Mixing
Determine the sequence generated by a generating function
$begingroup$
$A(z)=2z-1+frac{1}{2z-2z^2}$
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
asked Jan 26 at 22:58
GorossoGorosso
315
$endgroup$
add a comment |
$begingroup$
$A(z)=2z-1+frac{1}{2z-2z^2}$
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
asked Jan 26 at 22:58
GorossoGorosso
315
$endgroup$
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
add a comment |
$begingroup$
$A(z)=2z-1+frac{1}{2z-2z^2}$
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
asked Jan 26 at 22:58
GorossoGorosso
315
$endgroup$
$A(z)=2z-1+frac{1}{2z-2z^2}$
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
asked Jan 26 at 22:58
GorossoGorosso
315
asked Jan 26 at 22:58
GorossoGorosso
315
asked Jan 26 at 22:58
GorossoGorosso
315
asked Jan 26 at 22:58
GorossoGorosso
315
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
add a comment |
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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 $A$.
answered Jan 27 at 0:30
marty cohenmarty cohen
74.5k549129
$endgroup$
$begingroup$
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
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 $A$.
answered Jan 27 at 0:30
marty cohenmarty cohen
74.5k549129
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $A$.
answered Jan 27 at 0:30
marty cohenmarty cohen
74.5k549129
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $A$.
answered Jan 27 at 0:30
marty cohenmarty cohen
74.5k549129
$endgroup$
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 $A$.
answered Jan 27 at 0:30
marty cohenmarty cohen
74.5k549129
answered Jan 27 at 0:30
marty cohenmarty cohen
74.5k549129
answered Jan 27 at 0:30
marty cohenmarty cohen
74.5k549129
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
add a comment |
$begingroup$
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
add a comment |
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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