Fibonacci sequence/recurrence relation (limits)
Let $lbrace F_nrbrace_{n in mathbb{N_0}}$ be the Fibonacci sequence.
$F_{n+1}=F_{n-1}+F_{n-2}$ for $n in mathbb{N}$ with $n geq 2$ and start values $F_0:=0$ and $F_1:=1$.
How to determine:
$limlimits_{ntoinfty}frac{F_n}{F_{n+1}}$?
I used:
If $a=limlimits_{ntoinfty}frac{F_n}{F_{n+1}}$ then
$a=limlimits_{ntoinfty}frac{F_n}{F_{n+1}}=limlimits_{ntoinfty}frac{F_n}{F_n+F_{n-1}}$
Here I don't know how to continue.
limits recurrence-relations fibonacci-numbers
asked Nov 25 at 15:07
Nekarts
234
add a comment |
Let $lbrace F_nrbrace_{n in mathbb{N_0}}$ be the Fibonacci sequence.
$F_{n+1}=F_{n-1}+F_{n-2}$ for $n in mathbb{N}$ with $n geq 2$ and start values $F_0:=0$ and $F_1:=1$.
How to determine:
$limlimits_{ntoinfty}frac{F_n}{F_{n+1}}$?
I used:
If $a=limlimits_{ntoinfty}frac{F_n}{F_{n+1}}$ then
$a=limlimits_{ntoinfty}frac{F_n}{F_{n+1}}=limlimits_{ntoinfty}frac{F_n}{F_n+F_{n-1}}$
Here I don't know how to continue.
limits recurrence-relations fibonacci-numbers
asked Nov 25 at 15:07
Nekarts
234
add a comment |
Let $lbrace F_nrbrace_{n in mathbb{N_0}}$ be the Fibonacci sequence.
$F_{n+1}=F_{n-1}+F_{n-2}$ for $n in mathbb{N}$ with $n geq 2$ and start values $F_0:=0$ and $F_1:=1$.
How to determine:
$limlimits_{ntoinfty}frac{F_n}{F_{n+1}}$?
I used:
If $a=limlimits_{ntoinfty}frac{F_n}{F_{n+1}}$ then
$a=limlimits_{ntoinfty}frac{F_n}{F_{n+1}}=limlimits_{ntoinfty}frac{F_n}{F_n+F_{n-1}}$
Here I don't know how to continue.
limits recurrence-relations fibonacci-numbers
asked Nov 25 at 15:07
Nekarts
234
Let $lbrace F_nrbrace_{n in mathbb{N_0}}$ be the Fibonacci sequence.
$F_{n+1}=F_{n-1}+F_{n-2}$ for $n in mathbb{N}$ with $n geq 2$ and start values $F_0:=0$ and $F_1:=1$.
How to determine:
$limlimits_{ntoinfty}frac{F_n}{F_{n+1}}$?
I used:
If $a=limlimits_{ntoinfty}frac{F_n}{F_{n+1}}$ then
$a=limlimits_{ntoinfty}frac{F_n}{F_{n+1}}=limlimits_{ntoinfty}frac{F_n}{F_n+F_{n-1}}$
Here I don't know how to continue.
limits recurrence-relations fibonacci-numbers
limits recurrence-relations fibonacci-numbers
asked Nov 25 at 15:07
Nekarts
234
asked Nov 25 at 15:07
Nekarts
234
edited Nov 25 at 18:05
asked Nov 25 at 15:07
Nekarts
234
asked Nov 25 at 15:07
Nekarts
234
asked Nov 25 at 15:07
Nekarts
234
234
add a comment |
add a comment |
3 Answers
3
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oldest
votes
Assign a variable to the first limit, such as $x$.
$$lim_{ntoinfty}frac{F_n}{F_{n+1}} = x$$
By definition, $F_{n+1} = F_n+F_{n-1}$, so rewrite the limit.
$$lim_{ntoinfty}frac{F_n}{F_{n}+F_{n-1}} = x$$
$$lim_{ntoinfty}frac{F_{n}+F_{n-1}}{F_n} = frac{1}{x}$$
$$1+lim_{ntoinfty} frac{F_{n-1}}{F_n} = frac{1}{x}$$
If $lim_limits{ntoinfty}frac{F_n}{F_{n+1}} = x$, then $lim_limits{n to infty} frac{F_{n-1}}{F_n} = x$ as well.
$$1 + x =frac{1}{x}$$
$$x+x^2 = 1 implies x^2+x-1 = 0 implies x = frac{-1pmsqrt{5}}{2}$$
Now, it's clear that only $x = frac{-1+sqrt{5}}{2}$ applies here, so $x = frac{1}{φ}$, or the reciprocal of the golden ratio. The interesting thing is that this does not apply specifically to the Fibonacci sequence, but for Fibonacci-like sequences in general.
answered Nov 25 at 15:17
KM101
4,043417
add a comment |
HINT
Solve the recurrence analytically, the characteristic polynomial would be
$$
x^2-x-1=0
$$
and its roots $r,R$ would produce
$$
F_n = Ar^n + BR^n...
$$
answered Nov 25 at 15:12
gt6989b
32.8k22452
add a comment |
Using Binet's formula:
$$lim_{ntoinfty} frac{F_n}{F_{n+1}}=lim_{ntoinfty} frac{frac1{sqrt{5}}left(varphi^n-phi^nright)}{frac1{sqrt{5}}left(varphi^{n+1}-phi^{n+1}right)}=lim_{ntoinfty} frac{varphi^nleft(1-frac{phi^n}{varphi^n}right)}{varphi^{n+1}left(1-frac{phi^{n+1}}{varphi^{n+1}}right)}=frac1{varphi},$$
because:
$$lim_{ntoinfty} left(frac{phi}{varphi}right)^n=0,$$
because:
$$left|frac{phi}{varphi}right|=left|frac{frac{1-sqrt{5}}{2}}{frac{1+sqrt{5}}{2}}right|=left|frac{3-sqrt{5}}{-2}right|approx 0.38<1.$$
answered Nov 25 at 15:47
farruhota
18.9k2736
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
Assign a variable to the first limit, such as $x$.
$$lim_{ntoinfty}frac{F_n}{F_{n+1}} = x$$
By definition, $F_{n+1} = F_n+F_{n-1}$, so rewrite the limit.
$$lim_{ntoinfty}frac{F_n}{F_{n}+F_{n-1}} = x$$
$$lim_{ntoinfty}frac{F_{n}+F_{n-1}}{F_n} = frac{1}{x}$$
$$1+lim_{ntoinfty} frac{F_{n-1}}{F_n} = frac{1}{x}$$
If $lim_limits{ntoinfty}frac{F_n}{F_{n+1}} = x$, then $lim_limits{n to infty} frac{F_{n-1}}{F_n} = x$ as well.
$$1 + x =frac{1}{x}$$
$$x+x^2 = 1 implies x^2+x-1 = 0 implies x = frac{-1pmsqrt{5}}{2}$$
Now, it's clear that only $x = frac{-1+sqrt{5}}{2}$ applies here, so $x = frac{1}{φ}$, or the reciprocal of the golden ratio. The interesting thing is that this does not apply specifically to the Fibonacci sequence, but for Fibonacci-like sequences in general.
answered Nov 25 at 15:17
KM101
4,043417
add a comment |
Assign a variable to the first limit, such as $x$.
$$lim_{ntoinfty}frac{F_n}{F_{n+1}} = x$$
By definition, $F_{n+1} = F_n+F_{n-1}$, so rewrite the limit.
$$lim_{ntoinfty}frac{F_n}{F_{n}+F_{n-1}} = x$$
$$lim_{ntoinfty}frac{F_{n}+F_{n-1}}{F_n} = frac{1}{x}$$
$$1+lim_{ntoinfty} frac{F_{n-1}}{F_n} = frac{1}{x}$$
If $lim_limits{ntoinfty}frac{F_n}{F_{n+1}} = x$, then $lim_limits{n to infty} frac{F_{n-1}}{F_n} = x$ as well.
$$1 + x =frac{1}{x}$$
$$x+x^2 = 1 implies x^2+x-1 = 0 implies x = frac{-1pmsqrt{5}}{2}$$
Now, it's clear that only $x = frac{-1+sqrt{5}}{2}$ applies here, so $x = frac{1}{φ}$, or the reciprocal of the golden ratio. The interesting thing is that this does not apply specifically to the Fibonacci sequence, but for Fibonacci-like sequences in general.
answered Nov 25 at 15:17
KM101
4,043417
add a comment |
Assign a variable to the first limit, such as $x$.
$$lim_{ntoinfty}frac{F_n}{F_{n+1}} = x$$
By definition, $F_{n+1} = F_n+F_{n-1}$, so rewrite the limit.
$$lim_{ntoinfty}frac{F_n}{F_{n}+F_{n-1}} = x$$
$$lim_{ntoinfty}frac{F_{n}+F_{n-1}}{F_n} = frac{1}{x}$$
$$1+lim_{ntoinfty} frac{F_{n-1}}{F_n} = frac{1}{x}$$
If $lim_limits{ntoinfty}frac{F_n}{F_{n+1}} = x$, then $lim_limits{n to infty} frac{F_{n-1}}{F_n} = x$ as well.
$$1 + x =frac{1}{x}$$
$$x+x^2 = 1 implies x^2+x-1 = 0 implies x = frac{-1pmsqrt{5}}{2}$$
Now, it's clear that only $x = frac{-1+sqrt{5}}{2}$ applies here, so $x = frac{1}{φ}$, or the reciprocal of the golden ratio. The interesting thing is that this does not apply specifically to the Fibonacci sequence, but for Fibonacci-like sequences in general.
answered Nov 25 at 15:17
KM101
4,043417
Assign a variable to the first limit, such as $x$.
$$lim_{ntoinfty}frac{F_n}{F_{n+1}} = x$$
By definition, $F_{n+1} = F_n+F_{n-1}$, so rewrite the limit.
$$lim_{ntoinfty}frac{F_n}{F_{n}+F_{n-1}} = x$$
$$lim_{ntoinfty}frac{F_{n}+F_{n-1}}{F_n} = frac{1}{x}$$
$$1+lim_{ntoinfty} frac{F_{n-1}}{F_n} = frac{1}{x}$$
If $lim_limits{ntoinfty}frac{F_n}{F_{n+1}} = x$, then $lim_limits{n to infty} frac{F_{n-1}}{F_n} = x$ as well.
$$1 + x =frac{1}{x}$$
$$x+x^2 = 1 implies x^2+x-1 = 0 implies x = frac{-1pmsqrt{5}}{2}$$
Now, it's clear that only $x = frac{-1+sqrt{5}}{2}$ applies here, so $x = frac{1}{φ}$, or the reciprocal of the golden ratio. The interesting thing is that this does not apply specifically to the Fibonacci sequence, but for Fibonacci-like sequences in general.
answered Nov 25 at 15:17
KM101
4,043417
edited Nov 25 at 15:26
answered Nov 25 at 15:17
KM101
4,043417
answered Nov 25 at 15:17
KM101
4,043417
answered Nov 25 at 15:17
KM101
4,043417
4,043417
add a comment |
add a comment |
HINT
Solve the recurrence analytically, the characteristic polynomial would be
$$
x^2-x-1=0
$$
and its roots $r,R$ would produce
$$
F_n = Ar^n + BR^n...
$$
answered Nov 25 at 15:12
gt6989b
32.8k22452
add a comment |
HINT
Solve the recurrence analytically, the characteristic polynomial would be
$$
x^2-x-1=0
$$
and its roots $r,R$ would produce
$$
F_n = Ar^n + BR^n...
$$
answered Nov 25 at 15:12
gt6989b
32.8k22452
add a comment |
HINT
Solve the recurrence analytically, the characteristic polynomial would be
$$
x^2-x-1=0
$$
and its roots $r,R$ would produce
$$
F_n = Ar^n + BR^n...
$$
answered Nov 25 at 15:12
gt6989b
32.8k22452
HINT
Solve the recurrence analytically, the characteristic polynomial would be
$$
x^2-x-1=0
$$
and its roots $r,R$ would produce
$$
F_n = Ar^n + BR^n...
$$
answered Nov 25 at 15:12
gt6989b
32.8k22452
answered Nov 25 at 15:12
gt6989b
32.8k22452
answered Nov 25 at 15:12
gt6989b
32.8k22452
answered Nov 25 at 15:12
gt6989b
32.8k22452
32.8k22452
add a comment |
add a comment |
Using Binet's formula:
$$lim_{ntoinfty} frac{F_n}{F_{n+1}}=lim_{ntoinfty} frac{frac1{sqrt{5}}left(varphi^n-phi^nright)}{frac1{sqrt{5}}left(varphi^{n+1}-phi^{n+1}right)}=lim_{ntoinfty} frac{varphi^nleft(1-frac{phi^n}{varphi^n}right)}{varphi^{n+1}left(1-frac{phi^{n+1}}{varphi^{n+1}}right)}=frac1{varphi},$$
because:
$$lim_{ntoinfty} left(frac{phi}{varphi}right)^n=0,$$
because:
$$left|frac{phi}{varphi}right|=left|frac{frac{1-sqrt{5}}{2}}{frac{1+sqrt{5}}{2}}right|=left|frac{3-sqrt{5}}{-2}right|approx 0.38<1.$$
answered Nov 25 at 15:47
farruhota
18.9k2736
add a comment |
Using Binet's formula:
$$lim_{ntoinfty} frac{F_n}{F_{n+1}}=lim_{ntoinfty} frac{frac1{sqrt{5}}left(varphi^n-phi^nright)}{frac1{sqrt{5}}left(varphi^{n+1}-phi^{n+1}right)}=lim_{ntoinfty} frac{varphi^nleft(1-frac{phi^n}{varphi^n}right)}{varphi^{n+1}left(1-frac{phi^{n+1}}{varphi^{n+1}}right)}=frac1{varphi},$$
because:
$$lim_{ntoinfty} left(frac{phi}{varphi}right)^n=0,$$
because:
$$left|frac{phi}{varphi}right|=left|frac{frac{1-sqrt{5}}{2}}{frac{1+sqrt{5}}{2}}right|=left|frac{3-sqrt{5}}{-2}right|approx 0.38<1.$$
answered Nov 25 at 15:47
farruhota
18.9k2736
add a comment |
Using Binet's formula:
$$lim_{ntoinfty} frac{F_n}{F_{n+1}}=lim_{ntoinfty} frac{frac1{sqrt{5}}left(varphi^n-phi^nright)}{frac1{sqrt{5}}left(varphi^{n+1}-phi^{n+1}right)}=lim_{ntoinfty} frac{varphi^nleft(1-frac{phi^n}{varphi^n}right)}{varphi^{n+1}left(1-frac{phi^{n+1}}{varphi^{n+1}}right)}=frac1{varphi},$$
because:
$$lim_{ntoinfty} left(frac{phi}{varphi}right)^n=0,$$
because:
$$left|frac{phi}{varphi}right|=left|frac{frac{1-sqrt{5}}{2}}{frac{1+sqrt{5}}{2}}right|=left|frac{3-sqrt{5}}{-2}right|approx 0.38<1.$$
answered Nov 25 at 15:47
farruhota
18.9k2736
Using Binet's formula:
$$lim_{ntoinfty} frac{F_n}{F_{n+1}}=lim_{ntoinfty} frac{frac1{sqrt{5}}left(varphi^n-phi^nright)}{frac1{sqrt{5}}left(varphi^{n+1}-phi^{n+1}right)}=lim_{ntoinfty} frac{varphi^nleft(1-frac{phi^n}{varphi^n}right)}{varphi^{n+1}left(1-frac{phi^{n+1}}{varphi^{n+1}}right)}=frac1{varphi},$$
because:
$$lim_{ntoinfty} left(frac{phi}{varphi}right)^n=0,$$
because:
$$left|frac{phi}{varphi}right|=left|frac{frac{1-sqrt{5}}{2}}{frac{1+sqrt{5}}{2}}right|=left|frac{3-sqrt{5}}{-2}right|approx 0.38<1.$$
answered Nov 25 at 15:47
farruhota
18.9k2736
answered Nov 25 at 15:47
farruhota
18.9k2736
answered Nov 25 at 15:47
farruhota
18.9k2736
answered Nov 25 at 15:47
farruhota
18.9k2736
18.9k2736
add a comment |
add a comment |
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