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#1 Mariusz M

Mariusz M

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Napisano 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

 


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Afroman

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Napisano 25.09.2011 - 17:55

#2 Jarekzulus

Jarekzulus

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Napisano 22.06.2016 - 00:11

Ładne obliczenia :)     za mały Dyzio jestem chyba do takich rozważań :D   ale patrzyłem ostatnio na to zadanie z http://matma4u.pl/to...mentu-głównego/wybrem. Znalazłem jakiś psełdo kod ale z zapisaniem chwilowo nie idzie tak prosto. Jak się uda zapisać to podam.


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:wave: :wave: :wave: Jeśli rzuciłem choć promyczek światła na problem który postawiłeś - podziękuj. pre_1433974176__syg.jpgNad kreską


#3 Mariusz M

Mariusz M

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Napisano 22.06.2016 - 00:24

Jeżeli chodzi o funkcję odwrotną to Wolfram nie potrafi sobie poradzić

Program którego używam podaje że funkcja odwrotna jest postaci

 

f\left(x\right)=\frac{3}{2}x+\frac{29}{20}-\frac{\sqrt{3}}{30}\cos{\theta}

 

Problem w tym jak obliczyć \theta


Użytkownik Mariusz M edytował ten post 22.06.2016 - 00:24

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#4 karolbiegan

karolbiegan

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Napisano 25.06.2016 - 17:20

Niezłe rozwiązania niezły ze mnie głąb jak na to popatrze


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