{0}x+2a_{0})+(b_{1}x+b_{0})(x^2+2x+2)\\</p>\\<p>x=a_{1}x^2+2a_{1}x+2a_{1}-2a_{1}x^2-2a_{1}x-2a_{0}x-2a_{0}+b_{1}x^3+2b_{1}x^2+2b_{1}x+b_{0}x^2+2b_{0}x+2b_{0}\\</p>\\<p>x=b_{1}x^3+(2b_{1}+b_{0}-a_{1})x^2+(2b_{1}+2b_{0}-2a_{0})x+2b_{0}+2a_{1}-2a_{0}\\</p>\\<p>\begin{cases}b_{1}=0\\2b_{1}+b_{0}-a_{1}=0\\2b_{1}+2b_{0}-2a_{0}=1\\2b_{0}+2a_{1}-2a_{0}=0 \end{cases}\\<br>\\\begin{cases}b_{1}=0\\b_{0}=a_{1}\\2a_{1}-2a_{0}=1\\4a_{1}-2a_{0}=0 \end{cases}\\<br>\\\begin{cases}b_{1}=0\\b_{0}=a_{1}\\2a_{1}=-1\\a_{0}=2a_{1} \end{cases}\\<br>\\\begin{cases}b_{1}=0\\2b_{0}=-1\\2a_{1}=-1\\a_{0}=-1 \end{cases}\\<br>\\\int{\frac{x}{\left(x^2+2x+2\right)^2}\mbox{d}x}=-\frac{1}{2}\frac{x+2}{x^2+2x+2}-\frac{1}{2}\int{\frac{\mbox{d}x}{(x+1)^2+1}}\\</p>\\<p>\int{\frac{x}{\left(x^2+2x+2\right)^2}\mbox{d}x}=-\frac{1}{2}\frac{x+2}{x^2+2x+2}-\frac{1}{2}\arctan{(x+1)}+C\\<br>\\

Można jeszcze bawić się całkowaniem przez części