Mixing
equating coefficients in algebraic expansion
$begingroup$
If $displaystyle bigg(frac{1+x}{1-x}bigg)^n=1+b_{1}x+b_{2}x^2+b_{3}x^3+cdots +infty,$ then value of
$(1); displaystyle frac{3b_{3}-b_{1}}{b_{2}}$
$(2); displaystyle frac{2b_{4}-b_{2}}{b_{3}}$
$(3); displaystyle frac{3b_{6}-2b_{4}}{b_{5}}$
$(4); displaystyle frac{5b_{10}-4b_{8}}{b_{9}}$
Answers given in $n,2n,3n,4n$ formats
What I tried:
$displaystyle (1+x)^n(1-x)^{-n}=1+b_{1}x+b_{2}x^2+b_{3}x^3+cdots cdots infty$
$displaystyle bigg[1+nx+frac{n(n-1)}{2}x^2+frac{n(n-1)(n-2)}{6}cdots bigg]bigg[1+nx+frac{n(n+1)}{2}x^2+frac{n(n+1)(n+2)}{6}x^3+cdots bigg]=1+b_{1}x+b_{2}x^2+b_{3}x^3+cdots $
$displaystyle b_{1}=n,b_{2}=n,b_{3}=4n^2$
How do I find other coefficients? Help me, please.
binomial-theorem
edited Feb 1 at 5:37
J. W. Tanner
4,6441420
asked Feb 1 at 5:03
jackyjacky
1,344816
$endgroup$
add a comment |
$begingroup$
If $displaystyle bigg(frac{1+x}{1-x}bigg)^n=1+b_{1}x+b_{2}x^2+b_{3}x^3+cdots +infty,$ then value of
$(1); displaystyle frac{3b_{3}-b_{1}}{b_{2}}$
$(2); displaystyle frac{2b_{4}-b_{2}}{b_{3}}$
$(3); displaystyle frac{3b_{6}-2b_{4}}{b_{5}}$
$(4); displaystyle frac{5b_{10}-4b_{8}}{b_{9}}$
Answers given in $n,2n,3n,4n$ formats
What I tried:
$displaystyle (1+x)^n(1-x)^{-n}=1+b_{1}x+b_{2}x^2+b_{3}x^3+cdots cdots infty$
$displaystyle bigg[1+nx+frac{n(n-1)}{2}x^2+frac{n(n-1)(n-2)}{6}cdots bigg]bigg[1+nx+frac{n(n+1)}{2}x^2+frac{n(n+1)(n+2)}{6}x^3+cdots bigg]=1+b_{1}x+b_{2}x^2+b_{3}x^3+cdots $
$displaystyle b_{1}=n,b_{2}=n,b_{3}=4n^2$
How do I find other coefficients? Help me, please.
binomial-theorem
edited Feb 1 at 5:37
J. W. Tanner
4,6441420
asked Feb 1 at 5:03
jackyjacky
1,344816
$endgroup$
add a comment |
$begingroup$
If $displaystyle bigg(frac{1+x}{1-x}bigg)^n=1+b_{1}x+b_{2}x^2+b_{3}x^3+cdots +infty,$ then value of
$(1); displaystyle frac{3b_{3}-b_{1}}{b_{2}}$
$(2); displaystyle frac{2b_{4}-b_{2}}{b_{3}}$
$(3); displaystyle frac{3b_{6}-2b_{4}}{b_{5}}$
$(4); displaystyle frac{5b_{10}-4b_{8}}{b_{9}}$
Answers given in $n,2n,3n,4n$ formats
What I tried:
$displaystyle (1+x)^n(1-x)^{-n}=1+b_{1}x+b_{2}x^2+b_{3}x^3+cdots cdots infty$
$displaystyle bigg[1+nx+frac{n(n-1)}{2}x^2+frac{n(n-1)(n-2)}{6}cdots bigg]bigg[1+nx+frac{n(n+1)}{2}x^2+frac{n(n+1)(n+2)}{6}x^3+cdots bigg]=1+b_{1}x+b_{2}x^2+b_{3}x^3+cdots $
$displaystyle b_{1}=n,b_{2}=n,b_{3}=4n^2$
How do I find other coefficients? Help me, please.
binomial-theorem
edited Feb 1 at 5:37
J. W. Tanner
4,6441420
asked Feb 1 at 5:03
jackyjacky
1,344816
$endgroup$
If $displaystyle bigg(frac{1+x}{1-x}bigg)^n=1+b_{1}x+b_{2}x^2+b_{3}x^3+cdots +infty,$ then value of
$(1); displaystyle frac{3b_{3}-b_{1}}{b_{2}}$
$(2); displaystyle frac{2b_{4}-b_{2}}{b_{3}}$
$(3); displaystyle frac{3b_{6}-2b_{4}}{b_{5}}$
$(4); displaystyle frac{5b_{10}-4b_{8}}{b_{9}}$
Answers given in $n,2n,3n,4n$ formats
What I tried:
$displaystyle (1+x)^n(1-x)^{-n}=1+b_{1}x+b_{2}x^2+b_{3}x^3+cdots cdots infty$
$displaystyle bigg[1+nx+frac{n(n-1)}{2}x^2+frac{n(n-1)(n-2)}{6}cdots bigg]bigg[1+nx+frac{n(n+1)}{2}x^2+frac{n(n+1)(n+2)}{6}x^3+cdots bigg]=1+b_{1}x+b_{2}x^2+b_{3}x^3+cdots $
$displaystyle b_{1}=n,b_{2}=n,b_{3}=4n^2$
How do I find other coefficients? Help me, please.
binomial-theorem
binomial-theorem
edited Feb 1 at 5:37
J. W. Tanner
4,6441420
asked Feb 1 at 5:03
jackyjacky
1,344816
edited Feb 1 at 5:37
J. W. Tanner
4,6441420
asked Feb 1 at 5:03
jackyjacky
1,344816
edited Feb 1 at 5:37
J. W. Tanner
4,6441420
edited Feb 1 at 5:37
J. W. Tanner
4,6441420
edited Feb 1 at 5:37
J. W. Tanner
4,6441420
4,6441420
asked Feb 1 at 5:03
jackyjacky
1,344816
asked Feb 1 at 5:03
jackyjacky
1,344816
asked Feb 1 at 5:03
jackyjacky
1,344816
1,344816
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You made a mistake. It is not $displaystyle b_{1}=n,b_{2}=n,b_{3}=4n^2$ , but
$b_{1}=2n$
$b_{2}=2n^2$
$b_{3}=frac23(n+2n^3)$
To go up to $b_{10}$, the series has to be expended up to $x^{10}$.
By brut force :
$b_{4}=frac23(2n^2+n^4)$
$b_{5}=frac{2}{15}(3n+10n^3+2n^5)$
$b_{6}=frac{2}{45}(23n^2+20n^4+2n^6)$
$b_{7}=frac{2}{315}(45n+196n^3+70n^5+4n^7)$
$b_{8}=frac{2}{315}(132n^2+154n^4+28n^6+n^8)$
$b_{9}=frac{2}{2835}(315n+1636n^3+798n^5+84n^7+2n^8)$
$b_{10}=frac{2}{14175}(5167n^2+7180n^4+1806n^6+120n^8+2n^{10})$
$frac{3b_{3}-b_{1}}{b_{2}}=2n$
Then, a supposed recurrence :
$frac{1b_{2}-0b_{0}}{b_{1}}=n$
$frac{2b_{4}-1b_{2}}{b_{3}}=n$
$frac{3b_{6}-2b_{4}}{b_{5}}=n$
$frac{4b_{8}-3b_{6}}{b_{7}}=n$
$frac{5b_{10}-4b_{8}}{b_{9}}=n$
$...$
$frac{k,b_{2k}-(k-1)b_{2k-2}}{b_{2k-1}}=n$ , not proved.
answered Feb 1 at 7:27
JJacquelinJJacquelin
45.6k21857
$endgroup$
add a comment |
$begingroup$
Here is a slightly different method to find the coefficients. It is convenient to use the coefficient of operator $[x^j]$ to denote the coefficient of $x^j$ in a series.
We obtain
begin{align*}
color{blue}{[x^j]left(frac{1+x}{1-x}right)^n}
&=[x^j](1-x)^{-n}(1+x)^n\
&=[x^j]sum_{k=0}^inftybinom{-n}{k}(-x)^k(1+x)^ntag{1}\
&=[x^j]sum_{k=0}^inftybinom{n+k-1}{k}x^k(1+x)^ntag{2}\
&=sum_{k=0}^jbinom{n+k-1}{k}[x^{j-k}](1+x)^ntag{3}\
&,,color{blue}{=sum_{k=0}^jbinom{n+k-1}{k}binom{n}{j-k}}tag{4}
end{align*}
Comment:
In (1) we apply the binomial series expansion
In (2) we use the binomial identity $binom{-p}{q}=binom{p+q-1}{q}(-1)^q$
In (3) we use the linearity of the coefficient of operator and apply the rule $[x^{p-q}]A(x)=[x^p]x^qA(x)$. We also set the upper limit of the sum to $j$ since other values do not contribute.
In (4) we select the coefficient of $[x^{j-k}](1+x)^n=[x^{j-k}]sum_{t=0}^nbinom{n}{t}x^t$.
answered Feb 1 at 20:42
Markus ScheuerMarkus Scheuer
64.1k460152
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
You made a mistake. It is not $displaystyle b_{1}=n,b_{2}=n,b_{3}=4n^2$ , but
$b_{1}=2n$
$b_{2}=2n^2$
$b_{3}=frac23(n+2n^3)$
To go up to $b_{10}$, the series has to be expended up to $x^{10}$.
By brut force :
$b_{4}=frac23(2n^2+n^4)$
$b_{5}=frac{2}{15}(3n+10n^3+2n^5)$
$b_{6}=frac{2}{45}(23n^2+20n^4+2n^6)$
$b_{7}=frac{2}{315}(45n+196n^3+70n^5+4n^7)$
$b_{8}=frac{2}{315}(132n^2+154n^4+28n^6+n^8)$
$b_{9}=frac{2}{2835}(315n+1636n^3+798n^5+84n^7+2n^8)$
$b_{10}=frac{2}{14175}(5167n^2+7180n^4+1806n^6+120n^8+2n^{10})$
$frac{3b_{3}-b_{1}}{b_{2}}=2n$
Then, a supposed recurrence :
$frac{1b_{2}-0b_{0}}{b_{1}}=n$
$frac{2b_{4}-1b_{2}}{b_{3}}=n$
$frac{3b_{6}-2b_{4}}{b_{5}}=n$
$frac{4b_{8}-3b_{6}}{b_{7}}=n$
$frac{5b_{10}-4b_{8}}{b_{9}}=n$
$...$
$frac{k,b_{2k}-(k-1)b_{2k-2}}{b_{2k-1}}=n$ , not proved.
answered Feb 1 at 7:27
JJacquelinJJacquelin
45.6k21857
$endgroup$
add a comment |
$begingroup$
You made a mistake. It is not $displaystyle b_{1}=n,b_{2}=n,b_{3}=4n^2$ , but
$b_{1}=2n$
$b_{2}=2n^2$
$b_{3}=frac23(n+2n^3)$
To go up to $b_{10}$, the series has to be expended up to $x^{10}$.
By brut force :
$b_{4}=frac23(2n^2+n^4)$
$b_{5}=frac{2}{15}(3n+10n^3+2n^5)$
$b_{6}=frac{2}{45}(23n^2+20n^4+2n^6)$
$b_{7}=frac{2}{315}(45n+196n^3+70n^5+4n^7)$
$b_{8}=frac{2}{315}(132n^2+154n^4+28n^6+n^8)$
$b_{9}=frac{2}{2835}(315n+1636n^3+798n^5+84n^7+2n^8)$
$b_{10}=frac{2}{14175}(5167n^2+7180n^4+1806n^6+120n^8+2n^{10})$
$frac{3b_{3}-b_{1}}{b_{2}}=2n$
Then, a supposed recurrence :
$frac{1b_{2}-0b_{0}}{b_{1}}=n$
$frac{2b_{4}-1b_{2}}{b_{3}}=n$
$frac{3b_{6}-2b_{4}}{b_{5}}=n$
$frac{4b_{8}-3b_{6}}{b_{7}}=n$
$frac{5b_{10}-4b_{8}}{b_{9}}=n$
$...$
$frac{k,b_{2k}-(k-1)b_{2k-2}}{b_{2k-1}}=n$ , not proved.
answered Feb 1 at 7:27
JJacquelinJJacquelin
45.6k21857
$endgroup$
add a comment |
$begingroup$
You made a mistake. It is not $displaystyle b_{1}=n,b_{2}=n,b_{3}=4n^2$ , but
$b_{1}=2n$
$b_{2}=2n^2$
$b_{3}=frac23(n+2n^3)$
To go up to $b_{10}$, the series has to be expended up to $x^{10}$.
By brut force :
$b_{4}=frac23(2n^2+n^4)$
$b_{5}=frac{2}{15}(3n+10n^3+2n^5)$
$b_{6}=frac{2}{45}(23n^2+20n^4+2n^6)$
$b_{7}=frac{2}{315}(45n+196n^3+70n^5+4n^7)$
$b_{8}=frac{2}{315}(132n^2+154n^4+28n^6+n^8)$
$b_{9}=frac{2}{2835}(315n+1636n^3+798n^5+84n^7+2n^8)$
$b_{10}=frac{2}{14175}(5167n^2+7180n^4+1806n^6+120n^8+2n^{10})$
$frac{3b_{3}-b_{1}}{b_{2}}=2n$
Then, a supposed recurrence :
$frac{1b_{2}-0b_{0}}{b_{1}}=n$
$frac{2b_{4}-1b_{2}}{b_{3}}=n$
$frac{3b_{6}-2b_{4}}{b_{5}}=n$
$frac{4b_{8}-3b_{6}}{b_{7}}=n$
$frac{5b_{10}-4b_{8}}{b_{9}}=n$
$...$
$frac{k,b_{2k}-(k-1)b_{2k-2}}{b_{2k-1}}=n$ , not proved.
answered Feb 1 at 7:27
JJacquelinJJacquelin
45.6k21857
$endgroup$
You made a mistake. It is not $displaystyle b_{1}=n,b_{2}=n,b_{3}=4n^2$ , but
$b_{1}=2n$
$b_{2}=2n^2$
$b_{3}=frac23(n+2n^3)$
To go up to $b_{10}$, the series has to be expended up to $x^{10}$.
By brut force :
$b_{4}=frac23(2n^2+n^4)$
$b_{5}=frac{2}{15}(3n+10n^3+2n^5)$
$b_{6}=frac{2}{45}(23n^2+20n^4+2n^6)$
$b_{7}=frac{2}{315}(45n+196n^3+70n^5+4n^7)$
$b_{8}=frac{2}{315}(132n^2+154n^4+28n^6+n^8)$
$b_{9}=frac{2}{2835}(315n+1636n^3+798n^5+84n^7+2n^8)$
$b_{10}=frac{2}{14175}(5167n^2+7180n^4+1806n^6+120n^8+2n^{10})$
$frac{3b_{3}-b_{1}}{b_{2}}=2n$
Then, a supposed recurrence :
$frac{1b_{2}-0b_{0}}{b_{1}}=n$
$frac{2b_{4}-1b_{2}}{b_{3}}=n$
$frac{3b_{6}-2b_{4}}{b_{5}}=n$
$frac{4b_{8}-3b_{6}}{b_{7}}=n$
$frac{5b_{10}-4b_{8}}{b_{9}}=n$
$...$
$frac{k,b_{2k}-(k-1)b_{2k-2}}{b_{2k-1}}=n$ , not proved.
answered Feb 1 at 7:27
JJacquelinJJacquelin
45.6k21857
answered Feb 1 at 7:27
JJacquelinJJacquelin
45.6k21857
answered Feb 1 at 7:27
JJacquelinJJacquelin
45.6k21857
answered Feb 1 at 7:27
JJacquelinJJacquelin
45.6k21857
45.6k21857
add a comment |
add a comment |
$begingroup$
Here is a slightly different method to find the coefficients. It is convenient to use the coefficient of operator $[x^j]$ to denote the coefficient of $x^j$ in a series.
We obtain
begin{align*}
color{blue}{[x^j]left(frac{1+x}{1-x}right)^n}
&=[x^j](1-x)^{-n}(1+x)^n\
&=[x^j]sum_{k=0}^inftybinom{-n}{k}(-x)^k(1+x)^ntag{1}\
&=[x^j]sum_{k=0}^inftybinom{n+k-1}{k}x^k(1+x)^ntag{2}\
&=sum_{k=0}^jbinom{n+k-1}{k}[x^{j-k}](1+x)^ntag{3}\
&,,color{blue}{=sum_{k=0}^jbinom{n+k-1}{k}binom{n}{j-k}}tag{4}
end{align*}
Comment:
In (1) we apply the binomial series expansion
In (2) we use the binomial identity $binom{-p}{q}=binom{p+q-1}{q}(-1)^q$
In (3) we use the linearity of the coefficient of operator and apply the rule $[x^{p-q}]A(x)=[x^p]x^qA(x)$. We also set the upper limit of the sum to $j$ since other values do not contribute.
In (4) we select the coefficient of $[x^{j-k}](1+x)^n=[x^{j-k}]sum_{t=0}^nbinom{n}{t}x^t$.
answered Feb 1 at 20:42
Markus ScheuerMarkus Scheuer
64.1k460152
$endgroup$
add a comment |
$begingroup$
Here is a slightly different method to find the coefficients. It is convenient to use the coefficient of operator $[x^j]$ to denote the coefficient of $x^j$ in a series.
We obtain
begin{align*}
color{blue}{[x^j]left(frac{1+x}{1-x}right)^n}
&=[x^j](1-x)^{-n}(1+x)^n\
&=[x^j]sum_{k=0}^inftybinom{-n}{k}(-x)^k(1+x)^ntag{1}\
&=[x^j]sum_{k=0}^inftybinom{n+k-1}{k}x^k(1+x)^ntag{2}\
&=sum_{k=0}^jbinom{n+k-1}{k}[x^{j-k}](1+x)^ntag{3}\
&,,color{blue}{=sum_{k=0}^jbinom{n+k-1}{k}binom{n}{j-k}}tag{4}
end{align*}
Comment:
In (1) we apply the binomial series expansion
In (2) we use the binomial identity $binom{-p}{q}=binom{p+q-1}{q}(-1)^q$
In (3) we use the linearity of the coefficient of operator and apply the rule $[x^{p-q}]A(x)=[x^p]x^qA(x)$. We also set the upper limit of the sum to $j$ since other values do not contribute.
In (4) we select the coefficient of $[x^{j-k}](1+x)^n=[x^{j-k}]sum_{t=0}^nbinom{n}{t}x^t$.
answered Feb 1 at 20:42
Markus ScheuerMarkus Scheuer
64.1k460152
$endgroup$
add a comment |
$begingroup$
Here is a slightly different method to find the coefficients. It is convenient to use the coefficient of operator $[x^j]$ to denote the coefficient of $x^j$ in a series.
We obtain
begin{align*}
color{blue}{[x^j]left(frac{1+x}{1-x}right)^n}
&=[x^j](1-x)^{-n}(1+x)^n\
&=[x^j]sum_{k=0}^inftybinom{-n}{k}(-x)^k(1+x)^ntag{1}\
&=[x^j]sum_{k=0}^inftybinom{n+k-1}{k}x^k(1+x)^ntag{2}\
&=sum_{k=0}^jbinom{n+k-1}{k}[x^{j-k}](1+x)^ntag{3}\
&,,color{blue}{=sum_{k=0}^jbinom{n+k-1}{k}binom{n}{j-k}}tag{4}
end{align*}
Comment:
In (1) we apply the binomial series expansion
In (2) we use the binomial identity $binom{-p}{q}=binom{p+q-1}{q}(-1)^q$
In (3) we use the linearity of the coefficient of operator and apply the rule $[x^{p-q}]A(x)=[x^p]x^qA(x)$. We also set the upper limit of the sum to $j$ since other values do not contribute.
In (4) we select the coefficient of $[x^{j-k}](1+x)^n=[x^{j-k}]sum_{t=0}^nbinom{n}{t}x^t$.
answered Feb 1 at 20:42
Markus ScheuerMarkus Scheuer
64.1k460152
$endgroup$
Here is a slightly different method to find the coefficients. It is convenient to use the coefficient of operator $[x^j]$ to denote the coefficient of $x^j$ in a series.
We obtain
begin{align*}
color{blue}{[x^j]left(frac{1+x}{1-x}right)^n}
&=[x^j](1-x)^{-n}(1+x)^n\
&=[x^j]sum_{k=0}^inftybinom{-n}{k}(-x)^k(1+x)^ntag{1}\
&=[x^j]sum_{k=0}^inftybinom{n+k-1}{k}x^k(1+x)^ntag{2}\
&=sum_{k=0}^jbinom{n+k-1}{k}[x^{j-k}](1+x)^ntag{3}\
&,,color{blue}{=sum_{k=0}^jbinom{n+k-1}{k}binom{n}{j-k}}tag{4}
end{align*}
Comment:
In (1) we apply the binomial series expansion
In (2) we use the binomial identity $binom{-p}{q}=binom{p+q-1}{q}(-1)^q$
In (3) we use the linearity of the coefficient of operator and apply the rule $[x^{p-q}]A(x)=[x^p]x^qA(x)$. We also set the upper limit of the sum to $j$ since other values do not contribute.
In (4) we select the coefficient of $[x^{j-k}](1+x)^n=[x^{j-k}]sum_{t=0}^nbinom{n}{t}x^t$.
answered Feb 1 at 20:42
Markus ScheuerMarkus Scheuer
64.1k460152
answered Feb 1 at 20:42
Markus ScheuerMarkus Scheuer
64.1k460152
answered Feb 1 at 20:42
Markus ScheuerMarkus Scheuer
64.1k460152
answered Feb 1 at 20:42
Markus ScheuerMarkus Scheuer
64.1k460152
64.1k460152
add a comment |
add a comment |
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StackExchange.helpers.onClickDraftSave('#login-link');
});
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StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
