From 56b740565fd6467564be28b5465ad24cccf71109 Mon Sep 17 00:00:00 2001 From: Holden Rohrer Date: Thu, 16 Jan 2020 21:24:46 -0500 Subject: moved comb hw --- tech-math/hw5.tex | 61 ------------------------------------------------------- 1 file changed, 61 deletions(-) delete mode 100644 tech-math/hw5.tex (limited to 'tech-math/hw5.tex') diff --git a/tech-math/hw5.tex b/tech-math/hw5.tex deleted file mode 100644 index b20efb0..0000000 --- a/tech-math/hw5.tex +++ /dev/null @@ -1,61 +0,0 @@ -\def\pre#1{\leavevmode\llap{\hbox to \parindent{\hfil #1 \hfil}}} -\noindent {\bf Q1)} %#2 - -\pre{a.} -$$\sum_{i=0}^\infty x^{i+1} = {x \over 1-x}$$ - -\pre{c.} -$$\sum_{i=0}^\infty 2^ix^i = {1 \over 1-2x}$$ - -\pre{d.} -$$\sum_{i=0}^\infty x^{i+4} = {x^4 \over 1-x}$$ - -\pre{f.} -$$(2+x)^8$$ %%BUG BUG BUG - -\pre{i.} -$$3+2x+4x^2+x^3\sum_{i=0}^\infty x^i = 3 + 2x + 4x^2 + {x^3 \over 1 - x}$$ - -\noindent {\bf Q2)} %#7 - -The generating function for $x_1$ is $1 \over 1-x$ because it is possible for all $n$. For $x_2$, $x^2 \over 1-x$ because there is no way of satisfying the restriction for $n<2$. For $x_3$, $1 \over 1-x^4$ as it is impossible for $n \not\equiv 0 ({\rm mod} 4)$ and $1+x+x^2+x^3$ for $x_4$ (because it cannot be all of the partition for $n$ more than $4$). One must also consider a fictional $x_5$ to account for the solutions which sum to less than $n$. This has the generating function $1 \over 1-x$. The combined possible number of solutions $c_n$ is the value of the coefficient for $x^n$ in their product. The product is $$x^2(1+x+x^2+x^3) \over (1-x)^3(1-x^4),$$ but by the fact that $$1+x+x^2+x^3={1-x^4 \over 1-x},$$ this is equal to $$x^2 \over (1-x)^4.$$ - -This, in closed form, becomes $${x^2 \over 6}\sum^\infty_{n=0} n(n-1)(n-2)x^{n-2} -= \sum^\infty_{n=0} {n(n-1)(n-2) \over 6}x^n -= \sum^\infty_{n=0} {n \choose 3}x^{n-1} -= \sum^\infty_{n=-1} {n+1 \choose 3}x^n,$$ -giving a number of solutions $n+1 \choose 3$ - - - -\noindent {\bf Q3)} %#9 - -$$\sum^\infty_{n=0} {n \choose p}x^n$$ - -Note: in this problem and Q2, I used 0 and infinity as my bounds (mostly) because ${k \choose n}=0$ for $k<0$ and $k>n$. - -\noindent {\bf Q4)} %#21 - -\pre{a.} -$a_n = 7^n$ - -\pre{b.} -$$x^2e^{3x} -= x^2\sum^\infty_{n=0} {(3x)^n \over n!} -= x^2\sum^\infty_{n=0} {3^n x^n \over n!} -= \sum^\infty_{n=0} {3^n x^{n+2} \over n!} -= \sum^\infty_{n=2} {3^{n-2} x^n \over {n-2}!} -= \sum^\infty_{n=2} {3^{n-2} x^n \over {n-2}!} -= \sum^\infty_{n=2} {3^{n-2}(n)(n-1) x^n \over n!}.$$ - -This shows $a_n = n(n-1)3^n$. - -\pre{c.} -$a_n = (-1)^n n!$ - -\pre{d.} -\def\mod#1{\thinspace(\thinspace{\rm mod} #1)} -$$e^{x^4} = 1 + x^4 + {x^8 \over 2!} \dots $$ -$$a_n = \left\{{ 0 : n\not\equiv 0 \mod{4} \atop {n! \over {n\over 4}!}: n\equiv 0\mod{4} }\right\}$$ - -\bye -- cgit