High School Mathematics Extensions/Counting and Generating Functions/Solutions

At the moment, the main focus is on authoring the main content of each chapter. Therefore this exercise solutions section may be out of date and appear disorganised.

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These solutions were not written by the author of the rest of the book. They are simply the answers I thought were correct while doing the exercises. I hope these answers are useful for someone and that people will correct my work if I made some mistakes.

Generating functions exercises

(a)

S=1-z+z^{2}-z^{3}+z^{4}-z^{5}+...

zS=z-z^{2}+z^{3}-z^{4}+z^{5}-...

(1+z)S=1

S={\frac {1}{1+z}}

(b)

S=1+2z+4z^{2}+8z^{3}+16z^{4}+32z^{5}+...

2zS=2z+4z^{2}+8z^{3}+16z^{4}+32z^{5}+...

(1-2z)S=1

S={\frac {1}{1-2z}}

(c)

S=z+z^{2}+z^{3}+z^{4}+z^{5}+...

zS=z^{2}+z^{3}+z^{4}+z^{5}+...

(1-z)S=z

S={\frac {z}{1-z}}

(d)

S=3-4z+4z^{2}-4z^{3}+4z^{4}-4z^{5}+...

z(S+1)=4z-4z^{2}+4z^{3}-4z^{4}+4z^{5}-...

S+z(S+1)=3

S+zS+z=3

(1+z)S=3-z

S={\frac {3-z}{1+z}}

(a)

S={\frac {1}{1+z}}

S={\frac {1}{1--z}}

S=1-x+x^{2}-x^{3}+x^{4}-x^{5}+...

f(n)=(-1)^{n}

(b)

S={\frac {z^{3}}{1-z^{2}}}

(1-z^{2})S=z^{3}

S=z^{3}+z^{5}+z^{7}+z^{9}+...

f(n)=1;{\mbox{for n}}\geq 2{\mbox{ and even}}

f(n)=0;{\mbox{for n is odd}}

2c only contains the exercise and not the answer for the moment

(c)

{\frac {z^{2}-1}{1+3z^{3}}}

Linear Recurrence Relations exercises

This section only contains the incomplete answers because I couldn't figure out where to go from here.

{\begin{matrix}x_{n}&=&2x_{n-1}&-&1;\ {\mbox{for n}}\geq 1\\x_{0}&=&1\end{matrix}}

Let G(z) be the generating function of the sequence described above.

G(z)=x_{0}+x_{1}z+x_{2}z^{2}+...

(1-2z)G(z)=x_{0}+(x_{1}-2x_{0})z+(x_{2}-2x_{1})z^{2}+...

(1-2z)G(z)=1-z-z^{2}-z^{3}-z^{4}-...

(1-2z)G(z)=1-z(1+z+z^{2}+...)

(1-2z)G(z)=1-{\frac {z}{1-z}}

(1-2z)G(z)={\frac {1-2z}{1-z}}

G(z)={\frac {1}{1-z}}

x_{n}=1

{\begin{matrix}3x_{n}&=&-4x_{n-1}&+&x_{n-2};\ {\mbox{for n}}\geq 2\\x_{0}&=&1\\x_{1}&=&1\\\end{matrix}}

Let G(z) be the generating function of the sequence described above.

G(z)=x_{0}+x_{1}z+x_{2}z^{2}+...

(3+4z-z^{2})G(z)=3x_{0}+(3x_{1}+4x_{0})z+(3x_{2}+4x_{1}-x_{0})z^{2}+(3x_{3}+4x_{2}-x_{1})z^{3}+...

(3+4z-z^{2})G(z)=3x_{0}+(3x_{1}+4x_{0})z

(3+4z-z^{2})G(z)=3+7z

G(z)={\frac {3+7z}{-z^{2}+4z+3}}

3. Let G(z) be the generating function of the sequence described above.

G(z)=x_{0}+x_{1}z+x_{2}z^{2}+...

(1-z-z^{2})G(z)=x_{0}+(x_{1}-x_{0})z+(x_{2}-x_{1}-x_{0})z^{2}+(x_{3}-x_{2}-x_{1})z^{2}+...

(1-z-z^{2})G(z)=1

G(z)={\frac {1}{1-z-z^{2}}}

G(z)={\frac {-1}{z^{2}+z-1}}

We want to factorize

f(z)=z^{2}+z-1

into

(z-\alpha )(z-\beta )

, by the converse of factor theorem, if (z - p) is a factor of f(z), f(p)=0.

Hence α and β are the roots of the quadratic equation

z^{2}+z-1=0

Using the quadratic formula to find the roots:

\alpha ={\frac {{\sqrt {5}}-1}{2}},\beta =-{\frac {{\sqrt {5}}+1}{2}}

In fact, these two numbers are the faomus golden ratio and to make things simple, we use the greek symbols for golden ratio from now on.

Note:

{\frac {{\sqrt {5}}-1}{2}}

is denoted

\phi

and

{\frac {{\sqrt {5}}+1}{2}}

is denoted

\Phi

G(z)={\frac {-1}{(z-\phi )(z+\Phi )}}

By the method of partial fraction:

G(z)={\frac {1}{{\sqrt {5}}(z+\Phi )}}-{\frac {1}{{\sqrt {5}}(z-\phi )}}

G(z)={\frac {1}{\Phi {\sqrt {5}}({\frac {z}{\Phi }}+1)}}-{\frac {1}{\phi {\sqrt {5}}({\frac {z}{\phi }}-1)}}

G(z)={\frac {1}{\Phi {\sqrt {5}}(1--\phi z)}}+{\frac {1}{\phi {\sqrt {5}}(1-\Phi z)}}

x_{n}={\frac {\phi }{\sqrt {5}}}\times (-\phi )^{n}+{\frac {\Phi }{\sqrt {5}}}\times \Phi ^{n}

x_{n}={\frac {\Phi ^{n+1}-(-\phi )^{n+1}}{\sqrt {5}}}

Further Counting exercises

1. We know that

T(z)={\frac {1}{(1-z)^{2}}}=\sum _{i=0}^{\infty }{i+1 \choose i}z^{i}=\sum _{i=0}^{\infty }(i+1)z^{i}

therefore

T(z)={\frac {1}{(1+z)^{2}}}=\sum _{i=0}^{\infty }(i+1)(-1)^{i}z^{i}

Thus

T_{k}=(-1)^{k}(k+1)

2. $a+b+c=m$

T(z)={\frac {1}{(1-z)^{3}}}=\sum _{i=0}^{\infty }{i+2 \choose i}z^{i}

Thus

T_{k}={i+2 \choose i}

Differentiate from first principle exercises

f'(z)=\lim _{h\to 0}{\frac {1}{(1-(z+h))^{2}}}-{\frac {1}{(1-z)^{2}}}=

\lim _{h\to 0}{\frac {1}{h}}{\frac {(1-z)^{2}-(1-(z+h))^{2}}{(1-z-h)^{2}(1-z)^{2}}}=

\lim _{h\to 0}{\frac {1}{h}}{\frac {z^{2}-2z+1-(z+h)^{2}+2(z+h)-1}{(1-z-h)^{2}(1-z)^{2}}}=

\lim _{h\to 0}{\frac {1}{h}}{\frac {z^{2}-2z+1-z^{2}-h^{2}-2zh+2z+2h-1}{(1-z-h)^{2}(1-z)^{2}}}=

\lim _{h\to 0}{\frac {1}{h}}{\frac {-h^{2}-2zh+2h}{(1-z-h)^{2}(1-z)^{2}}}=

\lim _{h\to 0}{\frac {-h-2z+2}{(1-z-h)^{2}(1-z)^{2}}}=

{\frac {-2z+2}{(1-z)^{4}}}=

{\frac {-2}{(1-z)^{3}}}

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High School Mathematics Extensions/Counting and Generating Functions/Solutions

Counting and Generating Functions

Generating functions exercises

Linear Recurrence Relations exercises

Further Counting exercises

*Differentiate from first principle* exercises

Differentiate from first principle exercises