anonymous
  • anonymous
Define \[f:\mathbb{N}\rightarrow \mathbb{N}\] by \[f(1)=2, f(2)=-8 \], for \[n\geq 3\] \[f(n)=8f(n-1)-15f(n-2)+6*2^n\] Prove that for all n\in \mathbb{N}, \[f(n)=-5*3^n+5^{n-1}+2^{n+3}\] Any help. I am stuck on the inductive step
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
\[f(n)=-5*3^{k+1}+5^{k}+2^{k+4}\] \[f(n)=8f(k)-15f(k-1)+6*2^{k+1}\] Then I dont know where to go from here.
anonymous
  • anonymous
suppose $$f(n-1)=-5\cdot3^{n-1}+5^{n-2}+2^{n+2}\\f(n)=-5\cdot3^n+5^{n-1}+2^{n+3}$$now we have $$\begin{align*}f(n+1)&=8 f(n)-15 f(n-1)+6\cdot 2^{n+1}\\&=8\left(-5\cdot3^n+5^{n-1}+2^{n+3}\right)-15\left(-5\cdot 3^{n-1}+5^{n-2}+2^{n+2}\right)+6\cdot 2^{n+1}\\&=-40\cdot 3^n+8\cdot 5^{n-1}+8\cdot 2^{n+3}+75\cdot 3^{n-1}-15\cdot 5^{n-2}-15\cdot2^{n+2}+6\cdot 2^{n+1}\\&=(-40+25)\cdot 3^n+(8-3)\cdot 5^{n-1}+(8\cdot 2-15+3)\cdot 2^{n+2}\\&=-15\cdot 3^n+5\cdot 5^{n-1}+4\cdot 2^{n+2}\\&=-5\cdot 3^{n+1}+5^n+2^{n+4}\end{align*}$$
anonymous
  • anonymous
http://www.mathblog.dk/strong-induction/

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
Are you sure it's \(f(2)=8\)? @keynote
anonymous
  • anonymous
he meant \(f(2)=-8\)
anonymous
  • anonymous
That 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%5Bn-1%5D-15f%5Bn-2%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%5Bn-1%5D-15f%5Bn-2%5D%2B6+2%5En%2Cf%5B1%5D%3D%3D2%2Cf%5B2%5D%3D%3D-8%7D%2Cf%5Bn%5D%2Cn%5D
anonymous
  • anonymous
it's obvious to solve $$f(n)=8f(n-1)-15f(n-2)+6\cdot 2^n$$ suppose \(f(n)=k\cdot 2^n\), so $$k\cdot 2^n=8k\cdot 2^{n-1}-15k\cdot 2^{n-2}+6\cdot 2^n\\4k\cdot 2^{n-2}-16k\cdot 2^{n-2}+15k\cdot 2^{n-2}-24\cdot 2^{n-2}=0\\2^{n-2}\left(4k-16k+15k-24\right)=0\\3k-24=0\\k-8=0\\k=8$$ so $$f(n)=8\cdot 2^n=2^{n+3}$$ is the particular solution to the recurrence
anonymous
  • anonymous
and then we find the homogeneous solution with \(f(n)=r^n\) so $$r^n=8r^{n-1}-15r^{n-2}\\r^{n-2}(r^2-8r+15)=0\\(r-5)(r-3)=0\implies r\in \{3,5\}$$ 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
  • anonymous
so 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 $$f(n)=-5\cdot 3^n+\frac15\cdot 5^n+2^{n+3}=-5\cdot 3^n+5^{n-1}+2^{n+3}$$
anonymous
  • anonymous
@oldrin.bataku I updated it. I had made a typo
anonymous
  • anonymous
that is irrelevant, I already knew what you meant
anonymous
  • anonymous
Just 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. \[\begin{align*}F(x)&=\sum_{n=1}^\infty f(n)x^n\\[1ex] &=\sum_{n=3}^\infty f(n)x^n+2x-8x^2\\[1ex] &=\sum_{n=3}^\infty \bigg(8f(n-1)-15f(n-2)+6\times2^n\bigg)x^n+2x-8x^2\\[1ex] &=8\sum_{n=3}^\infty f(n-1)x^n-15\sum_{n=3}^\infty f(n-2)x^n+6\sum_{n=3}^\infty(2x)^n+2x-8x^2\\[1ex] &=8x\sum_{n=3}^\infty f(n-1)x^{n-1}-15x^2\sum_{n=3}^\infty f(n-2)x^{n-2}+\frac{48x^2}{1-2x}+2x-8x^2\\[1ex] &=8x\sum_{n=2}^\infty f(n)x^n-15x^2\sum_{n=1}^\infty f(n)x^n+\frac{48x^2}{1-2x}+2x-8x^2\\[1ex] &=8x\left(\sum_{n=1}^\infty f(n)x^n-2x\right)-15x^2\sum_{n=1}^\infty f(n)x^n+\frac{48x^2}{1-2x}+2x-8x^2\\[1ex] &=8x(F(x)-2x)-15x^2F(x)+\frac{48x^2}{1-2x}+2x-8x^2\\[1ex] F(x)&=\frac{\dfrac{48x^2}{1-2x}+2x-24x^2}{1-8x+15x^2} \end{align*}\]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. W|A agrees: http://www.wolframalpha.com/input/?i=McLaurin+series+of+%2848x%5E3%2F%281-2x%29%2B2x-24x%5E2%29%2F%281-8x%2B15x%5E2%29
anonymous
  • anonymous
yeah, that's also called the Z transform

Looking for something else?

Not the answer you are looking for? Search for more explanations.