Mixing
Largest coefficient in the power of a polynomial
How do I find the largest coefficient of any power of $x$ in an expansion such as $(1 + 2x + 2x^2)^n$, as a function of $n$?
In the case $(1+x)^n$, we know that the central coefficients are the largest, and for $(1+ax)^n$ I can take ratios of consecutive terms to find the largest one, but these methods fail in the case mentioned above.
Also, can we find the degree of the term for which this largest coefficient occurs, again as a function of $n$?
combinatorics algebra-precalculus power-series
asked Dec 30 '18 at 19:38
Sagnik BhattacharyaSagnik Bhattacharya
285
add a comment |
How do I find the largest coefficient of any power of $x$ in an expansion such as $(1 + 2x + 2x^2)^n$, as a function of $n$?
In the case $(1+x)^n$, we know that the central coefficients are the largest, and for $(1+ax)^n$ I can take ratios of consecutive terms to find the largest one, but these methods fail in the case mentioned above.
Also, can we find the degree of the term for which this largest coefficient occurs, again as a function of $n$?
combinatorics algebra-precalculus power-series
asked Dec 30 '18 at 19:38
Sagnik BhattacharyaSagnik Bhattacharya
285
For the example you have given try $((1+x)^2+x^2)^n$ from which you can compute the coefficients.
– Mark Bennet
Dec 30 '18 at 19:42
add a comment |
How do I find the largest coefficient of any power of $x$ in an expansion such as $(1 + 2x + 2x^2)^n$, as a function of $n$?
In the case $(1+x)^n$, we know that the central coefficients are the largest, and for $(1+ax)^n$ I can take ratios of consecutive terms to find the largest one, but these methods fail in the case mentioned above.
Also, can we find the degree of the term for which this largest coefficient occurs, again as a function of $n$?
combinatorics algebra-precalculus power-series
asked Dec 30 '18 at 19:38
Sagnik BhattacharyaSagnik Bhattacharya
285
How do I find the largest coefficient of any power of $x$ in an expansion such as $(1 + 2x + 2x^2)^n$, as a function of $n$?
In the case $(1+x)^n$, we know that the central coefficients are the largest, and for $(1+ax)^n$ I can take ratios of consecutive terms to find the largest one, but these methods fail in the case mentioned above.
Also, can we find the degree of the term for which this largest coefficient occurs, again as a function of $n$?
combinatorics algebra-precalculus power-series
combinatorics algebra-precalculus power-series
asked Dec 30 '18 at 19:38
Sagnik BhattacharyaSagnik Bhattacharya
285
asked Dec 30 '18 at 19:38
Sagnik BhattacharyaSagnik Bhattacharya
285
edited Dec 30 '18 at 20:09
Sagnik Bhattacharya
asked Dec 30 '18 at 19:38
Sagnik BhattacharyaSagnik Bhattacharya
285
asked Dec 30 '18 at 19:38
Sagnik BhattacharyaSagnik Bhattacharya
285
asked Dec 30 '18 at 19:38
Sagnik BhattacharyaSagnik Bhattacharya
285
285
For the example you have given try $((1+x)^2+x^2)^n$ from which you can compute the coefficients.
– Mark Bennet
Dec 30 '18 at 19:42
add a comment |
For the example you have given try $((1+x)^2+x^2)^n$ from which you can compute the coefficients.
– Mark Bennet
Dec 30 '18 at 19:42
For the example you have given try $((1+x)^2+x^2)^n$ from which you can compute the coefficients.
– Mark Bennet
Dec 30 '18 at 19:42
For the example you have given try $((1+x)^2+x^2)^n$ from which you can compute the coefficients.
– Mark Bennet
Dec 30 '18 at 19:42
add a comment |
2 Answers
2
active
oldest
votes
- Write
$$left(frac{1}{5} + frac{2x}{5} + frac{2x^2}{5}right)^n = sum_{l=0}^{2n} a_lx^l.$$
- Let $x_1,x_2,ldots, x_n$ be iids where each $x_i$ is 0 with probability $frac{1}{5}$; 1 with probability $frac{2}{5}$; and 2 with probability $frac{2}{5}$. Then for each $l$, note the following:
$${bf{P}}left[left(sum_{i=1}^n x_i right) = l right] = a_l$$
It turns out that (Chernoff bounds) that the value of $l$ that maximizes $a_l={bf{P}}left[left(sum_{i=1}^n x_i right) = l right]$ is $l approx {bf{E}}left[sum_{i=1}^n x_i right]$ which is $l approx frac{6n}{5}$.
So putting the above together, writing $left(frac{1}{5} + frac{2x}{5} + frac{2x^2}{5}right)^n$ as $sum_{l=0}^{2n} a_lx^l$, the value of $a_l$ such that $a_l$ is the largest is $l approx frac{6n}{5}$ and for such $l$, the coefficient $a_l$ has value $theta left(frac{1}{sqrt{n}}right)$.
So writing
$$(1+2x+2x^2) = sum_{l=0}^{2n} b_lx^l,$$
the value of $l$ that maximzes $b_l$ is $lapprox frac{6n}{5}$, and $b_l$ has value $theta(frac{5^n}{sqrt{n}})$.
answered Dec 31 '18 at 18:46
MikeMike
3,208311
add a comment |
For something like $(1+2x+2x^2)^n$, you can say that the expected power of $x$ for each factor is $frac 65$, so you would expect the maximum to be at $frac 65n$. For $n=100$ this would be the $x^{120}$ term and it truly is per Alpha. You have to click more terms a bunch to see this. It is basically using the normal approximation to the binomial distribution.
answered Dec 30 '18 at 20:32
Ross MillikanRoss Millikan
292k23197371
Thank you for your answer. I went and used Python to plot out the dependence on $n$ of the index of the largest coefficient, and it indeed is linear with the exact coefficient that you calculated. In fact, the pattern holds for other polynomials too. I will try to come up with a formal proof using the central limit theorem.
– Sagnik Bhattacharya
Dec 31 '18 at 16:16
For small powers the pattern can be different. $(1+x+x^8)^n$ has an expected power of $3$, but for small powers the maximum coefficient will be about $frac n2$. As $n$ gets larger the $3n$ will take over. Also there can be missing powers, like if you have $(1+x^3)^n$ will only have powers that are multiples of $3$.
– Ross Millikan
Dec 31 '18 at 16:55
It should be $l=7n/5$ (and not $l=6n/5$) though, and I am not sure what you mean by "expected power of $x$"; I would rather the reasoning be fleshed out more.
– Mike
Dec 31 '18 at 18:49
My mistake, it should be $l=frac{6n}{5}$ just as you said. But it still would be nice to see the reasoning fleshed out wrong.
– Mike
Dec 31 '18 at 18:58
*fleshed out more
– Mike
Jan 1 at 0:05
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
- Write
$$left(frac{1}{5} + frac{2x}{5} + frac{2x^2}{5}right)^n = sum_{l=0}^{2n} a_lx^l.$$
- Let $x_1,x_2,ldots, x_n$ be iids where each $x_i$ is 0 with probability $frac{1}{5}$; 1 with probability $frac{2}{5}$; and 2 with probability $frac{2}{5}$. Then for each $l$, note the following:
$${bf{P}}left[left(sum_{i=1}^n x_i right) = l right] = a_l$$
It turns out that (Chernoff bounds) that the value of $l$ that maximizes $a_l={bf{P}}left[left(sum_{i=1}^n x_i right) = l right]$ is $l approx {bf{E}}left[sum_{i=1}^n x_i right]$ which is $l approx frac{6n}{5}$.
So putting the above together, writing $left(frac{1}{5} + frac{2x}{5} + frac{2x^2}{5}right)^n$ as $sum_{l=0}^{2n} a_lx^l$, the value of $a_l$ such that $a_l$ is the largest is $l approx frac{6n}{5}$ and for such $l$, the coefficient $a_l$ has value $theta left(frac{1}{sqrt{n}}right)$.
So writing
$$(1+2x+2x^2) = sum_{l=0}^{2n} b_lx^l,$$
the value of $l$ that maximzes $b_l$ is $lapprox frac{6n}{5}$, and $b_l$ has value $theta(frac{5^n}{sqrt{n}})$.
answered Dec 31 '18 at 18:46
MikeMike
3,208311
add a comment |
- Write
$$left(frac{1}{5} + frac{2x}{5} + frac{2x^2}{5}right)^n = sum_{l=0}^{2n} a_lx^l.$$
- Let $x_1,x_2,ldots, x_n$ be iids where each $x_i$ is 0 with probability $frac{1}{5}$; 1 with probability $frac{2}{5}$; and 2 with probability $frac{2}{5}$. Then for each $l$, note the following:
$${bf{P}}left[left(sum_{i=1}^n x_i right) = l right] = a_l$$
It turns out that (Chernoff bounds) that the value of $l$ that maximizes $a_l={bf{P}}left[left(sum_{i=1}^n x_i right) = l right]$ is $l approx {bf{E}}left[sum_{i=1}^n x_i right]$ which is $l approx frac{6n}{5}$.
So putting the above together, writing $left(frac{1}{5} + frac{2x}{5} + frac{2x^2}{5}right)^n$ as $sum_{l=0}^{2n} a_lx^l$, the value of $a_l$ such that $a_l$ is the largest is $l approx frac{6n}{5}$ and for such $l$, the coefficient $a_l$ has value $theta left(frac{1}{sqrt{n}}right)$.
So writing
$$(1+2x+2x^2) = sum_{l=0}^{2n} b_lx^l,$$
the value of $l$ that maximzes $b_l$ is $lapprox frac{6n}{5}$, and $b_l$ has value $theta(frac{5^n}{sqrt{n}})$.
answered Dec 31 '18 at 18:46
MikeMike
3,208311
add a comment |
- Write
$$left(frac{1}{5} + frac{2x}{5} + frac{2x^2}{5}right)^n = sum_{l=0}^{2n} a_lx^l.$$
- Let $x_1,x_2,ldots, x_n$ be iids where each $x_i$ is 0 with probability $frac{1}{5}$; 1 with probability $frac{2}{5}$; and 2 with probability $frac{2}{5}$. Then for each $l$, note the following:
$${bf{P}}left[left(sum_{i=1}^n x_i right) = l right] = a_l$$
It turns out that (Chernoff bounds) that the value of $l$ that maximizes $a_l={bf{P}}left[left(sum_{i=1}^n x_i right) = l right]$ is $l approx {bf{E}}left[sum_{i=1}^n x_i right]$ which is $l approx frac{6n}{5}$.
So putting the above together, writing $left(frac{1}{5} + frac{2x}{5} + frac{2x^2}{5}right)^n$ as $sum_{l=0}^{2n} a_lx^l$, the value of $a_l$ such that $a_l$ is the largest is $l approx frac{6n}{5}$ and for such $l$, the coefficient $a_l$ has value $theta left(frac{1}{sqrt{n}}right)$.
So writing
$$(1+2x+2x^2) = sum_{l=0}^{2n} b_lx^l,$$
the value of $l$ that maximzes $b_l$ is $lapprox frac{6n}{5}$, and $b_l$ has value $theta(frac{5^n}{sqrt{n}})$.
answered Dec 31 '18 at 18:46
MikeMike
3,208311
- Write
$$left(frac{1}{5} + frac{2x}{5} + frac{2x^2}{5}right)^n = sum_{l=0}^{2n} a_lx^l.$$
- Let $x_1,x_2,ldots, x_n$ be iids where each $x_i$ is 0 with probability $frac{1}{5}$; 1 with probability $frac{2}{5}$; and 2 with probability $frac{2}{5}$. Then for each $l$, note the following:
$${bf{P}}left[left(sum_{i=1}^n x_i right) = l right] = a_l$$
It turns out that (Chernoff bounds) that the value of $l$ that maximizes $a_l={bf{P}}left[left(sum_{i=1}^n x_i right) = l right]$ is $l approx {bf{E}}left[sum_{i=1}^n x_i right]$ which is $l approx frac{6n}{5}$.
So putting the above together, writing $left(frac{1}{5} + frac{2x}{5} + frac{2x^2}{5}right)^n$ as $sum_{l=0}^{2n} a_lx^l$, the value of $a_l$ such that $a_l$ is the largest is $l approx frac{6n}{5}$ and for such $l$, the coefficient $a_l$ has value $theta left(frac{1}{sqrt{n}}right)$.
So writing
$$(1+2x+2x^2) = sum_{l=0}^{2n} b_lx^l,$$
the value of $l$ that maximzes $b_l$ is $lapprox frac{6n}{5}$, and $b_l$ has value $theta(frac{5^n}{sqrt{n}})$.
answered Dec 31 '18 at 18:46
MikeMike
3,208311
edited Jan 1 at 3:44
answered Dec 31 '18 at 18:46
MikeMike
3,208311
answered Dec 31 '18 at 18:46
MikeMike
3,208311
answered Dec 31 '18 at 18:46
MikeMike
3,208311
3,208311
add a comment |
add a comment |
For something like $(1+2x+2x^2)^n$, you can say that the expected power of $x$ for each factor is $frac 65$, so you would expect the maximum to be at $frac 65n$. For $n=100$ this would be the $x^{120}$ term and it truly is per Alpha. You have to click more terms a bunch to see this. It is basically using the normal approximation to the binomial distribution.
answered Dec 30 '18 at 20:32
Ross MillikanRoss Millikan
292k23197371
Thank you for your answer. I went and used Python to plot out the dependence on $n$ of the index of the largest coefficient, and it indeed is linear with the exact coefficient that you calculated. In fact, the pattern holds for other polynomials too. I will try to come up with a formal proof using the central limit theorem.
– Sagnik Bhattacharya
Dec 31 '18 at 16:16
For small powers the pattern can be different. $(1+x+x^8)^n$ has an expected power of $3$, but for small powers the maximum coefficient will be about $frac n2$. As $n$ gets larger the $3n$ will take over. Also there can be missing powers, like if you have $(1+x^3)^n$ will only have powers that are multiples of $3$.
– Ross Millikan
Dec 31 '18 at 16:55
It should be $l=7n/5$ (and not $l=6n/5$) though, and I am not sure what you mean by "expected power of $x$"; I would rather the reasoning be fleshed out more.
– Mike
Dec 31 '18 at 18:49
My mistake, it should be $l=frac{6n}{5}$ just as you said. But it still would be nice to see the reasoning fleshed out wrong.
– Mike
Dec 31 '18 at 18:58
*fleshed out more
– Mike
Jan 1 at 0:05
add a comment |
For something like $(1+2x+2x^2)^n$, you can say that the expected power of $x$ for each factor is $frac 65$, so you would expect the maximum to be at $frac 65n$. For $n=100$ this would be the $x^{120}$ term and it truly is per Alpha. You have to click more terms a bunch to see this. It is basically using the normal approximation to the binomial distribution.
answered Dec 30 '18 at 20:32
Ross MillikanRoss Millikan
292k23197371
Thank you for your answer. I went and used Python to plot out the dependence on $n$ of the index of the largest coefficient, and it indeed is linear with the exact coefficient that you calculated. In fact, the pattern holds for other polynomials too. I will try to come up with a formal proof using the central limit theorem.
– Sagnik Bhattacharya
Dec 31 '18 at 16:16
For small powers the pattern can be different. $(1+x+x^8)^n$ has an expected power of $3$, but for small powers the maximum coefficient will be about $frac n2$. As $n$ gets larger the $3n$ will take over. Also there can be missing powers, like if you have $(1+x^3)^n$ will only have powers that are multiples of $3$.
– Ross Millikan
Dec 31 '18 at 16:55
It should be $l=7n/5$ (and not $l=6n/5$) though, and I am not sure what you mean by "expected power of $x$"; I would rather the reasoning be fleshed out more.
– Mike
Dec 31 '18 at 18:49
My mistake, it should be $l=frac{6n}{5}$ just as you said. But it still would be nice to see the reasoning fleshed out wrong.
– Mike
Dec 31 '18 at 18:58
*fleshed out more
– Mike
Jan 1 at 0:05
add a comment |
For something like $(1+2x+2x^2)^n$, you can say that the expected power of $x$ for each factor is $frac 65$, so you would expect the maximum to be at $frac 65n$. For $n=100$ this would be the $x^{120}$ term and it truly is per Alpha. You have to click more terms a bunch to see this. It is basically using the normal approximation to the binomial distribution.
answered Dec 30 '18 at 20:32
Ross MillikanRoss Millikan
292k23197371
For something like $(1+2x+2x^2)^n$, you can say that the expected power of $x$ for each factor is $frac 65$, so you would expect the maximum to be at $frac 65n$. For $n=100$ this would be the $x^{120}$ term and it truly is per Alpha. You have to click more terms a bunch to see this. It is basically using the normal approximation to the binomial distribution.
answered Dec 30 '18 at 20:32
Ross MillikanRoss Millikan
292k23197371
answered Dec 30 '18 at 20:32
Ross MillikanRoss Millikan
292k23197371
answered Dec 30 '18 at 20:32
Ross MillikanRoss Millikan
292k23197371
answered Dec 30 '18 at 20:32
Ross MillikanRoss Millikan
292k23197371
292k23197371
Thank you for your answer. I went and used Python to plot out the dependence on $n$ of the index of the largest coefficient, and it indeed is linear with the exact coefficient that you calculated. In fact, the pattern holds for other polynomials too. I will try to come up with a formal proof using the central limit theorem.
– Sagnik Bhattacharya
Dec 31 '18 at 16:16
For small powers the pattern can be different. $(1+x+x^8)^n$ has an expected power of $3$, but for small powers the maximum coefficient will be about $frac n2$. As $n$ gets larger the $3n$ will take over. Also there can be missing powers, like if you have $(1+x^3)^n$ will only have powers that are multiples of $3$.
– Ross Millikan
Dec 31 '18 at 16:55
It should be $l=7n/5$ (and not $l=6n/5$) though, and I am not sure what you mean by "expected power of $x$"; I would rather the reasoning be fleshed out more.
– Mike
Dec 31 '18 at 18:49
My mistake, it should be $l=frac{6n}{5}$ just as you said. But it still would be nice to see the reasoning fleshed out wrong.
– Mike
Dec 31 '18 at 18:58
*fleshed out more
– Mike
Jan 1 at 0:05
add a comment |
Thank you for your answer. I went and used Python to plot out the dependence on $n$ of the index of the largest coefficient, and it indeed is linear with the exact coefficient that you calculated. In fact, the pattern holds for other polynomials too. I will try to come up with a formal proof using the central limit theorem.
– Sagnik Bhattacharya
Dec 31 '18 at 16:16
For small powers the pattern can be different. $(1+x+x^8)^n$ has an expected power of $3$, but for small powers the maximum coefficient will be about $frac n2$. As $n$ gets larger the $3n$ will take over. Also there can be missing powers, like if you have $(1+x^3)^n$ will only have powers that are multiples of $3$.
– Ross Millikan
Dec 31 '18 at 16:55
It should be $l=7n/5$ (and not $l=6n/5$) though, and I am not sure what you mean by "expected power of $x$"; I would rather the reasoning be fleshed out more.
– Mike
Dec 31 '18 at 18:49
My mistake, it should be $l=frac{6n}{5}$ just as you said. But it still would be nice to see the reasoning fleshed out wrong.
– Mike
Dec 31 '18 at 18:58
*fleshed out more
– Mike
Jan 1 at 0:05
Thank you for your answer. I went and used Python to plot out the dependence on $n$ of the index of the largest coefficient, and it indeed is linear with the exact coefficient that you calculated. In fact, the pattern holds for other polynomials too. I will try to come up with a formal proof using the central limit theorem.
– Sagnik Bhattacharya
Dec 31 '18 at 16:16
Thank you for your answer. I went and used Python to plot out the dependence on $n$ of the index of the largest coefficient, and it indeed is linear with the exact coefficient that you calculated. In fact, the pattern holds for other polynomials too. I will try to come up with a formal proof using the central limit theorem.
– Sagnik Bhattacharya
Dec 31 '18 at 16:16
For small powers the pattern can be different. $(1+x+x^8)^n$ has an expected power of $3$, but for small powers the maximum coefficient will be about $frac n2$. As $n$ gets larger the $3n$ will take over. Also there can be missing powers, like if you have $(1+x^3)^n$ will only have powers that are multiples of $3$.
– Ross Millikan
Dec 31 '18 at 16:55
For small powers the pattern can be different. $(1+x+x^8)^n$ has an expected power of $3$, but for small powers the maximum coefficient will be about $frac n2$. As $n$ gets larger the $3n$ will take over. Also there can be missing powers, like if you have $(1+x^3)^n$ will only have powers that are multiples of $3$.
– Ross Millikan
Dec 31 '18 at 16:55
It should be $l=7n/5$ (and not $l=6n/5$) though, and I am not sure what you mean by "expected power of $x$"; I would rather the reasoning be fleshed out more.
– Mike
Dec 31 '18 at 18:49
It should be $l=7n/5$ (and not $l=6n/5$) though, and I am not sure what you mean by "expected power of $x$"; I would rather the reasoning be fleshed out more.
– Mike
Dec 31 '18 at 18:49
My mistake, it should be $l=frac{6n}{5}$ just as you said. But it still would be nice to see the reasoning fleshed out wrong.
– Mike
Dec 31 '18 at 18:58
My mistake, it should be $l=frac{6n}{5}$ just as you said. But it still would be nice to see the reasoning fleshed out wrong.
– Mike
Dec 31 '18 at 18:58
*fleshed out more
– Mike
Jan 1 at 0:05
*fleshed out more
– Mike
Jan 1 at 0:05
add a comment |
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For the example you have given try $((1+x)^2+x^2)^n$ from which you can compute the coefficients.
– Mark Bennet
Dec 30 '18 at 19:42