Mixing
Maximum coefficient in the expansion of $(5+3x)^{10}$
$begingroup$
Though I know that I could simply just expand $(5+3x)^{10}$ with the binomial theorem for each power of x, is there a simpler and quicker method of finding out the largest coefficient? After manual expansion, I know that it wouldn't be the coefficient of $x^6$ since there are other coefficients that are larger, but how do I prove that there's a quick way to get to the largest coefficient possible?
binomial-theorem
edited Jan 10 at 14:58
Larry
2,39131129
asked Jan 10 at 14:47
Roo23Roo23
163
$endgroup$
add a comment |
$begingroup$
Though I know that I could simply just expand $(5+3x)^{10}$ with the binomial theorem for each power of x, is there a simpler and quicker method of finding out the largest coefficient? After manual expansion, I know that it wouldn't be the coefficient of $x^6$ since there are other coefficients that are larger, but how do I prove that there's a quick way to get to the largest coefficient possible?
binomial-theorem
edited Jan 10 at 14:58
Larry
2,39131129
asked Jan 10 at 14:47
Roo23Roo23
163
$endgroup$
2
$begingroup$
If you divide a general coefficient by its successor you should get a simpler expression which will tell you which of them is larger.
$endgroup$
– Mark Bennet
Jan 10 at 14:54
$begingroup$
math.stackexchange.com/questions/722952/…
$endgroup$
– lab bhattacharjee
Jan 10 at 15:10
add a comment |
$begingroup$
Though I know that I could simply just expand $(5+3x)^{10}$ with the binomial theorem for each power of x, is there a simpler and quicker method of finding out the largest coefficient? After manual expansion, I know that it wouldn't be the coefficient of $x^6$ since there are other coefficients that are larger, but how do I prove that there's a quick way to get to the largest coefficient possible?
binomial-theorem
edited Jan 10 at 14:58
Larry
2,39131129
asked Jan 10 at 14:47
Roo23Roo23
163
$endgroup$
Though I know that I could simply just expand $(5+3x)^{10}$ with the binomial theorem for each power of x, is there a simpler and quicker method of finding out the largest coefficient? After manual expansion, I know that it wouldn't be the coefficient of $x^6$ since there are other coefficients that are larger, but how do I prove that there's a quick way to get to the largest coefficient possible?
binomial-theorem
binomial-theorem
edited Jan 10 at 14:58
Larry
2,39131129
asked Jan 10 at 14:47
Roo23Roo23
163
edited Jan 10 at 14:58
Larry
2,39131129
asked Jan 10 at 14:47
Roo23Roo23
163
edited Jan 10 at 14:58
Larry
2,39131129
edited Jan 10 at 14:58
Larry
2,39131129
edited Jan 10 at 14:58
Larry
2,39131129
2,39131129
asked Jan 10 at 14:47
Roo23Roo23
163
asked Jan 10 at 14:47
Roo23Roo23
163
asked Jan 10 at 14:47
Roo23Roo23
163
163
2
$begingroup$
If you divide a general coefficient by its successor you should get a simpler expression which will tell you which of them is larger.
$endgroup$
– Mark Bennet
Jan 10 at 14:54
$begingroup$
math.stackexchange.com/questions/722952/…
$endgroup$
– lab bhattacharjee
Jan 10 at 15:10
add a comment |
2
$begingroup$
If you divide a general coefficient by its successor you should get a simpler expression which will tell you which of them is larger.
$endgroup$
– Mark Bennet
Jan 10 at 14:54
$begingroup$
math.stackexchange.com/questions/722952/…
$endgroup$
– lab bhattacharjee
Jan 10 at 15:10
2
2
$begingroup$
If you divide a general coefficient by its successor you should get a simpler expression which will tell you which of them is larger.
$endgroup$
– Mark Bennet
Jan 10 at 14:54
$begingroup$
If you divide a general coefficient by its successor you should get a simpler expression which will tell you which of them is larger.
$endgroup$
– Mark Bennet
Jan 10 at 14:54
$begingroup$
math.stackexchange.com/questions/722952/…
$endgroup$
– lab bhattacharjee
Jan 10 at 15:10
$begingroup$
math.stackexchange.com/questions/722952/…
$endgroup$
– lab bhattacharjee
Jan 10 at 15:10
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Write $$(5+3x)^{10}=sum_{i=0}^{10}a_ix^i.$$ So that
$$
a_i=binom{10}{i}3^i5^{10-i}.
$$ Then $$f(i):=a_i/a_{i+1}=frac{5}{3}cdot frac{i+1}{10-i}.$$ This is an increasing function of $i$ (for $0leq i<10$). The maximum of the $a_i$ is attained at the smallest $i$ for which $f(i)>1$. Solving $f(i)=1$ yields $i=25/8$, which is larger than $3$ but smaller than $4$. Therefore $a_4$ is the largest coefficient.
answered Jan 10 at 15:08
UncountableUncountable
3,040414
$endgroup$
add a comment |
$begingroup$
The coefficients have the form
$$dbinom{10}k5^k3^{10-k}$$
The ratio of two consecutive coefficients is:
$$frac{dbinom{10}k5^k3^{10-k}}{dbinom{10}{k+1}5^{k+1}3^{9-k}}=frac{3(k+1)}{5(10-k)}$$
This ratio is lesser than $1$ when
$$3k+3<50-5k$$
That is, when $kle 5$. This means that the maximum coefficient is $dbinom{10}65^6cdot3^4$.
answered Jan 10 at 15:09
ajotatxeajotatxe
53.8k23890
$endgroup$
$begingroup$
According to OP question, it would be more logical to define $k$ as the "power of $x$": $a_k x^k$
$endgroup$
– Damien
Jan 10 at 15:14
add a comment |
$begingroup$
With the result proved below, the largest coefficient is among the set $(a^n,b^n,C_p)$, where $C_p$ is the largest coefficient as $pin[1,n-1]$.
In this case, $a=5, b=3, n=10$, $displaystylefrac ab>0$. So by $(1)$,
$$frac{an-b}{a+b}<p<frac{an+a}{a+b}impliesfrac{47}{8}<p<frac{55}{8}implies p=6$$
$$C_p={10choose6}5^63^4>5^{10}>3^{10}$$
So the largest coefficient occurs at $k=6$.
Proof
By Binomial Formula,
$$(ax+by)^n=sum_{k=0}^n {nchoose k}(ax)^k (by)^{n-k}$$
The coefficients can be expressed as
$$C_k={nchoose k}a^k b^{n-k}$$
Consider for all $k<n$, one may prove that
$$frac{C_{k+1}}{C_k}=frac abcdotfrac{n-k}{k+1}$$
This ratio, $widetilde C$, indicates if $|C_{k+1}|>|C_k|$. This is true if $|widetilde C|>1$.
Now if there exists an integer $pin[1,n-1]$ such that $C_{p}>0$ is maximum, then
$$left|frac{C_{p}}{C_{p-1}}right|>1, left|frac{C_{p+1}}{C_{p}}right|<1$$
For $displaystylefrac ab>0$,
$$frac ab n-1<(1+frac ab)p<frac ab(n+1)$$
$$impliesfrac{an-b}{a+b}<p<frac{an+a}{a+b}tag1$$
For $displaystyle-1<frac ab<0$,
$$frac ab(n+1)<(1+frac ab)p<frac ab n-1$$
$$impliesfrac{an+a}{a+b}<p<frac{an-b}{a+b}tag2$$
For $displaystylefrac ab<-1$,
$$frac ab n-1>(1+frac ab)p; (frac ab-1)p<frac ab(n+1)$$
$$impliesfrac{an-b}{a+b}<p<frac{an+a}{a-b}tag3$$
The final step is to choose the largest coefficient among the set
$$(C_0=b^n, C_p, C_n=a^n)tag4$$
answered Jan 10 at 17:17
MythomorphicMythomorphic
5,3321733
$endgroup$
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
$begingroup$
Write $$(5+3x)^{10}=sum_{i=0}^{10}a_ix^i.$$ So that
$$
a_i=binom{10}{i}3^i5^{10-i}.
$$ Then $$f(i):=a_i/a_{i+1}=frac{5}{3}cdot frac{i+1}{10-i}.$$ This is an increasing function of $i$ (for $0leq i<10$). The maximum of the $a_i$ is attained at the smallest $i$ for which $f(i)>1$. Solving $f(i)=1$ yields $i=25/8$, which is larger than $3$ but smaller than $4$. Therefore $a_4$ is the largest coefficient.
answered Jan 10 at 15:08
UncountableUncountable
3,040414
$endgroup$
add a comment |
$begingroup$
Write $$(5+3x)^{10}=sum_{i=0}^{10}a_ix^i.$$ So that
$$
a_i=binom{10}{i}3^i5^{10-i}.
$$ Then $$f(i):=a_i/a_{i+1}=frac{5}{3}cdot frac{i+1}{10-i}.$$ This is an increasing function of $i$ (for $0leq i<10$). The maximum of the $a_i$ is attained at the smallest $i$ for which $f(i)>1$. Solving $f(i)=1$ yields $i=25/8$, which is larger than $3$ but smaller than $4$. Therefore $a_4$ is the largest coefficient.
answered Jan 10 at 15:08
UncountableUncountable
3,040414
$endgroup$
add a comment |
$begingroup$
Write $$(5+3x)^{10}=sum_{i=0}^{10}a_ix^i.$$ So that
$$
a_i=binom{10}{i}3^i5^{10-i}.
$$ Then $$f(i):=a_i/a_{i+1}=frac{5}{3}cdot frac{i+1}{10-i}.$$ This is an increasing function of $i$ (for $0leq i<10$). The maximum of the $a_i$ is attained at the smallest $i$ for which $f(i)>1$. Solving $f(i)=1$ yields $i=25/8$, which is larger than $3$ but smaller than $4$. Therefore $a_4$ is the largest coefficient.
answered Jan 10 at 15:08
UncountableUncountable
3,040414
$endgroup$
Write $$(5+3x)^{10}=sum_{i=0}^{10}a_ix^i.$$ So that
$$
a_i=binom{10}{i}3^i5^{10-i}.
$$ Then $$f(i):=a_i/a_{i+1}=frac{5}{3}cdot frac{i+1}{10-i}.$$ This is an increasing function of $i$ (for $0leq i<10$). The maximum of the $a_i$ is attained at the smallest $i$ for which $f(i)>1$. Solving $f(i)=1$ yields $i=25/8$, which is larger than $3$ but smaller than $4$. Therefore $a_4$ is the largest coefficient.
answered Jan 10 at 15:08
UncountableUncountable
3,040414
edited Jan 10 at 15:19
answered Jan 10 at 15:08
UncountableUncountable
3,040414
answered Jan 10 at 15:08
UncountableUncountable
3,040414
answered Jan 10 at 15:08
UncountableUncountable
3,040414
3,040414
add a comment |
add a comment |
$begingroup$
The coefficients have the form
$$dbinom{10}k5^k3^{10-k}$$
The ratio of two consecutive coefficients is:
$$frac{dbinom{10}k5^k3^{10-k}}{dbinom{10}{k+1}5^{k+1}3^{9-k}}=frac{3(k+1)}{5(10-k)}$$
This ratio is lesser than $1$ when
$$3k+3<50-5k$$
That is, when $kle 5$. This means that the maximum coefficient is $dbinom{10}65^6cdot3^4$.
answered Jan 10 at 15:09
ajotatxeajotatxe
53.8k23890
$endgroup$
$begingroup$
According to OP question, it would be more logical to define $k$ as the "power of $x$": $a_k x^k$
$endgroup$
– Damien
Jan 10 at 15:14
add a comment |
$begingroup$
The coefficients have the form
$$dbinom{10}k5^k3^{10-k}$$
The ratio of two consecutive coefficients is:
$$frac{dbinom{10}k5^k3^{10-k}}{dbinom{10}{k+1}5^{k+1}3^{9-k}}=frac{3(k+1)}{5(10-k)}$$
This ratio is lesser than $1$ when
$$3k+3<50-5k$$
That is, when $kle 5$. This means that the maximum coefficient is $dbinom{10}65^6cdot3^4$.
answered Jan 10 at 15:09
ajotatxeajotatxe
53.8k23890
$endgroup$
$begingroup$
According to OP question, it would be more logical to define $k$ as the "power of $x$": $a_k x^k$
$endgroup$
– Damien
Jan 10 at 15:14
add a comment |
$begingroup$
The coefficients have the form
$$dbinom{10}k5^k3^{10-k}$$
The ratio of two consecutive coefficients is:
$$frac{dbinom{10}k5^k3^{10-k}}{dbinom{10}{k+1}5^{k+1}3^{9-k}}=frac{3(k+1)}{5(10-k)}$$
This ratio is lesser than $1$ when
$$3k+3<50-5k$$
That is, when $kle 5$. This means that the maximum coefficient is $dbinom{10}65^6cdot3^4$.
answered Jan 10 at 15:09
ajotatxeajotatxe
53.8k23890
$endgroup$
The coefficients have the form
$$dbinom{10}k5^k3^{10-k}$$
The ratio of two consecutive coefficients is:
$$frac{dbinom{10}k5^k3^{10-k}}{dbinom{10}{k+1}5^{k+1}3^{9-k}}=frac{3(k+1)}{5(10-k)}$$
This ratio is lesser than $1$ when
$$3k+3<50-5k$$
That is, when $kle 5$. This means that the maximum coefficient is $dbinom{10}65^6cdot3^4$.
answered Jan 10 at 15:09
ajotatxeajotatxe
53.8k23890
edited Jan 10 at 15:19
answered Jan 10 at 15:09
ajotatxeajotatxe
53.8k23890
answered Jan 10 at 15:09
ajotatxeajotatxe
53.8k23890
answered Jan 10 at 15:09
ajotatxeajotatxe
53.8k23890
53.8k23890
$begingroup$
According to OP question, it would be more logical to define $k$ as the "power of $x$": $a_k x^k$
$endgroup$
– Damien
Jan 10 at 15:14
add a comment |
$begingroup$
According to OP question, it would be more logical to define $k$ as the "power of $x$": $a_k x^k$
$endgroup$
– Damien
Jan 10 at 15:14
$begingroup$
According to OP question, it would be more logical to define $k$ as the "power of $x$": $a_k x^k$
$endgroup$
– Damien
Jan 10 at 15:14
$begingroup$
According to OP question, it would be more logical to define $k$ as the "power of $x$": $a_k x^k$
$endgroup$
– Damien
Jan 10 at 15:14
add a comment |
$begingroup$
With the result proved below, the largest coefficient is among the set $(a^n,b^n,C_p)$, where $C_p$ is the largest coefficient as $pin[1,n-1]$.
In this case, $a=5, b=3, n=10$, $displaystylefrac ab>0$. So by $(1)$,
$$frac{an-b}{a+b}<p<frac{an+a}{a+b}impliesfrac{47}{8}<p<frac{55}{8}implies p=6$$
$$C_p={10choose6}5^63^4>5^{10}>3^{10}$$
So the largest coefficient occurs at $k=6$.
Proof
By Binomial Formula,
$$(ax+by)^n=sum_{k=0}^n {nchoose k}(ax)^k (by)^{n-k}$$
The coefficients can be expressed as
$$C_k={nchoose k}a^k b^{n-k}$$
Consider for all $k<n$, one may prove that
$$frac{C_{k+1}}{C_k}=frac abcdotfrac{n-k}{k+1}$$
This ratio, $widetilde C$, indicates if $|C_{k+1}|>|C_k|$. This is true if $|widetilde C|>1$.
Now if there exists an integer $pin[1,n-1]$ such that $C_{p}>0$ is maximum, then
$$left|frac{C_{p}}{C_{p-1}}right|>1, left|frac{C_{p+1}}{C_{p}}right|<1$$
For $displaystylefrac ab>0$,
$$frac ab n-1<(1+frac ab)p<frac ab(n+1)$$
$$impliesfrac{an-b}{a+b}<p<frac{an+a}{a+b}tag1$$
For $displaystyle-1<frac ab<0$,
$$frac ab(n+1)<(1+frac ab)p<frac ab n-1$$
$$impliesfrac{an+a}{a+b}<p<frac{an-b}{a+b}tag2$$
For $displaystylefrac ab<-1$,
$$frac ab n-1>(1+frac ab)p; (frac ab-1)p<frac ab(n+1)$$
$$impliesfrac{an-b}{a+b}<p<frac{an+a}{a-b}tag3$$
The final step is to choose the largest coefficient among the set
$$(C_0=b^n, C_p, C_n=a^n)tag4$$
answered Jan 10 at 17:17
MythomorphicMythomorphic
5,3321733
$endgroup$
add a comment |
$begingroup$
With the result proved below, the largest coefficient is among the set $(a^n,b^n,C_p)$, where $C_p$ is the largest coefficient as $pin[1,n-1]$.
In this case, $a=5, b=3, n=10$, $displaystylefrac ab>0$. So by $(1)$,
$$frac{an-b}{a+b}<p<frac{an+a}{a+b}impliesfrac{47}{8}<p<frac{55}{8}implies p=6$$
$$C_p={10choose6}5^63^4>5^{10}>3^{10}$$
So the largest coefficient occurs at $k=6$.
Proof
By Binomial Formula,
$$(ax+by)^n=sum_{k=0}^n {nchoose k}(ax)^k (by)^{n-k}$$
The coefficients can be expressed as
$$C_k={nchoose k}a^k b^{n-k}$$
Consider for all $k<n$, one may prove that
$$frac{C_{k+1}}{C_k}=frac abcdotfrac{n-k}{k+1}$$
This ratio, $widetilde C$, indicates if $|C_{k+1}|>|C_k|$. This is true if $|widetilde C|>1$.
Now if there exists an integer $pin[1,n-1]$ such that $C_{p}>0$ is maximum, then
$$left|frac{C_{p}}{C_{p-1}}right|>1, left|frac{C_{p+1}}{C_{p}}right|<1$$
For $displaystylefrac ab>0$,
$$frac ab n-1<(1+frac ab)p<frac ab(n+1)$$
$$impliesfrac{an-b}{a+b}<p<frac{an+a}{a+b}tag1$$
For $displaystyle-1<frac ab<0$,
$$frac ab(n+1)<(1+frac ab)p<frac ab n-1$$
$$impliesfrac{an+a}{a+b}<p<frac{an-b}{a+b}tag2$$
For $displaystylefrac ab<-1$,
$$frac ab n-1>(1+frac ab)p; (frac ab-1)p<frac ab(n+1)$$
$$impliesfrac{an-b}{a+b}<p<frac{an+a}{a-b}tag3$$
The final step is to choose the largest coefficient among the set
$$(C_0=b^n, C_p, C_n=a^n)tag4$$
answered Jan 10 at 17:17
MythomorphicMythomorphic
5,3321733
$endgroup$
add a comment |
$begingroup$
With the result proved below, the largest coefficient is among the set $(a^n,b^n,C_p)$, where $C_p$ is the largest coefficient as $pin[1,n-1]$.
In this case, $a=5, b=3, n=10$, $displaystylefrac ab>0$. So by $(1)$,
$$frac{an-b}{a+b}<p<frac{an+a}{a+b}impliesfrac{47}{8}<p<frac{55}{8}implies p=6$$
$$C_p={10choose6}5^63^4>5^{10}>3^{10}$$
So the largest coefficient occurs at $k=6$.
Proof
By Binomial Formula,
$$(ax+by)^n=sum_{k=0}^n {nchoose k}(ax)^k (by)^{n-k}$$
The coefficients can be expressed as
$$C_k={nchoose k}a^k b^{n-k}$$
Consider for all $k<n$, one may prove that
$$frac{C_{k+1}}{C_k}=frac abcdotfrac{n-k}{k+1}$$
This ratio, $widetilde C$, indicates if $|C_{k+1}|>|C_k|$. This is true if $|widetilde C|>1$.
Now if there exists an integer $pin[1,n-1]$ such that $C_{p}>0$ is maximum, then
$$left|frac{C_{p}}{C_{p-1}}right|>1, left|frac{C_{p+1}}{C_{p}}right|<1$$
For $displaystylefrac ab>0$,
$$frac ab n-1<(1+frac ab)p<frac ab(n+1)$$
$$impliesfrac{an-b}{a+b}<p<frac{an+a}{a+b}tag1$$
For $displaystyle-1<frac ab<0$,
$$frac ab(n+1)<(1+frac ab)p<frac ab n-1$$
$$impliesfrac{an+a}{a+b}<p<frac{an-b}{a+b}tag2$$
For $displaystylefrac ab<-1$,
$$frac ab n-1>(1+frac ab)p; (frac ab-1)p<frac ab(n+1)$$
$$impliesfrac{an-b}{a+b}<p<frac{an+a}{a-b}tag3$$
The final step is to choose the largest coefficient among the set
$$(C_0=b^n, C_p, C_n=a^n)tag4$$
answered Jan 10 at 17:17
MythomorphicMythomorphic
5,3321733
$endgroup$
With the result proved below, the largest coefficient is among the set $(a^n,b^n,C_p)$, where $C_p$ is the largest coefficient as $pin[1,n-1]$.
In this case, $a=5, b=3, n=10$, $displaystylefrac ab>0$. So by $(1)$,
$$frac{an-b}{a+b}<p<frac{an+a}{a+b}impliesfrac{47}{8}<p<frac{55}{8}implies p=6$$
$$C_p={10choose6}5^63^4>5^{10}>3^{10}$$
So the largest coefficient occurs at $k=6$.
Proof
By Binomial Formula,
$$(ax+by)^n=sum_{k=0}^n {nchoose k}(ax)^k (by)^{n-k}$$
The coefficients can be expressed as
$$C_k={nchoose k}a^k b^{n-k}$$
Consider for all $k<n$, one may prove that
$$frac{C_{k+1}}{C_k}=frac abcdotfrac{n-k}{k+1}$$
This ratio, $widetilde C$, indicates if $|C_{k+1}|>|C_k|$. This is true if $|widetilde C|>1$.
Now if there exists an integer $pin[1,n-1]$ such that $C_{p}>0$ is maximum, then
$$left|frac{C_{p}}{C_{p-1}}right|>1, left|frac{C_{p+1}}{C_{p}}right|<1$$
For $displaystylefrac ab>0$,
$$frac ab n-1<(1+frac ab)p<frac ab(n+1)$$
$$impliesfrac{an-b}{a+b}<p<frac{an+a}{a+b}tag1$$
For $displaystyle-1<frac ab<0$,
$$frac ab(n+1)<(1+frac ab)p<frac ab n-1$$
$$impliesfrac{an+a}{a+b}<p<frac{an-b}{a+b}tag2$$
For $displaystylefrac ab<-1$,
$$frac ab n-1>(1+frac ab)p; (frac ab-1)p<frac ab(n+1)$$
$$impliesfrac{an-b}{a+b}<p<frac{an+a}{a-b}tag3$$
The final step is to choose the largest coefficient among the set
$$(C_0=b^n, C_p, C_n=a^n)tag4$$
answered Jan 10 at 17:17
MythomorphicMythomorphic
5,3321733
answered Jan 10 at 17:17
MythomorphicMythomorphic
5,3321733
answered Jan 10 at 17:17
MythomorphicMythomorphic
5,3321733
answered Jan 10 at 17:17
MythomorphicMythomorphic
5,3321733
5,3321733
add a comment |
add a comment |
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2
$begingroup$
If you divide a general coefficient by its successor you should get a simpler expression which will tell you which of them is larger.
$endgroup$
– Mark Bennet
Jan 10 at 14:54
$begingroup$
math.stackexchange.com/questions/722952/…
$endgroup$
– lab bhattacharjee
Jan 10 at 15:10