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Prove ${|C_a^n|}$ is decreasing starting from some $k$ given $C_a^n = frac{a(a-1)(a-2)dots(a-n+1)}{n!}$
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Let $C_a^n$ be defined as:
$$
begin{cases}
C_a^n = frac{a(a-1)(a-2)dots(a-n+1)}{n!}\
bin mathbb N \
a in mathbb R \
C_a^0 = 1
end{cases}
$$
Prove that ${|C_a^n|}$ is decreasing starting from some index $k in mathbb N$.
Since we are dealing with absolute values lets find out when the given sequence is bounded. Consider consecutive terms:
$$
frac{C_{a}^{n+1}}{C_a^n} = frac{n!}{(n+1)!}cdot(a - n)
$$
Now inspect absolute values:
$$
frac{|C_{a}^{n+1}|}{|C_a^n|} = frac{1}{n+1}cdot|a-n| = frac{|n-a|}{n+1}
$$
So clearly the sequence is bounded only in case $a ge -1$. Before this one i've also shown that $C_a^n$ is bounded in case $a ge -1$ and unbounded in case $a < -1$ which was a generalization of this question.
My further thoughts were as follows. Let's take for instance $a = 3.5$ then the sequence becomes:
$$
C_{3.5}^n = frac{3.5(3.5-1)(3.5-2)(3.5-3)cdots(3.5-n+1)}{n!} = \
= frac{3.5}{1} cdot frac{2.5}{2} cdot frac{1.5}{3}cdot cdots cdot frac{3.5-n+1}{n}
$$
This instance of sequence is increasing for $n<3$ and then starts decreasing. After playing around with different cases of $a$ it seems like the index $k$ is in the form:
$$
k_0 =leftlceilfrac{a-1}{2}rightrceil
$$
To give more insights here is a visualization of the sequence.
My thoughts above do not feel like formal ones and $k_0$ is obtained by a guess so i would like to see a formal proof of what's in problem statement.
I know that $C_a^n$ has somewhat to deal with the Gamma function, but this question is in precalculus level. Thank you.
sequences-and-series algebra-precalculus upper-lower-bounds monotone-functions
asked yesterday
roman
9421815
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Let $C_a^n$ be defined as:
$$
begin{cases}
C_a^n = frac{a(a-1)(a-2)dots(a-n+1)}{n!}\
bin mathbb N \
a in mathbb R \
C_a^0 = 1
end{cases}
$$
Prove that ${|C_a^n|}$ is decreasing starting from some index $k in mathbb N$.
Since we are dealing with absolute values lets find out when the given sequence is bounded. Consider consecutive terms:
$$
frac{C_{a}^{n+1}}{C_a^n} = frac{n!}{(n+1)!}cdot(a - n)
$$
Now inspect absolute values:
$$
frac{|C_{a}^{n+1}|}{|C_a^n|} = frac{1}{n+1}cdot|a-n| = frac{|n-a|}{n+1}
$$
So clearly the sequence is bounded only in case $a ge -1$. Before this one i've also shown that $C_a^n$ is bounded in case $a ge -1$ and unbounded in case $a < -1$ which was a generalization of this question.
My further thoughts were as follows. Let's take for instance $a = 3.5$ then the sequence becomes:
$$
C_{3.5}^n = frac{3.5(3.5-1)(3.5-2)(3.5-3)cdots(3.5-n+1)}{n!} = \
= frac{3.5}{1} cdot frac{2.5}{2} cdot frac{1.5}{3}cdot cdots cdot frac{3.5-n+1}{n}
$$
This instance of sequence is increasing for $n<3$ and then starts decreasing. After playing around with different cases of $a$ it seems like the index $k$ is in the form:
$$
k_0 =leftlceilfrac{a-1}{2}rightrceil
$$
To give more insights here is a visualization of the sequence.
My thoughts above do not feel like formal ones and $k_0$ is obtained by a guess so i would like to see a formal proof of what's in problem statement.
I know that $C_a^n$ has somewhat to deal with the Gamma function, but this question is in precalculus level. Thank you.
sequences-and-series algebra-precalculus upper-lower-bounds monotone-functions
asked yesterday
roman
9421815
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $C_a^n$ be defined as:
$$
begin{cases}
C_a^n = frac{a(a-1)(a-2)dots(a-n+1)}{n!}\
bin mathbb N \
a in mathbb R \
C_a^0 = 1
end{cases}
$$
Prove that ${|C_a^n|}$ is decreasing starting from some index $k in mathbb N$.
Since we are dealing with absolute values lets find out when the given sequence is bounded. Consider consecutive terms:
$$
frac{C_{a}^{n+1}}{C_a^n} = frac{n!}{(n+1)!}cdot(a - n)
$$
Now inspect absolute values:
$$
frac{|C_{a}^{n+1}|}{|C_a^n|} = frac{1}{n+1}cdot|a-n| = frac{|n-a|}{n+1}
$$
So clearly the sequence is bounded only in case $a ge -1$. Before this one i've also shown that $C_a^n$ is bounded in case $a ge -1$ and unbounded in case $a < -1$ which was a generalization of this question.
My further thoughts were as follows. Let's take for instance $a = 3.5$ then the sequence becomes:
$$
C_{3.5}^n = frac{3.5(3.5-1)(3.5-2)(3.5-3)cdots(3.5-n+1)}{n!} = \
= frac{3.5}{1} cdot frac{2.5}{2} cdot frac{1.5}{3}cdot cdots cdot frac{3.5-n+1}{n}
$$
This instance of sequence is increasing for $n<3$ and then starts decreasing. After playing around with different cases of $a$ it seems like the index $k$ is in the form:
$$
k_0 =leftlceilfrac{a-1}{2}rightrceil
$$
To give more insights here is a visualization of the sequence.
My thoughts above do not feel like formal ones and $k_0$ is obtained by a guess so i would like to see a formal proof of what's in problem statement.
I know that $C_a^n$ has somewhat to deal with the Gamma function, but this question is in precalculus level. Thank you.
sequences-and-series algebra-precalculus upper-lower-bounds monotone-functions
asked yesterday
roman
9421815
Let $C_a^n$ be defined as:
$$
begin{cases}
C_a^n = frac{a(a-1)(a-2)dots(a-n+1)}{n!}\
bin mathbb N \
a in mathbb R \
C_a^0 = 1
end{cases}
$$
Prove that ${|C_a^n|}$ is decreasing starting from some index $k in mathbb N$.
Since we are dealing with absolute values lets find out when the given sequence is bounded. Consider consecutive terms:
$$
frac{C_{a}^{n+1}}{C_a^n} = frac{n!}{(n+1)!}cdot(a - n)
$$
Now inspect absolute values:
$$
frac{|C_{a}^{n+1}|}{|C_a^n|} = frac{1}{n+1}cdot|a-n| = frac{|n-a|}{n+1}
$$
So clearly the sequence is bounded only in case $a ge -1$. Before this one i've also shown that $C_a^n$ is bounded in case $a ge -1$ and unbounded in case $a < -1$ which was a generalization of this question.
My further thoughts were as follows. Let's take for instance $a = 3.5$ then the sequence becomes:
$$
C_{3.5}^n = frac{3.5(3.5-1)(3.5-2)(3.5-3)cdots(3.5-n+1)}{n!} = \
= frac{3.5}{1} cdot frac{2.5}{2} cdot frac{1.5}{3}cdot cdots cdot frac{3.5-n+1}{n}
$$
This instance of sequence is increasing for $n<3$ and then starts decreasing. After playing around with different cases of $a$ it seems like the index $k$ is in the form:
$$
k_0 =leftlceilfrac{a-1}{2}rightrceil
$$
To give more insights here is a visualization of the sequence.
My thoughts above do not feel like formal ones and $k_0$ is obtained by a guess so i would like to see a formal proof of what's in problem statement.
I know that $C_a^n$ has somewhat to deal with the Gamma function, but this question is in precalculus level. Thank you.
sequences-and-series algebra-precalculus upper-lower-bounds monotone-functions
sequences-and-series algebra-precalculus upper-lower-bounds monotone-functions
asked yesterday
roman
9421815
asked yesterday
roman
9421815
edited yesterday
asked yesterday
roman
9421815
asked yesterday
roman
9421815
asked yesterday
roman
9421815
9421815
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