Fibonacci elegance sought for $F_{f (n)} + F_{f(n)-1}$Solving Fibonaccis Term Using Golden Ratio...
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Fibonacci elegance sought for $F_{f (n)} + F_{f(n)-1}$
Solving Fibonaccis Term Using Golden Ratio ConverganceFibonacci nth termNeed formula for sequence related to Lucas/Fibonacci numbersFibonacci numbers and golden ratio: $Phi = lim sqrt[n]{F_n}$How is the Binet's formula for Fibonacci reversed in order to find the index for a given Fibonacci number?Fibonacci-related sumHow does one arrive at a certain expression for the Fibonacci Zeta function?Solve for n in golden ratio fibonacci equationWhat is the connection and the difference between the Golden Ratio and Fibonacci Sequence?Powers of the golden ratio
$begingroup$
Updated on Friday 15th March 2019 at 5 pm in the light of comments received over the last 24 hours.
The original question was; given the well known variation of Binet's Formula:
$$F_n = frac{phi^n - (-phi)^{-n}}{sqrt{5}}$$
Derive an elegant expression, should one exist, for;
$$F_{log (n)} + F_{log(n)-1}$$
Of course,
$$phi=frac{1+sqrt{5}}{2}$$
Wolfgang Kais has pointed out that this is going to run into technical issues with raising negative numbers to the power of $log(n)$.
Consequently, initially at least, I've been looking instead at
$$F_{f(n)} + F_{f(n)-1}$$
where the function $f$ is sufficiently well behaved to not run into such issues.
In this case,
$$ (sqrt{5})(F_{f(n)} + F_{f(n)-1})=phi^{f(n)}-(-phi)^{-f(n)}+ phi^{f(n)-1}-(-phi)^{-(f(n)-1)}$$
Using the two equivalent facts that
$$1-phi=- frac{1}{phi} :or: phi=1+frac{1}{phi}$$
I proceeded as follows;
$$ (sqrt{5})(F_{f(n)} + F_{f(n)-1})=phi^{f(n)}-big(-frac{1}{phi}big)^{f(n)}+ phi^{f(n)} times phi^{-1}-big(-frac{1}{phi}big)^{(f(n)-1)}$$
$$ =phi^{f(n)}-(1-phi)^{f(n)}+ frac{phi^{f(n)}}{phi}-big(-frac{1}{phi}big)^{f(n)} times big(-frac{1}{phi}big)^{-1}$$
$$ =phi^{f(n)}big(1+frac{1}{phi}big)-(1-phi)^{f(n)}-(1-phi)^{f(n)} times (-phi)$$
$$ =phi times phi^{f(n)} - (1-phi) times (1-phi)^{f(n)}$$
Wolfgang Kais suggested immediately starting with the Fibonacci recurrence relation
$$F_m=F_{m-1}+F_{m-2}$$
from which I deduce that
$$F_{f(n)}+F_{f(n)-1}=F_{f(n)+1}$$
Now, applying the variation on Binet's Formula,
$$F_{f(n)+1} = frac{phi^{f(n)+1} - (-phi)^{-(f(n)+1)}}{sqrt{5}}$$
$$(sqrt{5})(F_{f(n)+1}) = phi times phi^{f(n)} - big(-frac{1}{phi}big)^{f(n)+1}$$
$$ = phi times phi^{f(n)} - (1-phi)^{f(n)+1}$$
$$ =phi times phi^{f(n)} - (1-phi) times (1-phi)^{f(n)}$$
which is the same result as obtained previously which, if nothing else, shows that the variation on Binet's formula satisfies the Fibonacci Recurrence relation.
I'm curious to know if
$$ F_{f(n)+1} = frac{phi times phi^{f(n)}}{sqrt{5}} - frac{(1-phi) times (1-phi)^{f(n)}}{sqrt{5}}$$
is of any use, and is it the best that can be done ?
Possibly I've exhausted what can usefully be done with this question, as I can't see the point of introducing the $log(n)$ function, given the technical hurdles it immediately throws up.
However, any further thoughts are most welcome.
The question originally came from a friend yesterday.
The earlier ask is here : how can i place then values = F(logn)+F(logn-1) in golden ratio Fibonacci series?lognflogn-1-in-golden-ratio-fibonacci-series
elementary-number-theory fibonacci-numbers golden-ratio
asked Mar 14 at 22:19
Martin HansenMartin Hansen
707114
$endgroup$
|
show 5 more comments
$begingroup$
Updated on Friday 15th March 2019 at 5 pm in the light of comments received over the last 24 hours.
The original question was; given the well known variation of Binet's Formula:
$$F_n = frac{phi^n - (-phi)^{-n}}{sqrt{5}}$$
Derive an elegant expression, should one exist, for;
$$F_{log (n)} + F_{log(n)-1}$$
Of course,
$$phi=frac{1+sqrt{5}}{2}$$
Wolfgang Kais has pointed out that this is going to run into technical issues with raising negative numbers to the power of $log(n)$.
Consequently, initially at least, I've been looking instead at
$$F_{f(n)} + F_{f(n)-1}$$
where the function $f$ is sufficiently well behaved to not run into such issues.
In this case,
$$ (sqrt{5})(F_{f(n)} + F_{f(n)-1})=phi^{f(n)}-(-phi)^{-f(n)}+ phi^{f(n)-1}-(-phi)^{-(f(n)-1)}$$
Using the two equivalent facts that
$$1-phi=- frac{1}{phi} :or: phi=1+frac{1}{phi}$$
I proceeded as follows;
$$ (sqrt{5})(F_{f(n)} + F_{f(n)-1})=phi^{f(n)}-big(-frac{1}{phi}big)^{f(n)}+ phi^{f(n)} times phi^{-1}-big(-frac{1}{phi}big)^{(f(n)-1)}$$
$$ =phi^{f(n)}-(1-phi)^{f(n)}+ frac{phi^{f(n)}}{phi}-big(-frac{1}{phi}big)^{f(n)} times big(-frac{1}{phi}big)^{-1}$$
$$ =phi^{f(n)}big(1+frac{1}{phi}big)-(1-phi)^{f(n)}-(1-phi)^{f(n)} times (-phi)$$
$$ =phi times phi^{f(n)} - (1-phi) times (1-phi)^{f(n)}$$
Wolfgang Kais suggested immediately starting with the Fibonacci recurrence relation
$$F_m=F_{m-1}+F_{m-2}$$
from which I deduce that
$$F_{f(n)}+F_{f(n)-1}=F_{f(n)+1}$$
Now, applying the variation on Binet's Formula,
$$F_{f(n)+1} = frac{phi^{f(n)+1} - (-phi)^{-(f(n)+1)}}{sqrt{5}}$$
$$(sqrt{5})(F_{f(n)+1}) = phi times phi^{f(n)} - big(-frac{1}{phi}big)^{f(n)+1}$$
$$ = phi times phi^{f(n)} - (1-phi)^{f(n)+1}$$
$$ =phi times phi^{f(n)} - (1-phi) times (1-phi)^{f(n)}$$
which is the same result as obtained previously which, if nothing else, shows that the variation on Binet's formula satisfies the Fibonacci Recurrence relation.
I'm curious to know if
$$ F_{f(n)+1} = frac{phi times phi^{f(n)}}{sqrt{5}} - frac{(1-phi) times (1-phi)^{f(n)}}{sqrt{5}}$$
is of any use, and is it the best that can be done ?
Possibly I've exhausted what can usefully be done with this question, as I can't see the point of introducing the $log(n)$ function, given the technical hurdles it immediately throws up.
However, any further thoughts are most welcome.
The question originally came from a friend yesterday.
The earlier ask is here : how can i place then values = F(logn)+F(logn-1) in golden ratio Fibonacci series?lognflogn-1-in-golden-ratio-fibonacci-series
elementary-number-theory fibonacci-numbers golden-ratio
asked Mar 14 at 22:19
Martin HansenMartin Hansen
707114
$endgroup$
1
$begingroup$
Note the same basic question was asked about $2$ hours earlier at how can i place then values = F(logn)+F(logn-1) in golden ratio Fibonacci series?. I appreciate you leaving a comment on the other question about you asking it here, but you should also mention that somewhere in this new question. Thanks.
$endgroup$
– John Omielan
Mar 14 at 22:28
2
$begingroup$
As you can't take a negative number to a non-integer power, $log(n)$ will have to be an integer, and so $F_{log (n)} + F_{log(n)-1} = F_{log(n)+1}$.
$endgroup$
– Wolfgang Kais
Mar 15 at 0:28
1
$begingroup$
@Wolfgang Kais : Of course ! And if I use that in the Binet formula I can get to $$F_{log (n)} + F_{log(n)-1} = frac{ phi times phi^{log (n)}-(1-phi) times (1- phi)^{log(n)}}{sqrt{5}}$$ which is so similar to the expression I mention in my question that I think I have a single minus sign error and so have $phi^2$ instead of $(1 - phi)$. I'll try to reconcile the two methods tomorrow. Thanks !
$endgroup$
– Martin Hansen
Mar 15 at 1:21
1
$begingroup$
You can also try and use the identity $$x^{log a}=a^{log x}.$$ Unfortunately that requires careful interpretation of complex powers unless $a$ and $x$ are both positive real numbers. It does give you a simple estimate for the sum.
$endgroup$
– Jyrki Lahtonen
Mar 15 at 8:16
1
$begingroup$
@MartinHansen: $1-phi$ is also negative. How will you define $(-1)^{log 2}$?
$endgroup$
– Wolfgang Kais
Mar 15 at 8:56
|
show 5 more comments
$begingroup$
Updated on Friday 15th March 2019 at 5 pm in the light of comments received over the last 24 hours.
The original question was; given the well known variation of Binet's Formula:
$$F_n = frac{phi^n - (-phi)^{-n}}{sqrt{5}}$$
Derive an elegant expression, should one exist, for;
$$F_{log (n)} + F_{log(n)-1}$$
Of course,
$$phi=frac{1+sqrt{5}}{2}$$
Wolfgang Kais has pointed out that this is going to run into technical issues with raising negative numbers to the power of $log(n)$.
Consequently, initially at least, I've been looking instead at
$$F_{f(n)} + F_{f(n)-1}$$
where the function $f$ is sufficiently well behaved to not run into such issues.
In this case,
$$ (sqrt{5})(F_{f(n)} + F_{f(n)-1})=phi^{f(n)}-(-phi)^{-f(n)}+ phi^{f(n)-1}-(-phi)^{-(f(n)-1)}$$
Using the two equivalent facts that
$$1-phi=- frac{1}{phi} :or: phi=1+frac{1}{phi}$$
I proceeded as follows;
$$ (sqrt{5})(F_{f(n)} + F_{f(n)-1})=phi^{f(n)}-big(-frac{1}{phi}big)^{f(n)}+ phi^{f(n)} times phi^{-1}-big(-frac{1}{phi}big)^{(f(n)-1)}$$
$$ =phi^{f(n)}-(1-phi)^{f(n)}+ frac{phi^{f(n)}}{phi}-big(-frac{1}{phi}big)^{f(n)} times big(-frac{1}{phi}big)^{-1}$$
$$ =phi^{f(n)}big(1+frac{1}{phi}big)-(1-phi)^{f(n)}-(1-phi)^{f(n)} times (-phi)$$
$$ =phi times phi^{f(n)} - (1-phi) times (1-phi)^{f(n)}$$
Wolfgang Kais suggested immediately starting with the Fibonacci recurrence relation
$$F_m=F_{m-1}+F_{m-2}$$
from which I deduce that
$$F_{f(n)}+F_{f(n)-1}=F_{f(n)+1}$$
Now, applying the variation on Binet's Formula,
$$F_{f(n)+1} = frac{phi^{f(n)+1} - (-phi)^{-(f(n)+1)}}{sqrt{5}}$$
$$(sqrt{5})(F_{f(n)+1}) = phi times phi^{f(n)} - big(-frac{1}{phi}big)^{f(n)+1}$$
$$ = phi times phi^{f(n)} - (1-phi)^{f(n)+1}$$
$$ =phi times phi^{f(n)} - (1-phi) times (1-phi)^{f(n)}$$
which is the same result as obtained previously which, if nothing else, shows that the variation on Binet's formula satisfies the Fibonacci Recurrence relation.
I'm curious to know if
$$ F_{f(n)+1} = frac{phi times phi^{f(n)}}{sqrt{5}} - frac{(1-phi) times (1-phi)^{f(n)}}{sqrt{5}}$$
is of any use, and is it the best that can be done ?
Possibly I've exhausted what can usefully be done with this question, as I can't see the point of introducing the $log(n)$ function, given the technical hurdles it immediately throws up.
However, any further thoughts are most welcome.
The question originally came from a friend yesterday.
The earlier ask is here : how can i place then values = F(logn)+F(logn-1) in golden ratio Fibonacci series?lognflogn-1-in-golden-ratio-fibonacci-series
elementary-number-theory fibonacci-numbers golden-ratio
asked Mar 14 at 22:19
Martin HansenMartin Hansen
707114
$endgroup$
Updated on Friday 15th March 2019 at 5 pm in the light of comments received over the last 24 hours.
The original question was; given the well known variation of Binet's Formula:
$$F_n = frac{phi^n - (-phi)^{-n}}{sqrt{5}}$$
Derive an elegant expression, should one exist, for;
$$F_{log (n)} + F_{log(n)-1}$$
Of course,
$$phi=frac{1+sqrt{5}}{2}$$
Wolfgang Kais has pointed out that this is going to run into technical issues with raising negative numbers to the power of $log(n)$.
Consequently, initially at least, I've been looking instead at
$$F_{f(n)} + F_{f(n)-1}$$
where the function $f$ is sufficiently well behaved to not run into such issues.
In this case,
$$ (sqrt{5})(F_{f(n)} + F_{f(n)-1})=phi^{f(n)}-(-phi)^{-f(n)}+ phi^{f(n)-1}-(-phi)^{-(f(n)-1)}$$
Using the two equivalent facts that
$$1-phi=- frac{1}{phi} :or: phi=1+frac{1}{phi}$$
I proceeded as follows;
$$ (sqrt{5})(F_{f(n)} + F_{f(n)-1})=phi^{f(n)}-big(-frac{1}{phi}big)^{f(n)}+ phi^{f(n)} times phi^{-1}-big(-frac{1}{phi}big)^{(f(n)-1)}$$
$$ =phi^{f(n)}-(1-phi)^{f(n)}+ frac{phi^{f(n)}}{phi}-big(-frac{1}{phi}big)^{f(n)} times big(-frac{1}{phi}big)^{-1}$$
$$ =phi^{f(n)}big(1+frac{1}{phi}big)-(1-phi)^{f(n)}-(1-phi)^{f(n)} times (-phi)$$
$$ =phi times phi^{f(n)} - (1-phi) times (1-phi)^{f(n)}$$
Wolfgang Kais suggested immediately starting with the Fibonacci recurrence relation
$$F_m=F_{m-1}+F_{m-2}$$
from which I deduce that
$$F_{f(n)}+F_{f(n)-1}=F_{f(n)+1}$$
Now, applying the variation on Binet's Formula,
$$F_{f(n)+1} = frac{phi^{f(n)+1} - (-phi)^{-(f(n)+1)}}{sqrt{5}}$$
$$(sqrt{5})(F_{f(n)+1}) = phi times phi^{f(n)} - big(-frac{1}{phi}big)^{f(n)+1}$$
$$ = phi times phi^{f(n)} - (1-phi)^{f(n)+1}$$
$$ =phi times phi^{f(n)} - (1-phi) times (1-phi)^{f(n)}$$
which is the same result as obtained previously which, if nothing else, shows that the variation on Binet's formula satisfies the Fibonacci Recurrence relation.
I'm curious to know if
$$ F_{f(n)+1} = frac{phi times phi^{f(n)}}{sqrt{5}} - frac{(1-phi) times (1-phi)^{f(n)}}{sqrt{5}}$$
is of any use, and is it the best that can be done ?
Possibly I've exhausted what can usefully be done with this question, as I can't see the point of introducing the $log(n)$ function, given the technical hurdles it immediately throws up.
However, any further thoughts are most welcome.
The question originally came from a friend yesterday.
The earlier ask is here : how can i place then values = F(logn)+F(logn-1) in golden ratio Fibonacci series?lognflogn-1-in-golden-ratio-fibonacci-series
elementary-number-theory fibonacci-numbers golden-ratio
elementary-number-theory fibonacci-numbers golden-ratio
asked Mar 14 at 22:19
Martin HansenMartin Hansen
707114
asked Mar 14 at 22:19
Martin HansenMartin Hansen
707114
edited Mar 15 at 17:22
Martin Hansen
asked Mar 14 at 22:19
Martin HansenMartin Hansen
707114
asked Mar 14 at 22:19
Martin HansenMartin Hansen
707114
asked Mar 14 at 22:19
Martin HansenMartin Hansen
707114
707114
1
$begingroup$
Note the same basic question was asked about $2$ hours earlier at how can i place then values = F(logn)+F(logn-1) in golden ratio Fibonacci series?. I appreciate you leaving a comment on the other question about you asking it here, but you should also mention that somewhere in this new question. Thanks.
$endgroup$
– John Omielan
Mar 14 at 22:28
2
$begingroup$
As you can't take a negative number to a non-integer power, $log(n)$ will have to be an integer, and so $F_{log (n)} + F_{log(n)-1} = F_{log(n)+1}$.
$endgroup$
– Wolfgang Kais
Mar 15 at 0:28
1
$begingroup$
@Wolfgang Kais : Of course ! And if I use that in the Binet formula I can get to $$F_{log (n)} + F_{log(n)-1} = frac{ phi times phi^{log (n)}-(1-phi) times (1- phi)^{log(n)}}{sqrt{5}}$$ which is so similar to the expression I mention in my question that I think I have a single minus sign error and so have $phi^2$ instead of $(1 - phi)$. I'll try to reconcile the two methods tomorrow. Thanks !
$endgroup$
– Martin Hansen
Mar 15 at 1:21
1
$begingroup$
You can also try and use the identity $$x^{log a}=a^{log x}.$$ Unfortunately that requires careful interpretation of complex powers unless $a$ and $x$ are both positive real numbers. It does give you a simple estimate for the sum.
$endgroup$
– Jyrki Lahtonen
Mar 15 at 8:16
1
$begingroup$
@MartinHansen: $1-phi$ is also negative. How will you define $(-1)^{log 2}$?
$endgroup$
– Wolfgang Kais
Mar 15 at 8:56
|
show 5 more comments
1
$begingroup$
Note the same basic question was asked about $2$ hours earlier at how can i place then values = F(logn)+F(logn-1) in golden ratio Fibonacci series?. I appreciate you leaving a comment on the other question about you asking it here, but you should also mention that somewhere in this new question. Thanks.
$endgroup$
– John Omielan
Mar 14 at 22:28
2
$begingroup$
As you can't take a negative number to a non-integer power, $log(n)$ will have to be an integer, and so $F_{log (n)} + F_{log(n)-1} = F_{log(n)+1}$.
$endgroup$
– Wolfgang Kais
Mar 15 at 0:28
1
$begingroup$
@Wolfgang Kais : Of course ! And if I use that in the Binet formula I can get to $$F_{log (n)} + F_{log(n)-1} = frac{ phi times phi^{log (n)}-(1-phi) times (1- phi)^{log(n)}}{sqrt{5}}$$ which is so similar to the expression I mention in my question that I think I have a single minus sign error and so have $phi^2$ instead of $(1 - phi)$. I'll try to reconcile the two methods tomorrow. Thanks !
$endgroup$
– Martin Hansen
Mar 15 at 1:21
1
$begingroup$
You can also try and use the identity $$x^{log a}=a^{log x}.$$ Unfortunately that requires careful interpretation of complex powers unless $a$ and $x$ are both positive real numbers. It does give you a simple estimate for the sum.
$endgroup$
– Jyrki Lahtonen
Mar 15 at 8:16
1
$begingroup$
@MartinHansen: $1-phi$ is also negative. How will you define $(-1)^{log 2}$?
$endgroup$
– Wolfgang Kais
Mar 15 at 8:56
1
1
$begingroup$
Note the same basic question was asked about $2$ hours earlier at how can i place then values = F(logn)+F(logn-1) in golden ratio Fibonacci series?. I appreciate you leaving a comment on the other question about you asking it here, but you should also mention that somewhere in this new question. Thanks.
$endgroup$
– John Omielan
Mar 14 at 22:28
$begingroup$
Note the same basic question was asked about $2$ hours earlier at how can i place then values = F(logn)+F(logn-1) in golden ratio Fibonacci series?. I appreciate you leaving a comment on the other question about you asking it here, but you should also mention that somewhere in this new question. Thanks.
$endgroup$
– John Omielan
Mar 14 at 22:28
2
2
$begingroup$
As you can't take a negative number to a non-integer power, $log(n)$ will have to be an integer, and so $F_{log (n)} + F_{log(n)-1} = F_{log(n)+1}$.
$endgroup$
– Wolfgang Kais
Mar 15 at 0:28
$begingroup$
As you can't take a negative number to a non-integer power, $log(n)$ will have to be an integer, and so $F_{log (n)} + F_{log(n)-1} = F_{log(n)+1}$.
$endgroup$
– Wolfgang Kais
Mar 15 at 0:28
1
1
$begingroup$
@Wolfgang Kais : Of course ! And if I use that in the Binet formula I can get to $$F_{log (n)} + F_{log(n)-1} = frac{ phi times phi^{log (n)}-(1-phi) times (1- phi)^{log(n)}}{sqrt{5}}$$ which is so similar to the expression I mention in my question that I think I have a single minus sign error and so have $phi^2$ instead of $(1 - phi)$. I'll try to reconcile the two methods tomorrow. Thanks !
$endgroup$
– Martin Hansen
Mar 15 at 1:21
$begingroup$
@Wolfgang Kais : Of course ! And if I use that in the Binet formula I can get to $$F_{log (n)} + F_{log(n)-1} = frac{ phi times phi^{log (n)}-(1-phi) times (1- phi)^{log(n)}}{sqrt{5}}$$ which is so similar to the expression I mention in my question that I think I have a single minus sign error and so have $phi^2$ instead of $(1 - phi)$. I'll try to reconcile the two methods tomorrow. Thanks !
$endgroup$
– Martin Hansen
Mar 15 at 1:21
1
1
$begingroup$
You can also try and use the identity $$x^{log a}=a^{log x}.$$ Unfortunately that requires careful interpretation of complex powers unless $a$ and $x$ are both positive real numbers. It does give you a simple estimate for the sum.
$endgroup$
– Jyrki Lahtonen
Mar 15 at 8:16
$begingroup$
You can also try and use the identity $$x^{log a}=a^{log x}.$$ Unfortunately that requires careful interpretation of complex powers unless $a$ and $x$ are both positive real numbers. It does give you a simple estimate for the sum.
$endgroup$
– Jyrki Lahtonen
Mar 15 at 8:16
1
1
$begingroup$
@MartinHansen: $1-phi$ is also negative. How will you define $(-1)^{log 2}$?
$endgroup$
– Wolfgang Kais
Mar 15 at 8:56
$begingroup$
@MartinHansen: $1-phi$ is also negative. How will you define $(-1)^{log 2}$?
$endgroup$
– Wolfgang Kais
Mar 15 at 8:56
|
show 5 more comments
1 Answer
1
active
oldest
votes
$begingroup$
@MartinHansen
Sorry i cant comment as i have reputation less than 50
$$F_{log (n)} + F_{log(n)-1} = frac{ phi^{logn}-(-phi)^{-log(n)}}{sqrt5}+frac{ phi^{log(n)-1}-(-phi)^{-(log(n)-1)}}{sqrt5}$$
$${log (n)} + F_{log(n)-1} = frac{ phi^{log(n)}+(phi)^{(log(n)-1)}}{sqrt5}+frac{ phi^{-log(n)}+(phi)^{-(log(n)-1)}}{sqrt5}$$
as in asymptotic time complexity we tends to ignore constants
$${log (n)} + F_{log(n)-1} = { phi^{log(n)}+(phi)^{(log(n)-1)}}+{phi^{-log(n)}+(phi)^{-(log(n)-1)}}$$
How can i proceed further from here
answered Mar 18 at 12:06
S.OhanzeeS.Ohanzee
217
$endgroup$
add a comment |
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@MartinHansen
Sorry i cant comment as i have reputation less than 50
$$F_{log (n)} + F_{log(n)-1} = frac{ phi^{logn}-(-phi)^{-log(n)}}{sqrt5}+frac{ phi^{log(n)-1}-(-phi)^{-(log(n)-1)}}{sqrt5}$$
$${log (n)} + F_{log(n)-1} = frac{ phi^{log(n)}+(phi)^{(log(n)-1)}}{sqrt5}+frac{ phi^{-log(n)}+(phi)^{-(log(n)-1)}}{sqrt5}$$
as in asymptotic time complexity we tends to ignore constants
$${log (n)} + F_{log(n)-1} = { phi^{log(n)}+(phi)^{(log(n)-1)}}+{phi^{-log(n)}+(phi)^{-(log(n)-1)}}$$
How can i proceed further from here
answered Mar 18 at 12:06
S.OhanzeeS.Ohanzee
217
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add a comment |
$begingroup$
@MartinHansen
Sorry i cant comment as i have reputation less than 50
$$F_{log (n)} + F_{log(n)-1} = frac{ phi^{logn}-(-phi)^{-log(n)}}{sqrt5}+frac{ phi^{log(n)-1}-(-phi)^{-(log(n)-1)}}{sqrt5}$$
$${log (n)} + F_{log(n)-1} = frac{ phi^{log(n)}+(phi)^{(log(n)-1)}}{sqrt5}+frac{ phi^{-log(n)}+(phi)^{-(log(n)-1)}}{sqrt5}$$
as in asymptotic time complexity we tends to ignore constants
$${log (n)} + F_{log(n)-1} = { phi^{log(n)}+(phi)^{(log(n)-1)}}+{phi^{-log(n)}+(phi)^{-(log(n)-1)}}$$
How can i proceed further from here
answered Mar 18 at 12:06
S.OhanzeeS.Ohanzee
217
$endgroup$
add a comment |
$begingroup$
@MartinHansen
Sorry i cant comment as i have reputation less than 50
$$F_{log (n)} + F_{log(n)-1} = frac{ phi^{logn}-(-phi)^{-log(n)}}{sqrt5}+frac{ phi^{log(n)-1}-(-phi)^{-(log(n)-1)}}{sqrt5}$$
$${log (n)} + F_{log(n)-1} = frac{ phi^{log(n)}+(phi)^{(log(n)-1)}}{sqrt5}+frac{ phi^{-log(n)}+(phi)^{-(log(n)-1)}}{sqrt5}$$
as in asymptotic time complexity we tends to ignore constants
$${log (n)} + F_{log(n)-1} = { phi^{log(n)}+(phi)^{(log(n)-1)}}+{phi^{-log(n)}+(phi)^{-(log(n)-1)}}$$
How can i proceed further from here
answered Mar 18 at 12:06
S.OhanzeeS.Ohanzee
217
$endgroup$
@MartinHansen
Sorry i cant comment as i have reputation less than 50
$$F_{log (n)} + F_{log(n)-1} = frac{ phi^{logn}-(-phi)^{-log(n)}}{sqrt5}+frac{ phi^{log(n)-1}-(-phi)^{-(log(n)-1)}}{sqrt5}$$
$${log (n)} + F_{log(n)-1} = frac{ phi^{log(n)}+(phi)^{(log(n)-1)}}{sqrt5}+frac{ phi^{-log(n)}+(phi)^{-(log(n)-1)}}{sqrt5}$$
as in asymptotic time complexity we tends to ignore constants
$${log (n)} + F_{log(n)-1} = { phi^{log(n)}+(phi)^{(log(n)-1)}}+{phi^{-log(n)}+(phi)^{-(log(n)-1)}}$$
How can i proceed further from here
answered Mar 18 at 12:06
S.OhanzeeS.Ohanzee
217
answered Mar 18 at 12:06
S.OhanzeeS.Ohanzee
217
answered Mar 18 at 12:06
S.OhanzeeS.Ohanzee
217
answered Mar 18 at 12:06
S.OhanzeeS.Ohanzee
217
217
add a comment |
add a comment |
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Note the same basic question was asked about $2$ hours earlier at how can i place then values = F(logn)+F(logn-1) in golden ratio Fibonacci series?. I appreciate you leaving a comment on the other question about you asking it here, but you should also mention that somewhere in this new question. Thanks.
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– John Omielan
Mar 14 at 22:28
2
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As you can't take a negative number to a non-integer power, $log(n)$ will have to be an integer, and so $F_{log (n)} + F_{log(n)-1} = F_{log(n)+1}$.
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– Wolfgang Kais
Mar 15 at 0:28
1
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@Wolfgang Kais : Of course ! And if I use that in the Binet formula I can get to $$F_{log (n)} + F_{log(n)-1} = frac{ phi times phi^{log (n)}-(1-phi) times (1- phi)^{log(n)}}{sqrt{5}}$$ which is so similar to the expression I mention in my question that I think I have a single minus sign error and so have $phi^2$ instead of $(1 - phi)$. I'll try to reconcile the two methods tomorrow. Thanks !
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– Martin Hansen
Mar 15 at 1:21
1
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You can also try and use the identity $$x^{log a}=a^{log x}.$$ Unfortunately that requires careful interpretation of complex powers unless $a$ and $x$ are both positive real numbers. It does give you a simple estimate for the sum.
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– Jyrki Lahtonen
Mar 15 at 8:16
1
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@MartinHansen: $1-phi$ is also negative. How will you define $(-1)^{log 2}$?
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– Wolfgang Kais
Mar 15 at 8:56