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May 22[edit]

derivative of sin(ax)[edit]

I'm pretty sure that the derivative of sin(ax) for some constant a is a cos(ax), but I'm having trouble proving that. I keep getting stuck in a puddle of messed up Δxs. I have got so far:
${\frac {d}{dx}}\sin(ax)=\lim _{\Delta x\to 0}{\sin(ax+a\Delta x)-\sin(ax) \over \Delta x}$
$=\lim _{\Delta x\to 0}{\sin(ax)\cos(a\Delta x)+\cos(ax)\sin(a\Delta x)-\sin(ax) \over \Delta x}$
$=\lim _{\Delta x\to 0}{\sin(ax)(\cos(a\Delta x)-1)+\cos(ax)\sin(a\Delta x) \over \Delta x}$
$=\lim _{\Delta x\to 0}{\sin(ax)(\cos(a\Delta x)-1) \over \Delta x}+\lim _{\Delta x\to 0}{\cos(ax)\sin(a\Delta x) \over \Delta x}$ If I apply the limit so that it is cos(0)-1/0=0 and sin(0)/0=1, I end up with cos(ax), which is not right. I have a feeling how the limit is applied at b:Calculus/Derivatives of Trigonometric Functions is not right in the same way as I applied the limit not right. 76.229.239.223 (talk) 01:53, 22 May 2010 (UTC)[reply]

Each step appears to be correct to me, but your claim that you end up with cos(ax) is incorrect; you're just not evaluating the limit correctly. Hint: try multiplying by a/a, putting the denominator a inside the limit, and then doing some substitution. --COVIZAPIBETEFOKY (talk) 02:12, 22 May 2010 (UTC)[reply]

The traditional way to prove this is with the chain rule. -- Meni Rosenfeld (talk) 17:46, 22 May 2010 (UTC)[reply]

First you need to unlearn the rule that says sin 0/0 = 1 and learn this:

{\begin{cases}\lim _{u\to 0}{\dfrac {\sin u}{u}}=1&({\text{right}})\\[12pt]\lim _{u\to 0}{\dfrac {\sin(au)}{u}}=1&({\text{wrong}})\end{cases}}

Then try a simple substitution:

\lim _{u\to 0}{\frac {\sin(au)}{u}}=a\lim _{u\to 0}{\frac {\sin(au)}{au}}{\text{ (since the }}a\;{\text{s cancel)}}=a\lim _{u\to 0}{\frac {\sin w}{w}},

and then instead of

u\to 0\,

you need to write

w\to {\text{something}}.\,

Since w = au, it's not hard to figure out what should go in the role of "something", and then use one of the limits mentioned above. Michael Hardy (talk) 22:25, 22 May 2010 (UTC)[reply]

Stable ranges for Albedo and Emissivity of Earth[edit]

The Earth has a Global Annual Average Temperature of 14°C.
The Global Average Albedo is 0.30 and the concomitant Emissivity is 0.622.
Hence:
$T_{p}={\frac {((6.955\times 10^{8})^{2})\times (5778^{4})\times (1-0.3)}{(4\times 0.622\times 149597876600)^{2}}}={14}$
(derived from Luminosity)

So I was wondering how far that 14°C can shift and the planet still sustain "liquid water,"
what range is possible, and also what is range for the Albedo, and the range for the Emissivity,
for a stable habitable temperature for Earth life?
For clarity, this question is only about the Earth, as a reference point, not exo-moons.
Earth's temperature range is 60°C to -90°C, or 150 degrees.
There must be limits.
Global Annual Average Temperature Maximum: ??
Global Annual Average Temperature Minimum: ??
Minimum Albedo: ??
Minimum Emissivity: ??
Maximum Albedo: ??
Maximum Emissivity: ??
24.78.178.147 (talk) 01:30, 23 May 2010 (UTC)[reply]

Average doesn't tell you that water is not liquid in Antarctica, if this is anything to do with the real world the answer is rather complicated. That's part of the problem about climate change and why it needs so many scientists working hard to try and get predictions more accurate. Even albedo and emissivity doesn't tell you everything even if the whole world was treated the same, it differs at different wavelengths and light can get transformed to heat when striking things. Besides that there may be other sources of heat like Europa (moon) which may have a huge ocean of water beneath its thick icy crust. The list of special factors goes on, I suppose one could set wide margins but even then there's microbes living at temperatures above the boiling point of water or below its freezing point on earth. Dmcq (talk) 12:22, 22 May 2010 (UTC)[reply]

If diameter divided Circumference, what we will get?[edit]

If diameter divided Circumference, what we will get? —Preceding unsigned comment added by Svenand496 (talk • contribs) 13:54, 22 May 2010 (UTC)[reply]

The circumference C of a circle is related to the diameter D via the formula

C=\pi D

. Therefore, the value of

{\frac {D}{C}}

is equal to

{\frac {1}{\pi }}

. PS T 15:45, 22 May 2010 (UTC)[reply]

The diameter dividing the circumference (of the same circle) is the inverse of the diameter being divided by the circumference (of the same circle).

"C/D = pi" is the inverse of "D/C = 1/pi". -- Wavelength (talk) 16:40, 22 May 2010 (UTC)[reply]

See Active Voice, Passive Voice. -- Wavelength (talk) 17:58, 22 May 2010 (UTC)[reply]

See simple:Circle, simple:Circumference, simple:Diameter, and simple:Pi. -- Wavelength (talk) 18:28, 22 May 2010 (UTC)[reply]

In general a diameter is a longest line across a figure so the diameters of a square are its two diagonals. And dividing can mean cutting. So here a diameter could divide the circumference into two lots of two equal lines joined at right angles. Just being pedantic. Though I guess you mean the value pi as explained above. Dmcq (talk) 12:04, 23 May 2010 (UTC)[reply]

Quick query on index 5 subgroups of A5[edit]

Hi all,

I'm looking at order 12 subgroups of the alternating group A5, and I'm aware that the only such groups are those which fix 1 element, i.e. are isomorphic to A4: however, I'm wondering if anyone could suggest a fairly clean, fast way of proving that these are the only such subgroups: i.e. that if a subgroup of A5 moves every element {1, 2, ..., 5}, it cannot be of order 12.

Cauchy's Theorem tells us any such group of order 12 must have an element of order 2 in (i.e. a product of 2 disjoint transpositions) and an element of order 3 in (a 3-cycle), and cannot have an element of order 5, a 5-cycle. I've tried something like showing that if H moves all 5 elements then it must contain an element of order 5, but there are loads of possible combinations of 2x2-cycles and 3-cycles which contain all 5 elements in {1,...,5}, so I didn't really get anywhere with that, and it certainly wasn't neat - if possible, I'd like to try to find a quick approach which doesn't require testing many different cases of permutations (for example, I used Cayley's theorem to eliminate all nontrivial subgroups of index n<5 in 2 lines) - could anyone suggest anything?

I'm allowed to use any general results I choose, but obviously the less machinery I need to complete the problem, the better - if there's no way to do this concisely without using any well-known results however then that's fine too! Thanks very much in advance, 131.111.185.68 (talk) 14:06, 22 May 2010 (UTC)[reply]

Subgroups of S5 come in a few varieties: (1) those that fix a point are conjugate to a subgroup of S1 x S4 (so the only one of order 12 is A4), (2) those that are not transitive but move all 5 points are contained in S2 x S3 (so the only one of order 12 is S2 x S3), (3) those that are transitive, and so have order divisible by 5 (so none of order 12). The groups of type 2 are not contained in A5, so that leaves only the conjugates of A4 by elements of S5, that is, the subgroups you described. This is a standard first step in analyzing a permutation group: it is a subgroup of a direct product of symmetric groups on its orbits, also known as examining its transitive constituents. JackSchmidt (talk) 14:49, 22 May 2010 (UTC)[reply]

(Edit Conflict) Let G be the alternating group on five letters and let S be an arbitrary Sylow 2-subgroup of G. Then the normalizer, N, of S in G must have order 12. (Proof: The subgroup N must have index either 1, 3, 5 or 15 in G. Neither 1 nor 3 is possible since that would contradict the simplicity of G, and if N had index 15 in G, S would be a self-normalizing self-centralizing Sylow subgroup of G. However, this would imply that there is no fusion in S (it is said that there is no fusion in S if any two elements of S conjugate in G are already conjugate in in S), so that (by the focal subgroup theorem) $G'\cap S=S'={1}$ , and $G'<G$ ; a contradiction of the simplicity of G. Therefore, N has index 5 in G.) Conversely, if N is any subgroup of G of index 5, N contains a Sylow 2-subgroup S, which must also be a Sylow 2-subgroup of G. Furthermore, S must be normal in N (Proof: If S is not normal in N, let T be a Sylow 3-subgroup of N and note that T must be normal in N. Let P be the core of S in N and note that P has order 2. It follows that the product PT has order 6 and hence is normal in N. In particular, $N'=PT$ (where, N' is the commutator subgroup of N), and since there is no fusion in S, it follows from the focal subgroup theorem that $G'\cap S=S'={1}$ . This is a contradiction since $G'\cap S=P$ .) In particular, a subgroup N of G has order 12 if and only if it is the normalizer of some Sylow 2-subgroup of G. However, we know that for any Sylow 2-subgroup S of G, there exists $\alpha \in \{1,2,3,4,5\}$ such that S is the subgroup of elements of G of order 2 that fixes $\alpha$ . Such an S must be normal in the subgroup, N, of G that fixes $\alpha$ (since the Klein group may be naturally viewed as a normal subgroup of the alternating group on 4 letters). The result now follows. (I apologize for the rather complex argument; there is almost certainly a much simpler argument than this, but I am a bit tired at the moment and this was the first argument that came to my mind (for instance, I do not think that one needs to use either transfer or the focal subgroup theorem to prove basic facts about groups of order 12!).) PS T 15:43, 22 May 2010 (UTC)[reply]

Looking back over my above post again, note that I actually made an error: the commutator subgroup of a group of order 12 that possesses a normal subgroup of order 3, is the normal subgroup of order 3 (since the quotient would be a group of order 4 and hence would be abelian). However, I have provided an alternative proof of the result in question below, which is free of errors. PS T 03:20, 23 May 2010 (UTC)[reply]

Thankyou both, that's been very helpful! (And I'll be sure to read up on fusion and the focal subgroup theorem ;-)) 131.111.185.68 (talk) 16:33, 22 May 2010 (UTC)[reply]

If you wish to learn about fusion and the focal subgroup theorem, note that they are subsumed within the general theory of transfer. I would recommend Martin Isaacs' Algebra: A Graduate Course; the first 10 chapters of the book cover some of the fundamentals of finite group theory, and the ninth chapter covers transfer. I think the prerequisite for the ninth chapter of this book would probably be chapters 1-5 and chaper 8 (i.e., basic group theory, group actions and Sylow theory, solvable and nilpotent groups etc.); from the looks of your post, you already seem to know most of this so you could probably read chapter 9 without going through the previous chapters. The same author also has a more specialized book on finite groups, entitled Finite Group Theory by Martin Isaacs, so if you do intend to pursue finite group theory in greater depth at some point in the future, you might find that book useful as well.

Anyway, I thought that I would mention a (much) simpler (though considerably longer) proof of the fact that a subgroup, N, of G (the alternating group on five letters) of order 12 must be the normalizer of some Sylow 2-subgroup of G.

Proof: If N is a subgroup of G of order 12, N contains a Sylow 2-subgroup, S, of G and we will show that N is in fact the normalizer of S in G. Note that if this is false, S is not normal in N, and consequently, a Sylow 3-subgroup, T, of G must be normal in N. (To see this, note that if T were not normal in N, there would have to be four Sylow 3-subgroups of N (each of order 3). If you use the "inclusion-exclusion principle" to count the number of elements of order 3 in N, we would obtain

4(3-1)+1=9

(since the intersection of two distnct Sylow 3-subgroups of N is trivial), and this would imply that there is only room for one Sylow 2-subgroup of N (N has only 12 elements!), and thus S has to be normal in N (since S would be the unique Sylow 2-subgroup of N).)

Now, since T is normal in N, its normalizer in N is, of course, N. We may ask: what is the centralizer, C, of T in N? Note that the quotient N/C is naturally a subgroup of the automorphism group of T (also known as the "N/C theorem"). Since T is cyclic of order 3, its automorphism group has order 2, and hence C has order at least 6. Thus either C has order 6 or C has order 12 (and equals N).

If the former, note that C will have a subgroup P of order 2 by Cauchy's theorem, and P would be normal in N since it would have to be the core of S in G. Since T is also normal in N, and hence normal in C, C would be the internal direct product of two cyclic groups of relatively prime orders, and hence would be cyclic of order 6. However, the alternating group on 5 letters cannot have either a 6-cycle, or a product of a 2-cycle and a 3-cycle, and hence cannot have any element of order 6; a contradiction.

If the latter, that is, if the centralizer, C, were N, T would be the center of the group N. However, we know that N decomposes as a product of S and T (i.e.,

N=ST

since the order of the product is 12). Therfore, for any n in N,

n=st

for some s in S and t in T. Conjugating S by n yields:

S^{n}=(S^{st})=(S^{s})^{t}=S^{t}=S

where the last equality follows from the assumption that t is central in N (T is the center of N). This implies that S is normal in N; a contradiction.

Since no matter what we assume, given the hypothesis that S is not normal in N, we are led to a cotradiction, S must be normal in N. The normalizer of S in G now has to be N since N is maximal in G (since it has prime index 5), and the normalizer contains N (if it were not N, it would have to be G by the maximality of N and this contradicts the simplicity of G).

Phew! The proof is long; there are some relatively simple facts about permutation groups in general that can be used to derive this result in a much simpler manner, but this proof is almost purely "permutation group theory free" (with the exception of our use of the fact that the alternating group on five letters cannot have an element of order 6). I still used a couple of non-trivial facts (mostly about group actions), such as the notion of "the core of a subgroup", which I think you might already be familiar with (if not, I do not mind explaining it if you note it here).

Finally, as I mentioned in my previous post, if any subgroup, N, of order 12 is the normalizer of some subgroup of order 4 (i.e., a Sylow 2-subgroup), it readily follows that N is the subgroup of the alternating group on 5 letters, fixing a letter. (We actually do not need the fact that the normalizer of a Sylow 2-subgroup of G has order 12 (which I proved in my last post); we only really need to know that any subgroup of order 12 is a normalizer of this form.) PS T 03:14, 23 May 2010 (UTC)[reply]

Euler's constant again[edit]

Some times ago I asked what is the use of Euler's gamma. As I get no comments, then I evaluated the following new constants using Euler's reasoning:( changing dx value)

dx-----------------------(SUM-Ln)----------new constants

0,001--------------0,000507602
0,01----------------0,005015955
0,1------------------0,050840136
1---------------------0,577223298----Euler( confirmed when dx=1)
2---------------------1,270370479
3,850601754----2,718281828----e
4,361917092----3,141592654----Pi
10-------------------8,121177481
16,17709104----13,87338721----TOP=2*ln

According these evaluations, Euler's gamma has not a practical use,if not an "artistic math show".TASDELEN (talk) 19:50, 22 May 2010 (UTC)[reply]

Again? And you call this "no comments"? DVdm (talk) 19:59, 22 May 2010 (UTC)[reply]

Yes and what's your practical use, apart from the artistic one? --pm a 20:13, 22 May 2010 (UTC)[reply]

At a basic level - to save time and ink, just as any shorthand representation of a value, concept or idea does. The same reason poeple write USA instead of United States of America etc.. 20:23, 22 May 2010 (UTC) —Preceding unsigned comment added by 87.102.18.191 (talk)

Gentlemen,if these are "comments",then you have time to comment my "artistic values".As you see,except for dx=1,Euler's constant is not confirmed.Thanks.TASDELEN (talk) 21:17, 22 May 2010 (UTC)[reply]

An appendix,from an earlier talk about this topic

All these mathematics may be correct.What I want to say is,where for example this difference is involved in our naive evaluations.Suppose I search for an integral,but I am unable to solve it ,then I am approaching the integral by summing Delta(y).I understand, Euler says:A difference between integral and sum will be existing.So,shall we say that (a sum-an integral= this constant).Or shall we say that ONLY for the case of logarithmic evaluaton,Euler constant is valid.Or ,this constant is correct when we consider ONLY 1/2,1/3,1/4,....1/n when n tends to infinity.I am looking for a practical use.Mathematics should not be considered as an artistic art like painting or singingTASDELEN (talk) 20:45, 12 May 2010 (UTC)TASDELEN (talk) 21:33, 22 May 2010 (UTC)[reply]

You need to go and learn some calculus before trying to have a discussion about the Euler–Mascheroni constant. The expression "dx=1" makes absolutely no sense and shows that you don't have a clue about calculus. There is one part of your confusion I may be able to help with, though: There are infinitely many constants in maths, but we only give names to the interesting and/or useful ones (they may not be obviously useful in real life, but they are useful in maths). That doesn't mean the others are any less real, we just don't see the need to talk about them. --Tango (talk) 21:48, 22 May 2010 (UTC)[reply]

You seem to be suggesting that using steps of 1 to is some sort of inflated conceit on the part of mathematicians; yet the numerous applications of the constant obtained by using that suggest that it's an entirely practicable and useful (at least theoretically) conceit.

Perhaps an alternative set of constants should be proposed, lets call it Tasdelen's number(s), actually a function, T(n) , where T(1)=Euler's constant. Would that silence you? 87.102.18.191 (talk) 21:59, 22 May 2010 (UTC)[reply]

Actually, T(1)=Euler's constant, only when dx=1. --84.221.208.86 (talk) 22:15, 22 May 2010 (UTC) [reply]

That's what I meant T(n)=T(dx)87.102.18.191 (talk) 22:20, 22 May 2010 (UTC)[reply]

In response to our original poster's earlier question, I linked to a number of Wikipedia articles that mention various uses for the Euler–Mascheroni constant. Michael Hardy (talk) 22:29, 22 May 2010 (UTC)[reply]

This is a reference desk not a soapbox. You stated no question. You should give a question about mathematics if you wish to use this reference desk. Dmcq (talk) 22:35, 22 May 2010 (UTC)[reply]

Thanks for the comments.The stated question was:"What is the practical use of Euler's constant?".I appreciate "There are infinitely many constants in maths, but we only give names to the interesting and/or useful ones (they may not be obviously useful in real life, but they are useful in maths)".To give a "sense" to (dx=1),you may consider SUM(dx/x) when dx=1 and x=1+n*dx when (n) tends to infinity.TASDELEN (talk) 11:16, 23 May 2010 (UTC)[reply]

The practical use of the constant is that it makes it easier to work with all the formulae in which it appears, which themselves have practical uses (eg. in cryptography, as already mentioned). I think know what you are trying to say with dx=1 - you are talking about the size of step used in the integration - however "dx" means the size of step after you've let it tend to zero, so it is infinitesimal (well, it's usually thought of as just abstract notation, but it can be thought of as an infinitesimal). You need to use a different notation to describe the step size prior to taking the limit - I like

\Delta x

, personally. --Tango (talk) 16:03, 23 May 2010 (UTC)[reply]

BTW, the notation

\Delta x

for (usually small) increments of the variable x maybe sometimes could have some use (so e.g one knows that

\Delta y

refers to an increments of the variable y and so on. But the counter-indications are so many... What if I need to exchange things and consider permutations like f(y+

\Delta x

, x+

\Delta y

) + f(x+

\Delta y

, y+

\Delta x

)+..? a mess, IMO. Besides,

\Delta x

is misleading because it looks like an operator

\Delta

applied to the object x, and it is not. Lastly,

\Delta

is already the symbol for the Laplacian. I can't see what's wrong in any symbol as h (or whatever) to indicate the increment from x to x+h. To summarize: I hate the notation

\Delta x

for increments! I wish it were banned forever from maths! ;-) --pm a 16:58, 23 May 2010 (UTC)[reply]

Is dx(x,Δx) = Δx as in differential of a function okay by you then or does that fill you with revulsion? :) Dmcq (talk) 18:23, 23 May 2010 (UTC)[reply]

By SUM(dx/x), I meant SUM(Delta.x/x) where x=1+n*Delta.x and n=0 to infinity, with any arbitrarily chosen 0<Delta.x<infinity. When I choose Delta.x=1, Euler's constant is confirmed, otherwise not confirmed.I prefer to write Delta.x=hTASDELEN (talk) 22:01, 23 May 2010 (UTC)[reply]

What do you mean by "confirmed"? Euler's constant is what it is. There is no reason to expect to get the same answer if you change the formula. --Tango (talk) 22:05, 23 May 2010 (UTC)[reply]

The formula remain the same. The value of Delta.x (or h) is changing. When h=1, Euler gamma is what it is. When h=0,001, gamma value=0,000507602. When h=3,850601754 gamma value=2,718281828. For h=1, Euler's constant is existing.For other values of (h), it does not exist. Existence=confirmed; non-existence=non-confirmed.TASDELEN (talk) 08:07, 24 May 2010 (UTC)[reply]

Our article gives two equivalent definitions of the constant:

\gamma =\lim _{n\rightarrow \infty }\left(\sum _{k=1}^{n}{\frac {1}{k}}-\ln(n)\right)=\int _{1}^{\infty }\left({1 \over \lfloor x\rfloor }-{1 \over x}\right)\,dx.

I don't see any

\Delta x

's or h's in either of those expressions. What are you talking about? --Tango (talk) 00:29, 25 May 2010 (UTC)[reply]

Consider k=1+z*h with (z=0;1;2;3;4.....n..) and use the above expression.

When h=0,001 (Sum-Ln)=0,000507602...

When h=1 (Sum-Ln)=Euler's constant

When h=3,850601754.. (Sum-Ln)=2,718281828..

Only for h=1 Euler's constant is existing.Otherwise it does not exist.(SUM-Ln) is a function of (h).TASDELEN (talk) 21:11, 25 May 2010 (UTC)[reply]

\sum _{k=1}^{n}{\frac {1}{k}}

appears frequently in mathematics.

\sum _{j=0}^{n}\left(1+{\frac {1}{1000j}}\right)

does not. (See Harmonic number.) That's why Euler's constant gets a name, and the other constants don't. —Bkell (talk) 18:03, 26 May 2010 (UTC)[reply]

In other words: consider the steps (h)

SUM [h/(1+0*h)+h/(1+1*h)+h/(1+2*h)+h/(1+3*h)+h/(1+4*h)+h/(1+5*h)+.....+h/(1+n*h)+....]-Ln(n)= ???

When h=0,000000....1 (Sum-Ln)=0 (Zero)= (Integral-Integral).Rigth or wrong?

When h=0,001 (Sum-Ln)=0,000507602...

When h=1 (Sum-Ln)=Euler's constant

When h=3,850601754.. (Sum-Ln)=2,718281828..

Only for h=1 Euler's constant is existing.Otherwise it does not exist.(SUM-Ln) is a function of (h).Is that wrong? Probabily this will be considered by mathematicians for new discoveries. Why to consider only Harmonic number where h=1 ?. TASDELEN (talk) 18:12, 27 May 2010 (UTC)[reply]

Oops, I messed up the second summation above. Anyway, the point stands—harmonic numbers come up quite often and quite naturally in many places in mathematics, and so they have been rather extensively studied. See harmonic number for examples. The other sums you are referring to do not come up nearly as often. This does not mean they are not interesting; they simply do not appear to have as many applications. The Euler–Mascheroni constant gets a name because it has an intimate connection with the harmonic numbers and the natural logarithm and consequently is quite useful in many applications; that's all. If you are interested in studying these other constants that arise for different values of h, then by all means you should do so. They are generalizations of the Euler–Mascheroni constant, and studying generalizations is what mathematics is all about. —Bkell (talk) 20:58, 27 May 2010 (UTC)[reply]

Thanks Bkell. I should be a mathematician to go further, but I am a mechanic.TASDELEN (talk) 22:26, 31 May 2010 (UTC)[reply]