Mixing
A sum involving Fibonacci numbers, $sum_{k=1}^infty F_k/k!$
Let $F_k$ be Fibonacci numbers. I am looking for a closed form of the sum $sum_{k=1}^infty F_k/k!$.
I tried to use Wolfram Alpha, but it is not doing the sum Fibonacci[k]/k! , k=1 to infinity.
Can someone tell what is the problem with WA and what this sum equals to?
sequences-and-series factorial fibonacci-numbers
edited Aug 8 '14 at 14:36
Cleric
3,15442466
asked Nov 14 '13 at 11:45
Waqar Ahmad
204
add a comment |
Let $F_k$ be Fibonacci numbers. I am looking for a closed form of the sum $sum_{k=1}^infty F_k/k!$.
I tried to use Wolfram Alpha, but it is not doing the sum Fibonacci[k]/k! , k=1 to infinity.
Can someone tell what is the problem with WA and what this sum equals to?
sequences-and-series factorial fibonacci-numbers
edited Aug 8 '14 at 14:36
Cleric
3,15442466
asked Nov 14 '13 at 11:45
Waqar Ahmad
204
use mathworld.wolfram.com/BinetsFibonacciNumberFormula.html, proofwiki.org/wiki/Euler-Binet_Formula
– lab bhattacharjee
Nov 14 '13 at 11:51
Appendix: Also need en.wikipedia.org/wiki/Exponential_function#Formal_definition
– lab bhattacharjee
Nov 14 '13 at 12:00
add a comment |
Let $F_k$ be Fibonacci numbers. I am looking for a closed form of the sum $sum_{k=1}^infty F_k/k!$.
I tried to use Wolfram Alpha, but it is not doing the sum Fibonacci[k]/k! , k=1 to infinity.
Can someone tell what is the problem with WA and what this sum equals to?
sequences-and-series factorial fibonacci-numbers
edited Aug 8 '14 at 14:36
Cleric
3,15442466
asked Nov 14 '13 at 11:45
Waqar Ahmad
204
Let $F_k$ be Fibonacci numbers. I am looking for a closed form of the sum $sum_{k=1}^infty F_k/k!$.
I tried to use Wolfram Alpha, but it is not doing the sum Fibonacci[k]/k! , k=1 to infinity.
Can someone tell what is the problem with WA and what this sum equals to?
sequences-and-series factorial fibonacci-numbers
sequences-and-series factorial fibonacci-numbers
edited Aug 8 '14 at 14:36
Cleric
3,15442466
asked Nov 14 '13 at 11:45
Waqar Ahmad
204
edited Aug 8 '14 at 14:36
Cleric
3,15442466
asked Nov 14 '13 at 11:45
Waqar Ahmad
204
edited Aug 8 '14 at 14:36
Cleric
3,15442466
edited Aug 8 '14 at 14:36
Cleric
3,15442466
edited Aug 8 '14 at 14:36
Cleric
3,15442466
3,15442466
asked Nov 14 '13 at 11:45
Waqar Ahmad
204
asked Nov 14 '13 at 11:45
Waqar Ahmad
204
asked Nov 14 '13 at 11:45
Waqar Ahmad
204
204
use mathworld.wolfram.com/BinetsFibonacciNumberFormula.html, proofwiki.org/wiki/Euler-Binet_Formula
– lab bhattacharjee
Nov 14 '13 at 11:51
Appendix: Also need en.wikipedia.org/wiki/Exponential_function#Formal_definition
– lab bhattacharjee
Nov 14 '13 at 12:00
add a comment |
use mathworld.wolfram.com/BinetsFibonacciNumberFormula.html, proofwiki.org/wiki/Euler-Binet_Formula
– lab bhattacharjee
Nov 14 '13 at 11:51
Appendix: Also need en.wikipedia.org/wiki/Exponential_function#Formal_definition
– lab bhattacharjee
Nov 14 '13 at 12:00
use mathworld.wolfram.com/BinetsFibonacciNumberFormula.html, proofwiki.org/wiki/Euler-Binet_Formula
– lab bhattacharjee
Nov 14 '13 at 11:51
use mathworld.wolfram.com/BinetsFibonacciNumberFormula.html, proofwiki.org/wiki/Euler-Binet_Formula
– lab bhattacharjee
Nov 14 '13 at 11:51
Appendix: Also need en.wikipedia.org/wiki/Exponential_function#Formal_definition
– lab bhattacharjee
Nov 14 '13 at 12:00
Appendix: Also need en.wikipedia.org/wiki/Exponential_function#Formal_definition
– lab bhattacharjee
Nov 14 '13 at 12:00
add a comment |
4 Answers
4
active
oldest
votes
A hint:
You can prove for yourself or find in a book a formula of the form
$${rm fib}(k)=alambda^k + bmu^k$$
with certain constants $a$, $b$, $lambda$, $mu$. The requested sum can then be easily written as a sum of two exponentials.
answered Nov 14 '13 at 12:15
Christian Blatter
172k7112326
add a comment |
Im getting the solution:
$$ frac{e^phi - e^varphi}{sqrt{5}} $$
Since it is known that $F_n = frac{phi^n - varphi^n}{sqrt{5}}$.
$$sum_{k=0}^infty frac{F_k}{k!} = sum_{k=0}^infty frac{phi^k - varphi^k}{k!sqrt{5}} = frac{1}{sqrt{5}}left[sum_{k=0}^infty frac{phi^k}{k!} - sum_{k=0}^inftyfrac{varphi^k}{k!}right]$$
Youre welcome.
answered Dec 6 '14 at 21:56
CogitoErgoCogitoSum
1
1
Im sure the down voting was well-deserved.
– CogitoErgoCogitoSum
Dec 6 '14 at 22:09
1
Why bump an old question?
– Cyclohexanol.
Dec 6 '14 at 22:22
1
I dont check the dates on these questions. It was near the top of the list already when I clicked on it. Besides, how does a legitimate and valid answer deserve a down thumb, regardless of how old it is?
– CogitoErgoCogitoSum
Dec 6 '14 at 22:25
1
I did not downvote, just stating a possible reason.
– Cyclohexanol.
Dec 6 '14 at 22:27
add a comment |
Using Mma, Limit[Sum[Fibonacci[k]/k!, {k, 1, p}], p -> Infinity] gives an answer and
Sum[Fibonacci[k]/k!, {k, 1, Infinity}] gives the same. Is your problem with Wolfram Alpha ?
answered Nov 14 '13 at 11:52
Claude Leibovici
119k1157132
computation time out every time. Now please tell the numerical value now.
– Waqar Ahmad
Nov 14 '13 at 12:00
The numerical value is 2.014322733458315736581346. But I think you should establish the analytical value using what lab bhattacharjee sent you. The formula is quite simple.
– Claude Leibovici
Nov 14 '13 at 12:07
According to Maple, with: evalf(sum(fibonacci(k)/k!,k=1..infinity)); Giving: 2.014322733. But, does there exist an "exact" solution? Hmm, "the formula is quite simple". I'm quite curious ..
– Han de Bruijn
Nov 14 '13 at 12:09
@HandeBruijn. Yes and it is (-1 + E**Sqrt(5))/(Sqrt(5)*E**(2/(1 + Sqrt(5))))
– Claude Leibovici
Nov 14 '13 at 12:10
@Claude_Leibovici: Numerically the same, OK, but how to prove it?
– Han de Bruijn
Nov 14 '13 at 12:16
|
show 1 more comment
Solve y'' = y' + y both via power series and then by elementary methods (substitution of exponentials) and setting boundary conditions y(0) = 0, and y'(0) = 1. The result is the following identity:
$sum_{n=1}^infty frac{F_n}{n!} x^n = frac{2}{sqrt{5}} e^frac{x}{2} sinh(frac{sqrt{5}}{2}x)$
If you don't feel like solving the differential equation yourself you can easily verify by substituting each side of the above equation into the differential equation and boundary conditions and then remind yourself via the existence and uniqueness theorem that the two solutions must be the same. Meaning they are the same at the point x=1.
Setting x=1 gives
$sum_{n=1}^infty frac{F_n}{n!} = 2 sqrt{frac{e}{5}} sinh(frac{sqrt{5}}{2}) = frac{e^{phi_+} - e^{phi_-}}{sqrt{5}}$
where $phi_pm = frac{1 pm sqrt{5}}{2}$ are the golden ratio and it's congugate. I.e. the two solutions to the quadratic equation $y^2 = y + 1$. (Note that this is the same as the original differential equation with the order of the derivative replaced by the power of y: $y^{(n)} rightarrow y^n$. Is there a reason for this or is it just an interesting coincidence?)
The numerical value to 1234 decimal places is:
2.0143227334583157365813462554697591356591114695811241821088403766742128397097006637111011319457016312404491456095258793427009006861303485673596801638909160410193717669927023995171048184677900815088560097082872535199871890392172995661754732113194156588018029328948474213043876788216048899918772417664402571617571139330772358975323915258959445759063347851074795879633549994051398333466312388233898626203203269110709326570399986964305691732913351786602533868094239267790638117143928091599301761319009602970110828764189441445889515565022310301247296449293897072832251096862983584326320448058898294430874162784325966040344781269902152316671151601869355616381547811631999794833349897492383827197553543970970099927748899115338963618621987298237221140308693191390458816782367508072038660426129042190625209844154130995652380767395098326152648319830337117521840226032419661899959469674966107384253745331849405956049149602323282511659984227035172708526550586977824786544373977008821755560614931363394267293413160342932047952358802051132151029521708819143029550619292413657149885241166291353201988752249793933504301075491208793989947801376838994380077603990476851019102426649383266999848219352544560978465856984179911553255421975337456890201316269
A fun exercise is to write a C program to generate this (not as easy as you might assume due to the limits of native C data types).
answered Nov 17 '18 at 17:07
1o_o7
11
1
Simply stating a solution is not helpful in order to understand the problem in more detail. Please add you whole evaluation.
– mrtaurho
Nov 17 '18 at 17:33
I have added further detail which should satisfy your complaint but I won't give more than that. It is a fun problem which should be solvable by an undergraduate differential equations student and I'm not about to ruin their fun. I'm sure their professor expects them to solve it themselves and not simply copy a solution off the internet. If it were my class I would also require them to write a computer program to calculate the sum of the series to 15 decimal places (i.e. size of double precision floating point). A student getting 200 decimal places from a program written in C would impress me.
– 1o_o7
Nov 19 '18 at 10:31
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
A hint:
You can prove for yourself or find in a book a formula of the form
$${rm fib}(k)=alambda^k + bmu^k$$
with certain constants $a$, $b$, $lambda$, $mu$. The requested sum can then be easily written as a sum of two exponentials.
answered Nov 14 '13 at 12:15
Christian Blatter
172k7112326
add a comment |
A hint:
You can prove for yourself or find in a book a formula of the form
$${rm fib}(k)=alambda^k + bmu^k$$
with certain constants $a$, $b$, $lambda$, $mu$. The requested sum can then be easily written as a sum of two exponentials.
answered Nov 14 '13 at 12:15
Christian Blatter
172k7112326
add a comment |
A hint:
You can prove for yourself or find in a book a formula of the form
$${rm fib}(k)=alambda^k + bmu^k$$
with certain constants $a$, $b$, $lambda$, $mu$. The requested sum can then be easily written as a sum of two exponentials.
answered Nov 14 '13 at 12:15
Christian Blatter
172k7112326
A hint:
You can prove for yourself or find in a book a formula of the form
$${rm fib}(k)=alambda^k + bmu^k$$
with certain constants $a$, $b$, $lambda$, $mu$. The requested sum can then be easily written as a sum of two exponentials.
answered Nov 14 '13 at 12:15
Christian Blatter
172k7112326
answered Nov 14 '13 at 12:15
Christian Blatter
172k7112326
answered Nov 14 '13 at 12:15
Christian Blatter
172k7112326
answered Nov 14 '13 at 12:15
Christian Blatter
172k7112326
172k7112326
add a comment |
add a comment |
Im getting the solution:
$$ frac{e^phi - e^varphi}{sqrt{5}} $$
Since it is known that $F_n = frac{phi^n - varphi^n}{sqrt{5}}$.
$$sum_{k=0}^infty frac{F_k}{k!} = sum_{k=0}^infty frac{phi^k - varphi^k}{k!sqrt{5}} = frac{1}{sqrt{5}}left[sum_{k=0}^infty frac{phi^k}{k!} - sum_{k=0}^inftyfrac{varphi^k}{k!}right]$$
Youre welcome.
answered Dec 6 '14 at 21:56
CogitoErgoCogitoSum
1
1
Im sure the down voting was well-deserved.
– CogitoErgoCogitoSum
Dec 6 '14 at 22:09
1
Why bump an old question?
– Cyclohexanol.
Dec 6 '14 at 22:22
1
I dont check the dates on these questions. It was near the top of the list already when I clicked on it. Besides, how does a legitimate and valid answer deserve a down thumb, regardless of how old it is?
– CogitoErgoCogitoSum
Dec 6 '14 at 22:25
1
I did not downvote, just stating a possible reason.
– Cyclohexanol.
Dec 6 '14 at 22:27
add a comment |
Im getting the solution:
$$ frac{e^phi - e^varphi}{sqrt{5}} $$
Since it is known that $F_n = frac{phi^n - varphi^n}{sqrt{5}}$.
$$sum_{k=0}^infty frac{F_k}{k!} = sum_{k=0}^infty frac{phi^k - varphi^k}{k!sqrt{5}} = frac{1}{sqrt{5}}left[sum_{k=0}^infty frac{phi^k}{k!} - sum_{k=0}^inftyfrac{varphi^k}{k!}right]$$
Youre welcome.
answered Dec 6 '14 at 21:56
CogitoErgoCogitoSum
1
1
Im sure the down voting was well-deserved.
– CogitoErgoCogitoSum
Dec 6 '14 at 22:09
1
Why bump an old question?
– Cyclohexanol.
Dec 6 '14 at 22:22
1
I dont check the dates on these questions. It was near the top of the list already when I clicked on it. Besides, how does a legitimate and valid answer deserve a down thumb, regardless of how old it is?
– CogitoErgoCogitoSum
Dec 6 '14 at 22:25
1
I did not downvote, just stating a possible reason.
– Cyclohexanol.
Dec 6 '14 at 22:27
add a comment |
Im getting the solution:
$$ frac{e^phi - e^varphi}{sqrt{5}} $$
Since it is known that $F_n = frac{phi^n - varphi^n}{sqrt{5}}$.
$$sum_{k=0}^infty frac{F_k}{k!} = sum_{k=0}^infty frac{phi^k - varphi^k}{k!sqrt{5}} = frac{1}{sqrt{5}}left[sum_{k=0}^infty frac{phi^k}{k!} - sum_{k=0}^inftyfrac{varphi^k}{k!}right]$$
Youre welcome.
answered Dec 6 '14 at 21:56
CogitoErgoCogitoSum
1
Im getting the solution:
$$ frac{e^phi - e^varphi}{sqrt{5}} $$
Since it is known that $F_n = frac{phi^n - varphi^n}{sqrt{5}}$.
$$sum_{k=0}^infty frac{F_k}{k!} = sum_{k=0}^infty frac{phi^k - varphi^k}{k!sqrt{5}} = frac{1}{sqrt{5}}left[sum_{k=0}^infty frac{phi^k}{k!} - sum_{k=0}^inftyfrac{varphi^k}{k!}right]$$
Youre welcome.
answered Dec 6 '14 at 21:56
CogitoErgoCogitoSum
1
edited Dec 6 '14 at 22:03
answered Dec 6 '14 at 21:56
CogitoErgoCogitoSum
1
answered Dec 6 '14 at 21:56
CogitoErgoCogitoSum
1
answered Dec 6 '14 at 21:56
CogitoErgoCogitoSum
1
1
1
Im sure the down voting was well-deserved.
– CogitoErgoCogitoSum
Dec 6 '14 at 22:09
1
Why bump an old question?
– Cyclohexanol.
Dec 6 '14 at 22:22
1
I dont check the dates on these questions. It was near the top of the list already when I clicked on it. Besides, how does a legitimate and valid answer deserve a down thumb, regardless of how old it is?
– CogitoErgoCogitoSum
Dec 6 '14 at 22:25
1
I did not downvote, just stating a possible reason.
– Cyclohexanol.
Dec 6 '14 at 22:27
add a comment |
1
Im sure the down voting was well-deserved.
– CogitoErgoCogitoSum
Dec 6 '14 at 22:09
1
Why bump an old question?
– Cyclohexanol.
Dec 6 '14 at 22:22
1
I dont check the dates on these questions. It was near the top of the list already when I clicked on it. Besides, how does a legitimate and valid answer deserve a down thumb, regardless of how old it is?
– CogitoErgoCogitoSum
Dec 6 '14 at 22:25
1
I did not downvote, just stating a possible reason.
– Cyclohexanol.
Dec 6 '14 at 22:27
1
1
Im sure the down voting was well-deserved.
– CogitoErgoCogitoSum
Dec 6 '14 at 22:09
Im sure the down voting was well-deserved.
– CogitoErgoCogitoSum
Dec 6 '14 at 22:09
1
1
Why bump an old question?
– Cyclohexanol.
Dec 6 '14 at 22:22
Why bump an old question?
– Cyclohexanol.
Dec 6 '14 at 22:22
1
1
I dont check the dates on these questions. It was near the top of the list already when I clicked on it. Besides, how does a legitimate and valid answer deserve a down thumb, regardless of how old it is?
– CogitoErgoCogitoSum
Dec 6 '14 at 22:25
I dont check the dates on these questions. It was near the top of the list already when I clicked on it. Besides, how does a legitimate and valid answer deserve a down thumb, regardless of how old it is?
– CogitoErgoCogitoSum
Dec 6 '14 at 22:25
1
1
I did not downvote, just stating a possible reason.
– Cyclohexanol.
Dec 6 '14 at 22:27
I did not downvote, just stating a possible reason.
– Cyclohexanol.
Dec 6 '14 at 22:27
add a comment |
Using Mma, Limit[Sum[Fibonacci[k]/k!, {k, 1, p}], p -> Infinity] gives an answer and
Sum[Fibonacci[k]/k!, {k, 1, Infinity}] gives the same. Is your problem with Wolfram Alpha ?
answered Nov 14 '13 at 11:52
Claude Leibovici
119k1157132
computation time out every time. Now please tell the numerical value now.
– Waqar Ahmad
Nov 14 '13 at 12:00
The numerical value is 2.014322733458315736581346. But I think you should establish the analytical value using what lab bhattacharjee sent you. The formula is quite simple.
– Claude Leibovici
Nov 14 '13 at 12:07
According to Maple, with: evalf(sum(fibonacci(k)/k!,k=1..infinity)); Giving: 2.014322733. But, does there exist an "exact" solution? Hmm, "the formula is quite simple". I'm quite curious ..
– Han de Bruijn
Nov 14 '13 at 12:09
@HandeBruijn. Yes and it is (-1 + E**Sqrt(5))/(Sqrt(5)*E**(2/(1 + Sqrt(5))))
– Claude Leibovici
Nov 14 '13 at 12:10
@Claude_Leibovici: Numerically the same, OK, but how to prove it?
– Han de Bruijn
Nov 14 '13 at 12:16
|
show 1 more comment
Using Mma, Limit[Sum[Fibonacci[k]/k!, {k, 1, p}], p -> Infinity] gives an answer and
Sum[Fibonacci[k]/k!, {k, 1, Infinity}] gives the same. Is your problem with Wolfram Alpha ?
answered Nov 14 '13 at 11:52
Claude Leibovici
119k1157132
computation time out every time. Now please tell the numerical value now.
– Waqar Ahmad
Nov 14 '13 at 12:00
The numerical value is 2.014322733458315736581346. But I think you should establish the analytical value using what lab bhattacharjee sent you. The formula is quite simple.
– Claude Leibovici
Nov 14 '13 at 12:07
According to Maple, with: evalf(sum(fibonacci(k)/k!,k=1..infinity)); Giving: 2.014322733. But, does there exist an "exact" solution? Hmm, "the formula is quite simple". I'm quite curious ..
– Han de Bruijn
Nov 14 '13 at 12:09
@HandeBruijn. Yes and it is (-1 + E**Sqrt(5))/(Sqrt(5)*E**(2/(1 + Sqrt(5))))
– Claude Leibovici
Nov 14 '13 at 12:10
@Claude_Leibovici: Numerically the same, OK, but how to prove it?
– Han de Bruijn
Nov 14 '13 at 12:16
|
show 1 more comment
Using Mma, Limit[Sum[Fibonacci[k]/k!, {k, 1, p}], p -> Infinity] gives an answer and
Sum[Fibonacci[k]/k!, {k, 1, Infinity}] gives the same. Is your problem with Wolfram Alpha ?
answered Nov 14 '13 at 11:52
Claude Leibovici
119k1157132
Using Mma, Limit[Sum[Fibonacci[k]/k!, {k, 1, p}], p -> Infinity] gives an answer and
Sum[Fibonacci[k]/k!, {k, 1, Infinity}] gives the same. Is your problem with Wolfram Alpha ?
answered Nov 14 '13 at 11:52
Claude Leibovici
119k1157132
answered Nov 14 '13 at 11:52
Claude Leibovici
119k1157132
answered Nov 14 '13 at 11:52
Claude Leibovici
119k1157132
answered Nov 14 '13 at 11:52
Claude Leibovici
119k1157132
119k1157132
computation time out every time. Now please tell the numerical value now.
– Waqar Ahmad
Nov 14 '13 at 12:00
The numerical value is 2.014322733458315736581346. But I think you should establish the analytical value using what lab bhattacharjee sent you. The formula is quite simple.
– Claude Leibovici
Nov 14 '13 at 12:07
According to Maple, with: evalf(sum(fibonacci(k)/k!,k=1..infinity)); Giving: 2.014322733. But, does there exist an "exact" solution? Hmm, "the formula is quite simple". I'm quite curious ..
– Han de Bruijn
Nov 14 '13 at 12:09
@HandeBruijn. Yes and it is (-1 + E**Sqrt(5))/(Sqrt(5)*E**(2/(1 + Sqrt(5))))
– Claude Leibovici
Nov 14 '13 at 12:10
@Claude_Leibovici: Numerically the same, OK, but how to prove it?
– Han de Bruijn
Nov 14 '13 at 12:16
|
show 1 more comment
computation time out every time. Now please tell the numerical value now.
– Waqar Ahmad
Nov 14 '13 at 12:00
The numerical value is 2.014322733458315736581346. But I think you should establish the analytical value using what lab bhattacharjee sent you. The formula is quite simple.
– Claude Leibovici
Nov 14 '13 at 12:07
According to Maple, with: evalf(sum(fibonacci(k)/k!,k=1..infinity)); Giving: 2.014322733. But, does there exist an "exact" solution? Hmm, "the formula is quite simple". I'm quite curious ..
– Han de Bruijn
Nov 14 '13 at 12:09
@HandeBruijn. Yes and it is (-1 + E**Sqrt(5))/(Sqrt(5)*E**(2/(1 + Sqrt(5))))
– Claude Leibovici
Nov 14 '13 at 12:10
@Claude_Leibovici: Numerically the same, OK, but how to prove it?
– Han de Bruijn
Nov 14 '13 at 12:16
computation time out every time. Now please tell the numerical value now.
– Waqar Ahmad
Nov 14 '13 at 12:00
computation time out every time. Now please tell the numerical value now.
– Waqar Ahmad
Nov 14 '13 at 12:00
The numerical value is 2.014322733458315736581346. But I think you should establish the analytical value using what lab bhattacharjee sent you. The formula is quite simple.
– Claude Leibovici
Nov 14 '13 at 12:07
The numerical value is 2.014322733458315736581346. But I think you should establish the analytical value using what lab bhattacharjee sent you. The formula is quite simple.
– Claude Leibovici
Nov 14 '13 at 12:07
According to Maple, with: evalf(sum(fibonacci(k)/k!,k=1..infinity)); Giving: 2.014322733. But, does there exist an "exact" solution? Hmm, "the formula is quite simple". I'm quite curious ..
– Han de Bruijn
Nov 14 '13 at 12:09
According to Maple, with: evalf(sum(fibonacci(k)/k!,k=1..infinity)); Giving: 2.014322733. But, does there exist an "exact" solution? Hmm, "the formula is quite simple". I'm quite curious ..
– Han de Bruijn
Nov 14 '13 at 12:09
@HandeBruijn. Yes and it is (-1 + E**Sqrt(5))/(Sqrt(5)*E**(2/(1 + Sqrt(5))))
– Claude Leibovici
Nov 14 '13 at 12:10
@HandeBruijn. Yes and it is (-1 + E**Sqrt(5))/(Sqrt(5)*E**(2/(1 + Sqrt(5))))
– Claude Leibovici
Nov 14 '13 at 12:10
@Claude_Leibovici: Numerically the same, OK, but how to prove it?
– Han de Bruijn
Nov 14 '13 at 12:16
@Claude_Leibovici: Numerically the same, OK, but how to prove it?
– Han de Bruijn
Nov 14 '13 at 12:16
|
show 1 more comment
Solve y'' = y' + y both via power series and then by elementary methods (substitution of exponentials) and setting boundary conditions y(0) = 0, and y'(0) = 1. The result is the following identity:
$sum_{n=1}^infty frac{F_n}{n!} x^n = frac{2}{sqrt{5}} e^frac{x}{2} sinh(frac{sqrt{5}}{2}x)$
If you don't feel like solving the differential equation yourself you can easily verify by substituting each side of the above equation into the differential equation and boundary conditions and then remind yourself via the existence and uniqueness theorem that the two solutions must be the same. Meaning they are the same at the point x=1.
Setting x=1 gives
$sum_{n=1}^infty frac{F_n}{n!} = 2 sqrt{frac{e}{5}} sinh(frac{sqrt{5}}{2}) = frac{e^{phi_+} - e^{phi_-}}{sqrt{5}}$
where $phi_pm = frac{1 pm sqrt{5}}{2}$ are the golden ratio and it's congugate. I.e. the two solutions to the quadratic equation $y^2 = y + 1$. (Note that this is the same as the original differential equation with the order of the derivative replaced by the power of y: $y^{(n)} rightarrow y^n$. Is there a reason for this or is it just an interesting coincidence?)
The numerical value to 1234 decimal places is:
2.0143227334583157365813462554697591356591114695811241821088403766742128397097006637111011319457016312404491456095258793427009006861303485673596801638909160410193717669927023995171048184677900815088560097082872535199871890392172995661754732113194156588018029328948474213043876788216048899918772417664402571617571139330772358975323915258959445759063347851074795879633549994051398333466312388233898626203203269110709326570399986964305691732913351786602533868094239267790638117143928091599301761319009602970110828764189441445889515565022310301247296449293897072832251096862983584326320448058898294430874162784325966040344781269902152316671151601869355616381547811631999794833349897492383827197553543970970099927748899115338963618621987298237221140308693191390458816782367508072038660426129042190625209844154130995652380767395098326152648319830337117521840226032419661899959469674966107384253745331849405956049149602323282511659984227035172708526550586977824786544373977008821755560614931363394267293413160342932047952358802051132151029521708819143029550619292413657149885241166291353201988752249793933504301075491208793989947801376838994380077603990476851019102426649383266999848219352544560978465856984179911553255421975337456890201316269
A fun exercise is to write a C program to generate this (not as easy as you might assume due to the limits of native C data types).
answered Nov 17 '18 at 17:07
1o_o7
11
1
Simply stating a solution is not helpful in order to understand the problem in more detail. Please add you whole evaluation.
– mrtaurho
Nov 17 '18 at 17:33
I have added further detail which should satisfy your complaint but I won't give more than that. It is a fun problem which should be solvable by an undergraduate differential equations student and I'm not about to ruin their fun. I'm sure their professor expects them to solve it themselves and not simply copy a solution off the internet. If it were my class I would also require them to write a computer program to calculate the sum of the series to 15 decimal places (i.e. size of double precision floating point). A student getting 200 decimal places from a program written in C would impress me.
– 1o_o7
Nov 19 '18 at 10:31
add a comment |
Solve y'' = y' + y both via power series and then by elementary methods (substitution of exponentials) and setting boundary conditions y(0) = 0, and y'(0) = 1. The result is the following identity:
$sum_{n=1}^infty frac{F_n}{n!} x^n = frac{2}{sqrt{5}} e^frac{x}{2} sinh(frac{sqrt{5}}{2}x)$
If you don't feel like solving the differential equation yourself you can easily verify by substituting each side of the above equation into the differential equation and boundary conditions and then remind yourself via the existence and uniqueness theorem that the two solutions must be the same. Meaning they are the same at the point x=1.
Setting x=1 gives
$sum_{n=1}^infty frac{F_n}{n!} = 2 sqrt{frac{e}{5}} sinh(frac{sqrt{5}}{2}) = frac{e^{phi_+} - e^{phi_-}}{sqrt{5}}$
where $phi_pm = frac{1 pm sqrt{5}}{2}$ are the golden ratio and it's congugate. I.e. the two solutions to the quadratic equation $y^2 = y + 1$. (Note that this is the same as the original differential equation with the order of the derivative replaced by the power of y: $y^{(n)} rightarrow y^n$. Is there a reason for this or is it just an interesting coincidence?)
The numerical value to 1234 decimal places is:
2.0143227334583157365813462554697591356591114695811241821088403766742128397097006637111011319457016312404491456095258793427009006861303485673596801638909160410193717669927023995171048184677900815088560097082872535199871890392172995661754732113194156588018029328948474213043876788216048899918772417664402571617571139330772358975323915258959445759063347851074795879633549994051398333466312388233898626203203269110709326570399986964305691732913351786602533868094239267790638117143928091599301761319009602970110828764189441445889515565022310301247296449293897072832251096862983584326320448058898294430874162784325966040344781269902152316671151601869355616381547811631999794833349897492383827197553543970970099927748899115338963618621987298237221140308693191390458816782367508072038660426129042190625209844154130995652380767395098326152648319830337117521840226032419661899959469674966107384253745331849405956049149602323282511659984227035172708526550586977824786544373977008821755560614931363394267293413160342932047952358802051132151029521708819143029550619292413657149885241166291353201988752249793933504301075491208793989947801376838994380077603990476851019102426649383266999848219352544560978465856984179911553255421975337456890201316269
A fun exercise is to write a C program to generate this (not as easy as you might assume due to the limits of native C data types).
answered Nov 17 '18 at 17:07
1o_o7
11
1
Simply stating a solution is not helpful in order to understand the problem in more detail. Please add you whole evaluation.
– mrtaurho
Nov 17 '18 at 17:33
I have added further detail which should satisfy your complaint but I won't give more than that. It is a fun problem which should be solvable by an undergraduate differential equations student and I'm not about to ruin their fun. I'm sure their professor expects them to solve it themselves and not simply copy a solution off the internet. If it were my class I would also require them to write a computer program to calculate the sum of the series to 15 decimal places (i.e. size of double precision floating point). A student getting 200 decimal places from a program written in C would impress me.
– 1o_o7
Nov 19 '18 at 10:31
add a comment |
Solve y'' = y' + y both via power series and then by elementary methods (substitution of exponentials) and setting boundary conditions y(0) = 0, and y'(0) = 1. The result is the following identity:
$sum_{n=1}^infty frac{F_n}{n!} x^n = frac{2}{sqrt{5}} e^frac{x}{2} sinh(frac{sqrt{5}}{2}x)$
If you don't feel like solving the differential equation yourself you can easily verify by substituting each side of the above equation into the differential equation and boundary conditions and then remind yourself via the existence and uniqueness theorem that the two solutions must be the same. Meaning they are the same at the point x=1.
Setting x=1 gives
$sum_{n=1}^infty frac{F_n}{n!} = 2 sqrt{frac{e}{5}} sinh(frac{sqrt{5}}{2}) = frac{e^{phi_+} - e^{phi_-}}{sqrt{5}}$
where $phi_pm = frac{1 pm sqrt{5}}{2}$ are the golden ratio and it's congugate. I.e. the two solutions to the quadratic equation $y^2 = y + 1$. (Note that this is the same as the original differential equation with the order of the derivative replaced by the power of y: $y^{(n)} rightarrow y^n$. Is there a reason for this or is it just an interesting coincidence?)
The numerical value to 1234 decimal places is:
2.0143227334583157365813462554697591356591114695811241821088403766742128397097006637111011319457016312404491456095258793427009006861303485673596801638909160410193717669927023995171048184677900815088560097082872535199871890392172995661754732113194156588018029328948474213043876788216048899918772417664402571617571139330772358975323915258959445759063347851074795879633549994051398333466312388233898626203203269110709326570399986964305691732913351786602533868094239267790638117143928091599301761319009602970110828764189441445889515565022310301247296449293897072832251096862983584326320448058898294430874162784325966040344781269902152316671151601869355616381547811631999794833349897492383827197553543970970099927748899115338963618621987298237221140308693191390458816782367508072038660426129042190625209844154130995652380767395098326152648319830337117521840226032419661899959469674966107384253745331849405956049149602323282511659984227035172708526550586977824786544373977008821755560614931363394267293413160342932047952358802051132151029521708819143029550619292413657149885241166291353201988752249793933504301075491208793989947801376838994380077603990476851019102426649383266999848219352544560978465856984179911553255421975337456890201316269
A fun exercise is to write a C program to generate this (not as easy as you might assume due to the limits of native C data types).
answered Nov 17 '18 at 17:07
1o_o7
11
Solve y'' = y' + y both via power series and then by elementary methods (substitution of exponentials) and setting boundary conditions y(0) = 0, and y'(0) = 1. The result is the following identity:
$sum_{n=1}^infty frac{F_n}{n!} x^n = frac{2}{sqrt{5}} e^frac{x}{2} sinh(frac{sqrt{5}}{2}x)$
If you don't feel like solving the differential equation yourself you can easily verify by substituting each side of the above equation into the differential equation and boundary conditions and then remind yourself via the existence and uniqueness theorem that the two solutions must be the same. Meaning they are the same at the point x=1.
Setting x=1 gives
$sum_{n=1}^infty frac{F_n}{n!} = 2 sqrt{frac{e}{5}} sinh(frac{sqrt{5}}{2}) = frac{e^{phi_+} - e^{phi_-}}{sqrt{5}}$
where $phi_pm = frac{1 pm sqrt{5}}{2}$ are the golden ratio and it's congugate. I.e. the two solutions to the quadratic equation $y^2 = y + 1$. (Note that this is the same as the original differential equation with the order of the derivative replaced by the power of y: $y^{(n)} rightarrow y^n$. Is there a reason for this or is it just an interesting coincidence?)
The numerical value to 1234 decimal places is:
2.0143227334583157365813462554697591356591114695811241821088403766742128397097006637111011319457016312404491456095258793427009006861303485673596801638909160410193717669927023995171048184677900815088560097082872535199871890392172995661754732113194156588018029328948474213043876788216048899918772417664402571617571139330772358975323915258959445759063347851074795879633549994051398333466312388233898626203203269110709326570399986964305691732913351786602533868094239267790638117143928091599301761319009602970110828764189441445889515565022310301247296449293897072832251096862983584326320448058898294430874162784325966040344781269902152316671151601869355616381547811631999794833349897492383827197553543970970099927748899115338963618621987298237221140308693191390458816782367508072038660426129042190625209844154130995652380767395098326152648319830337117521840226032419661899959469674966107384253745331849405956049149602323282511659984227035172708526550586977824786544373977008821755560614931363394267293413160342932047952358802051132151029521708819143029550619292413657149885241166291353201988752249793933504301075491208793989947801376838994380077603990476851019102426649383266999848219352544560978465856984179911553255421975337456890201316269
A fun exercise is to write a C program to generate this (not as easy as you might assume due to the limits of native C data types).
answered Nov 17 '18 at 17:07
1o_o7
11
edited Nov 21 '18 at 11:40
answered Nov 17 '18 at 17:07
1o_o7
11
answered Nov 17 '18 at 17:07
1o_o7
11
answered Nov 17 '18 at 17:07
1o_o7
11
11
1
Simply stating a solution is not helpful in order to understand the problem in more detail. Please add you whole evaluation.
– mrtaurho
Nov 17 '18 at 17:33
I have added further detail which should satisfy your complaint but I won't give more than that. It is a fun problem which should be solvable by an undergraduate differential equations student and I'm not about to ruin their fun. I'm sure their professor expects them to solve it themselves and not simply copy a solution off the internet. If it were my class I would also require them to write a computer program to calculate the sum of the series to 15 decimal places (i.e. size of double precision floating point). A student getting 200 decimal places from a program written in C would impress me.
– 1o_o7
Nov 19 '18 at 10:31
add a comment |
1
Simply stating a solution is not helpful in order to understand the problem in more detail. Please add you whole evaluation.
– mrtaurho
Nov 17 '18 at 17:33
I have added further detail which should satisfy your complaint but I won't give more than that. It is a fun problem which should be solvable by an undergraduate differential equations student and I'm not about to ruin their fun. I'm sure their professor expects them to solve it themselves and not simply copy a solution off the internet. If it were my class I would also require them to write a computer program to calculate the sum of the series to 15 decimal places (i.e. size of double precision floating point). A student getting 200 decimal places from a program written in C would impress me.
– 1o_o7
Nov 19 '18 at 10:31
1
1
Simply stating a solution is not helpful in order to understand the problem in more detail. Please add you whole evaluation.
– mrtaurho
Nov 17 '18 at 17:33
Simply stating a solution is not helpful in order to understand the problem in more detail. Please add you whole evaluation.
– mrtaurho
Nov 17 '18 at 17:33
I have added further detail which should satisfy your complaint but I won't give more than that. It is a fun problem which should be solvable by an undergraduate differential equations student and I'm not about to ruin their fun. I'm sure their professor expects them to solve it themselves and not simply copy a solution off the internet. If it were my class I would also require them to write a computer program to calculate the sum of the series to 15 decimal places (i.e. size of double precision floating point). A student getting 200 decimal places from a program written in C would impress me.
– 1o_o7
Nov 19 '18 at 10:31
I have added further detail which should satisfy your complaint but I won't give more than that. It is a fun problem which should be solvable by an undergraduate differential equations student and I'm not about to ruin their fun. I'm sure their professor expects them to solve it themselves and not simply copy a solution off the internet. If it were my class I would also require them to write a computer program to calculate the sum of the series to 15 decimal places (i.e. size of double precision floating point). A student getting 200 decimal places from a program written in C would impress me.
– 1o_o7
Nov 19 '18 at 10:31
add a comment |
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use mathworld.wolfram.com/BinetsFibonacciNumberFormula.html, proofwiki.org/wiki/Euler-Binet_Formula
– lab bhattacharjee
Nov 14 '13 at 11:51
Appendix: Also need en.wikipedia.org/wiki/Exponential_function#Formal_definition
– lab bhattacharjee
Nov 14 '13 at 12:00