Wikipedia:Reference desk/Archives/Mathematics/2007 February 20

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February 20

Fibonacci Sequences

Alrighty, I've noticed something interesting with this generator thing. When you increase b by one (in that generator), starting at the second term you're just adding the standard Fibonacci 1,1,2,3,5...

Ex. (a,b)

(10,11)- 10 11 21 32 53 85 138

(10,12)- 10 12 22 34 56 90 146

(10,13)- 10 13 23 36 59 95 154

Is there any specific name for this or is it just there? Deltacom1515 00:37, 20 February 2007 (UTC)

OEIS kindly gives the name, Fibonacci sequence beginning 1 10.  ;) As for number 2, you get Fibonacci sequence beginning 2 10. So unless you consider these specific, I guess... No? x42bn6 Talk 01:18, 20 February 2007 (UTC)

Well, I noticed that as I increase that second number in the parenthesis, b, the number sequence starting at the second term, is increased by 1,1,2,3,5,8,13, etc. which is what I guess you would cal la normal Fibonacci sequence. I just wondered if there was a name/reason for this occurring in this fashion. Deltacom1515 02:16, 20 February 2007 (UTC)

Of course it would, and for the same reason the Fibonacci numbers exist in the first place. Each is the sum of the previous two. I'll take the second entry in your list. At 10, you've added zero to the base sequence, at 12 you've added one. To get the next number, it's (base 0 + Fib 0) + (base 1 + Fib 1), which is (base 0 + base 1) + (Fib 0 + Fib 1), which is (base 2) + (Fib 2). To get the next number, it's (base 1 + Fib 1) + (base 2 + Fib 2), which is (base 1 + base 2) + (Fib 1 + Fib 2), which is (base 3) + (Fib 3). In other words, you're doing exactly what you've described: adding up the base case as before, with the Fibonacci numbers layered on top. Black Carrot 07:40, 20 February 2007 (UTC)

This works because the Fibonacci sequence is generated by a linear recurrence relation. If F(a,b) is the Fibonacci sequence generated by initial terms a and b, then you have noticed that F(10,11)+F(0,1)=F(10,12). This is a special case of the general relationship F(a,b)+F(c,d)=F(a+c,b+d). This will not work for a non-linear relation - for example, if each term is the product of the previous two terms, instead of their sum. Gandalf61 11:31, 20 February 2007 (UTC)

primitive of 1/ln(x)

When investigating the number of primes less than x, i came across the function li(x), which is the integral from 0 to x of 1/ln(x). My computer can estimate this function, but I was wondering if a primitive existed. Does anyone know if a primitive exists or weather it does, the article on this does not seem to give an answer either way.Thepalm 06:25, 20 February 2007 (UTC)

Surely you don't mean $\int _{0}^{x}{\frac {1}{\ln x}}\ dx$? There's too many x's in there. Maybe $\int _{0}^{x}{\frac {1}{\ln t}}\ dt$? Anyway, I don't know of a primitive; something tells me such a thing is hard to find. At least, I hope it is. I'm teaching Calc II at a university right now, and my students would probably celebrate in the streets if they found an integral I couldn't do. =) –King Bee ^{(T ‚Ä¢ C)} 06:30, 20 February 2007 (UTC)

I'm not exactly sure but i think that $\int _{0}^{x}{\frac {1}{\ln t}}\ dt$ is the more correct version. Basically i was looking for a primitive of 1/ln(x). Maybe that clarifies things.Thepalm 07:08, 20 February 2007 (UTC)

Did you check the logarithmic integral function article? It has a few ways to calculate it, but not exactly what you're looking for I think. One thing that's very close that they don't mention is x/lnx. Black Carrot 07:29, 20 February 2007 (UTC)

I was wondering if there was a way to calculate it without using an infinite series, which is what most of the methods there suggest. that article does mention the x/ln(x) approximation, but the values that they give are quite different as x gets very large Thepalm 09:39, 20 February 2007 (UTC)

1/ln(x) is an example of an elementary function whose antiderivative cannot itself be expressed in terms of elementary functions. And even if you expand your list of "elementary functions" to include functions such as li(x), then there will be functions on your new list whose antiderivatives are still not on your new list - see differential Galois theory for a more precise statement of this fact. So you should be pleasantly surprised when a function does turn out to have an antiderivative that you recognise. Of course, the standard examples in introductory calculus courses are carefully chosen to disguise this inconvenient truth ! Gandalf61 11:02, 20 February 2007 (UTC)

Examples in every subject are carefully selected. I remember as a very young student being struck by the realization that answers to assigned problems always came out too cleanly to be accidental. Much later I discovered that research questions often have a different character, and may defy any answer, much less a pretty clean one. We publish the stuff that works out well, and quietly discard the rest. Worse still, we hide the usually messy path of discovery. Consider Hamilton's well-documented route to quaternions; he had fifteen years of failure before the answer suddenly became "obvious". A more modern example is the ordeal Andrew Wiles went through with Fermat's Last Theorem. :-)

Aside from deliberate selection, we also see something else, more intriguing. We can prove that most real numbers cannot be computed, most real-valued functions are wild, and so on. We can also prove that theorems can be true yet not provable. We could say that our entire mathematical playground is a set of measure zero compared to what's really out there. --KSmrq^T 14:33, 20 February 2007 (UTC)

standard integrals

1.A free electron in a metal in a state .¢=sin¶x÷a where 0<x<a or = 0 elsewhere.Determine the average value of x,<x>.Hence,find the uncertainty in position.

a mile

How many steps in a mile68.12.114.162 17:49, 20 February 2007 (UTC)

This would depend on the how big your stride is, but according to this site, the average is about 2000 steps. - Akamad 21:00, 20 February 2007 (UTC)

There are 8 furlongs to a mile. Next determine how many steps you can take in a minute. Then determine how many furlongs you can travel in a fortnight. The mathematics from here onwards is quite simple. Convert minutes to fortnight and divide the steps by the furlongs and further divide by 8. 202.168.50.40 23:29, 20 February 2007 (UTC)

.3(9)

Since, .(9) = 1 and 2.(9) = 3, does .3(9) = 4? Jac _roe 23:33, 20 February 2007 (UTC)

If you mean 3.999..., then yes. Splintercellguy 23:41, 20 February 2007 (UTC)

0.3999999999... = 0.4 202.168.50.40 00:04, 21 February 2007 (UTC)

We have a simple general procedure for converting any repeating decimal to a quotient of integers. To illustrate, take

$r=3.6{\overline {142857}}.\,\!$

The repetition involves six digits, so we multiply six times by ten, or once by 10⁶ (which is 1,000,000).

$1\,000\,000r=3614285.7{\overline {142857}}\,\!$

Since we have shifted by one complete repetition, subtraction eliminates it all.

$1\,000\,000r-r=3614282.1\,\!$

Now it is a simple matter to finish.

${\begin{aligned}999\,999r&{}={\frac {36142821}{10}}\\r&{}={\frac {36142821}{9\,999\,990}}\\&{}={\frac {253}{70}}\end{aligned}}$

The only hard part is the last step, reducing to lowest terms, which is not that hard. --KSmrq^T 11:30, 21 February 2007 (UTC)