Mixing
Develop Spread Polynomials Recursively
$begingroup$
I have the following recursive formula:
$$ S_0 = 0$$
$$ S_1 = alpha$$
$$ S_{n+1} = 2(1 - 2alpha)S_n - S_{n - 1} + 2alpha$$
So $S_2$ is given by:
$$S_{1 + 1} = 2(1 - 2alpha)S_1 - S_{1 - 1} + 2alpha$$
Then:
$$S_2 = (2 - 4alpha)alpha - 0 + 2alpha = 4alpha(1 - alpha)$$
And we are tasked to show that:
$$S_2(S_2(x)) = S_4(x)$$
I tried:
$$S_2(2(1 - 2alpha)S_1 - S_{1 - 1} + 2alpha) = (4alpha - 4alpha^2)^2 = 16alpha^2 - 32alpha^3 + 16alpha^4$$
Which is not equal to:
$$S_4 = 16alpha(1 - alpha)(1 - 2alpha)^2 = 16alpha(1 - 5alpha + 8alpha^2 - 4alpha^3)$$
Any suggestion or pointers in the direction that I should be focusing my attention. My feeling is that I am staring at the solution but its subtlety is just our of my reach.
linear-algebra
asked Jan 27 at 2:52
hlogomahlogoma
112
$endgroup$
add a comment |
$begingroup$
I have the following recursive formula:
$$ S_0 = 0$$
$$ S_1 = alpha$$
$$ S_{n+1} = 2(1 - 2alpha)S_n - S_{n - 1} + 2alpha$$
So $S_2$ is given by:
$$S_{1 + 1} = 2(1 - 2alpha)S_1 - S_{1 - 1} + 2alpha$$
Then:
$$S_2 = (2 - 4alpha)alpha - 0 + 2alpha = 4alpha(1 - alpha)$$
And we are tasked to show that:
$$S_2(S_2(x)) = S_4(x)$$
I tried:
$$S_2(2(1 - 2alpha)S_1 - S_{1 - 1} + 2alpha) = (4alpha - 4alpha^2)^2 = 16alpha^2 - 32alpha^3 + 16alpha^4$$
Which is not equal to:
$$S_4 = 16alpha(1 - alpha)(1 - 2alpha)^2 = 16alpha(1 - 5alpha + 8alpha^2 - 4alpha^3)$$
Any suggestion or pointers in the direction that I should be focusing my attention. My feeling is that I am staring at the solution but its subtlety is just our of my reach.
linear-algebra
asked Jan 27 at 2:52
hlogomahlogoma
112
$endgroup$
$begingroup$
There seems to be a mix of $x$s and $alpha$s; it appears that you're using them interchangeably, so I'll use $x$ to save typing. Since you've already shown that $S_2(x)=4x(1-x)$, you can write simply $$S_2(S_2(x)) = 4cdot 4x(1-x)cdot(1-4x(1-x))=16x(1-x)(1-4x+4x^2)=16x(1-x)(1-2x)^2$$ The last expression is, indeed, $S_4(x)$. So, you simply erred somewhere in expanding $S_2(S_2(x))$.
$endgroup$
– Blue
Jan 27 at 4:00
$begingroup$
Thanks you. Your answer highlights the elegance of mathematics and thereby underlines its attraction to many. I marvel but unfortunately find myself still looking in from the outside. I hope you have the time to unpack $4⋅4x(1−x)⋅(1−4x(1−x))$. I see the term $(4x(1-x))^2$ but I do not see where we get the $4$ times quantity nor the term $(1 - 4x(1-x))$ from.
$endgroup$
– hlogoma
Jan 27 at 14:39
$begingroup$
In determining $S_2(S_2(x))$, we simply replace every $x$ in $4x(1-x)$ with $S_2(x)$, which is $4x(1-x)$. $$4cdotcolor{red}{x}cdot(1-color{red}{x}) quadtoquad 4cdotcolor{red}{S_2(x)}cdot (;1-color{red}{S_2(x)};)quadtoquad 4cdotcolor{red}{4x(1-x)}cdot (;1-color{red}{4x(1-x)};)$$
$endgroup$
– Blue
Jan 27 at 19:41
1
$begingroup$
Wow ... I got it now ... thanks for the patience. It is like a double shot of expresso, I need a chaser. Thanks again.
$endgroup$
– hlogoma
Jan 27 at 20:13
add a comment |
$begingroup$
I have the following recursive formula:
$$ S_0 = 0$$
$$ S_1 = alpha$$
$$ S_{n+1} = 2(1 - 2alpha)S_n - S_{n - 1} + 2alpha$$
So $S_2$ is given by:
$$S_{1 + 1} = 2(1 - 2alpha)S_1 - S_{1 - 1} + 2alpha$$
Then:
$$S_2 = (2 - 4alpha)alpha - 0 + 2alpha = 4alpha(1 - alpha)$$
And we are tasked to show that:
$$S_2(S_2(x)) = S_4(x)$$
I tried:
$$S_2(2(1 - 2alpha)S_1 - S_{1 - 1} + 2alpha) = (4alpha - 4alpha^2)^2 = 16alpha^2 - 32alpha^3 + 16alpha^4$$
Which is not equal to:
$$S_4 = 16alpha(1 - alpha)(1 - 2alpha)^2 = 16alpha(1 - 5alpha + 8alpha^2 - 4alpha^3)$$
Any suggestion or pointers in the direction that I should be focusing my attention. My feeling is that I am staring at the solution but its subtlety is just our of my reach.
linear-algebra
asked Jan 27 at 2:52
hlogomahlogoma
112
$endgroup$
I have the following recursive formula:
$$ S_0 = 0$$
$$ S_1 = alpha$$
$$ S_{n+1} = 2(1 - 2alpha)S_n - S_{n - 1} + 2alpha$$
So $S_2$ is given by:
$$S_{1 + 1} = 2(1 - 2alpha)S_1 - S_{1 - 1} + 2alpha$$
Then:
$$S_2 = (2 - 4alpha)alpha - 0 + 2alpha = 4alpha(1 - alpha)$$
And we are tasked to show that:
$$S_2(S_2(x)) = S_4(x)$$
I tried:
$$S_2(2(1 - 2alpha)S_1 - S_{1 - 1} + 2alpha) = (4alpha - 4alpha^2)^2 = 16alpha^2 - 32alpha^3 + 16alpha^4$$
Which is not equal to:
$$S_4 = 16alpha(1 - alpha)(1 - 2alpha)^2 = 16alpha(1 - 5alpha + 8alpha^2 - 4alpha^3)$$
Any suggestion or pointers in the direction that I should be focusing my attention. My feeling is that I am staring at the solution but its subtlety is just our of my reach.
linear-algebra
linear-algebra
asked Jan 27 at 2:52
hlogomahlogoma
112
asked Jan 27 at 2:52
hlogomahlogoma
112
asked Jan 27 at 2:52
hlogomahlogoma
112
asked Jan 27 at 2:52
hlogomahlogoma
112
asked Jan 27 at 2:52
hlogomahlogoma
112
112
$begingroup$
There seems to be a mix of $x$s and $alpha$s; it appears that you're using them interchangeably, so I'll use $x$ to save typing. Since you've already shown that $S_2(x)=4x(1-x)$, you can write simply $$S_2(S_2(x)) = 4cdot 4x(1-x)cdot(1-4x(1-x))=16x(1-x)(1-4x+4x^2)=16x(1-x)(1-2x)^2$$ The last expression is, indeed, $S_4(x)$. So, you simply erred somewhere in expanding $S_2(S_2(x))$.
$endgroup$
– Blue
Jan 27 at 4:00
$begingroup$
Thanks you. Your answer highlights the elegance of mathematics and thereby underlines its attraction to many. I marvel but unfortunately find myself still looking in from the outside. I hope you have the time to unpack $4⋅4x(1−x)⋅(1−4x(1−x))$. I see the term $(4x(1-x))^2$ but I do not see where we get the $4$ times quantity nor the term $(1 - 4x(1-x))$ from.
$endgroup$
– hlogoma
Jan 27 at 14:39
$begingroup$
In determining $S_2(S_2(x))$, we simply replace every $x$ in $4x(1-x)$ with $S_2(x)$, which is $4x(1-x)$. $$4cdotcolor{red}{x}cdot(1-color{red}{x}) quadtoquad 4cdotcolor{red}{S_2(x)}cdot (;1-color{red}{S_2(x)};)quadtoquad 4cdotcolor{red}{4x(1-x)}cdot (;1-color{red}{4x(1-x)};)$$
$endgroup$
– Blue
Jan 27 at 19:41
1
$begingroup$
Wow ... I got it now ... thanks for the patience. It is like a double shot of expresso, I need a chaser. Thanks again.
$endgroup$
– hlogoma
Jan 27 at 20:13
add a comment |
$begingroup$
There seems to be a mix of $x$s and $alpha$s; it appears that you're using them interchangeably, so I'll use $x$ to save typing. Since you've already shown that $S_2(x)=4x(1-x)$, you can write simply $$S_2(S_2(x)) = 4cdot 4x(1-x)cdot(1-4x(1-x))=16x(1-x)(1-4x+4x^2)=16x(1-x)(1-2x)^2$$ The last expression is, indeed, $S_4(x)$. So, you simply erred somewhere in expanding $S_2(S_2(x))$.
$endgroup$
– Blue
Jan 27 at 4:00
$begingroup$
Thanks you. Your answer highlights the elegance of mathematics and thereby underlines its attraction to many. I marvel but unfortunately find myself still looking in from the outside. I hope you have the time to unpack $4⋅4x(1−x)⋅(1−4x(1−x))$. I see the term $(4x(1-x))^2$ but I do not see where we get the $4$ times quantity nor the term $(1 - 4x(1-x))$ from.
$endgroup$
– hlogoma
Jan 27 at 14:39
$begingroup$
In determining $S_2(S_2(x))$, we simply replace every $x$ in $4x(1-x)$ with $S_2(x)$, which is $4x(1-x)$. $$4cdotcolor{red}{x}cdot(1-color{red}{x}) quadtoquad 4cdotcolor{red}{S_2(x)}cdot (;1-color{red}{S_2(x)};)quadtoquad 4cdotcolor{red}{4x(1-x)}cdot (;1-color{red}{4x(1-x)};)$$
$endgroup$
– Blue
Jan 27 at 19:41
1
$begingroup$
Wow ... I got it now ... thanks for the patience. It is like a double shot of expresso, I need a chaser. Thanks again.
$endgroup$
– hlogoma
Jan 27 at 20:13
$begingroup$
There seems to be a mix of $x$s and $alpha$s; it appears that you're using them interchangeably, so I'll use $x$ to save typing. Since you've already shown that $S_2(x)=4x(1-x)$, you can write simply $$S_2(S_2(x)) = 4cdot 4x(1-x)cdot(1-4x(1-x))=16x(1-x)(1-4x+4x^2)=16x(1-x)(1-2x)^2$$ The last expression is, indeed, $S_4(x)$. So, you simply erred somewhere in expanding $S_2(S_2(x))$.
$endgroup$
– Blue
Jan 27 at 4:00
$begingroup$
There seems to be a mix of $x$s and $alpha$s; it appears that you're using them interchangeably, so I'll use $x$ to save typing. Since you've already shown that $S_2(x)=4x(1-x)$, you can write simply $$S_2(S_2(x)) = 4cdot 4x(1-x)cdot(1-4x(1-x))=16x(1-x)(1-4x+4x^2)=16x(1-x)(1-2x)^2$$ The last expression is, indeed, $S_4(x)$. So, you simply erred somewhere in expanding $S_2(S_2(x))$.
$endgroup$
– Blue
Jan 27 at 4:00
$begingroup$
Thanks you. Your answer highlights the elegance of mathematics and thereby underlines its attraction to many. I marvel but unfortunately find myself still looking in from the outside. I hope you have the time to unpack $4⋅4x(1−x)⋅(1−4x(1−x))$. I see the term $(4x(1-x))^2$ but I do not see where we get the $4$ times quantity nor the term $(1 - 4x(1-x))$ from.
$endgroup$
– hlogoma
Jan 27 at 14:39
$begingroup$
Thanks you. Your answer highlights the elegance of mathematics and thereby underlines its attraction to many. I marvel but unfortunately find myself still looking in from the outside. I hope you have the time to unpack $4⋅4x(1−x)⋅(1−4x(1−x))$. I see the term $(4x(1-x))^2$ but I do not see where we get the $4$ times quantity nor the term $(1 - 4x(1-x))$ from.
$endgroup$
– hlogoma
Jan 27 at 14:39
$begingroup$
In determining $S_2(S_2(x))$, we simply replace every $x$ in $4x(1-x)$ with $S_2(x)$, which is $4x(1-x)$. $$4cdotcolor{red}{x}cdot(1-color{red}{x}) quadtoquad 4cdotcolor{red}{S_2(x)}cdot (;1-color{red}{S_2(x)};)quadtoquad 4cdotcolor{red}{4x(1-x)}cdot (;1-color{red}{4x(1-x)};)$$
$endgroup$
– Blue
Jan 27 at 19:41
$begingroup$
In determining $S_2(S_2(x))$, we simply replace every $x$ in $4x(1-x)$ with $S_2(x)$, which is $4x(1-x)$. $$4cdotcolor{red}{x}cdot(1-color{red}{x}) quadtoquad 4cdotcolor{red}{S_2(x)}cdot (;1-color{red}{S_2(x)};)quadtoquad 4cdotcolor{red}{4x(1-x)}cdot (;1-color{red}{4x(1-x)};)$$
$endgroup$
– Blue
Jan 27 at 19:41
1
1
$begingroup$
Wow ... I got it now ... thanks for the patience. It is like a double shot of expresso, I need a chaser. Thanks again.
$endgroup$
– hlogoma
Jan 27 at 20:13
$begingroup$
Wow ... I got it now ... thanks for the patience. It is like a double shot of expresso, I need a chaser. Thanks again.
$endgroup$
– hlogoma
Jan 27 at 20:13
add a comment |
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$begingroup$
There seems to be a mix of $x$s and $alpha$s; it appears that you're using them interchangeably, so I'll use $x$ to save typing. Since you've already shown that $S_2(x)=4x(1-x)$, you can write simply $$S_2(S_2(x)) = 4cdot 4x(1-x)cdot(1-4x(1-x))=16x(1-x)(1-4x+4x^2)=16x(1-x)(1-2x)^2$$ The last expression is, indeed, $S_4(x)$. So, you simply erred somewhere in expanding $S_2(S_2(x))$.
$endgroup$
– Blue
Jan 27 at 4:00
$begingroup$
Thanks you. Your answer highlights the elegance of mathematics and thereby underlines its attraction to many. I marvel but unfortunately find myself still looking in from the outside. I hope you have the time to unpack $4⋅4x(1−x)⋅(1−4x(1−x))$. I see the term $(4x(1-x))^2$ but I do not see where we get the $4$ times quantity nor the term $(1 - 4x(1-x))$ from.
$endgroup$
– hlogoma
Jan 27 at 14:39
$begingroup$
In determining $S_2(S_2(x))$, we simply replace every $x$ in $4x(1-x)$ with $S_2(x)$, which is $4x(1-x)$. $$4cdotcolor{red}{x}cdot(1-color{red}{x}) quadtoquad 4cdotcolor{red}{S_2(x)}cdot (;1-color{red}{S_2(x)};)quadtoquad 4cdotcolor{red}{4x(1-x)}cdot (;1-color{red}{4x(1-x)};)$$
$endgroup$
– Blue
Jan 27 at 19:41
1
$begingroup$
Wow ... I got it now ... thanks for the patience. It is like a double shot of expresso, I need a chaser. Thanks again.
$endgroup$
– hlogoma
Jan 27 at 20:13