Timeline for Fibonacci Sequence proof by induction
Current License: CC BY-SA 4.0
21 events
| when toggle format | what | by | license | comment | |
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| Jul 21, 2019 at 20:26 | vote | accept | EtherealMist | ||
| Jul 21, 2019 at 13:11 | history | edited | Sil |
edited tags
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| Jul 21, 2019 at 13:01 | answer | added | Sil | timeline score: 1 | |
| Jul 20, 2019 at 23:49 | comment | added | EtherealMist | @Sil I see. Unfortunately we don't deal with infinite sums in this course so it would have to be by induction. I just can't see it, though. Thanks for the help. | |
| Jul 20, 2019 at 22:33 | comment | added | Sil | Not by induction (induction is generally bad when you want to compare inequality of form $f(n)<c$ with a constant $c$ (there is no reason why if $f(n)=0.999999$ should imply that $f(n+1)<1$ as well). This is usually resolved by proving stronger statement instead, something like $f(n)<c-\frac{1}{n}$, which in this case might be tricky to find. Without induction, there is quite an easy proof though, you can prove that the infinite sum $\sum_{i=0}^{\infty} \frac{F_i}{2^{2+i}}=1$, and since individual terms are non-negative, the statement follows. | |
| Jul 20, 2019 at 22:27 | comment | added | EtherealMist | @Sil I don't know. The question was just given that way. Any tips or hints? I'm really having trouble figuring it out. | |
| Jul 20, 2019 at 22:25 | comment | added | Sil | Why $n+2$? This is true even with $\sum_{i=0}^{n} \frac{F_i}{2^{2+i}} < 1$ | |
| Jul 20, 2019 at 20:01 | comment | added | EtherealMist | Does anyone have any hints to solve this? Some additional context: this is a first year intro to proofs course. | |
| Jul 20, 2019 at 19:47 | comment | added | EtherealMist | It is $2^{2+i}$ | |
| Jul 20, 2019 at 9:10 | answer | added | Witold | timeline score: 1 | |
| Jul 20, 2019 at 7:04 | comment | added | DanielWainfleet | @GerryMyerson/ I assumed that $2^2+i$ was a typo and edited it. | |
| Jul 20, 2019 at 7:03 | comment | added | DanielWainfleet | It is more common to define $F_0=0$ and $F_1=F_2=1.$ | |
| Jul 20, 2019 at 6:52 | history | edited | DanielWainfleet | CC BY-SA 4.0 |
typo. punctuation.
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| Jul 20, 2019 at 4:38 | comment | added | Gerry Myerson | You have $2^{2+i}$ in one place, $2^2+i$ in another. They are different. Which (if either) do you want? | |
| Jul 20, 2019 at 2:47 | history | reopened |
mechanodroid dantopa Michael Rozenberg YuiTo Cheng John Omielan |
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| Jul 19, 2019 at 23:20 | review | Reopen votes | |||
| Jul 20, 2019 at 2:47 | |||||
| Jul 19, 2019 at 23:07 | comment | added | EtherealMist | @MarkFischler I edited the question to add more details. Could you clarify your comment? | |
| Jul 19, 2019 at 23:02 | history | edited | EtherealMist | CC BY-SA 4.0 |
Added more information/context.
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| Jul 19, 2019 at 22:53 | comment | added | Mark Fischler | Evidently he means the second of those definitions; otherwise $\frac12$ is an upper bound. | |
| Jul 19, 2019 at 22:43 | history | closed |
Martin R Peter Foreman Henno Brandsma ArsenBerk Sil |
Not suitable for this site | |
| Jul 19, 2019 at 22:36 | history | asked | EtherealMist | CC BY-SA 4.0 |