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anonymous
 one year ago
Define \[f:\mathbb{N}\rightarrow \mathbb{N}\] by \[f(1)=2, f(2)=8 \], for \[n\geq 3\]
\[f(n)=8f(n1)15f(n2)+6*2^n\]
Prove that for all n\in \mathbb{N}, \[f(n)=5*3^n+5^{n1}+2^{n+3}\]
Any help. I am stuck on the inductive step
anonymous
 one year ago
Define \[f:\mathbb{N}\rightarrow \mathbb{N}\] by \[f(1)=2, f(2)=8 \], for \[n\geq 3\] \[f(n)=8f(n1)15f(n2)+6*2^n\] Prove that for all n\in \mathbb{N}, \[f(n)=5*3^n+5^{n1}+2^{n+3}\] Any help. I am stuck on the inductive step

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[f(n)=5*3^{k+1}+5^{k}+2^{k+4}\] \[f(n)=8f(k)15f(k1)+6*2^{k+1}\] Then I dont know where to go from here.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0suppose $$f(n1)=5\cdot3^{n1}+5^{n2}+2^{n+2}\\f(n)=5\cdot3^n+5^{n1}+2^{n+3}$$now we have $$\begin{align*}f(n+1)&=8 f(n)15 f(n1)+6\cdot 2^{n+1}\\&=8\left(5\cdot3^n+5^{n1}+2^{n+3}\right)15\left(5\cdot 3^{n1}+5^{n2}+2^{n+2}\right)+6\cdot 2^{n+1}\\&=40\cdot 3^n+8\cdot 5^{n1}+8\cdot 2^{n+3}+75\cdot 3^{n1}15\cdot 5^{n2}15\cdot2^{n+2}+6\cdot 2^{n+1}\\&=(40+25)\cdot 3^n+(83)\cdot 5^{n1}+(8\cdot 215+3)\cdot 2^{n+2}\\&=15\cdot 3^n+5\cdot 5^{n1}+4\cdot 2^{n+2}\\&=5\cdot 3^{n+1}+5^n+2^{n+4}\end{align*}$$

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Are you sure it's \(f(2)=8\)? @keynote

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0he meant \(f(2)=8\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0That would definitely make more sense. Solution with \(f(2)=\color{red}+8\): http://www.wolframalpha.com/input/?i=RSolve%5B%7Bf%5Bn%5D%3D%3D8f%5Bn1%5D15f%5Bn2%5D%2B6+2%5En%2Cf%5B1%5D%3D%3D2%2Cf%5B2%5D%3D%3D8%7D%2Cf%5Bn%5D%2Cn%5D Solution with \(f(2)=\color{red}8\): http://www.wolframalpha.com/input/?i=RSolve%5B%7Bf%5Bn%5D%3D%3D8f%5Bn1%5D15f%5Bn2%5D%2B6+2%5En%2Cf%5B1%5D%3D%3D2%2Cf%5B2%5D%3D%3D8%7D%2Cf%5Bn%5D%2Cn%5D

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0it's obvious to solve $$f(n)=8f(n1)15f(n2)+6\cdot 2^n$$ suppose \(f(n)=k\cdot 2^n\), so $$k\cdot 2^n=8k\cdot 2^{n1}15k\cdot 2^{n2}+6\cdot 2^n\\4k\cdot 2^{n2}16k\cdot 2^{n2}+15k\cdot 2^{n2}24\cdot 2^{n2}=0\\2^{n2}\left(4k16k+15k24\right)=0\\3k24=0\\k8=0\\k=8$$ so $$f(n)=8\cdot 2^n=2^{n+3}$$ is the particular solution to the recurrence

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and then we find the homogeneous solution with \(f(n)=r^n\) so $$r^n=8r^{n1}15r^{n2}\\r^{n2}(r^28r+15)=0\\(r5)(r3)=0\implies r\in \{3,5\}$$ so \(f(n)=A\cdot 3^n+B\cdot 5^n\) is the general solution to the homogeneous problem and \(f(n)=A\cdot 3^n+B\cdot 5^n+2^{n+3}\) where \(A,B\) are determined using initial conditions

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so here we want \(f(1)=2,f(2)=8\) giving $$f(1)=A\cdot 3+B\cdot 5+16\\f(2)=A\cdot 9+B\cdot 25+32$$ giving a system in \(A,B\): $$3A+5B+16=2\\9A+25B+32=8\\\text{rewriting gives:}\\ 3A+5B=14\\9A+25B=40$$ we can cancel so: $$9A+15B=42\\9A+25B=40$$ subtracting the first from the second: $$10B=40+42=2\\B=\frac15$$and similarly $$15A+25B=70\\9A+25B=40$$ giving $$6A=30\\A=5$$ which yields $$f(n)=5\cdot 3^n+\frac15\cdot 5^n+2^{n+3}=5\cdot 3^n+5^{n1}+2^{n+3}$$

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@oldrin.bataku I updated it. I had made a typo

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0that is irrelevant, I already knew what you meant

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Just to offer another method of solving, you can try finding the generating function \(F(x)=\sum\limits_{n=1}^\infty f(n)x^n\). Far less efficient, but it's a pretty useful technique nonetheless. \[\begin{align*}F(x)&=\sum_{n=1}^\infty f(n)x^n\\[1ex] &=\sum_{n=3}^\infty f(n)x^n+2x8x^2\\[1ex] &=\sum_{n=3}^\infty \bigg(8f(n1)15f(n2)+6\times2^n\bigg)x^n+2x8x^2\\[1ex] &=8\sum_{n=3}^\infty f(n1)x^n15\sum_{n=3}^\infty f(n2)x^n+6\sum_{n=3}^\infty(2x)^n+2x8x^2\\[1ex] &=8x\sum_{n=3}^\infty f(n1)x^{n1}15x^2\sum_{n=3}^\infty f(n2)x^{n2}+\frac{48x^2}{12x}+2x8x^2\\[1ex] &=8x\sum_{n=2}^\infty f(n)x^n15x^2\sum_{n=1}^\infty f(n)x^n+\frac{48x^2}{12x}+2x8x^2\\[1ex] &=8x\left(\sum_{n=1}^\infty f(n)x^n2x\right)15x^2\sum_{n=1}^\infty f(n)x^n+\frac{48x^2}{12x}+2x8x^2\\[1ex] &=8x(F(x)2x)15x^2F(x)+\frac{48x^2}{12x}+2x8x^2\\[1ex] F(x)&=\frac{\dfrac{48x^2}{12x}+2x24x^2}{18x+15x^2} \end{align*}\]where \(f(n)\) is the coefficient of the \(n\)th term of the Taylor expansion of \(F\) around \(x=0\). Finding a closed form for the coefficient is tedious, but it can be done. WA agrees: http://www.wolframalpha.com/input/?i=McLaurin+series+of+%2848x%5E3%2F%2812x%29%2B2x24x%5E2%29%2F%2818x%2B15x%5E2%29

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah, that's also called the Z transform
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