Mixing
Finding coefficient of $x^{100}$
$begingroup$
What will be solution of this function for coefficient of $x^{100}$?
$$displaystylefrac{1}{left ( 1-x^{10} right )(1-x^{20})(1-x^{50})}$$
My solution:
$[x^{100}] $$displaystyle frac{1}{left ( 1-x^{10} right )(1-x^{20})(1-x^{50})}$
$[x^{100}] $ $(1-x^{10})^{-1}$$(1-x^{20})^{-1}$$(1-x^{50})^{-1}$
$begin{aligned}displaystyle &binom{-n}{k} = (-1)^rbinom{n + k - 1}{k}\
&=displaystyle color{red} {(-1)^{10}binom{-1}{10}binom{-1}{0}binom{-1}{0} + (-1)^5 binom{-1}{0}binom{-1}{5}binom{-1}{0} + (-1)^2binom{-1}{0}binom{-1}{0}binom{-1}{2}} \ & + color{blue}{ (-1)^8 (-1)^1 binom{-1}{8}binom{-1}{1}binom{-1}{0} + (-1)^6 (-1)^2binom{-1}{6}binom{-1}{2}binom{-1}{0} + (-1)^4 (-1)^3 binom{-1}{4}binom{-1}{3}binom{-1}{0} + (-1)^2 (-1)^4binom{-1}{2}binom{-1}{4}binom{-1}{0} }\ & + color{Orange}{ (-1)^5 (-1)^1 binom{-1}{5}binom{-1}{0}binom{-1}{1} } \ & + color{maroon}{(-1)^3(-1)^1(-1)^1binom{-1}{3}binom{-1}{1}binom{-1}{1} + (-1)^1 (-1)^2(-1)^3binom{-1}{1}binom{-1}{2}binom{-1}{1} }end{aligned} $
$ 1 – 1 + 1 + 1 – 1 + 1 – 1 – 1 + 1 – 1 = 0$
So coefficient of $x^{100}$ is $0$
There are many terms so to differentiate among them distinct colors are used
$color{red}{text{Red}}$ - Selecting $x$ from any one of $(1-x^{10})^{-1}$$(1-x^{20})^{-1}$$(1-x^{50})^{-1}$
$color{Blue}{text{Blue}}$ - Selecting from both $(1-x^{10})^{-1}$$(1-x^{20})^{-1}$
$color{Orange}{text{Orange}}$ - Selecting from both $(1-x^{10})^{-1}$$(1-x^{50})^{-1}$
$color{maroon}{text{Maroon}}$ - Selecting from all $(1-x^{10})^{-1}$$(1-x^{20})^{-1}$$(1-x^{50})^{-1}$
But answer is $10$,
is anything wrong with my solution?
combinatorics proof-verification generating-functions
edited Jan 29 at 5:05
Yadati Kiran
2,1121622
asked Jan 29 at 4:37
Mk UtkarshMk Utkarsh
94110
$endgroup$
add a comment |
$begingroup$
What will be solution of this function for coefficient of $x^{100}$?
$$displaystylefrac{1}{left ( 1-x^{10} right )(1-x^{20})(1-x^{50})}$$
My solution:
$[x^{100}] $$displaystyle frac{1}{left ( 1-x^{10} right )(1-x^{20})(1-x^{50})}$
$[x^{100}] $ $(1-x^{10})^{-1}$$(1-x^{20})^{-1}$$(1-x^{50})^{-1}$
$begin{aligned}displaystyle &binom{-n}{k} = (-1)^rbinom{n + k - 1}{k}\
&=displaystyle color{red} {(-1)^{10}binom{-1}{10}binom{-1}{0}binom{-1}{0} + (-1)^5 binom{-1}{0}binom{-1}{5}binom{-1}{0} + (-1)^2binom{-1}{0}binom{-1}{0}binom{-1}{2}} \ & + color{blue}{ (-1)^8 (-1)^1 binom{-1}{8}binom{-1}{1}binom{-1}{0} + (-1)^6 (-1)^2binom{-1}{6}binom{-1}{2}binom{-1}{0} + (-1)^4 (-1)^3 binom{-1}{4}binom{-1}{3}binom{-1}{0} + (-1)^2 (-1)^4binom{-1}{2}binom{-1}{4}binom{-1}{0} }\ & + color{Orange}{ (-1)^5 (-1)^1 binom{-1}{5}binom{-1}{0}binom{-1}{1} } \ & + color{maroon}{(-1)^3(-1)^1(-1)^1binom{-1}{3}binom{-1}{1}binom{-1}{1} + (-1)^1 (-1)^2(-1)^3binom{-1}{1}binom{-1}{2}binom{-1}{1} }end{aligned} $
$ 1 – 1 + 1 + 1 – 1 + 1 – 1 – 1 + 1 – 1 = 0$
So coefficient of $x^{100}$ is $0$
There are many terms so to differentiate among them distinct colors are used
$color{red}{text{Red}}$ - Selecting $x$ from any one of $(1-x^{10})^{-1}$$(1-x^{20})^{-1}$$(1-x^{50})^{-1}$
$color{Blue}{text{Blue}}$ - Selecting from both $(1-x^{10})^{-1}$$(1-x^{20})^{-1}$
$color{Orange}{text{Orange}}$ - Selecting from both $(1-x^{10})^{-1}$$(1-x^{50})^{-1}$
$color{maroon}{text{Maroon}}$ - Selecting from all $(1-x^{10})^{-1}$$(1-x^{20})^{-1}$$(1-x^{50})^{-1}$
But answer is $10$,
is anything wrong with my solution?
combinatorics proof-verification generating-functions
edited Jan 29 at 5:05
Yadati Kiran
2,1121622
asked Jan 29 at 4:37
Mk UtkarshMk Utkarsh
94110
$endgroup$
add a comment |
$begingroup$
What will be solution of this function for coefficient of $x^{100}$?
$$displaystylefrac{1}{left ( 1-x^{10} right )(1-x^{20})(1-x^{50})}$$
My solution:
$[x^{100}] $$displaystyle frac{1}{left ( 1-x^{10} right )(1-x^{20})(1-x^{50})}$
$[x^{100}] $ $(1-x^{10})^{-1}$$(1-x^{20})^{-1}$$(1-x^{50})^{-1}$
$begin{aligned}displaystyle &binom{-n}{k} = (-1)^rbinom{n + k - 1}{k}\
&=displaystyle color{red} {(-1)^{10}binom{-1}{10}binom{-1}{0}binom{-1}{0} + (-1)^5 binom{-1}{0}binom{-1}{5}binom{-1}{0} + (-1)^2binom{-1}{0}binom{-1}{0}binom{-1}{2}} \ & + color{blue}{ (-1)^8 (-1)^1 binom{-1}{8}binom{-1}{1}binom{-1}{0} + (-1)^6 (-1)^2binom{-1}{6}binom{-1}{2}binom{-1}{0} + (-1)^4 (-1)^3 binom{-1}{4}binom{-1}{3}binom{-1}{0} + (-1)^2 (-1)^4binom{-1}{2}binom{-1}{4}binom{-1}{0} }\ & + color{Orange}{ (-1)^5 (-1)^1 binom{-1}{5}binom{-1}{0}binom{-1}{1} } \ & + color{maroon}{(-1)^3(-1)^1(-1)^1binom{-1}{3}binom{-1}{1}binom{-1}{1} + (-1)^1 (-1)^2(-1)^3binom{-1}{1}binom{-1}{2}binom{-1}{1} }end{aligned} $
$ 1 – 1 + 1 + 1 – 1 + 1 – 1 – 1 + 1 – 1 = 0$
So coefficient of $x^{100}$ is $0$
There are many terms so to differentiate among them distinct colors are used
$color{red}{text{Red}}$ - Selecting $x$ from any one of $(1-x^{10})^{-1}$$(1-x^{20})^{-1}$$(1-x^{50})^{-1}$
$color{Blue}{text{Blue}}$ - Selecting from both $(1-x^{10})^{-1}$$(1-x^{20})^{-1}$
$color{Orange}{text{Orange}}$ - Selecting from both $(1-x^{10})^{-1}$$(1-x^{50})^{-1}$
$color{maroon}{text{Maroon}}$ - Selecting from all $(1-x^{10})^{-1}$$(1-x^{20})^{-1}$$(1-x^{50})^{-1}$
But answer is $10$,
is anything wrong with my solution?
combinatorics proof-verification generating-functions
edited Jan 29 at 5:05
Yadati Kiran
2,1121622
asked Jan 29 at 4:37
Mk UtkarshMk Utkarsh
94110
$endgroup$
What will be solution of this function for coefficient of $x^{100}$?
$$displaystylefrac{1}{left ( 1-x^{10} right )(1-x^{20})(1-x^{50})}$$
My solution:
$[x^{100}] $$displaystyle frac{1}{left ( 1-x^{10} right )(1-x^{20})(1-x^{50})}$
$[x^{100}] $ $(1-x^{10})^{-1}$$(1-x^{20})^{-1}$$(1-x^{50})^{-1}$
$begin{aligned}displaystyle &binom{-n}{k} = (-1)^rbinom{n + k - 1}{k}\
&=displaystyle color{red} {(-1)^{10}binom{-1}{10}binom{-1}{0}binom{-1}{0} + (-1)^5 binom{-1}{0}binom{-1}{5}binom{-1}{0} + (-1)^2binom{-1}{0}binom{-1}{0}binom{-1}{2}} \ & + color{blue}{ (-1)^8 (-1)^1 binom{-1}{8}binom{-1}{1}binom{-1}{0} + (-1)^6 (-1)^2binom{-1}{6}binom{-1}{2}binom{-1}{0} + (-1)^4 (-1)^3 binom{-1}{4}binom{-1}{3}binom{-1}{0} + (-1)^2 (-1)^4binom{-1}{2}binom{-1}{4}binom{-1}{0} }\ & + color{Orange}{ (-1)^5 (-1)^1 binom{-1}{5}binom{-1}{0}binom{-1}{1} } \ & + color{maroon}{(-1)^3(-1)^1(-1)^1binom{-1}{3}binom{-1}{1}binom{-1}{1} + (-1)^1 (-1)^2(-1)^3binom{-1}{1}binom{-1}{2}binom{-1}{1} }end{aligned} $
$ 1 – 1 + 1 + 1 – 1 + 1 – 1 – 1 + 1 – 1 = 0$
So coefficient of $x^{100}$ is $0$
There are many terms so to differentiate among them distinct colors are used
$color{red}{text{Red}}$ - Selecting $x$ from any one of $(1-x^{10})^{-1}$$(1-x^{20})^{-1}$$(1-x^{50})^{-1}$
$color{Blue}{text{Blue}}$ - Selecting from both $(1-x^{10})^{-1}$$(1-x^{20})^{-1}$
$color{Orange}{text{Orange}}$ - Selecting from both $(1-x^{10})^{-1}$$(1-x^{50})^{-1}$
$color{maroon}{text{Maroon}}$ - Selecting from all $(1-x^{10})^{-1}$$(1-x^{20})^{-1}$$(1-x^{50})^{-1}$
But answer is $10$,
is anything wrong with my solution?
combinatorics proof-verification generating-functions
combinatorics proof-verification generating-functions
edited Jan 29 at 5:05
Yadati Kiran
2,1121622
asked Jan 29 at 4:37
Mk UtkarshMk Utkarsh
94110
edited Jan 29 at 5:05
Yadati Kiran
2,1121622
asked Jan 29 at 4:37
Mk UtkarshMk Utkarsh
94110
edited Jan 29 at 5:05
Yadati Kiran
2,1121622
edited Jan 29 at 5:05
Yadati Kiran
2,1121622
edited Jan 29 at 5:05
Yadati Kiran
2,1121622
2,1121622
asked Jan 29 at 4:37
Mk UtkarshMk Utkarsh
94110
asked Jan 29 at 4:37
Mk UtkarshMk Utkarsh
94110
asked Jan 29 at 4:37
Mk UtkarshMk Utkarsh
94110
94110
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
The problem is equivalent to that of how many ways of making one pound out of
ten-pence, twenty-pence and fifty-pence pieces, an just writing out all possibilities rapidly gives ten of them.
I'll try to explain in monochrome. For any $t$,
$$frac1{1-t}=sum_{n=0}^infty t^n$$
at least if $t$ is near zero. Then
begin{align}
frac1{(1-x^{10})(1-x^{20})(1-x^{50})}
&=(1+x^{10}+x^{20}+x^{30}+cdots)\
×
(1+x^{20}+x^{40}+x^{60}+cdots)\
×
(1+x^{50}+x^{100}+x^{150}+cdots)
end{align}
and we see the coefficients of all powers of $x^{10}$ in this expansion
are all positive. More formally,
begin{align}
frac1{(1-x^{10})(1-x^{20})(1-x^{50})}
&=sum_{a=0}^infty x^{10a}sum_{b=0}^infty x^{20b}sum_{c=0}^infty x^{50c}\
&=sum_{a,b,c=0}^infty x^{10a+20b+50c}
end{align}
so the coefficient of $x^{100}$ is the number of solutions
to $10a+20b+50c=100$ in non-negative integers.
edited Jan 29 at 5:43
Mk Utkarsh
94110
answered Jan 29 at 4:51
Lord Shark the UnknownLord Shark the Unknown
107k1162135
$endgroup$
$begingroup$
I understood your explanation, its crystal clear but still what's wrong with my solution, if I ignore all powers of -1 then I get correct answer however in extended binomial theorem there is power to (-1)
$endgroup$
– Mk Utkarsh
Jan 29 at 5:46
add a comment |
$begingroup$
Hint:
Set $y=x^{10},$
Now for $|y|<1, (1-y)^{-1}=1+y+y^2+cdots$
we need the coefficient of $y^{10}$ in $$(1-y^5)^{-1}(1-y^2)^{-1}(1-y)^{-1}$$
$$=(1+y^5+y^{10}+cdots)(1-y^2)^{-1}(1-y)^{-1}$$
So, the coefficient of $y^{10}$ in $(1-y^2)^{-1}(1-y)^{-1}$(which is $6$ due to $1cdot y^{10},y^2cdot y^8,y^4cdot y^6,y^6cdot y^4,y^8cdot y^2,,y^{10}cdot y^0)$
$+$ the coefficient of $y^{10-5}$ in $(1-y^2)^{-1}(1-y)^{-1}$ (which is $3,$ how ? )
$+$ the coefficient of $y^{10-10}$ in $(1-y^2)^{-1}(1-y)^{-1}$ (which is $1$ )
answered Jan 29 at 4:46
lab bhattacharjeelab bhattacharjee
228k15158279
$endgroup$
add a comment |
$begingroup$
Your mistake is in calculating $binom{-1}k$. You had
$$
(-1)^{10}binom{-1}{10}+(-1)^5binom{-1}{5}+dots=(-1)^{10}cdot 1+(-1)^5cdot color{red}1+dots
$$
The one in red is a mistake. $binom{-1}{5}=frac{(-1)(-2)cdots(-5)}{5!}=(-1)^5=-1$. In fact, your summation should just be a summation of ten ones, because each summand is a product of numbers of the form
$$
(-1)^kbinom{-1}{k}=(-1)^k(-1)^k=(-1)^{2k}=1
$$
The answer is then
$1+1+1+1+1+1+1+1+1+1=10$.
answered Jan 29 at 17:56
Mike EarnestMike Earnest
26.3k22151
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The problem is equivalent to that of how many ways of making one pound out of
ten-pence, twenty-pence and fifty-pence pieces, an just writing out all possibilities rapidly gives ten of them.
I'll try to explain in monochrome. For any $t$,
$$frac1{1-t}=sum_{n=0}^infty t^n$$
at least if $t$ is near zero. Then
begin{align}
frac1{(1-x^{10})(1-x^{20})(1-x^{50})}
&=(1+x^{10}+x^{20}+x^{30}+cdots)\
×
(1+x^{20}+x^{40}+x^{60}+cdots)\
×
(1+x^{50}+x^{100}+x^{150}+cdots)
end{align}
and we see the coefficients of all powers of $x^{10}$ in this expansion
are all positive. More formally,
begin{align}
frac1{(1-x^{10})(1-x^{20})(1-x^{50})}
&=sum_{a=0}^infty x^{10a}sum_{b=0}^infty x^{20b}sum_{c=0}^infty x^{50c}\
&=sum_{a,b,c=0}^infty x^{10a+20b+50c}
end{align}
so the coefficient of $x^{100}$ is the number of solutions
to $10a+20b+50c=100$ in non-negative integers.
edited Jan 29 at 5:43
Mk Utkarsh
94110
answered Jan 29 at 4:51
Lord Shark the UnknownLord Shark the Unknown
107k1162135
$endgroup$
$begingroup$
I understood your explanation, its crystal clear but still what's wrong with my solution, if I ignore all powers of -1 then I get correct answer however in extended binomial theorem there is power to (-1)
$endgroup$
– Mk Utkarsh
Jan 29 at 5:46
add a comment |
$begingroup$
The problem is equivalent to that of how many ways of making one pound out of
ten-pence, twenty-pence and fifty-pence pieces, an just writing out all possibilities rapidly gives ten of them.
I'll try to explain in monochrome. For any $t$,
$$frac1{1-t}=sum_{n=0}^infty t^n$$
at least if $t$ is near zero. Then
begin{align}
frac1{(1-x^{10})(1-x^{20})(1-x^{50})}
&=(1+x^{10}+x^{20}+x^{30}+cdots)\
×
(1+x^{20}+x^{40}+x^{60}+cdots)\
×
(1+x^{50}+x^{100}+x^{150}+cdots)
end{align}
and we see the coefficients of all powers of $x^{10}$ in this expansion
are all positive. More formally,
begin{align}
frac1{(1-x^{10})(1-x^{20})(1-x^{50})}
&=sum_{a=0}^infty x^{10a}sum_{b=0}^infty x^{20b}sum_{c=0}^infty x^{50c}\
&=sum_{a,b,c=0}^infty x^{10a+20b+50c}
end{align}
so the coefficient of $x^{100}$ is the number of solutions
to $10a+20b+50c=100$ in non-negative integers.
edited Jan 29 at 5:43
Mk Utkarsh
94110
answered Jan 29 at 4:51
Lord Shark the UnknownLord Shark the Unknown
107k1162135
$endgroup$
$begingroup$
I understood your explanation, its crystal clear but still what's wrong with my solution, if I ignore all powers of -1 then I get correct answer however in extended binomial theorem there is power to (-1)
$endgroup$
– Mk Utkarsh
Jan 29 at 5:46
add a comment |
$begingroup$
The problem is equivalent to that of how many ways of making one pound out of
ten-pence, twenty-pence and fifty-pence pieces, an just writing out all possibilities rapidly gives ten of them.
I'll try to explain in monochrome. For any $t$,
$$frac1{1-t}=sum_{n=0}^infty t^n$$
at least if $t$ is near zero. Then
begin{align}
frac1{(1-x^{10})(1-x^{20})(1-x^{50})}
&=(1+x^{10}+x^{20}+x^{30}+cdots)\
×
(1+x^{20}+x^{40}+x^{60}+cdots)\
×
(1+x^{50}+x^{100}+x^{150}+cdots)
end{align}
and we see the coefficients of all powers of $x^{10}$ in this expansion
are all positive. More formally,
begin{align}
frac1{(1-x^{10})(1-x^{20})(1-x^{50})}
&=sum_{a=0}^infty x^{10a}sum_{b=0}^infty x^{20b}sum_{c=0}^infty x^{50c}\
&=sum_{a,b,c=0}^infty x^{10a+20b+50c}
end{align}
so the coefficient of $x^{100}$ is the number of solutions
to $10a+20b+50c=100$ in non-negative integers.
edited Jan 29 at 5:43
Mk Utkarsh
94110
answered Jan 29 at 4:51
Lord Shark the UnknownLord Shark the Unknown
107k1162135
$endgroup$
The problem is equivalent to that of how many ways of making one pound out of
ten-pence, twenty-pence and fifty-pence pieces, an just writing out all possibilities rapidly gives ten of them.
I'll try to explain in monochrome. For any $t$,
$$frac1{1-t}=sum_{n=0}^infty t^n$$
at least if $t$ is near zero. Then
begin{align}
frac1{(1-x^{10})(1-x^{20})(1-x^{50})}
&=(1+x^{10}+x^{20}+x^{30}+cdots)\
×
(1+x^{20}+x^{40}+x^{60}+cdots)\
×
(1+x^{50}+x^{100}+x^{150}+cdots)
end{align}
and we see the coefficients of all powers of $x^{10}$ in this expansion
are all positive. More formally,
begin{align}
frac1{(1-x^{10})(1-x^{20})(1-x^{50})}
&=sum_{a=0}^infty x^{10a}sum_{b=0}^infty x^{20b}sum_{c=0}^infty x^{50c}\
&=sum_{a,b,c=0}^infty x^{10a+20b+50c}
end{align}
so the coefficient of $x^{100}$ is the number of solutions
to $10a+20b+50c=100$ in non-negative integers.
edited Jan 29 at 5:43
Mk Utkarsh
94110
answered Jan 29 at 4:51
Lord Shark the UnknownLord Shark the Unknown
107k1162135
edited Jan 29 at 5:43
Mk Utkarsh
94110
edited Jan 29 at 5:43
Mk Utkarsh
94110
edited Jan 29 at 5:43
Mk Utkarsh
94110
94110
answered Jan 29 at 4:51
Lord Shark the UnknownLord Shark the Unknown
107k1162135
answered Jan 29 at 4:51
Lord Shark the UnknownLord Shark the Unknown
107k1162135
answered Jan 29 at 4:51
Lord Shark the UnknownLord Shark the Unknown
107k1162135
107k1162135
$begingroup$
I understood your explanation, its crystal clear but still what's wrong with my solution, if I ignore all powers of -1 then I get correct answer however in extended binomial theorem there is power to (-1)
$endgroup$
– Mk Utkarsh
Jan 29 at 5:46
add a comment |
$begingroup$
I understood your explanation, its crystal clear but still what's wrong with my solution, if I ignore all powers of -1 then I get correct answer however in extended binomial theorem there is power to (-1)
$endgroup$
– Mk Utkarsh
Jan 29 at 5:46
$begingroup$
I understood your explanation, its crystal clear but still what's wrong with my solution, if I ignore all powers of -1 then I get correct answer however in extended binomial theorem there is power to (-1)
$endgroup$
– Mk Utkarsh
Jan 29 at 5:46
$begingroup$
I understood your explanation, its crystal clear but still what's wrong with my solution, if I ignore all powers of -1 then I get correct answer however in extended binomial theorem there is power to (-1)
$endgroup$
– Mk Utkarsh
Jan 29 at 5:46
add a comment |
$begingroup$
Hint:
Set $y=x^{10},$
Now for $|y|<1, (1-y)^{-1}=1+y+y^2+cdots$
we need the coefficient of $y^{10}$ in $$(1-y^5)^{-1}(1-y^2)^{-1}(1-y)^{-1}$$
$$=(1+y^5+y^{10}+cdots)(1-y^2)^{-1}(1-y)^{-1}$$
So, the coefficient of $y^{10}$ in $(1-y^2)^{-1}(1-y)^{-1}$(which is $6$ due to $1cdot y^{10},y^2cdot y^8,y^4cdot y^6,y^6cdot y^4,y^8cdot y^2,,y^{10}cdot y^0)$
$+$ the coefficient of $y^{10-5}$ in $(1-y^2)^{-1}(1-y)^{-1}$ (which is $3,$ how ? )
$+$ the coefficient of $y^{10-10}$ in $(1-y^2)^{-1}(1-y)^{-1}$ (which is $1$ )
answered Jan 29 at 4:46
lab bhattacharjeelab bhattacharjee
228k15158279
$endgroup$
add a comment |
$begingroup$
Hint:
Set $y=x^{10},$
Now for $|y|<1, (1-y)^{-1}=1+y+y^2+cdots$
we need the coefficient of $y^{10}$ in $$(1-y^5)^{-1}(1-y^2)^{-1}(1-y)^{-1}$$
$$=(1+y^5+y^{10}+cdots)(1-y^2)^{-1}(1-y)^{-1}$$
So, the coefficient of $y^{10}$ in $(1-y^2)^{-1}(1-y)^{-1}$(which is $6$ due to $1cdot y^{10},y^2cdot y^8,y^4cdot y^6,y^6cdot y^4,y^8cdot y^2,,y^{10}cdot y^0)$
$+$ the coefficient of $y^{10-5}$ in $(1-y^2)^{-1}(1-y)^{-1}$ (which is $3,$ how ? )
$+$ the coefficient of $y^{10-10}$ in $(1-y^2)^{-1}(1-y)^{-1}$ (which is $1$ )
answered Jan 29 at 4:46
lab bhattacharjeelab bhattacharjee
228k15158279
$endgroup$
add a comment |
$begingroup$
Hint:
Set $y=x^{10},$
Now for $|y|<1, (1-y)^{-1}=1+y+y^2+cdots$
we need the coefficient of $y^{10}$ in $$(1-y^5)^{-1}(1-y^2)^{-1}(1-y)^{-1}$$
$$=(1+y^5+y^{10}+cdots)(1-y^2)^{-1}(1-y)^{-1}$$
So, the coefficient of $y^{10}$ in $(1-y^2)^{-1}(1-y)^{-1}$(which is $6$ due to $1cdot y^{10},y^2cdot y^8,y^4cdot y^6,y^6cdot y^4,y^8cdot y^2,,y^{10}cdot y^0)$
$+$ the coefficient of $y^{10-5}$ in $(1-y^2)^{-1}(1-y)^{-1}$ (which is $3,$ how ? )
$+$ the coefficient of $y^{10-10}$ in $(1-y^2)^{-1}(1-y)^{-1}$ (which is $1$ )
answered Jan 29 at 4:46
lab bhattacharjeelab bhattacharjee
228k15158279
$endgroup$
Hint:
Set $y=x^{10},$
Now for $|y|<1, (1-y)^{-1}=1+y+y^2+cdots$
we need the coefficient of $y^{10}$ in $$(1-y^5)^{-1}(1-y^2)^{-1}(1-y)^{-1}$$
$$=(1+y^5+y^{10}+cdots)(1-y^2)^{-1}(1-y)^{-1}$$
So, the coefficient of $y^{10}$ in $(1-y^2)^{-1}(1-y)^{-1}$(which is $6$ due to $1cdot y^{10},y^2cdot y^8,y^4cdot y^6,y^6cdot y^4,y^8cdot y^2,,y^{10}cdot y^0)$
$+$ the coefficient of $y^{10-5}$ in $(1-y^2)^{-1}(1-y)^{-1}$ (which is $3,$ how ? )
$+$ the coefficient of $y^{10-10}$ in $(1-y^2)^{-1}(1-y)^{-1}$ (which is $1$ )
answered Jan 29 at 4:46
lab bhattacharjeelab bhattacharjee
228k15158279
answered Jan 29 at 4:46
lab bhattacharjeelab bhattacharjee
228k15158279
answered Jan 29 at 4:46
lab bhattacharjeelab bhattacharjee
228k15158279
answered Jan 29 at 4:46
lab bhattacharjeelab bhattacharjee
228k15158279
228k15158279
add a comment |
add a comment |
$begingroup$
Your mistake is in calculating $binom{-1}k$. You had
$$
(-1)^{10}binom{-1}{10}+(-1)^5binom{-1}{5}+dots=(-1)^{10}cdot 1+(-1)^5cdot color{red}1+dots
$$
The one in red is a mistake. $binom{-1}{5}=frac{(-1)(-2)cdots(-5)}{5!}=(-1)^5=-1$. In fact, your summation should just be a summation of ten ones, because each summand is a product of numbers of the form
$$
(-1)^kbinom{-1}{k}=(-1)^k(-1)^k=(-1)^{2k}=1
$$
The answer is then
$1+1+1+1+1+1+1+1+1+1=10$.
answered Jan 29 at 17:56
Mike EarnestMike Earnest
26.3k22151
$endgroup$
add a comment |
$begingroup$
Your mistake is in calculating $binom{-1}k$. You had
$$
(-1)^{10}binom{-1}{10}+(-1)^5binom{-1}{5}+dots=(-1)^{10}cdot 1+(-1)^5cdot color{red}1+dots
$$
The one in red is a mistake. $binom{-1}{5}=frac{(-1)(-2)cdots(-5)}{5!}=(-1)^5=-1$. In fact, your summation should just be a summation of ten ones, because each summand is a product of numbers of the form
$$
(-1)^kbinom{-1}{k}=(-1)^k(-1)^k=(-1)^{2k}=1
$$
The answer is then
$1+1+1+1+1+1+1+1+1+1=10$.
answered Jan 29 at 17:56
Mike EarnestMike Earnest
26.3k22151
$endgroup$
add a comment |
$begingroup$
Your mistake is in calculating $binom{-1}k$. You had
$$
(-1)^{10}binom{-1}{10}+(-1)^5binom{-1}{5}+dots=(-1)^{10}cdot 1+(-1)^5cdot color{red}1+dots
$$
The one in red is a mistake. $binom{-1}{5}=frac{(-1)(-2)cdots(-5)}{5!}=(-1)^5=-1$. In fact, your summation should just be a summation of ten ones, because each summand is a product of numbers of the form
$$
(-1)^kbinom{-1}{k}=(-1)^k(-1)^k=(-1)^{2k}=1
$$
The answer is then
$1+1+1+1+1+1+1+1+1+1=10$.
answered Jan 29 at 17:56
Mike EarnestMike Earnest
26.3k22151
$endgroup$
Your mistake is in calculating $binom{-1}k$. You had
$$
(-1)^{10}binom{-1}{10}+(-1)^5binom{-1}{5}+dots=(-1)^{10}cdot 1+(-1)^5cdot color{red}1+dots
$$
The one in red is a mistake. $binom{-1}{5}=frac{(-1)(-2)cdots(-5)}{5!}=(-1)^5=-1$. In fact, your summation should just be a summation of ten ones, because each summand is a product of numbers of the form
$$
(-1)^kbinom{-1}{k}=(-1)^k(-1)^k=(-1)^{2k}=1
$$
The answer is then
$1+1+1+1+1+1+1+1+1+1=10$.
answered Jan 29 at 17:56
Mike EarnestMike Earnest
26.3k22151
edited Jan 29 at 18:56
answered Jan 29 at 17:56
Mike EarnestMike Earnest
26.3k22151
answered Jan 29 at 17:56
Mike EarnestMike Earnest
26.3k22151
answered Jan 29 at 17:56
Mike EarnestMike Earnest
26.3k22151
26.3k22151
add a comment |
add a comment |
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