Why the value of this kind of sequence always get bigger and then smaller
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I tried to use $1$, $2$ and $3$ into this formula, and for all the answer will get bigger first than smaller. I want to know the reason for it.
sequences-and-series
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I tried to use $1$, $2$ and $3$ into this formula, and for all the answer will get bigger first than smaller. I want to know the reason for it.
sequences-and-series
The sequences are recursive sequences, as you might see. For example, in the first case, $$ x_n+1 = 1 + frac1x_n qquad textwith x_0 = 1 $$ So which one is bigger, $x_n$ or $x_n+1$? The limit of the sequence is $$ lim_x to infty x_n = frac1+sqrt52 = phi $$ and if $x_n < phi Rightarrow x_n+1 > phi$ etc ... You can see this by comparing the functions $$ f_1(x) = x $$ and $$ f_2(x) = 1+ frac1x $$ and figuring out when $f_1(x) > f_2(x)$.
â Matti P.
Sep 7 at 10:00
The sequence contains the convergents of the (simple) continued fraction of the golden ratio $ varphi $. Such a sequence always alternates between values above and values below the given number.
â Peter
Sep 7 at 11:08
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up vote
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up vote
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down vote
favorite
I tried to use $1$, $2$ and $3$ into this formula, and for all the answer will get bigger first than smaller. I want to know the reason for it.
sequences-and-series
I tried to use $1$, $2$ and $3$ into this formula, and for all the answer will get bigger first than smaller. I want to know the reason for it.
sequences-and-series
sequences-and-series
edited Sep 7 at 10:30
user220178
1768
1768
asked Sep 7 at 9:40
teng
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42
The sequences are recursive sequences, as you might see. For example, in the first case, $$ x_n+1 = 1 + frac1x_n qquad textwith x_0 = 1 $$ So which one is bigger, $x_n$ or $x_n+1$? The limit of the sequence is $$ lim_x to infty x_n = frac1+sqrt52 = phi $$ and if $x_n < phi Rightarrow x_n+1 > phi$ etc ... You can see this by comparing the functions $$ f_1(x) = x $$ and $$ f_2(x) = 1+ frac1x $$ and figuring out when $f_1(x) > f_2(x)$.
â Matti P.
Sep 7 at 10:00
The sequence contains the convergents of the (simple) continued fraction of the golden ratio $ varphi $. Such a sequence always alternates between values above and values below the given number.
â Peter
Sep 7 at 11:08
add a comment |Â
The sequences are recursive sequences, as you might see. For example, in the first case, $$ x_n+1 = 1 + frac1x_n qquad textwith x_0 = 1 $$ So which one is bigger, $x_n$ or $x_n+1$? The limit of the sequence is $$ lim_x to infty x_n = frac1+sqrt52 = phi $$ and if $x_n < phi Rightarrow x_n+1 > phi$ etc ... You can see this by comparing the functions $$ f_1(x) = x $$ and $$ f_2(x) = 1+ frac1x $$ and figuring out when $f_1(x) > f_2(x)$.
â Matti P.
Sep 7 at 10:00
The sequence contains the convergents of the (simple) continued fraction of the golden ratio $ varphi $. Such a sequence always alternates between values above and values below the given number.
â Peter
Sep 7 at 11:08
The sequences are recursive sequences, as you might see. For example, in the first case, $$ x_n+1 = 1 + frac1x_n qquad textwith x_0 = 1 $$ So which one is bigger, $x_n$ or $x_n+1$? The limit of the sequence is $$ lim_x to infty x_n = frac1+sqrt52 = phi $$ and if $x_n < phi Rightarrow x_n+1 > phi$ etc ... You can see this by comparing the functions $$ f_1(x) = x $$ and $$ f_2(x) = 1+ frac1x $$ and figuring out when $f_1(x) > f_2(x)$.
â Matti P.
Sep 7 at 10:00
The sequences are recursive sequences, as you might see. For example, in the first case, $$ x_n+1 = 1 + frac1x_n qquad textwith x_0 = 1 $$ So which one is bigger, $x_n$ or $x_n+1$? The limit of the sequence is $$ lim_x to infty x_n = frac1+sqrt52 = phi $$ and if $x_n < phi Rightarrow x_n+1 > phi$ etc ... You can see this by comparing the functions $$ f_1(x) = x $$ and $$ f_2(x) = 1+ frac1x $$ and figuring out when $f_1(x) > f_2(x)$.
â Matti P.
Sep 7 at 10:00
The sequence contains the convergents of the (simple) continued fraction of the golden ratio $ varphi $. Such a sequence always alternates between values above and values below the given number.
â Peter
Sep 7 at 11:08
The sequence contains the convergents of the (simple) continued fraction of the golden ratio $ varphi $. Such a sequence always alternates between values above and values below the given number.
â Peter
Sep 7 at 11:08
add a comment |Â
1 Answer
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These are simple continued fractions. This from wikipedia tells you the approximation errors alternate in sign:
Corollary 1: The even convergents continually increase, but are
always less than $x$.
Corollary 2: The odd convergents continually decrease, but are always
greater than $x$ .
(Good for you for noticing this.)
https://en.wikipedia.org/wiki/Continued_fraction#Properties
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
These are simple continued fractions. This from wikipedia tells you the approximation errors alternate in sign:
Corollary 1: The even convergents continually increase, but are
always less than $x$.
Corollary 2: The odd convergents continually decrease, but are always
greater than $x$ .
(Good for you for noticing this.)
https://en.wikipedia.org/wiki/Continued_fraction#Properties
add a comment |Â
up vote
0
down vote
These are simple continued fractions. This from wikipedia tells you the approximation errors alternate in sign:
Corollary 1: The even convergents continually increase, but are
always less than $x$.
Corollary 2: The odd convergents continually decrease, but are always
greater than $x$ .
(Good for you for noticing this.)
https://en.wikipedia.org/wiki/Continued_fraction#Properties
add a comment |Â
up vote
0
down vote
up vote
0
down vote
These are simple continued fractions. This from wikipedia tells you the approximation errors alternate in sign:
Corollary 1: The even convergents continually increase, but are
always less than $x$.
Corollary 2: The odd convergents continually decrease, but are always
greater than $x$ .
(Good for you for noticing this.)
https://en.wikipedia.org/wiki/Continued_fraction#Properties
These are simple continued fractions. This from wikipedia tells you the approximation errors alternate in sign:
Corollary 1: The even convergents continually increase, but are
always less than $x$.
Corollary 2: The odd convergents continually decrease, but are always
greater than $x$ .
(Good for you for noticing this.)
https://en.wikipedia.org/wiki/Continued_fraction#Properties
edited Sep 7 at 12:17
answered Sep 7 at 12:05
Ethan Bolker
36.7k54299
36.7k54299
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The sequences are recursive sequences, as you might see. For example, in the first case, $$ x_n+1 = 1 + frac1x_n qquad textwith x_0 = 1 $$ So which one is bigger, $x_n$ or $x_n+1$? The limit of the sequence is $$ lim_x to infty x_n = frac1+sqrt52 = phi $$ and if $x_n < phi Rightarrow x_n+1 > phi$ etc ... You can see this by comparing the functions $$ f_1(x) = x $$ and $$ f_2(x) = 1+ frac1x $$ and figuring out when $f_1(x) > f_2(x)$.
â Matti P.
Sep 7 at 10:00
The sequence contains the convergents of the (simple) continued fraction of the golden ratio $ varphi $. Such a sequence always alternates between values above and values below the given number.
â Peter
Sep 7 at 11:08