Determine the sequence generated by a generating function












1












$begingroup$


$A(z)=2z-1+frac{1}{2z-2z^2}$



I have no clue how to solve this, I tried looking at other examples but I am just stuck, could anyone be so kind and explain how to solve this step by step?






discrete-mathematics







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    What do you mean by solve? You only wrote down a function.
    $endgroup$
    – Klaus
    Jan 26 at 23:06










  • $begingroup$
    Do you mean finding the inverse?
    $endgroup$
    – Math Lover
    Jan 26 at 23:11










  • $begingroup$
    I guess thats finding the inverse. Sorry, wasn't sure how to translate it from my language.
    $endgroup$
    – Gorosso
    Jan 26 at 23:30


















1












$begingroup$


$A(z)=2z-1+frac{1}{2z-2z^2}$



I have no clue how to solve this, I tried looking at other examples but I am just stuck, could anyone be so kind and explain how to solve this step by step?






discrete-mathematics







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    What do you mean by solve? You only wrote down a function.
    $endgroup$
    – Klaus
    Jan 26 at 23:06










  • $begingroup$
    Do you mean finding the inverse?
    $endgroup$
    – Math Lover
    Jan 26 at 23:11










  • $begingroup$
    I guess thats finding the inverse. Sorry, wasn't sure how to translate it from my language.
    $endgroup$
    – Gorosso
    Jan 26 at 23:30
















1












1








1





$begingroup$


$A(z)=2z-1+frac{1}{2z-2z^2}$



I have no clue how to solve this, I tried looking at other examples but I am just stuck, could anyone be so kind and explain how to solve this step by step?






discrete-mathematics







share|cite|improve this question









$endgroup$




$A(z)=2z-1+frac{1}{2z-2z^2}$



I have no clue how to solve this, I tried looking at other examples but I am just stuck, could anyone be so kind and explain how to solve this step by step?






discrete-mathematics




discrete-mathematics






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 26 at 22:58









GorossoGorosso

315




315








  • 1




    $begingroup$
    What do you mean by solve? You only wrote down a function.
    $endgroup$
    – Klaus
    Jan 26 at 23:06










  • $begingroup$
    Do you mean finding the inverse?
    $endgroup$
    – Math Lover
    Jan 26 at 23:11










  • $begingroup$
    I guess thats finding the inverse. Sorry, wasn't sure how to translate it from my language.
    $endgroup$
    – Gorosso
    Jan 26 at 23:30
















  • 1




    $begingroup$
    What do you mean by solve? You only wrote down a function.
    $endgroup$
    – Klaus
    Jan 26 at 23:06










  • $begingroup$
    Do you mean finding the inverse?
    $endgroup$
    – Math Lover
    Jan 26 at 23:11










  • $begingroup$
    I guess thats finding the inverse. Sorry, wasn't sure how to translate it from my language.
    $endgroup$
    – Gorosso
    Jan 26 at 23:30










1




1




$begingroup$
What do you mean by solve? You only wrote down a function.
$endgroup$
– Klaus
Jan 26 at 23:06




$begingroup$
What do you mean by solve? You only wrote down a function.
$endgroup$
– Klaus
Jan 26 at 23:06












$begingroup$
Do you mean finding the inverse?
$endgroup$
– Math Lover
Jan 26 at 23:11




$begingroup$
Do you mean finding the inverse?
$endgroup$
– Math Lover
Jan 26 at 23:11












$begingroup$
I guess thats finding the inverse. Sorry, wasn't sure how to translate it from my language.
$endgroup$
– Gorosso
Jan 26 at 23:30






$begingroup$
I guess thats finding the inverse. Sorry, wasn't sure how to translate it from my language.
$endgroup$
– Gorosso
Jan 26 at 23:30












1 Answer
1






active

oldest

votes


















1












$begingroup$

If you want to see the generating function:



$begin{array}\
A(z)
&=2z-1+frac{1}{2z-2z^2}\
&=2z-1+frac1{2z}frac{1}{1-z}\
&=2z-1+frac1{2z}sum_{n=0}^{infty} z^n\
&=2z-1+frac1{2z}+sum_{n=1}^{infty} frac{z^{n-1}}{2}\
&=2z-1+frac1{2z}+sum_{n=0}^{infty} frac{z^{n}}{2}\
&=2z-1+frac1{2z}+frac12+frac{z}{2}+sum_{n=2}^{infty} frac{z^{n}}{2}\
&=frac53 z-frac12+frac1{2z}+sum_{n=2}^{infty} frac{z^{n}}{2}\
end{array}
$



If you want to invert the function,
that is,
find $z$ in terms of $A$:



$begin{array}\
A(z)
&=2z-1+frac{1}{2z-2z^2}\
&=frac{(2z-1)(2z-2z^2)+1}{2z-2z^2}\
&=-frac{4 z^3 - 6 z^2 + 2 z - 1
}{2z-2z^2}\
end{array}
$



so
$-A(2z-2z^2)
=4 z^3 - 6 z^2 + 2 z - 1
$
or
$4 z^3 - (2A+6) z^2 + (2a+2) z - 1
=0
$.



Wolfy gives some very complicated expressions
for $z$ in terms of $A$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Marty, do you mean Wolfram when you say Wolfy? can you please provide a link to their app.
    $endgroup$
    – user25406
    Jan 27 at 13:11






  • 1




    $begingroup$
    Yep. Look for Wolfram Alpha.
    $endgroup$
    – marty cohen
    Jan 27 at 17:45










  • $begingroup$
    How does $z^n$ turn into $frac{z^{n-1}}{2}$ in fourth row?
    $endgroup$
    – Gorosso
    Feb 6 at 20:15










  • $begingroup$
    $z^n(frac1{2z})$.
    $endgroup$
    – marty cohen
    Feb 6 at 22:31













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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

If you want to see the generating function:



$begin{array}\
A(z)
&=2z-1+frac{1}{2z-2z^2}\
&=2z-1+frac1{2z}frac{1}{1-z}\
&=2z-1+frac1{2z}sum_{n=0}^{infty} z^n\
&=2z-1+frac1{2z}+sum_{n=1}^{infty} frac{z^{n-1}}{2}\
&=2z-1+frac1{2z}+sum_{n=0}^{infty} frac{z^{n}}{2}\
&=2z-1+frac1{2z}+frac12+frac{z}{2}+sum_{n=2}^{infty} frac{z^{n}}{2}\
&=frac53 z-frac12+frac1{2z}+sum_{n=2}^{infty} frac{z^{n}}{2}\
end{array}
$



If you want to invert the function,
that is,
find $z$ in terms of $A$:



$begin{array}\
A(z)
&=2z-1+frac{1}{2z-2z^2}\
&=frac{(2z-1)(2z-2z^2)+1}{2z-2z^2}\
&=-frac{4 z^3 - 6 z^2 + 2 z - 1
}{2z-2z^2}\
end{array}
$



so
$-A(2z-2z^2)
=4 z^3 - 6 z^2 + 2 z - 1
$
or
$4 z^3 - (2A+6) z^2 + (2a+2) z - 1
=0
$.



Wolfy gives some very complicated expressions
for $z$ in terms of $A$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Marty, do you mean Wolfram when you say Wolfy? can you please provide a link to their app.
    $endgroup$
    – user25406
    Jan 27 at 13:11






  • 1




    $begingroup$
    Yep. Look for Wolfram Alpha.
    $endgroup$
    – marty cohen
    Jan 27 at 17:45










  • $begingroup$
    How does $z^n$ turn into $frac{z^{n-1}}{2}$ in fourth row?
    $endgroup$
    – Gorosso
    Feb 6 at 20:15










  • $begingroup$
    $z^n(frac1{2z})$.
    $endgroup$
    – marty cohen
    Feb 6 at 22:31


















1












$begingroup$

If you want to see the generating function:



$begin{array}\
A(z)
&=2z-1+frac{1}{2z-2z^2}\
&=2z-1+frac1{2z}frac{1}{1-z}\
&=2z-1+frac1{2z}sum_{n=0}^{infty} z^n\
&=2z-1+frac1{2z}+sum_{n=1}^{infty} frac{z^{n-1}}{2}\
&=2z-1+frac1{2z}+sum_{n=0}^{infty} frac{z^{n}}{2}\
&=2z-1+frac1{2z}+frac12+frac{z}{2}+sum_{n=2}^{infty} frac{z^{n}}{2}\
&=frac53 z-frac12+frac1{2z}+sum_{n=2}^{infty} frac{z^{n}}{2}\
end{array}
$



If you want to invert the function,
that is,
find $z$ in terms of $A$:



$begin{array}\
A(z)
&=2z-1+frac{1}{2z-2z^2}\
&=frac{(2z-1)(2z-2z^2)+1}{2z-2z^2}\
&=-frac{4 z^3 - 6 z^2 + 2 z - 1
}{2z-2z^2}\
end{array}
$



so
$-A(2z-2z^2)
=4 z^3 - 6 z^2 + 2 z - 1
$
or
$4 z^3 - (2A+6) z^2 + (2a+2) z - 1
=0
$.



Wolfy gives some very complicated expressions
for $z$ in terms of $A$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Marty, do you mean Wolfram when you say Wolfy? can you please provide a link to their app.
    $endgroup$
    – user25406
    Jan 27 at 13:11






  • 1




    $begingroup$
    Yep. Look for Wolfram Alpha.
    $endgroup$
    – marty cohen
    Jan 27 at 17:45










  • $begingroup$
    How does $z^n$ turn into $frac{z^{n-1}}{2}$ in fourth row?
    $endgroup$
    – Gorosso
    Feb 6 at 20:15










  • $begingroup$
    $z^n(frac1{2z})$.
    $endgroup$
    – marty cohen
    Feb 6 at 22:31
















1












1








1





$begingroup$

If you want to see the generating function:



$begin{array}\
A(z)
&=2z-1+frac{1}{2z-2z^2}\
&=2z-1+frac1{2z}frac{1}{1-z}\
&=2z-1+frac1{2z}sum_{n=0}^{infty} z^n\
&=2z-1+frac1{2z}+sum_{n=1}^{infty} frac{z^{n-1}}{2}\
&=2z-1+frac1{2z}+sum_{n=0}^{infty} frac{z^{n}}{2}\
&=2z-1+frac1{2z}+frac12+frac{z}{2}+sum_{n=2}^{infty} frac{z^{n}}{2}\
&=frac53 z-frac12+frac1{2z}+sum_{n=2}^{infty} frac{z^{n}}{2}\
end{array}
$



If you want to invert the function,
that is,
find $z$ in terms of $A$:



$begin{array}\
A(z)
&=2z-1+frac{1}{2z-2z^2}\
&=frac{(2z-1)(2z-2z^2)+1}{2z-2z^2}\
&=-frac{4 z^3 - 6 z^2 + 2 z - 1
}{2z-2z^2}\
end{array}
$



so
$-A(2z-2z^2)
=4 z^3 - 6 z^2 + 2 z - 1
$
or
$4 z^3 - (2A+6) z^2 + (2a+2) z - 1
=0
$.



Wolfy gives some very complicated expressions
for $z$ in terms of $A$.






share|cite|improve this answer









$endgroup$



If you want to see the generating function:



$begin{array}\
A(z)
&=2z-1+frac{1}{2z-2z^2}\
&=2z-1+frac1{2z}frac{1}{1-z}\
&=2z-1+frac1{2z}sum_{n=0}^{infty} z^n\
&=2z-1+frac1{2z}+sum_{n=1}^{infty} frac{z^{n-1}}{2}\
&=2z-1+frac1{2z}+sum_{n=0}^{infty} frac{z^{n}}{2}\
&=2z-1+frac1{2z}+frac12+frac{z}{2}+sum_{n=2}^{infty} frac{z^{n}}{2}\
&=frac53 z-frac12+frac1{2z}+sum_{n=2}^{infty} frac{z^{n}}{2}\
end{array}
$



If you want to invert the function,
that is,
find $z$ in terms of $A$:



$begin{array}\
A(z)
&=2z-1+frac{1}{2z-2z^2}\
&=frac{(2z-1)(2z-2z^2)+1}{2z-2z^2}\
&=-frac{4 z^3 - 6 z^2 + 2 z - 1
}{2z-2z^2}\
end{array}
$



so
$-A(2z-2z^2)
=4 z^3 - 6 z^2 + 2 z - 1
$
or
$4 z^3 - (2A+6) z^2 + (2a+2) z - 1
=0
$.



Wolfy gives some very complicated expressions
for $z$ in terms of $A$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 27 at 0:30









marty cohenmarty cohen

74.5k549129




74.5k549129












  • $begingroup$
    Marty, do you mean Wolfram when you say Wolfy? can you please provide a link to their app.
    $endgroup$
    – user25406
    Jan 27 at 13:11






  • 1




    $begingroup$
    Yep. Look for Wolfram Alpha.
    $endgroup$
    – marty cohen
    Jan 27 at 17:45










  • $begingroup$
    How does $z^n$ turn into $frac{z^{n-1}}{2}$ in fourth row?
    $endgroup$
    – Gorosso
    Feb 6 at 20:15










  • $begingroup$
    $z^n(frac1{2z})$.
    $endgroup$
    – marty cohen
    Feb 6 at 22:31




















  • $begingroup$
    Marty, do you mean Wolfram when you say Wolfy? can you please provide a link to their app.
    $endgroup$
    – user25406
    Jan 27 at 13:11






  • 1




    $begingroup$
    Yep. Look for Wolfram Alpha.
    $endgroup$
    – marty cohen
    Jan 27 at 17:45










  • $begingroup$
    How does $z^n$ turn into $frac{z^{n-1}}{2}$ in fourth row?
    $endgroup$
    – Gorosso
    Feb 6 at 20:15










  • $begingroup$
    $z^n(frac1{2z})$.
    $endgroup$
    – marty cohen
    Feb 6 at 22:31


















$begingroup$
Marty, do you mean Wolfram when you say Wolfy? can you please provide a link to their app.
$endgroup$
– user25406
Jan 27 at 13:11




$begingroup$
Marty, do you mean Wolfram when you say Wolfy? can you please provide a link to their app.
$endgroup$
– user25406
Jan 27 at 13:11




1




1




$begingroup$
Yep. Look for Wolfram Alpha.
$endgroup$
– marty cohen
Jan 27 at 17:45




$begingroup$
Yep. Look for Wolfram Alpha.
$endgroup$
– marty cohen
Jan 27 at 17:45












$begingroup$
How does $z^n$ turn into $frac{z^{n-1}}{2}$ in fourth row?
$endgroup$
– Gorosso
Feb 6 at 20:15




$begingroup$
How does $z^n$ turn into $frac{z^{n-1}}{2}$ in fourth row?
$endgroup$
– Gorosso
Feb 6 at 20:15












$begingroup$
$z^n(frac1{2z})$.
$endgroup$
– marty cohen
Feb 6 at 22:31






$begingroup$
$z^n(frac1{2z})$.
$endgroup$
– marty cohen
Feb 6 at 22:31




















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