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
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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:
$$F_n = fracphi^n - (-phi)^-nsqrt5$$
Derive an elegant expression, should one exist, for;
$$F_log (n) + F_log(n)-1$$
Of course,
$$phi=frac1+sqrt52$$
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,
$$ (sqrt5)(F_f(n) + F_f(n)-1)=phi^f(n)-(-phi)^-f(n)+ phi^f(n)-1-(-phi)^-(f(n)-1)$$
Using the two equivalent facts that
$$1-phi=- frac1phi :or: phi=1+frac1phi$$
I proceeded as follows;
$$ (sqrt5)(F_f(n) + F_f(n)-1)=phi^f(n)-big(-frac1phibig)^f(n)+ phi^f(n) times phi^-1-big(-frac1phibig)^(f(n)-1)$$
$$ =phi^f(n)-(1-phi)^f(n)+ fracphi^f(n)phi-big(-frac1phibig)^f(n) times big(-frac1phibig)^-1$$
$$ =phi^f(n)big(1+frac1phibig)-(1-phi)^f(n)-(1-phi)^f(n) times (-phi)$$
$$ =phi times phi^f(n) - (1-phi) times (1-phi)^f(n)$$
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,
$$F_f(n)+1 = fracphi^f(n)+1 - (-phi)^-(f(n)+1)sqrt5$$
$$(sqrt5)(F_f(n)+1) = phi times phi^f(n) - big(-frac1phibig)^f(n)+1$$
$$ = phi times phi^f(n) - (1-phi)^f(n)+1$$
$$ =phi times phi^f(n) - (1-phi) times (1-phi)^f(n)$$
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
$$ F_f(n)+1 = fracphi times phi^f(n)sqrt5 - frac(1-phi) times (1-phi)^f(n)sqrt5$$
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
720114
$endgroup$
|
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:
$$F_n = fracphi^n - (-phi)^-nsqrt5$$
Derive an elegant expression, should one exist, for;
$$F_log (n) + F_log(n)-1$$
Of course,
$$phi=frac1+sqrt52$$
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,
$$ (sqrt5)(F_f(n) + F_f(n)-1)=phi^f(n)-(-phi)^-f(n)+ phi^f(n)-1-(-phi)^-(f(n)-1)$$
Using the two equivalent facts that
$$1-phi=- frac1phi :or: phi=1+frac1phi$$
I proceeded as follows;
$$ (sqrt5)(F_f(n) + F_f(n)-1)=phi^f(n)-big(-frac1phibig)^f(n)+ phi^f(n) times phi^-1-big(-frac1phibig)^(f(n)-1)$$
$$ =phi^f(n)-(1-phi)^f(n)+ fracphi^f(n)phi-big(-frac1phibig)^f(n) times big(-frac1phibig)^-1$$
$$ =phi^f(n)big(1+frac1phibig)-(1-phi)^f(n)-(1-phi)^f(n) times (-phi)$$
$$ =phi times phi^f(n) - (1-phi) times (1-phi)^f(n)$$
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,
$$F_f(n)+1 = fracphi^f(n)+1 - (-phi)^-(f(n)+1)sqrt5$$
$$(sqrt5)(F_f(n)+1) = phi times phi^f(n) - big(-frac1phibig)^f(n)+1$$
$$ = phi times phi^f(n) - (1-phi)^f(n)+1$$
$$ =phi times phi^f(n) - (1-phi) times (1-phi)^f(n)$$
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
$$ F_f(n)+1 = fracphi times phi^f(n)sqrt5 - frac(1-phi) times (1-phi)^f(n)sqrt5$$
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
720114
$endgroup$
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 $$F_log (n) + F_log(n)-1 = frac phi times phi^log (n)-(1-phi) times (1- phi)^log(n)sqrt5$$ 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:
$$F_n = fracphi^n - (-phi)^-nsqrt5$$
Derive an elegant expression, should one exist, for;
$$F_log (n) + F_log(n)-1$$
Of course,
$$phi=frac1+sqrt52$$
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,
$$ (sqrt5)(F_f(n) + F_f(n)-1)=phi^f(n)-(-phi)^-f(n)+ phi^f(n)-1-(-phi)^-(f(n)-1)$$
Using the two equivalent facts that
$$1-phi=- frac1phi :or: phi=1+frac1phi$$
I proceeded as follows;
$$ (sqrt5)(F_f(n) + F_f(n)-1)=phi^f(n)-big(-frac1phibig)^f(n)+ phi^f(n) times phi^-1-big(-frac1phibig)^(f(n)-1)$$
$$ =phi^f(n)-(1-phi)^f(n)+ fracphi^f(n)phi-big(-frac1phibig)^f(n) times big(-frac1phibig)^-1$$
$$ =phi^f(n)big(1+frac1phibig)-(1-phi)^f(n)-(1-phi)^f(n) times (-phi)$$
$$ =phi times phi^f(n) - (1-phi) times (1-phi)^f(n)$$
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,
$$F_f(n)+1 = fracphi^f(n)+1 - (-phi)^-(f(n)+1)sqrt5$$
$$(sqrt5)(F_f(n)+1) = phi times phi^f(n) - big(-frac1phibig)^f(n)+1$$
$$ = phi times phi^f(n) - (1-phi)^f(n)+1$$
$$ =phi times phi^f(n) - (1-phi) times (1-phi)^f(n)$$
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
$$ F_f(n)+1 = fracphi times phi^f(n)sqrt5 - frac(1-phi) times (1-phi)^f(n)sqrt5$$
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
720114
$endgroup$
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:
$$F_n = fracphi^n - (-phi)^-nsqrt5$$
Derive an elegant expression, should one exist, for;
$$F_log (n) + F_log(n)-1$$
Of course,
$$phi=frac1+sqrt52$$
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,
$$ (sqrt5)(F_f(n) + F_f(n)-1)=phi^f(n)-(-phi)^-f(n)+ phi^f(n)-1-(-phi)^-(f(n)-1)$$
Using the two equivalent facts that
$$1-phi=- frac1phi :or: phi=1+frac1phi$$
I proceeded as follows;
$$ (sqrt5)(F_f(n) + F_f(n)-1)=phi^f(n)-big(-frac1phibig)^f(n)+ phi^f(n) times phi^-1-big(-frac1phibig)^(f(n)-1)$$
$$ =phi^f(n)-(1-phi)^f(n)+ fracphi^f(n)phi-big(-frac1phibig)^f(n) times big(-frac1phibig)^-1$$
$$ =phi^f(n)big(1+frac1phibig)-(1-phi)^f(n)-(1-phi)^f(n) times (-phi)$$
$$ =phi times phi^f(n) - (1-phi) times (1-phi)^f(n)$$
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,
$$F_f(n)+1 = fracphi^f(n)+1 - (-phi)^-(f(n)+1)sqrt5$$
$$(sqrt5)(F_f(n)+1) = phi times phi^f(n) - big(-frac1phibig)^f(n)+1$$
$$ = phi times phi^f(n) - (1-phi)^f(n)+1$$
$$ =phi times phi^f(n) - (1-phi) times (1-phi)^f(n)$$
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
$$ F_f(n)+1 = fracphi times phi^f(n)sqrt5 - frac(1-phi) times (1-phi)^f(n)sqrt5$$
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
720114
asked Mar 14 at 22:19
Martin HansenMartin Hansen
720114
edited Mar 15 at 17:22
Martin Hansen
asked Mar 14 at 22:19
Martin HansenMartin Hansen
720114
asked Mar 14 at 22:19
Martin HansenMartin Hansen
720114
asked Mar 14 at 22:19
Martin HansenMartin Hansen
720114
720114
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 $$F_log (n) + F_log(n)-1 = frac phi times phi^log (n)-(1-phi) times (1- phi)^log(n)sqrt5$$ 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 $$F_log (n) + F_log(n)-1 = frac phi times phi^log (n)-(1-phi) times (1- phi)^log(n)sqrt5$$ 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 $$F_log (n) + F_log(n)-1 = frac phi times phi^log (n)-(1-phi) times (1- phi)^log(n)sqrt5$$ 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 $$F_log (n) + F_log(n)-1 = frac phi times phi^log (n)-(1-phi) times (1- phi)^log(n)sqrt5$$ 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
116
New contributor
S.Ohanzee is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
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
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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
116
New contributor
S.Ohanzee is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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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
116
New contributor
S.Ohanzee is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
@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
116
New contributor
S.Ohanzee is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Mar 18 at 12:06
S.OhanzeeS.Ohanzee
116
New contributor
S.Ohanzee is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Mar 18 at 12:06
S.OhanzeeS.Ohanzee
116
answered Mar 18 at 12:06
S.OhanzeeS.Ohanzee
116
116
New contributor
S.Ohanzee is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
S.Ohanzee is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
S.Ohanzee is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
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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.
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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 $$F_log (n) + F_log(n)-1 = frac phi times phi^log (n)-(1-phi) times (1- phi)^log(n)sqrt5$$ 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