Skocz do zawartości

  •  
  • Mini kompendium
  • MimeTeX
  • Regulamin

- zdjęcie

Mariusz M

Rejestracja: 11 Sep 2010
Offline Ostatnio: Dec 17 2016 00:11
*****

#128184 całka10

Napisane przez Mariusz M w 01.10.2016 - 00:53

\int{\frac{x^2}{\sqrt{13+11x^2}}\mbox{d}x}\\</p>\\<p>\sqrt{13+11x^2}=t-\sqrt{11}x\\</p>\\<p>13+11x^2=t^2-2\sqrt{11}tx+11x^2\\</p>\\<p>13=t^2-2\sqrt{11}tx\\</p>\\<p>2\sqrt{11}tx=t^2-13</p>\\<p>x=\frac{t^2-13}{2\sqrt{11}t}\\</p>\\<p>\mbox{d}x=\frac{2t\cdot 2\sqrt{11}t-2\sqrt{11}\left(t^2-13\right)}{44t^2}\mbox{d}t\\</p>\\<p>\mbox{d}x=\sqrt{11}\frac{t^2+13}{22t^2}\mbox{d}t\\</p>\\<p>t-\sqrt{11}x=\frac{2\sqrt{11}t^2-\sqrt{11}t^2+13\sqrt{11}}{2\sqrt{11}t}\\</p>\\<p>t-\sqrt{11}x=\frac{t^2+13}{2t}\\</p>\\<p>\int{\frac{\left(t^2-13\right)^2}{44t^2}\cdot\frac{2t}{t^2+13}\cdot\sqrt{11}\frac{t^2+13}{22t^2}\mbox{d}t}\\</p>\\<p>\frac{\sqrt{11}}{484}\int{\frac{\left(t^2-13\right)^2}{t^3}\mbox{d}t}\\</p>\\<p>\frac{\sqrt{11}}{484}\int{\frac{t^4-26t^2+169}{t^3}\mbox{d}t}\\</p>\\<p>\frac{\sqrt{11}}{484}\left(\int{t\mbox{d}t}+169\int{\frac{\mbox{d}t}{t^3}}-26\int{\frac{\mbox{d}t}{t}}\right)\\</p>\\<p>\frac{\sqrt{11}}{484}\left(\frac{t^2}{2}-\frac{169}{2}\frac{1}{t^2}-26\ln{\left|t\right|}\right)+C\\</p>\\<p>\frac{\sqrt{11}}{242}\left(\frac{t^4-169}{4t^2}-13\ln{\left|t\right|}\right)+C\\</p>\\<p>\frac{\sqrt{11}}{242}\left(\frac{\left(t^2-13\right)\left(t^2+13\right)}{4t^2}-13\ln{\left|t\right|}\right)+C\\</p>\\<p>\frac{\sqrt{11}}{242}\left(\sqrt{11}x\sqrt{11x^2+13}-13\ln{\left|\sqrt{11}x+\sqrt{11x^2+13}\right|}\right)+C\\</p>\\<p>

 


  • 1


#127944 całka12

Napisane przez Mariusz M w 13.08.2016 - 04:32

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


  • 1


#127841 Całka funkcji trygonometrycznej 6

Napisane przez Mariusz M w 27.07.2016 - 18:58

Jarek podstawienie za tangens połowy kąta jest dobre ale ja proponowałbym złożenie podstawienia Eulera z podstawieniem za sinus

 

\sqrt{\sin^{2}{x}+2}=t-\sin{x}

 

Później podstawienie  u=t^2

aby mieć nieco mniej współczynników w rozkładzie na sumę ułamków prostych


  • 1


#127840 Całka funkcji trygonometrycznej 4

Napisane przez Mariusz M w 27.07.2016 - 18:28

Jarek mógłbyś pokazać jak tę całkę policzyć   Na oko wyjdzie eliptyczna

a zdaje się że ty takie całki liczyłeś

\int{\sqrt{sin{x}}\mbox{d}x}\\</p>\\<p>\sin{x}=t\\</p>\\<p>\mbox{d}t=\cos{x}\mbox{d}x\\</p>\\<p>\mbox{d}t=\sqrt{1-t^2}\mbox{d}x\\</p>\\<p>\int{\sqrt{t-t^3}\mbox{d}t}</p>\\<p>


  • 1


#127808 Całka niewymierna 4

Napisane przez Mariusz M w 23.07.2016 - 23:11

\int{\frac{\mbox{d}x}{\sqrt{x^3-x}}}\\</p>\\<p>=\int{\frac{\mbox{d}x}{\sqrt{x\left(x^2-1\right)}}}\\</p>\\<p>x=t^2\\</p>\\<p>\mbox{d}x=2t\mbox{d}t\\</p>\\<p>\int{\frac{2t\mbox{d}t}{\sqrt{t^2\left(t^4-1\right)}}}\\</p>\\<p>=2\int{\frac{\mbox{d}t}{\sqrt{-\left(1-t^2\right)\left(1+t^2\right)}}}\\</p>\\<p>t=\frac{1}{\sqrt{1-z^2}}\\</p>\\<p>t^2=\frac{1}{1-z^2}\\</p>\\<p>\frac{1}{t^2}=1-z^2</p>\\<p>\frac{1-t^2}{t^2}=-z^2\\</p>\\<p>\frac{1+t^2}{t^2}=2-z^2\\<br>\\1-t^2=-\frac{z^2}{1-z^2}\\</p>\\<p>1+t^2=-\frac{2-z^2}{1-z^2}\\</p>\\<p>\mbox{d}t=\frac{z}{\left(1-z^2\right)\sqrt{1-z^2}}\mbox{d}z\\</p>\\<p>2\int{\frac{1}{\sqrt{\frac{z^2}{1-z^2}\cdot\frac{2-z^2}{1-z^2}}}\cdot\frac{z}{\left(1-z^2\right)\sqrt{1-z^2}}\mbox{d}z}\\</p>\\<p>2\int{\frac{\mbox{d}z}{\sqrt{2-z^2}\sqrt{1-z^2}}}\\</p>\\<p>=\sqrt{2}\int{\frac{\mbox{d}z}{\sqrt{1-\frac{1}{2}z^2}\sqrt{1-z^2}}}\\</p>\\<p>=\sqrt{2}F\left(z,\frac{\sqrt{2}}{2}\right)+C\\</p>\\<p>=\sqrt{2}F\left(\frac{sqrt{t^2-1}}{t},\frac{\sqrt{2}}{2}\right)+C\\</p>\\<p>=\sqrt{2}F\left(\frac{\sqrt{x^2-x}}{x},\frac{\sqrt{2}}{2}\right)+C\\</p>\\<p>

 

F to całka eliptyczna

 

 

 

 


  • 2


#127799 Całka funkcji niewymiernej 7

Napisane przez Mariusz M w 23.07.2016 - 21:15

Spróbuj pierwszego lub drugiego podstawienia Eulera

 

\sqrt{x^2+2x+5}=t-x\\</p>\\<p>x^2+2x+5=t^2-2tx+x^2\\</p>\\<p>2x+5=t^2-2tx\\</p>\\<p>t^2-5=2tx+2x\\</p>\\<p>x\left(2t+2\right)=t^2-5\\</p>\\<p>x=\frac{t^2-5}{2t+2}\\</p>\\<p>x-3=\frac{t^2-6t-11}{2t+2}\\</p>\\<p>x-4=\frac{t^2-8t-13}{2t+2}\\</p>\\<p>t-x=\frac{2t^2+2t-t^2+5}{2t+2}=\frac{t^2+2t+5}{2t+2}\\</p>\\<p>\mbox{d}x=\frac{2t\left(2t+2\right)-2\left(t^2-5\right)}{\left(2t+2\right)^2}\mbox{d}t\\</p>\\<p>\mbox{d}x=\frac{2t^2+4t+10}{\left(2t+2\right)^2}\mbox{d}t\\</p>\\<p>\int{\frac{t^2+2t+5}{2t+2}\cdot\frac{2t+2}{t^2-6t-11}\cdot\frac{2t+2}{t^2-8t-13}\cdot\frac{2\left(t^2+2t+5\right)}{\left(2t+2\right)^2}\mbox{d}t}\\</p>\\<p>\int{\frac{\left(t^2+2t+5\right)^2}{\left(t^2-6t-11\right)\left(t^2-8t-13\right)\left(t+1\right)}\mbox{d}t}\\</p>\\<p>\int{\frac{\left(t^2+2t+5\right)^2}{\left(t-3-2\sqrt{5}\right)\left(t-3+2\sqrt{5}\right)\left(t-4-\sqrt{29}\right)\left(t-4+\sqrt{29}\right)\left(t+1\right)}\mbox{d}t}\\</p>\\<p>


  • 2


#127798 Całka niewymierna 2

Napisane przez Mariusz M w 23.07.2016 - 20:47

</p>\\<p></p>\\<p>\frac{1}{3}x^2-\frac{2}{3}x+\frac{5}{9}</p>\\<p>x^4-5x:3x^2+6x+7</p>\\<p>x^4+2x^3+\frac{7}{3}x^2</p>\\<p>-2x^3-\frac{7}{3}x^2-5x</p>\\<p>-2x^3-4x^2-\frac{14}{3}x</p>\\<p>\frac{5}{3}x^2-\frac{1}{3}x</p>\\<p>\frac{5}{3}x^2+\frac{10}{3}x+\frac{35}{9}</p>\\<p>-\frac{11}{3}x-\frac{35}{9}</p>\\<p>\int{\left(\frac{1}{3}x^2-\frac{2}{3}x+\frac{5}{9}\right)\mbox{d}x}-\frac{1}{9}\int{\frac{33x+35}{3x^2+6x+7}\mbox{d}x}\\</p>\\<p>=\frac{1}{9}x^3-\frac{1}{3}x^2+\frac{5}{9}x-\frac{1}{9}\int{\frac{33x+35}{3\left(x^2+2x+1\right)+4}\mbox{d}x}\\</p>\\<p>=\frac{1}{9}x^3-\frac{1}{3}x^2+\frac{5}{9}x-\frac{1}{9}\int{\frac{33x+35}{3\left(x+1\right)^2+4}\mbox{d}x}\\</p>\\<p>\sqrt{3}\left(x+1\right)=2t\\</p>\\<p>\sqrt{3}\mbox{d}x=2\mbox{d}t\\</p>\\<p>\mbox{d}x=\frac{2\sqrt{3}}{3}\mbox{d}t\\</p>\\<p>-\frac{2\sqrt{3}}{27}\int{\frac{22\sqrt{3}t+2}{4t^2+4}\mbox{d}t}\\</p>\\<p>-\frac{\sqrt{3}}{54}\int{\frac{22\sqrt{3}t+2}{t^2+1}\mbox{d}t}\\</p>\\<p>-\frac{\sqrt{3}}{54}\left(11\sqrt{3}\ln{\left|t^2+1\right|}+2\arctan{t}\right)+C\\</p>\\<p>-\frac{11}{18}\ln{\left|t^2+1\right|}-\frac{\sqrt{3}}{27}\arctan{t}+C\\</p>\\<p>-\frac{11}{18}\ln{\left|\frac{3x^2+6x+3+4}{4}\right|}-\frac{\sqrt{3}}{27}\arctan{\left(\frac{\sqrt{3}\left(x+1\right)}{2}\right)}+C\\</p>\\<p>-\frac{11}{18}\ln{\left|3x^2+6x+7\right|}-\frac{\sqrt{3}}{27}\arctan{\left(\frac{\sqrt{3}\left(x+1\right)}{2}\right)}+C\\</p>\\<p></p>\\<p>=\frac{1}{9}x^3-\frac{1}{3}x^2+\frac{5}{9}x-\frac{11}{18}\ln{\left|3x^2+6x+7\right|}-\frac{\sqrt{3}}{27}\arctan{\left(\frac{\sqrt{3}\left(x+1\right)}{2}\right)}+C\\</p>\\<p>


  • 1


#127635 Równanie różniczkowe Bernulliego

Napisane przez Mariusz M w 01.07.2016 - 18:45

\frac{\mbox{d}y}{\mbox{d}x}+xy=xy^3\\</p>\\<p>\frac{\mbox{d}y}{\mbox{d}x}=xy^3-xy\\</p>\\<p>\frac{\mbox{d}y}{\mbox{d}x}=x\left(y^3-y\right)\\</p>\\<p>\frac{\mbox{d}y}{y\left(y^2-1\right)}=x\mbox{d}x\\</p>\\<p>\frac{2\mbox{d}y}{y\left(y^2-1\right)}=2x\mbox{d}x\\</p>\\<p>\frac{2-2y^2+2y^2}{y\left(y^2-1\right)}\mbox{d}y=2x\mbox{d}x\\</p>\\<p>\left(-\frac{2}{y}+\frac{2y}{y^2-1}\right)\mbox{d}y=2x\mbox{d}x\\</p>\\<p>-2\ln{\left|y\right|}+\ln{\left|y^2-1\right|}=x^2+C\\</p>\\<p>\frac{y^2-1}{y^2}=Ce^{x^2}\\</p>\\<p>1-\frac{1}{y^2}=Ce^{x^2}\\</p>\\<p>1-Ce^{x^2}=\frac{1}{y^2}\\</p>\\<p>y^{2}=\frac{1}{1-Ce^{x^2}}\\</p>\\<p>


  • 1


#127633 Równanie różniczkowe Bernulliego

Napisane przez Mariusz M w 01.07.2016 - 18:14

Kiniu  próbowałaś przyjąć y za zmienną niezależną ?

 

Wtedy zdaje się że będzie liniowe

 

\left(x-2xy-y^2\right)\frac{\mbox{d}y}{\mbox{d}x}+y^2=0\\</p>\\<p>\left(x-2xy-y^2\right)+y^2\frac{\mbox{d}x}{\mbox{d}y}=0\\</p>\\<p>y^{2}x'+x-2xy-y^2=0\\</p>\\<p>y^{2}x'+\left(1-2y\right)x-y^2=0\\</p>\\<p>y^{2}x'+\left(1-2y\right)x=y^2\\</p>\\<p>y^{2}x'+\left(1-2y\right)x=0\\</p>\\<p>y^{2}x'=\left(2y-1\right)x</p>\\<p>\frac{x'}{x}=\frac{2y-1}{y^2}\\</p>\\<p>\frac{\mbox{d}x}{x}=\frac{2y-1}{y^2}\mbox{d}y\\</p>\\<p>\ln{\left|x\right|}=2\ln{\left|y\right|}+\frac{1}{y}+C\\</p>\\<p>x=Cy^2e^{\frac{1}{y}}\\</p>\\<p>x\left(y\right)=C\left(y\right)y^2e^{\frac{1}{y}}\\</p>\\<p>y^2\left(C'\left(y\right)y^2e^{\frac{1}{y}}+C\left(y\right)\left(2y-1\right)e^{\frac{1}{y}}\right)+\left(1-2y\right)C\left(y\right)y^2e^{\frac{1}{y}}=y^2\\</p>\\<p>C'\left(y\right)y^4e^{\frac{1}{y}}=y^2</p>\\<p>C'\left(y\right)=\frac{1}{y^2}e^{-\frac{1}{y}}\\</p>\\<p>t=-\frac{1}{y}\\</p>\\<p>\mbox{d}t=\frac{1}{y^2}\mbox{d}y\\</p>\\<p>\int{e^{t}\mbox{d}t}=e^{t}+C\\</p>\\<p>x\left(y\right)=\left(e^{-\frac{1}{y}}+C\right)y^2e^{\frac{1}{y}}\\</p>\\<p>x\left(y\right)=y^2+Cy^2e^{\frac{1}{y}}\\</p>\\<p>


  • 1


#127509 całka z pierwiastkiem

Napisane przez Mariusz M w 23.06.2016 - 18:11

Można przez części ale jeśli już korzystać z podstawień

 

\int{\sqrt{4x-x^2}\mbox{d}x}\\</p>\\<p>\sqrt{4x-x^2}=xt\\</p>\\<p>4x-x^2=x^2t^2\\</p>\\<p>4-x=xt^2\\</p>\\<p>4=x+xt^2\\</p>\\<p>x\left(1+t^2\right)=4\\</p>\\<p>x=\frac{4}{1+t^2}\\</p>\\<p>xt=\frac{4t}{1+t^2}\\</p>\\<p>\mbox{d}x=4\left(-1\right)\left(1+t^2\right)^{-2}\cdot 2t\mbox{d}t\\</p>\\<p>\mbox{d}x=-\frac{8t}{\left(1+t^2\right)^2}\mbox{d}t\\</p>\\<p>\int{\frac{-32t^2}{\left(1+t^2\right)^3}\mbox{d}t}\\</p>\\<p>

 

\int{\frac{-32t^2}{\left(1+t^2\right)^3}\mbox{d}t}=\frac{a_{3}t^3+a_{2}t^2+a_{1}t+a_{0}}{\left(1+t^2\right)^2}+\int{\frac{b_{1}t+b_{0}}{1+t^2}\mbox{d}t}\\</p>\\<p>\frac{-32t^2}{\left(1+t^2\right)^3}=\frac{\left(3a_{3}t^2+2a_{2}t+a_{1}\right)\left(1+t^2\right)^2-\left(a_{3}t^3+a_{2}t^2+a_{1}t+a_{0}\right)\left(1+t^2\right)4t}{\left(1+t^2\right)^4}+\frac{b_{1}t+b_{0}}{1+t^2}\\</p>\\<p>\frac{-32t^2}{\left(1+t^2\right)^3}=\frac{\left(3a_{3}t^2+2a_{2}t+a_{1}\right)\left(1+t^2\right)-4t\left(a_{3}t^3+a_{2}t^2+a_{1}t+a_{0}\right)+\left(b_{1}t+b_{0}\right)\left(1+2t^2+t^4\right)}{\left(1+t^2\right)^3}\\</p>\\<p>-32t^2=\left(3a_{3}t^2+2a_{2}t+a_{1}\right)\left(1+t^2\right)-4t\left(a_{3}t^3+a_{2}t^2+a_{1}t+a_{0}\right)+\left(b_{1}t+b_{0}\right)\left(1+2t^2+t^4\right)\\</p>\\<p>-32t^2=\left(3a_{3}t^2+2a_{2}t+a_{1}+3a_{3}t^4+2a_{2}t^3+a_{1}t^2\right)-\left(4a_{3}t^4+4a_{2}t^3+4a_{1}t^2+4a_{0}t\right)+\left(b_{1}t+2b_{1}t^3+b_{1}t^5+b_{0}+2b_{0}t^2+b_{0}t^4\right)\\</p>\\<p>-32t^2=b_{1}t^5+\left(b_{0}-a_{3}\right)t^4+\left(2b_{1}-2a_{2}\right)t^3+\left(2b_{0}+3a_{3}-3a_{1}\right)t^2+\left(b_{1}+2a_{2}-4a_{0}\right)+a_{1}+b_{0}\\</p>\\<p>\begin{cases}b_{1}=0\\b_{0}-a_{3}=0\\2b_{1}-2a_{2}=0\\2b_{0}+3a_{3}-3a_{1}=-32\\b_{1}+2a_{2}-4a_{0}=0\\a_{1}+b_{0}=0\end{cases}\\</p>\\<p>\begin{cases}b_{1}=0\\b_{0}=a_{3}\\a_{2}=0\\5a_{3}-3a_{1}=-32\\a_{0}=0\\a_{1}+a_{3}=0\end{cases}\\</p>\\<p>\begin{cases}b_{1}=0\\b_{0}=a_{3}\\a_{2}=0\\a_{3}=-4\\a_{0}=0\\a_{1}=-a_{3}\end{cases}\\</p>\\<p>

 

\int{\frac{-32t^2}{\left(1+t^2\right)^3}\mbox{d}t}=\frac{-4t^3+4t}{\left(1+t^2\right)^2}-4\int{\frac{\mbox{d}t}{1+t^2}}\\</p>\\<p>\int{\frac{-32t^2}{\left(1+t^2\right)^3}\mbox{d}t}=\frac{4t}{1+t^2}\cdot\frac{1-t^2}{1+t^2}-4\arctan{\left(t\right)}+C\\</p>\\<p>\int{\frac{-32t^2}{\left(1+t^2\right)^3}\mbox{d}t}=\frac{4t}{1+t^2}\cdot\frac{-1-t^2+2}{1+t^2}-4\arctan{\left(t\right)}+C\\</p>\\<p>\int{\frac{-32t^2}{\left(1+t^2\right)^3}\mbox{d}t}=\frac{4t}{1+t^2}\cdot\left(-1+\frac{2}{1+t^2}\right)-4\arctan{\left(t\right)}+C\\</p>\\<p>\int{\sqrt{4x-x^2}\mbox{d}x}=\sqrt{4x-x^2}\left(-1+\frac{1}{2}x\right)-4\arctan{\left(\frac{\sqrt{4x-x^2}}{x}\right)}+C\\</p>\\<p>\int{\sqrt{4x-x^2}\mbox{d}x}=\frac{1}{2}\left(x-2\right)\sqrt{4x-x^2}-4\arctan{\left(\frac{\sqrt{4x-x^2}}{x}\right)}+C\\</p>\\<p>

 

 

 


  • 2


#127485 Znajdź funkcję odwrotną

Napisane przez Mariusz M w 21.06.2016 - 23:41

Mając dane wartości pewnej funkcji w punktach zauważyłem pewien wzór rekurencyjny

 

</p>\\<p>a_{n}=a_{n-3}+1\\<br>\\A\left(x\right)=\sum_{n=0}^{\infty}{a_{n}x^{n}}\\<br>\\\sum_{n=3}^{\infty}{a_{n}x^{n}}=\sum_{n=3}^{\infty}{a_{n-3}x^{n}}+\sum_{n=3}^{\infty}{x^{n}}\\<br>\\\sum_{n=3}^{\infty}{a_{n}x^{n}}=x^3\left(\sum_{n=3}^{\infty}{a_{n-3}x^{n-3}}\right)+\frac{x^3}{1-x}\\<br>\\\sum_{n=0}^{\infty}{a_{n}x^{n}}-a_{0}-a_{1}x-a_{2}x^2=x^3\left(\sum_{n=0}^{\infty}{a_{n}x^{n}}\right)+\frac{x^3}{1-x}\\<br>\\A\left(x\right)-a_{0}-a_{1}x-a_{2}x^2=x^3A\left(x\right)+\frac{x^3}{1-x}\\<br>\\A\left(x\right)\left(1-x^3\right)=a_{0}+a_{1}x+a_{2}x^2+\frac{x^3}{1-x}\\<br>\\A\left(x\right)\left(1-x^3\right)=\frac{\left(a_{0}+a_{1}x+a_{2}x^2\right)\left(1-x\right)+x^3}{1-x}\\<br>\\A\left(x\right)\left(1-x^3\right)=\frac{\left(1-a_{2}\right)x^3+\left(a_{2}-a_{1}\right)x^2+\left(a_{1}-a_{0}\right)x+a_{0}}{1-x}\\<br>\\A\left(x\right)=\frac{\left(1-a_{2}\right)x^3+\left(a_{2}-a_{1}\right)x^2+\left(a_{1}-a_{0}\right)x+a_{0}}{\left(1-x^3\right)\left(1-x\right)}\\<br>\\A\left(x\right)=\frac{\left(1-a_{2}\right)x^3+\left(a_{2}-a_{1}\right)x^2+\left(a_{1}-a_{0}\right)x+a_{0}}{\left(1-x\right)^2\left(1+x+x^2\right)}\\<br>\\A\left(x\right)=\frac{\left(1-a_{2}\right)x^3+\left(a_{2}-a_{1}\right)x^2+\left(a_{1}-a_{0}\right)x+a_{0}}{\left(1-x\right)^2\left(1-\lambda_{1}x\right)\left(1-\lambda_{2}x\right)}\\<br>\\\left(1-\lambda_{1}x\right)\left(1-\lambda_{2}x\right)=1+x+x^2\\<br>\\\frac{\left(1-a_{2}\right)x^3+\left(a_{2}-a_{1}\right)x^2+\left(a_{1}-a_{0}\right)x+a_{0}}{\left(1-x\right)^2\left(1-\lambda_{1}x\right)\left(1-\lambda_{2}x\right)}=\frac{B_{1}}{1-x}+\frac{B_{2}}{\left(1-x\right)^2}+\frac{B_{3}}{1-\lambda_{1}x}+\frac{B_{4}}{1-\lambda_{2}x}\\<br>\\\left(1-a_{2}\right)x^3+\left(a_{2}-a_{1}\right)x^2+\left(a_{1}-a_{0}\right)x+a_{0}=B_{1}\left(1-x^3\right)+B_{2}\left(1+x+x^2\right)+B_{3}\left(1-\lambda_{2}x\right)\left(1-2x+x^2\right)+B_{4}\left(1-\lambda_{1}x\right)\left(1-2x+x^2\right)\\<br>\\\left(1-a_{2}\right)x^3+\left(a_{2}-a_{1}\right)x^2+\left(a_{1}-a_{0}\right)x+a_{0}=B_{1}\left(1-x^3\right)+B_{2}\left(1+x+x^2\right)+B_{3}\left(1-2x+x^2-\lambda_{2}x+2\lambda_{2}x^2-\lambda_{2}x^3\right)+B_{4}\left(1-2x+x^2-\lambda_{1}x+2\lambda_{1}x^2-\lambda_{1}x^3\right)\\<br>\\\left(1-a_{2}\right)x^3+\left(a_{2}-a_{1}\right)x^2+\left(a_{1}-a_{0}\right)x+a_{0}=B_{1}\left(1-x^3\right)+B_{2}\left(1+x+x^2\right)+B_{3}\left(1-\left(2+\lambda_{2}\right)x+\left(1+2\lambda_{2}\right)x^2-\lambda_{2}x^3\right)+B_{4}\left(1-\left(2+\lambda_{1}\right)x+\left(1+2\lambda_{1}\right)x^2-\lambda_{1}x^3\right)\\<br>\\\begin{cases}-B_{1}-\lambda_{2}B_{3}-\lambda_{1}B_{4}=1-a_{2}\\B_{2}+\left(1+2\lambda_{2}\right)B_{3}+\left(1+2\lambda_{1}\right)B_{4}=a_{2}-a_{1}\\B_{2}-\left(2+\lambda_{2}\right)B_{3}-\left(2+\lambda_{1}\right)B_{4}=a_{1}-a_{0}\\B_{1}+B_{2}+B_{3}+B_{4}=a_{0}\end{cases}\\</p>\\<p>

 

</p>\\<p>\begin{bmatrix}-1&0&-\lambda_{2}&-\lambda_{1}\\0&1&1+2\lambda_{2}&1+2\lambda_{1}\\0&1&-2-\lambda_{2}&-2-\lambda_{1}\\1&1&1&1\end{bmatrix}\cdot\begin{bmatrix}B_{1}\\B_{2}\\B_{3}\\B_{4}\end{bmatrix}=\begin{bmatrix}1-a_{2}\\a_{2}-a_{1}\\a_{1}-a_{0}\\a_{0}\end{bmatrix}\\<br>\\\begin{bmatrix}-1&0&-\lambda_{2}&-\lambda_{1}&\qquad &1&0&0&0\\0&1&1+2\lambda_{2}&1+2\lambda_{1}&\qquad&0&1&0&0\\0&1&-2-\lambda_{2}&-2-\lambda_{1}&\qquad &0&0&1&0\\1&1&1&1&\qquad&0&0&0&1\end{bmatrix}\\<br>\\\begin{bmatrix}1&0&\lambda_{2}&\lambda_{1}&\qquad &-1&0&0&0\\0&1&1+2\lambda_{2}&1+2\lambda_{1}&\qquad&0&1&0&0\\0&1&-2-\lambda_{2}&-2-\lambda_{1}&\qquad &0&0&1&0\\0&1&1-\lambda_{2}&1-\lambda_{1}&\qquad&1&0&0&1\end{bmatrix}\\<br>\\\begin{bmatrix}1&0&\lambda_{2}&\lambda_{1}&\qquad &-1&0&0&0\\0&1&1+2\lambda_{2}&1+2\lambda_{1}&\qquad&0&1&0&0\\0&0&-3-3\lambda_{2}&-3-3\lambda_{1}&\qquad &0&-1&1&0\\0&0&-3\lambda_{2}&-3\lambda_{1}&\qquad&1&-1&0&1\end{bmatrix}\\<br>\\\begin{bmatrix}1&0&\lambda_{2}&\lambda_{1}&\qquad &-1&0&0&0\\0&3&3+6\lambda_{2}&3+6\lambda_{1}&\qquad&0&3&0&0\\0&0&3&3&\qquad &1&0&-1&1\\0&0&-3\lambda_{2}&-3\lambda_{1}&\qquad&1&-1&0&1\end{bmatrix}\\<br>\\\begin{bmatrix}1&0&\lambda_{2}&\lambda_{1}&\qquad &-1&0&0&0\\0&3&3&3&\qquad&2&1&0&2\\0&0&3&3&\qquad &1&0&-1&1\\0&0&-3\lambda_{2}&-3\lambda_{1}&\qquad&1&-1&0&1\end{bmatrix}\\<br>\\\begin{bmatrix}1&0&\lambda_{2}&\lambda_{1}&\qquad &-1&0&0&0\\0&3&0&0&\qquad&1&1&1&1\\0&0&3&3&\qquad &1&0&-1&1\\0&0&0&-3\lambda_{1}+3\lambda_{2}&\qquad&1+\lambda_{2}&-1&-\lambda_{2}&1+\lambda_{2}\end{bmatrix}\\<br>\\\begin{bmatrix}3\left(\lambda_{1}-\lambda_{2}\right)&0&3\lambda_{2}\left(\lambda_{1}-\lambda_{2}\right)&3\lambda_{1}\left(\lambda_{1}-\lambda_{2}\right)&\qquad &-3\left(\lambda_{1}-\lambda_{2}\right)&0&0&0\\0&3&0&0&\qquad&1&1&1&1\\0&0&3\left(\lambda_{1}-\lambda_{2}\right)&3\left(\lambda_{1}-\lambda_{2}\right)&\qquad &\left(\lambda_{1}-\lambda_{2}\right)&0&-\left(\lambda_{1}-\lambda_{2}\right)&\left(\lambda_{1}-\lambda_{2}\right)\\0&0&0&-3\lambda_{1}+3\lambda_{2}&\qquad&1+\lambda_{2}&-1&-\lambda_{2}&1+\lambda_{2}\end{bmatrix}\\<br>\\\begin{bmatrix}3\left(\lambda_{1}-\lambda_{2}\right)&0&3\lambda_{2}\left(\lambda_{1}-\lambda_{2}\right)&0&\qquad &-2\lambda_{1}+3\lambda_{2}+\lambda_{2}\lambda_{1}&-\lambda_{1}&-\lambda_{2}\lambda_{1}&\lambda_{1}+\lambda_{2}\lambda_{1}\\0&3&0&0&\qquad&1&1&1&1\\0&0&3\left(\lambda_{1}-\lambda_{2}\right)&0&\qquad &1+\lambda_{1}&-1&-\lambda_{1}&1+\lambda_{1}\\0&0&0&-3\lambda_{1}+3\lambda_{2}&\qquad&1+\lambda_{2}&-1&-\lambda_{2}&1+\lambda_{2}\end{bmatrix}\\<br>\\\begin{bmatrix}3\left(\lambda_{1}-\lambda_{2}\right)&0&0&0&\qquad &-2\lambda_{1}+2\lambda_{2}&-\lambda_{1}+\lambda_{2}&0&\lambda_{1}-\lambda_{2}\\0&3\left(\lambda_{1}-\lambda_{2}\right)&0&0&\qquad&\lambda_{1}-\lambda_{2}&\lambda_{1}-\lambda_{2}&\lambda_{1}-\lambda_{2}&\lambda_{1}-\lambda_{2}\\0&0&3\left(\lambda_{1}-\lambda_{2}\right)&0&\qquad &1+\lambda_{1}&-1&-\lambda_{1}&1+\lambda_{1}\\0&0&0&3\left(\lambda_{1}-\lambda_{2}\right)&\qquad&-1-\lambda_{2}&1&\lambda_{2}&-1-\lambda_{2}\end{bmatrix}\\<br>\\

 

</p>\\<p>B^{-1}=\frac{1}{3\left(\lambda_{1}-\lambda_{2}\right)}\begin{bmatrix}-2\lambda_{1}+2\lambda_{2}&-\lambda_{1}+\lambda_{2}&0&\lambda_{1}-\lambda_{2}\\\lambda_{1}-\lambda_{2}&\lambda_{1}-\lambda_{2}&\lambda_{1}-\lambda_{2}&\lambda_{1}-\lambda_{2}\\1+\lambda_{1}&-1&-\lambda_{1}&1+\lambda_{1}\\-1-\lambda_{2}&1&\lambda_{2}&-1-\lambda_{2}\end{bmatrix}\\<br>\\X=\frac{1}{3\left(\lambda_{1}-\lambda_{2}\right)}\begin{bmatrix}-2\lambda_{1}+2\lambda_{2}&-\lambda_{1}+\lambda_{2}&0&\lambda_{1}-\lambda_{2}\\\lambda_{1}-\lambda_{2}&\lambda_{1}-\lambda_{2}&\lambda_{1}-\lambda_{2}&\lambda_{1}-\lambda_{2}\\1+\lambda_{1}&-1&-\lambda_{1}&1+\lambda_{1}\\-1-\lambda_{2}&1&\lambda_{2}&-1-\lambda_{2}\end{bmatrix}\cdot\begin{bmatrix}1-a_{2}\\a_{2}-a_{1}\\a_{1}-a_{0}\\a_{0}\end{bmatrix}\\<br>\\X=\begin{bmatrix}-\frac{2}{3}+\frac{1}{3}a_{2}+\frac{1}{3}a_{1}+\frac{1}{3}a_{0}\\\frac{1}{3}\\\frac{-\left(2+\lambda_{1}\right)a_{2}+\left(1-\lambda_{1}\right)a_{1}+\left(1+2\lambda_{1}\right)a_{0}+\left(1+\lambda_{1}\right)}{3\left(\lambda_{1}-\lambda_{2}\right)}\\\frac{\left(2+\lambda_{2}\right)a_{2}+\left(-1+\lambda_{2}\right)a_{1}+\left(-1-2\lambda_{2}\right)a_{0}-1-\lambda_{2}}{3\left(\lambda_{1}-\lambda_{2}\right)}\end{bmatrix}\\<br>\\\left(1-\lambda_{1}x\right)\left(1-\lambda_{2}x\right)=1+x+x^2\\<br>\\1-\left(\lambda_{1}+\lambda_{2}\right)x+\lambda_{1}\lambda_{2}x^2=1+x+x^2\\<br>\\\begin{cases}\lambda_{1}+\lambda_{2}=-1\\\lambda_{1}\lambda_{2}=1\end{cases}\\<br>\\t^2+t+1=0\\<br>\\\lambda_{1}=\frac{-1-sqrt{3}i}{2}=\cos{\left(\frac{4\pi}{3}\right)}+i\sin{\left(\frac{4\pi}{3}\right)}\\<br>\\\lambda_{2}=\frac{-1+sqrt{3}i}{2}=\cos{\left(\frac{4\pi}{3}\right)}-i\sin{\left(\frac{4\pi}{3}\right)}\\<br>\\\sum_{n=0}^{\infty}{x^{n}}=\frac{1}{1-x}\\<br>\\\frac{\mbox{d}}{\mbox{d}x}\left(\sum_{n=0}^{\infty}{x^{n}}\right)=\frac{\mbox{d}}{\mbox{d}x}\left(\frac{1}{1-x}\right)\\<br>\\\sum_{n=0}^{\infty}{nx^{n-1}}=-\frac{1}{\left(1-x\right)^2}\left(-1\right)\\<br>\\\sum_{n=1}^{\infty}{nx^{n-1}}=\frac{1}{\left(1-x\right)^2}\\<br>\\\sum_{n=0}^{\infty}{\left(n+1\right)x^{n}}=\frac{1}{\left(1-x\right)^2}\\<br>\\a\cos{\left(x\right)}+b\sin{\left(x\right)}=\sqrt{a^2+b^2}\left(\frac{a}{\sqrt{a^2+b^2}}\cos{\left(x\right)}+\frac{b}{\sqrt{a^2+b^2}}\sin{\left(x\right)}\right)\\<br>\\a\cos{\left(x\right)}+b\sin{\left(x\right)}=\sqrt{a^2+b^2}\cos{\left(x-\varphi\right)}\\<br>\\\begin{cases}\cos{\left(\varphi\right)}=\frac{a}{\sqrt{a^2+b^2}}\\\sin{\left(\varphi\right)}=\frac{b}{\sqrt{a^2+b^2}}\end{cases}\\<br>\\X=\begin{bmatrix}-\frac{2}{3}+\frac{1}{3}a_{2}+\frac{1}{3}a_{1}+\frac{1}{3}a_{0}\\\frac{1}{3}\\\left(-\frac{1}{6}-\frac{\sqrt{3}i}{6}\right)a_{2}+\left(-\frac{1}{6}+\frac{\sqrt{3}i}{6}\right)a_{1}+\frac{1}{3}a_{0}+\frac{1}{6}+\frac{\sqrt{3}i}{18}\\\left(-\frac{1}{6}+\frac{\sqrt{3}i}{6}\right)a_{2}+\left(-\frac{1}{6}-\frac{\sqrt{3}i}{6}\right)a_{1}+\frac{1}{3}a_{0}+\frac{1}{6}-\frac{\sqrt{3}i}{18}\end{bmatrix}\\</p>\\<p>

 

</p>\\<p>A\left(x\right)=\left(-\frac{2}{3}+\frac{1}{3}a_{2}+\frac{1}{3}a_{1}+\frac{1}{3}a_{0}\right)\cdot\frac{1}{1-x}+\frac{1}{3}\cdot\frac{1}{\left(1-x\right)^2}+\left(\left(-\frac{1}{6}-\frac{\sqrt{3}i}{6}\right)a_{2}+\left(-\frac{1}{6}+\frac{\sqrt{3}i}{6}\right)a_{1}+\frac{1}{3}a_{0}+\frac{1}{6}+\frac{\sqrt{3}i}{18}\right)\cdot\frac{1}{1-\lambda_{1}x}+\left(\left(-\frac{1}{6}+\frac{\sqrt{3}i}{6}\right)a_{2}+\left(-\frac{1}{6}-\frac{\sqrt{3}i}{6}\right)a_{1}+\frac{1}{3}a_{0}+\frac{1}{6}-\frac{\sqrt{3}i}{18}\right)\cdot\frac{1}{1-\lambda_{2}x}\\<br>\\A\left(x\right)=\left(-\frac{2}{3}+\frac{1}{3}a_{2}+\frac{1}{3}a_{1}+\frac{1}{3}a_{0}\right)\cdot\left(\sum_{n=0}^{\infty}x^n\right)+\frac{1}{3}\left(\sum_{n=0}^{\infty}{\left(n+1\right)x^n}\right)+\left(\left(-\frac{1}{6}-\frac{\sqrt{3}i}{6}\right)a_{2}+\left(-\frac{1}{6}+\frac{\sqrt{3}i}{6}\right)a_{1}+\frac{1}{3}a_{0}+\frac{1}{6}+\frac{\sqrt{3}i}{18}\right)\cdot\left(\sum_{n=0}^{\infty}\lambda_{1}^nx^n\right)+\left(\left(-\frac{1}{6}+\frac{\sqrt{3}i}{6}\right)a_{2}+\left(-\frac{1}{6}-\frac{\sqrt{3}i}{6}\right)a_{1}+\frac{1}{3}a_{0}+\frac{1}{6}-\frac{\sqrt{3}i}{18}\right)\cdot\left(\sum_{n=0}^{\infty}\lambda_{2}^nx^n\right)\\<br>\\a_{n}=\left(-\frac{2}{3}+\frac{1}{3}a_{2}+\frac{1}{3}a_{1}+\frac{1}{3}a_{0}\right)+\frac{1}{3}\left(n+1\right)+\left(\left(-\frac{1}{6}-\frac{\sqrt{3}i}{6}\right)a_{2}+\left(-\frac{1}{6}+\frac{\sqrt{3}i}{6}\right)a_{1}+\frac{1}{3}a_{0}+\frac{1}{6}+\frac{\sqrt{3}i}{18}\right)\cdot \lambda_{1}^{n}+\left(\left(-\frac{1}{6}+\frac{\sqrt{3}i}{6}\right)a_{2}+\left(-\frac{1}{6}-\frac{\sqrt{3}i}{6}\right)a_{1}+\frac{1}{3}a_{0}+\frac{1}{6}-\frac{\sqrt{3}i}{18}\right)\cdot \lambda_{2}^{n}\\<br>\\a_{n}=\left(-\frac{2}{3}+\frac{1}{3}a_{2}+\frac{1}{3}a_{1}+\frac{1}{3}a_{0}\right)+\frac{1}{3}\left(n+1\right)+\left(\left(-\frac{1}{6}-\frac{\sqrt{3}i}{6}\right)a_{2}+\left(-\frac{1}{6}+\frac{\sqrt{3}i}{6}\right)a_{1}+\frac{1}{3}a_{0}+\frac{1}{6}+\frac{\sqrt{3}i}{18}\right)\cdot \left(\cos{\left(\frac{4}{3}n\pi\right)}+i\sin{\left(\frac{4}{3}n\pi\right)}\right)+\left(\left(-\frac{1}{6}+\frac{\sqrt{3}i}{6}\right)a_{2}+\left(-\frac{1}{6}-\frac{\sqrt{3}i}{6}\right)a_{1}+\frac{1}{3}a_{0}+\frac{1}{6}-\frac{\sqrt{3}i}{18}\right)\cdot \left(\cos{\left(\frac{4}{3}n\pi\right)}-i\sin{\left(\frac{4}{3}n\pi\right)}\right)\\<br>\\a_{n}=\frac{1}{3}\left(n-1+a_{2}+a_{1}+a_{0}\right)-\frac{1}{3}\left(a_{2}+a_{1}-2a_{0}-1\right)\cos{\left(\frac{4}{3}n\pi\right)}+\frac{\sqrt{3}}{9}\left(3a_{2}-3a_{1}-1\right)\sin{\left(\frac{4}{3}n\pi\right)}\\</p>\\<p>

 

Dla zadanych warunków początkowych np

a_{0}=-1\\a_{1}=-\frac{3}{5}\\a_{2}=-\frac{3}{10}

znajdź funkcję odwrotną

 


  • 1


#127483 Oblicz granicę funkcji

Napisane przez Mariusz M w 21.06.2016 - 23:04

Gdyby wykorzystać granicę \lim_{x\to 0}\frac{\sin{x}}{x}=1

to nawet raz by de l'Hospital starczył

 

\lim_{x\to 0}\left(\frac{1}{x}-\cot{x}\right)\\</p>\\<p>\lim_{x\to 0}\left(\frac{1}{x}-\frac{\cos{x}}{\sin{x}}\right)\\</p>\\<p>\lim_{x\to 0}\left(\frac{\sin{x}-x\cos{x}}{x\sin{x}}\right)\\</p>\\<p>\lim_{x\to 0}\left(\frac{\sin{x}-x\cos{x}}{x^2}\right)\cdot\frac{x}{\sin{x}}\\</p>\\<p>\lim_{x\to 0}\left(\frac{\sin{x}-x\cos{x}}{x^2}\right)\lim_{x\to 0}\frac{x}{\sin{x}}\\</p>\\<p>\lim_{x\to 0}\left(\frac{\sin{x}-x\cos{x}}{x^2}\right)=H=</p>\\<p>\lim_{x\to 0}\left(\frac{\cos{x}-\left(\cos{x}-x\sin{x}\right)}{2x}\right)\\</p>\\<p>\lim_{x\to 0}\left(\frac{x\sin{x}}{2x}\right)\\</p>\\<p>\lim_{x\to 0}\left(\frac{\sin{x}}{2}\right)=0\\</p>\\<p>

 

Jednak wolałbym bez de l'Hospitala


  • 1


#127479 Oblicz granicę funkcji

Napisane przez Mariusz M w 21.06.2016 - 21:19

\lim_{x\to 0}{\left(\frac{1}{x}-\cot{x}\right)}\\</p>\\<p>=\lim_{x\to 0}{\left(\frac{1}{x}-\frac{1}{\tan{x}}\right)}\\</p>\\<p>=\lim_{x\to 0}\frac{\tan{x}-x}{x\tan{x}}\\</p>\\<p>

 

\lim_{\Delta x\to 0}\frac{\left(\tan{\left(x+\Delta x\right)}-\tan{x}\right)-\left(\left(x+\Delta x\right)-x\right)}{\Delta x}\\</p>\\<p>\lim_{\Delta x\to 0}\frac{\frac{\tan{x}+\tan{\Delta x}}{1-\tan{x}\tan{\Delta x}}-\tan{x}-\Delta x}{\Delta x}\\</p>\\<p>\lim_{\Delta x\to 0}\frac{\frac{\tan{x}+\tan{\Delta x}-\tan{x}\left(1-\tan{x}\tan{\Delta x}\right)}{1-\tan{x}\tan{\Delta x}}-\Delta x}{\Delta x}\\</p>\\<p>\lim_{\Delta x\to 0}\frac{\frac{\tan{x}+\tan{\Delta x}-\tan{x}+\tan^2{x}\tan{\Delta x}}{1-\tan{x}\tan{\Delta x}}}{\Delta x}-\lim_{\Delta x\to 0}\frac{\Delta x}{\Delta x}\\</p>\\<p>\lim_{\Delta x\to 0}\frac{\frac{\tan{\Delta x}+\tan^2{x}\tan{\Delta x}}{1-\tan{x}\tan{\Delta x}}}{\Delta x}-1\\</p>\\<p>\lim_{\Delta x\to 0}\frac{\tan{\Delta x}}{\Delta x}\cdot\lim_{\Delta x\to 0}\frac{1+\tan^{2}{x}}{1-\tan{x}\tan{\Delta x}}-1\\</p>\\<p>1+\tan^{2}{x}-1\\</p>\\<p>=\tan^{2}{x}\\</p>\\<p>\lim_{\Delta x\to 0}\frac{\left(x+\Delta x\right)\tan{\left(x+\Delta x\right)}-x\tan{x}}{\Delta x}\\</p>\\<p>\lim_{\Delta x\to 0}\frac{\left(x+\Delta x\right)\tan{\left(x+\Delta x\right)}-x\tan{\left(x+\Delta x\right)}+x\tan{\left(x+\Delta x\right)}-x\tan{x}}{\Delta x}\\</p>\\<p>\lim_{\Delta x\to 0}\frac{\left(x+\Delta x\right)-x}{\Delta x}\lim_{\Delta x\to 0}\tan{\left(x+\Delta x\right)}+x\lim_{\Delta x\to 0}\frac{\tan{\left(x+\Delta x\right)}-\tan{x}}{\Delta x}\\</p>\\<p>\tan{x}+x\left(\lim_{\Delta x\to 0}\frac{\frac{\tan{x}+\tan{\Delta x}}{1-\tan{x}\tan{\Delta x}}-\tan{x}}{\Delta x}\right)\\</p>\\<p>\tan{x}+x\left(\lim_{\Delta x\to 0}\frac{\frac{\tan{x}+\tan{\Delta x}-\tan{x}+\tan^{2}{x}\tan{\Delta x}}{1-\tan{x}\tan{\Delta x}}}{\Delta x}\right)\\</p>\\<p>\tan{x}+x\left(\lim_{\Delta x\to 0}\frac{\frac{\tan{\Delta x}+\tan^{2}{x}\tan{\Delta x}}{1-\tan{x}\tan{\Delta x}}}{\Delta x}\right)\\</p>\\<p>\tan{x}+x\left(\lim_{\Delta x\to 0}\frac{\tan{\Delta x}}{\Delta x}\cdot\lim_{\Delta x\to 0}\frac{1+\tan^{2}{x}}{1-\tan{x}\tan{\Delta x}}\right)\\</p>\\<p>\tan{x}+x\left(1+\tan^{2}{x}\right)\\<br>\\

 

 

 

 

\lim_{x\to 0}{\left(\frac{1}{x}-\cot{x}\right)}=\lim_{x\to 0}\frac{\tan^{2}{x}}{\tan{x}+x\left(1+\tan^{2}{x}\right)}\\</p>\\<p>\left(\tan^{2}{x}\right)'</p>\\<p>\lim_{\Delta x\to 0}\frac{\tan^{2}{\left(x+\Delta x\right)-\tan^{2}{x}}}{\Delta x}\\</p>\\<p>\lim_{\Delta x\to 0}\frac{\tan{\left(x+\Delta x\right)}-\tan{x}}{\Delta x}\cdot\lim_{\Delta x\to 0}\left(\tan{\left(x+\Delta x\right)}+\tan{x}\right)\\</p>\\<p>=2\tan{x}\lim_{\Delta x\to 0}\frac{\tan{\left(x+\Delta x\right)}-\tan{x}}{\Delta x}\\</p>\\<p>=2\tan{x}\lim_{\Delta x\to 0}\frac{\frac{\tan{x}+\tan{\Delta x}}{1-\tan{x}\tan{\Delta x}}-\tan{x}}{\Delta x}\\</p>\\<p>=2\tan{x}\lim_{\Delta x\to 0}\frac{\frac{\tan{x}+\tan{\Delta x}-\tan{x}+\tan^{2}{x}\tan{\Delta x}}{1-\tan{x}\tan{\Delta x}}}{\Delta x}\\</p>\\<p>=2\tan{x}\lim_{\Delta x\to 0}\frac{\frac{\tan{\Delta x}+\tan^{2}{x}\tan{\Delta x}}{1-\tan{x}\tan{\Delta x}}}{\Delta x}\\</p>\\<p>=2\tan{x}\lim_{\Delta x\to 0}\frac{\tan{\Delta x}}{\Delta x}\lim_{\Delta x\to 0}\frac{1+\tan^{2}{x}}{1-\tan{x}\tan{\Delta x}}\\</p>\\<p>=2\tan{x}\left(1+\tan^{2}{x}\right)\\</p>\\<p>\left(\tan{x}+x+x\tan^2{x}\right)'\\</p>\\<p>\lim_{\Delta x\to 0}\frac{\tan{\left(x+\Delta x\right)}-\tan{x}+\left(x+\Delta x\right)-x+\left(x+\Delta x\right)\tan^{2}{\left(x+\Delta x\right)-x\tan^{2}{x}}}{\Delta x}\\</p>\\<p>=\lim_{\Delta x\to 0}\frac{\tan{\left(x+\Delta x\right)}-\tan{x}}{\Delta x}+\lim_{\Delta x\to 0}\frac{\left(x+\Delta x\right)-x}{\Delta x}+\lim_{\Delta x\to 0}\frac{\left(x+\Delta x\right)\tan^{2}{\left(x+\Delta x\right)}-x\tan^{2}{x}}{\Delta x}\\</p>\\<p>=\lim_{\Delta x\to 0}\frac{\frac{\tan{x}+\tan{\Delta x}}{1-\tan{x}\tan{\Delta x}}-\tan{x}}{\Delta x}+1+\lim_{\Delta x\to 0}\frac{\left(x+\Delta x\right)\tan^{2}{\left(x+\Delta x\right)}-x\tan^{2}{\left(x+\Delta x\right)}+x\tan^{2}{\left(x+\Delta x\right)}-x\tan^{2}{x}}{\Delta x} \\</p>\\<p>=\lim_{\Delta x\to 0}\frac{\frac{\tan{x}+\tan{\Delta x}-\tan{x}+\tan^{2}{x}\tan{\Delta x}}{1-\tan{x}\tan{\Delta x}}}{\Delta x}+1+\lim_{\Delta x\to 0}\frac{\left(x+\Delta x\right)-x}{\Delta x}\cdot\lim_{\Delta x\to 0}\tan^{2}{\left(x+\Delta x\right)}+x\lim_{\Delta x\to 0}\frac{\tan^{2}{\left(x+\Delta x\right)}-\tan^{2}{x}}{\Delta x}\\</p>\\<p>=\lim_{\Delta x\to 0}\frac{\frac{\tan{\Delta x}+\tan^{2}{x}\tan{\Delta x}}{1-\tan{x}\tan{\Delta x}}}{\Delta x}+1+\tan^{2}{x}+x\left(\lim_{\Delta x\to 0}\frac{\tan{\left(x+\Delta x\right)}-\tan{x}}{\Delta x}\cdot\lim_{\Delta x\to 0}\left(\tan{\left(x+\Delta x\right)}+\tan{x}\right)\right)\\</p>\\<p>=\lim_{\Delta x\to 0}\frac{\tan{\Delta x}}{\Delta x}\cdot\lim_{\Delta x\to 0}\frac{1+\tan^2{x}}{1-\tan{x}\tan{\Delta x}}+1+\tan^{2}{x}+x\left(2\tan{x}\lim_{\Delta x\to 0}\frac{\frac{\tan{x}+\tan{\Delta x}}{1-\tan{x}\tan{\Delta x}}-\tan{x}}{\Delta x}\right)\\</p>\\<p>=1+\tan^{2}{x}+1+\tan^{2}{x}+2x\tan{x}\left(\lim_{\Delta x\to 0}\frac{\frac{\tan{x}+\tan{\Delta x}-\tan{x}+\tan^{2}{x}\tan{\Delta x}}{1-\tan{x}\tan{\Delta x}}}{\Delta x}\right)\\</p>\\<p>=2+2\tan^{2}{x}+2x\tan{x}\left(\lim_{\Delta x\to 0}\frac{\frac{\tan{\Delta x}+\tan^{2}{x}\tan{\Delta x}}{1-\tan{x}\tan{\Delta x}}}{\Delta x}\right)\\</p>\\<p>=2+2\tan^{2}{x}+2x\tan{x}\left(\lim_{\Delta x\to 0}\frac{\tan{\Delta x}}{\Delta x}\lim_{\Delta x\to 0}\frac{1+\tan^{2}{x}}{1-\tan{x}\tan{\Delta x}}\right)\\</p>\\<p>=2+2\tan^{2}{x}+2x\tan{x}\left(1+\tan^{2}{x}\right)\\</p>\\<p>

 

 

\lim_{x\to 0}{\left(\frac{1}{x}-\cot{x}\right)}=\lim_{x\to 0}\frac{2\tan{x}\left(1+\tan^{2}{x}\right)}{2+2\tan^{2}{x}+2x\tan{x}\left(1+\tan^{2}{x}\right)}\\</p>\\<p>=\lim_{x\to 0}\frac{2\left(1+\tan^{2}{x}\right)\tan{x}}{2\left(1+\tan^2{x}\right)\left(1+x\tan{x}\right)}\\</p>\\<p>=\lim_{x\to 0}\frac{\tan{x}}{1+x\tan{x}}=0\\</p>\\<p>

 

 

Ciekaw jestem jak to policzyć bez reguły de l'Hospitala

 

 


  • 1


#127395 Nietabelkowe wartości funkcji trygonometrycznych

Napisane przez Mariusz M w 12.06.2016 - 20:19

Jeżeli miara w stopniach jest podzielna przez 3 to da się policzyć z użyciem czterech działań i wyciągania pierwiastka

Implikacja w drugą stronę nie zachodzi


  • 1


#127273 całka krzywoliniowa

Napisane przez Mariusz M w 22.05.2016 - 12:57

\int_{0}^{1}{\left(\left(t+2+2\cdot 4t\right)\cdot 1+\left(\left(t+2\right)^2+4\left(t+2\right)\cdot 4t\right)\cdot 4\right)\mbox{d}t}\\</p>\\<p>\int_{0}^{1}{\left(\left(9t+2\right)+4\left(\left(t^2+4t+4\right)+16t^2+32t\right)\right)\mbox{d}t}\\</p>\\<p>\int_{0}^{1}{\left(68t^2+153t+18\right)\mbox{d}t}=\frac{68}{3}+\frac{153}{2}+18-0-0-0\\</p>\\<p>=\frac{136+459+108}{6}=\frac{703}{6}</p>\\<p>

 

Ups! Coś poszło nie tak!

 

Wyniki mamy różne


  • 1