Wikipedia:Reference desk/Archives/Mathematics/2007 July 11

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July 11

Leap Years - What Would We Do Without Them?

I have been wondering for quite some time- what would happen if we did not observe leap years? Would our whole calendar just get one day ahead every four years? Would it affect our daylight hours?

Ultimately, yes. There are 97 leap days every 400 years. If we didn't have them, today's date 11 July for example, which currently occurs in winter in the southern hemisphere, would progressively occur earlier and earlier. It would at some stage fall in autumn, then in summer, then in spring, then back to winter, and so on. It would take approximately 400*365/97 = 1505 years for the whole cycle to occur. -- JackofOz 13:31, 11 July 2007 (UTC)

As for daylight hours - those depend, of course, on the time of solar year and not on our calendars. So the longest day will still be in the beginning of summer, but it won't be the middle of June but a different calendar day every time. -- Meni Rosenfeld (talk) 13:43, 11 July 2007 (UTC)

By "summer", I assume you're talking about "natural" or climatic summer, as opposed to official summer. In my country, we're now in winter, which is officially defined as the period 1 June-31 August. This roughly coincides with the natural winter - the times when it's cold, rainy, snowy, foggy etc - which is a good thing. However, if the 103-day scenario you worked out below applied, and seasonal designations were not officially changed, today would now be called 18 October and we'd be calling the season "spring" (officially), but climatically it would still be mid-winter. To correct this, the official start and end dates of the seasons would need to be moved one further day ahead every year that would otherwise have been a leap year. -- JackofOz 10:43, 12 July 2007 (UTC)

Now someone who REALLY likes math figure out exactly what time of the year it would be if there were no leap years starting from the time the Gregorian calendar was established --L^augh! 13:44, 11 July 2007 (UTC)

That's easy. From 1582 to today there have been 106 years which are multiples of 4. Among these, 1700, 1800 and 1900 are not leap years. So we would be 103 days ahead of today, which is October 18. -- Meni Rosenfeld (talk) 13:55, 11 July 2007 (UTC)

In the interest of historical accuracy, I would just like to point out that (a) the Gregorian calendar was not adopted at a single date - Spain and Portugal adopted it first in 1582; Great Britain in 1752; Greece did not adopt it until 1923; (b) the Gregorian calendar did not introduce leap years - in fact, it removed 3 leap years in every 400 years from the previous Julian calendar, which had a leap year every 4 years with no exceptions; (c) if we had stayed with the Julian calendar we would now be 13 days behind the current date. Gandalf61 14:17, 11 July 2007 (UTC)

For the record, I was thinking about mentioning those points, but ultimately concluded that they aren't relevant for the question or the answer. -- Meni Rosenfeld (talk) 14:25, 11 July 2007 (UTC)

Indeed I was being Mr. Sad & Pedantic - I just didn't want anyone to walk away from the RDs with the misguided impression that there were no leap years before the 16th century. Gandalf61 14:46, 11 July 2007 (UTC)

Note that the Muslim calendar is purely lunar with no adjustments to bring it into sync with the solar year, so their dates rotate through the seasons. Gzuckier 15:29, 11 July 2007 (UTC)

Leap years are a fun topic, involving mathematics, astronomy, agronomy, religion, and politics — at least!

The Earth spins around its axis, and the rate at which is does so is fairly constant. It also orbits the Sun, again completing each orbit with almost no timing variation from the one before. In this case, these two periods are — for most practical purposes — not only constant but independent. Not so with the Moon orbiting the Earth; not so with Mercury orbiting the Sun. One fine point (of many): taking the distant stars as a fixed reference, the time for one complete spin is not the same as the time between sun facings (noons), and the latter varies depending on position in orbit. Again, in this case, the difference is small. Although Earth's orbit is nearly a perfect circle, so the distance to the Sun is nearly constant, the Earth's axis of spin is tilted with respect to the plane of the orbit; thus we have seasons.

Sunlight is a major factor in human activity; so is temperature, and so are other seasonal variations. In today's world, when so few people in the developed world engage in agriculture, it is hard to appreciate the critical importance of calendars to survival.

We empirically discover that the length of a day and the length of a year and the length of a month exhibit no exact simple numeric relationships. However, a year (suitably defined) contains 365 days (suitably defined) and a little more. That fraction is our problem. What exactly is it, and how do we accomodate it? And, as requested, what happens if we ignore it? The fraction is almost exactly ¹⁄₄, but a little less. If we ignore it, then a calendar based on days bears no fixed relationship with the seasons.

We cannot ignore religion and politics. (Consider the Islamic calendar, which makes life complicated indeed!) The common Western calendar, the Gregorian calendar, was motivated by an inaccurate use of leap years (the Julian calendar) causing the Roman Catholic Church problems with its yearly major event: Easter. It was supposed to be in the Spring, but over hundreds of years the drift in the calendar became too much to ignore. So the church dictated a new calendar. Not surprisingly, various governments had their own ideas.

All timekeeping, not just calendars, must juggle science and commerce and politics and so on. Consider daylight saving time, with it's shifting of noon. And although few people know about them, leap seconds were recently a cause for internation debate. Ever notice that Wikipedia uses UTC times? With contributors around the world, we certainly couldn't use local time! --KSmrq^T 10:13, 12 July 2007 (UTC)

Cubic equation

I have just discovered Cardano's method for solving cubics and have tested it out, to no avail.

I would appreciate someone telling me where I have gone wrong or what to do next.

I start with the equation

$(x-2)(x-3)(x-4)=0\,\!$

Hence my solutions are obviously 2, 3 and 4 respectively. Expanding this results in

$x^{3}-9x^{2}+26x-24=0\,\!$

Then I have to use the substitution of

$x=t-{a \over 3}$

where a is the co-efficent of $x^{2}\,\!$ and in this case is therefore -9.

Substiuting in gives $x=t+3\,\!$

Substituting this into the original equation leaves

$(t+3)^{3}-9(t+3)^{2}+26(t+3)-24=0\,\!$

This simplifies to

$t^{3}-t=0\,\!$

Which can then simplify to

$t(t+1)(t-1)=0\,\!$

This last stage obviously confirms that the three solutions are consecutive but gives them as -1,0, and 1, which is incorrect.

Thanks. Algebra man 15:49, 11 July 2007 (UTC)

If $t$ takes the values -1, 0 and 1, $x$ takes the values 2, 3 and 4 because $x=t+3$, right? —Bromskloss 16:18, 11 July 2007 (UTC)

Wow, is my face red. Many thanks Algebra man 16:20, 11 July 2007 (UTC)

Hehe, you're welcome. I was a little afraid I was missing some difficult part. —Bromskloss 16:22, 11 July 2007 (UTC)

That's a big mistake my students made whenever they did substitutions. Donald Hosek 17:51, 11 July 2007 (UTC)

Nice to know I'm not alone! Algebra man 17:55, 11 July 2007 (UTC)

In theory can you solve any polynomial

$x^{n}+a_{1}x^{n-1}+a_{2}x^{n-2}...=0\,\!$

by making the substitution

$x=t-{a_{1} \over n}\,\!$

When I say this I suppose I really mean could you solve something relatively large, say n=9 by constantly applying the appropriate substitution? Algebra man

Not so fast. The Abel–Ruffini theorem states that for $n\geq 5$, a solution to a general polynomial equation of nth degree cannot be found using only arithmetic operations and radicals (roots). The substitution you suggest can eliminate the term of degree $n-1$, and other tricks might be able to eliminate other terms as well, but you will almost always end up with an essentially difficult equation. In this particular example, you were lucky enough to have one of the roots equal to the average of the other two, simplifying the equation for t. Otherwise you would have to solve something like $t^{3}+t+3=0$ which is a little harder. In the $n=5$ case you might have to solve something like $t^{5}+t+3=0$ which is much harder. -- Meni Rosenfeld (talk) 21:40, 11 July 2007 (UTC)

For quintics there are Bring Radicals--Cronholm¹⁴⁴ 23:22, 11 July 2007 (UTC)

Of course, there is a general method to find all the rational roots of any polynomial, if for any reason those are of interest. -- Meni Rosenfeld (talk) 21:43, 11 July 2007 (UTC)

If you're offering to explain or point me in the right direction then yes, I am interested. Algebra man 21:47, 11 July 2007 (UTC)

The short version is: Every rational root of a polynomial (with integer coefficients, of course) can be expressed as a divisor of the free term divided by a divisor of the leading term (the coefficient of $x^{n}$). Since there are only finitely many such divisors, you can find them all in finite (and I believe polylog) time. In your example, all roots are rational, and they are all a divisor of 24 divided by a divisor of 1. -- Meni Rosenfeld (talk) 21:51, 11 July 2007 (UTC)

Apparently, this is called the rational root theorem. -- Meni Rosenfeld (talk) 22:09, 11 July 2007 (UTC)

In case you think this is a major result, it isn't: it's a trivial application of the fundamental theorem of arithmetic. Algebraist 23:10, 11 July 2007 (UTC)

Meni Rosenfeld in your above explanation you mentioned 'other tricks' for eliminating other terms of a polynomial. Could you expand on 'other tricks' or show me where to find out about them? Algebra man 00:09, 12 July 2007 (UTC)

Here's a substitution to get rid of the linear term in a cubic ax^3 + bx^2 + cx + d = 0

Let x = y + (√(b^2 - 3ac) - 6ac)/3a

this should produce a cubic of the form My^3 + Ny^2 + S = 0

dunno if that's a "trick" you'd want to see or not..

This is indeed an example, but not a particularly useful one as it doesn't give something simpler than $x^{3}+ax+b=0$. A better example I can suggest appears in Bring radical, which describes some (relatively complicated) transformations which can be used to transform a general quintic equation to one of the form $x^{5}-5x-4t=0$, or equivalently, $x^{5}+x+a=0$. -- Meni Rosenfeld (talk) 12:37, 12 July 2007 (UTC)

Riemann Zeta Function

Many different sources, including wikipedia, state that the riemann zeta function have trivial zeroes at negative integers. However, this argument states that at the negative even integers, (in the example, s=-2,) the function diverges. $\zeta {(-2)}=\sum _{n=1}^{\infty }{\frac {1}{n^{-2}}}=\sum _{n=1}^{\infty }n^{2}=1+4+9+16+25+36...={\frac {\infty (\infty +1)(2\infty +1)}{6}}=\infty$ What's wrong with this argument? Indeed123 18:51, 11 July 2007 (UTC)

The sum form of the function is not valid for all values. For $\zeta (s),s\in \mathbb {Z} ^{-}$, you need to use the functional equation. Donald Hosek 19:02, 11 July 2007 (UTC)

Indeed, that series only defines the zeta function for z with real part >1. The function is extended to the complex plane (except 1) by analytic continuation. Algebraist 20:17, 11 July 2007 (UTC)

Ah, when we have an expression like this, which is only meaningful in a certain region, but can be analytically continued to form the function we really want, doesn't it feel like we (mankind) are missing something? It's like our concepts and notation can only reach a very small part of all the mathematical object out there, and some only if we bend over backward like in this case. Like we are a one-dimensional creature, living on a string, beginning to suspect that that is not all, there must be more dimensions out there! —Bromskloss 20:36, 11 July 2007 (UTC)

If you prefer, Riemann zeta function gives a global definition:

$\zeta (s)={\frac {1}{1-2^{1-s}}}\sum _{n=0}^{\infty }{\frac {1}{2^{n+1}}}\sum _{k=0}^{n}(-1)^{k}{n \choose k}(k+1)^{-s}.$

Algebraist 21:02, 11 July 2007 (UTC)

But it's much easier to see that there are trivial zeros of the zeta function at negative even values from the functional equation, with that $\sin {\frac {\pi s}{2}}$ factor in there plus the fact that no other factors are unbounded (which is why you don't have trivial zeroes at nonnegative even integers). Donald Hosek 22:11, 11 July 2007 (UTC)

Sure, only a madman would actually use that global definition, I just wanted to point out it's there for anyone who dislikes analytic continuation. Algebraist 11:36, 12 July 2007 (UTC)

Is this expression not a relatively efficient way of evaluating the function numerically? -- Meni Rosenfeld (talk) 12:55, 12 July 2007 (UTC)

According to the article, Borwein's method is much better. Algebraist 13:42, 12 July 2007 (UTC)