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May 11

How to derive asymptotic expansion of error function?

Resolved

In wiki's page on the error function it gives a useful asymptotic expansion of the complementary error function.

$\mathrm {erfc} (x)={\frac {e^{-x^{2}}}{x{\sqrt {\pi }}}}\left[1+\sum _{n=1}^{\infty }(-1)^{n}{\frac {1\cdot 3\cdot 5\cdots (2n-1)}{(2x^{2})^{n}}}\right]={\frac {e^{-x^{2}}}{x{\sqrt {\pi }}}}\sum _{n=0}^{\infty }(-1)^{n}{\frac {(2n)!}{n!(2x)^{2n}}}.\,$

How is this series derived? I think I understand how an asymptotic series could be derived in general: You have your original function $f(x)$, and make a substitution $x\rightarrow 1/y$ and you have a function of y. Now you do a Taylor series of this function in terms of y, then back-substitute $y\rightarrow 1/x$ and you have a power series in ${\frac {1}{x}}$. Is this how it is always done? How would you go about doing this for the (complementary) error function? Thanks mislih 00:32, 11 May 2010 (UTC)

It's not just a term by term integration of the appropriate series expansion involving exp(-x^2)? 69.228.170.24 (talk) 02:47, 11 May 2010 (UTC)

Doesn't look like it, since the x appears in the denominators. Michael Hardy (talk) 05:05, 11 May 2010 (UTC)

Right, this would give the convergent expansion around 0, while the OP asked about the asymptotic expansion around ∞. -- Meni Rosenfeld (talk) 06:39, 11 May 2010 (UTC)

First note that asymptotic expansion is not the same as an expansion which has x in the denominator and thus is good for large values of x. The latter is known as expansions around ∞. There's a technical definition for asymptotic expansions, the interesting part of which is that they don't necessarily converge. Whether the expansion is convergent or asymptotic is completely unrelated to whether it is around ∞ or any finite number.

I think asymptotic expansions are usually found by repeated integration by parts. You might find more details in the asymptotic theory literature.

The technique you've described is good for finding convergent expansions around ∞. In theory it can also be used to find asymptotic expansions if you factor by the right terms. That is, if you let $f(x)=e^{1/x^{2}}\mathrm {erfc} (1/x)$ and calculate the Taylor series of that around 0, you'll get a series with radius of convergence 0 which is an asymptotic expansion of f around 0. From that you can find an asymptotic expansion of erfc around ∞. However, the expressions involved (in this case at least) are unwieldy, so this may not be practical.

See also Laurent series for expansions having negative powers (which again has nothing to do with being asymptotic). -- Meni Rosenfeld (talk) 06:31, 11 May 2010 (UTC)

Thanks Meni. In this case, repeated application of integration by parts does the trick. mislih 16:39, 11 May 2010 (UTC)

...Added some details in the link ;) --pma 17:34, 11 May 2010 (UTC)

You have removed a correct (and important) statement. The slow convergence part referred to the Taylor series around 0 which was given earlier in the article, and does converge. Also the "in practice" part refers to x which is large enough, not "not too large". Remember, for expansions around ∞, the larger x is, the better. -- Meni Rosenfeld (talk) 17:49, 11 May 2010 (UTC)

OK then the reference to the series had to be stated clearly, because I (as a random user) found it ambiguous. "large enough" not "not too large", of course, sorry, my fault. Of course you can modify and improve everything as you like. Cheers, --pma 07:28, 12 May 2010 (UTC)

Fasten your seatbelts then. The reason why the divergent asymptotic series is more useful than the convergent Taylor expansion is explained by the even more confusing heuristic Carrier's rule which says: “Divergent series converge faster than convergent series because they don't have to converge.” :) Count Iblis (talk) 00:31, 13 May 2010 (UTC)

Closed level curves

It seems reasonable to think that if we have a C^∞ real function on the plane and a closed level curve C without singular points then there is another closed curve near C which has greater lenght. How could we build a proof?--Pokipsy76 (talk) 09:29, 11 May 2010 (UTC)

I've this idea: let v be the hamiltonian vector field of F, then the lenght of a closed orbit starting from x with period T(x) is

$L(x)=\int _{0}^{T(x)}|v(\phi ^{t}(x))|dt$

we have to show that L has nonsingular gradient when x is on a closed orbit (in such case L would be locally like a linear map and the result is proved). Indeed we have

$\langle \nabla L(x),h\rangle =\int _{0}^{T(x)}\langle {\frac {v(\phi ^{t}(x))}{|v(\phi ^{t}(x))|}},Dv(\phi ^{t}(x))D\phi ^{t}(x)h\rangle dt+\langle \nabla T(x),h\rangle$

now we should prove that on a closed orbit this quantity turns out to be not identically 0...--Pokipsy76 (talk) 16:33, 11 May 2010 (UTC)

...then there is another closed curve: you mean another level curve, I suppose? --pma 18:01, 11 May 2010 (UTC)

You are right, however now I think that my statement above is false...--Pokipsy76 (talk) 18:12, 11 May 2010 (UTC)

Yes it should be easy to make an example of a smooth function on the plane with banana-shaped regular level sets {f=t} having the same length for all t in an interval a<t<b. --pma 18:47, 11 May 2010 (UTC)

This fact implies that my proof Here is incorrect. Now I have to find another proof.--Pokipsy76 (talk) 18:52, 11 May 2010 (UTC)

4 digit number

Resolved

 – Shahab (talk) 14:13, 13 May 2010 (UTC)

There is a four digit number whose digits when reversed give us a number 4 times the previous one. What is the number? I'd apreciate any help. Thanks-Shahab (talk) 12:47, 11 May 2010 (UTC)

$ perl -le 'reverse($_) == 4 * $_ and print for 1000 .. 9999'

Spoiler

2178

HTH, HAND. :) —Ilmari Karonen (talk) 13:11, 11 May 2010 (UTC)

Here are some hints (let's call the original number N):

1. Since N and 4N are both four digit numbers, N must be between 1000 and 2499.
2. 4N is an even number, so ends in 0,2,4,6 or 8. But the last digit of 4N is the first digit of N. Bearing in mind the limits from hint 1, this means that the last digit of 4N and the first digit of N are both 2.
3. Since the first digit of N is 2, the first digit of 4N is 8 or 9. So the last digit of N is 8 or 9 ...
4. ... and to make last digit of 4N equal to 2 (which we know from hint 2), the last digit of N must be 8 - and so must the first digit of 4N.
5. To make the first digit of 4N equal to 8, N must be between 2000 and 2249.

So now you know that N has the form 2ab8, and is no greater then 2249. This leaves you with 25 possibile values for N, which you can check one by one. Gandalf61 (talk) 13:20, 11 May 2010 (UTC)

6. Since N reversed, 8ba2, is divisible by four, a is odd. As N is at most 2249, that makes a = 1, and N = 21b8.
7. The second digit of 4N, i.e., b, is thus between 4 and 7, so we are down to 4 possible values.—Emil J. 13:34, 11 May 2010 (UTC)

Thanks-Shahab (talk) 14:13, 13 May 2010 (UTC)

Calculating distance to a cloud on the horizon

According to the "intersecting chords" theorem, it should be a simple calculation. Given the Earth's radius r and cloud height h, distance d should be:

$d={\sqrt {(2r+h)*h}}$

However, that is giving me an unlikely answer. For r=6378100m and h=2000m, d=112952m. It seems unlikely to me that a cloud on the horizon would be as much as 112km away. Is that right?

— PhilHibbs | talk 15:29, 11 May 2010 (UTC)

The formula is correct, but I got 159km for the values you gave (did you remember to multiply r by 2?). However, other factors, such as the oblateness of Earth and the elevation of the observer can affect the distance in practice. -- Meni Rosenfeld (talk) 15:36, 11 May 2010 (UTC)

• (ec) If the cloud seems to be on the horizon, then you're looking at it along the horizontal line. That is You, the Cloud and the Earth center form a right triangle with the right angle at Y. So, by Pythagorean theorem, |EY|² + |YC|² = |EC|², in other symbols $r^{2}+d^{2}=(r+h)^{2}$. Subtract r² from both sides and you'll get $d={\sqrt {2rh+h^{2}}}={\sqrt {(2r+h)h}}$.
  For values you gave $d=159.74{\textrm {\,km}}$ --CiaPan (talk) 15:43, 11 May 2010 (UTC)

For some reason I was halving the product before square-rooting it, a hang-over from a previous calculation that was still in the formula that I was executing. I now agree with 159km. — PhilHibbs | talk 16:30, 11 May 2010 (UTC)

If you want to use three-digit approximation then it should be 160, rather than 159. Note that 159 is about 0.74 off from the precise value while 160 is only about 0.26 off. --CiaPan (talk) 10:44, 12 May 2010 (UTC)

Calculating distance to a cloud not on the horizon

How can the distance to a cloud at any given elevation from the horizon be calculated? If the calculation in my prior question is correct, how could I amend that calculation to tell me how far away a cloud that is 30° above the horizon is, given a guess of the cloud's height? — PhilHibbs | talk 15:32, 11 May 2010 (UTC)

Law of cosines relates the triangle sides and one of its angles. --CiaPan (talk) 15:45, 11 May 2010 (UTC)

Ah, so for a triangle between me, the centre of the earth, and the cloud that I am looking at, the angle where I am will be 90° plus the angle of elevation, and from that I can get the length of the side that I am looking along. Hm, I could also use that to calculate how far away the ground underneath the cloud is. — PhilHibbs | talk 16:49, 11 May 2010 (UTC)

Curvature not intrinsic for curves???

On the page curvature i read

In contrast to curves, which do not have intrinsic curvature, but do have extrinsic curvature (they only have a curvature given an embedding), surfaces can have intrinsic curvature, independent of an embedding.

now I have a problem with this because curvature is also defined trought the radius of the osculating circle (which is a property which depend only by the shape of the curve) so can anybody clarify to me in which sense the curvature is not an intrinsic property for a curve?

--Pokipsy76 (talk) 17:43, 11 May 2010 (UTC)

I'm a bit over my head here, but I think that the shape of a curve is not an intrinsic property. That is, given any two smooth curves having the same length, you can map one to the other in a way which preserves the local structure. -- Meni Rosenfeld (talk) 17:57, 11 May 2010 (UTC)

(edit conflict) Any two simple curves of the same length are isomorphic Riemannian manifolds, hence curvature of the curve is not derivable from its Riemannian manifold structure. It is not intrinsic in the sense that it cannot be computed from inside the curve itself, you have to go outside and look at how the curve is embedded in the ambient Euclidean space. You cannot see the shape of the curve when you are a 1D being confined to the curve.—Emil J. 18:03, 11 May 2010 (UTC)

They don't even need to be the same length, just have the same openness. --Tango (talk) 18:06, 11 May 2010 (UTC)

No, the length of the curve comes intrinsically from the Riemannian metric. Algebraist 18:29, 11 May 2010 (UTC)

Of course, but does an isomorphism of Riemannian manifolds have to be an isometry? I thought there was some degree of flexibility, although I can't remember exactly what (but something at least allowing scaling the metric by a constant, which doesn't really change anything interesting). --Tango (talk) 22:45, 11 May 2010 (UTC)

In simple language, if you were a one-dimensional creature living on a circle you would have no way of telling it's a circle without traveling all the way around. On the other hand if you were a two dimensional creature living on a sphere you could tell that your universe is curved by adding the angles in a triangle and getting more than 180 degrees.--RDBury (talk) 02:54, 12 May 2010 (UTC)

Does this result have a name?

Comparing formulas from curvature and winding number i notice that the integral of the curvature of a closed curve is equal to the winding number of the velocity vector and so it is equal to 2*pi. Is there a name for this result? --Pokipsy76 (talk) 17:50, 11 May 2010 (UTC)

See total curvature.--RDBury (talk) 02:47, 12 May 2010 (UTC)

Terminology

I'm looking for a term for a function f(x) such that f(f(x))=x. Is there a "nice" term describing such a function? --Lucas Brown 19:17, 11 May 2010 (UTC)

Involution (mathematics) Algebraist 19:25, 11 May 2010 (UTC)

It's also a functional square root of the identity function. -- Meni Rosenfeld (talk) 19:27, 11 May 2010 (UTC)

Thanks! --72.197.202.36 (talk) 22:31, 11 May 2010 (UTC)

It is often called a "self-inverse" function. Dbfirs 18:47, 13 May 2010 (UTC)

Triple functions

On a fairly basic calculator, one can easily do operations that cycle between two answers, for example Ans^-1. Are there any, however, that go between three? If my calculator could do matrices, then I could use a matrix representing a 120 degree turn and then that would work, but unfortunately it can't. 92.8.31.181 (talk) 20:59, 11 May 2010 (UTC)

If your calculator supports complex numbers, you can multiply by a primitive ${\sqrt[{3}]{1}}$.

If your calculator can give the floor function, you can use $f(x)=x+1/3+\lfloor x\rfloor -\lfloor x+1/3\rfloor$.

Both of these can be generalized to any desired cycle length. -- Meni Rosenfeld (talk) 08:26, 12 May 2010 (UTC)

In excel the latter method ends up switching cycle once if the number is an integer between 16 and 512. Presumably due to round off errors. Starting at a fractional value other than x/3 will avoid this.Taemyr (talk) 08:45, 12 May 2010 (UTC)

Or, to avoid the roundoff errors for typical inputs, you can change the phase of the function. Choose some number r and let $f(x)=x+1/3+\lfloor x+r\rfloor -\lfloor x+r+1/3\rfloor$. Then this is also a functional cube root of unity, and errors should occur only if the input is a multiple of 1/3 plus r (which is unlikely if r is a weird number like 0.118229347). -- Meni Rosenfeld (talk) 09:09, 12 May 2010 (UTC)

Iterating (1−x)^-1 has a cycle length of 3. Gandalf61 (talk) 09:09, 12 May 2010 (UTC)

Nice! This gave me another idea, $f(x)=\tan(\arctan x+\pi /n)$, where n is the cycle length. In fact you can do the same with any function that maps the real line to a circle. If the mapping is not one-to-one you'll also need a function that tells you the period (as in the fractional part\floor function example). -- Meni Rosenfeld (talk) 09:22, 12 May 2010 (UTC)

Euler's constant

Hi,

What is the meaning of Euler's constant=0,577215665….Where it is necessary to use this constant? What is the purpose?Do someone would like to explain it? Any example,any application? ThanksTASDELEN (talk) 21:29, 11 May 2010 (UTC)

We have an article Euler–Mascheroni constant. Algebraist 21:31, 11 May 2010 (UTC)

Okey! But this doesn't give any application area.I am looking to where it is used?TASDELEN (talk) 21:47, 11 May 2010 (UTC)

As Euclid said, 'Give him a coin, since he must profit by what he learns.' What use is a painting? How about just that it occurs all over the place so it is interesting? Dmcq (talk) 22:41, 11 May 2010 (UTC)

At Gamma function, we read this:

$\Gamma (z)={\frac {e^{-\gamma z}}{z}}\prod _{n=1}^{\infty }\left(1+{\frac {z}{n}}\right)^{-1}e^{z/n}$

where γ is the Euler–Mascheroni constant.

...and:

For positive integer m the derivative of Gamma function can be calculated as follows (here $\gamma$ is the Euler–Mascheroni constant):

$\Gamma '(m+1)=m!\cdot \left(-\gamma +\sum _{k=1}^{m}{\frac {1}{k}}\right).$

At Riemann zeta function it says:

$\gamma _{n}=\lim _{m\rightarrow \infty }{\left(\left(\sum _{k=1}^{m}{\frac {(\log k)^{n}}{k}}\right)-{\frac {(\log m)^{n+1}}{n+1}}\right)}.$

The constant term γ₀ is the Euler-Mascheroni constant.

and

$\zeta (s)={\frac {e^{(\log(2\pi )-1-\gamma /2)s}}{2(s-1)\Gamma (1+s/2)}}\prod _{\rho }\left(1-{\frac {s}{\rho }}\right)e^{s/\rho },\!$

where the product is over the non-trivial zeros ρ of ζ and the letter γ again denotes the Euler-Mascheroni constant.

At Digamma function it says:

$\gamma _{n}=\lim _{m\rightarrow \infty }{\left(\left(\sum _{k=1}^{m}{\frac {(\log k)^{n}}{k}}\right)-{\frac {(\log m)^{n+1}}{n+1}}\right)}.$

The constant term γ₀ is the Euler-Mascheroni constant.

and

$\Psi (n)\ =\ H_{n-1}-\gamma$

where $\gamma \,$ is the Euler-Mascheroni constant.

At Mertens' theorem it says:

Mertens' 3rd theorem:

$\lim _{n\to \infty }\ln n\prod _{p\leq n}\left(1-{\frac {1}{p}}\right)=e^{-\gamma },$

where γ is the Euler–Mascheroni constant.

At Exponential integral, it says:

$\mathrm {E_{1}} (z)=-\gamma -\ln z+\sum _{k=1}^{\infty }{\frac {(-1)^{k+1}z^{k}}{k\;k!}}\qquad (|\mathrm {Arg} (z)|<\pi )$

where $\gamma$ is the Euler–Mascheroni constant.

At Meissel–Mertens constant, we read that

$M=\lim _{n\rightarrow \infty }\left(\sum _{p\leq n}{\frac {1}{p}}-\ln(\ln(n))\right)=\gamma +\sum _{p}\left[\ln \left(1-{\frac {1}{p}}\right)+{\frac {1}{p}}\right].$

Here γ is the famous Euler–Mascheroni constant,

At Laplace transform it appears in a table:

$-{t_{0} \over s}\ [\ \ln(t_{0}s)+\gamma \ ]$

At Barnes G-function:

Formally, the Barnes G-function is defined (in the form of a Weierstrass product) as

$G(z+1)=(2\pi )^{z/2}\exp(-(z(z+1)+\gamma z^{2})/2)\ \times \ \prod _{n=1}^{\infty }\left[\left(1+{\frac {z}{n}}\right)^{n}\exp(-z+z^{2}/(2n))\right],$

where γ is the Euler-Mascheroni constant,

Stieltjes constants:

Cauchy's differentiation formula leads the integral representation

$\gamma _{n}={\frac {(-1)^{n}n!}{2\pi }}\int _{0}^{2\pi }e^{-nix}\zeta \left(e^{ix}+1\right)dx.$

The zero'th constant $\gamma _{0}=\gamma =0.577\dots$ is known as the Euler–Mascheroni constant.

At Gumbel distribution we see an application to statistics:

The mean is $\mu +\gamma \beta$ where $\gamma$ = Euler–Mascheroni constant

At Gauss–Kuzmin–Wirsing operator:

$t_{n}=1-\gamma +\sum _{k=1}^{n}(-1)^{n}{n \choose k}\left[{\frac {1}{k}}+{\frac {\zeta (k+1)}{k+1}}\right]$

where $\gamma$ is the Euler–Mascheroni constant.

...and there are others. Michael Hardy (talk) 00:11, 12 May 2010 (UTC)

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)

If you want a "practical" use, some of the number theory cited above may be applicable to cryptography, and you should notice that I mentioned an application to statistics. However, contrary to your last assertion, mathematics is often done years or decades, or sometimes centuries, before practical applications are found, and is normally treated as (as you put it) "an artistic art". Michael Hardy (talk) 00:21, 13 May 2010 (UTC)

Thanks Hardy.I have the following comments for Euler's constant: Euler says: when (s) tends to infinity, the value of the following expression reaches to: (1 to s)SUM(1/s)-ln(s)=0,577215665....But,if we write (ds to s)SUM(ds/s)-(ds to s)INTEGRAL(ds/s)=K value?

This K value=0,577215665...if ds=1 (confirmation for Euler)

When (ds=0,1) this K value=0,050861.. and when (ds=0,000000.....1),this K value= 0 (no confirmation for Euler) Therefore I deduct: SUM or INTEGRAL give the same value if the evaluation steps are not JUMPING,are continious.So,Euler's constant has not an interesting practical use.Choosing ds= Hardy's ds,you may invent your constant.TASDELEN (talk) 20:43, 13 May 2010 (UTC)

We already know that an integral is the limit of a sum as the step size goes to 0 (for well-behaved functions).

And also, for any step size you choose, the expression will involve Euler's constant. For example, $\lim _{n\to \infty }\left(\sum _{k=10}^{n}{\frac {1}{k}}-\int _{10}^{n}{\frac {dk}{k}}\right)=\gamma +\ln 10-{\frac {7129}{2520}}$. -- Meni Rosenfeld (talk) 08:03, 14 May 2010 (UTC)

Hi Meni.In your expression the steps are not "any step" but 1,2,3,4....n. When you choose steps 0,1;0,2;0,3,0,4;....n do we get Euler's gamma? I think not!This why,gamma constant is a special case.You may choose your steps and discover a different gamma.Gamma is v-ariable according the choose of the steps.Then why it is said "constant"TASDELEN (talk) 19:04, 14 May 2010 (UTC)

Are you asking something or are you trying to make some sort of point? You seem to have veered in tone and approach drastically from your original question. Dmcq (talk) 20:32, 14 May 2010 (UTC)

Yes, even when we choose steps 0,1;0,2;0,3,0,4;....n we do get Euler's gamma.

$\lim _{n\to \infty }\left(\sum _{k=1}^{n}{\frac {1}{0.1k}}-\int _{1}^{n}{\frac {dk}{0.1k}}\right)=10\lim _{n\to \infty }\left(\sum _{k=1}^{n}{\frac {1}{k}}-\int _{1}^{n}{\frac {dk}{k}}\right)=10\gamma$

Bo Jacoby (talk) 10:14, 15 May 2010 (UTC).

When I consider the following expression,where (ds=0,1 or 0,2 or 0,3.....or 0,9) I do not find Euler's gamma, [(1 to s)SUM(ds/s)-(1 to s)INTEGRAL(ds/s)]=Differents gamma.

I find: Euler's gamma=0,577215665...if ds=1 .When (ds=0,1) this value=0,050861.. .And this should be normal as "We already know that an integral is the limit of a sum as the step size goes to 0".This why,I do not appreciate gamma as constant,and I say every body may invent its personal constant according the choose of the steps (ds)TASDELEN (talk) 20:53, 15 May 2010 (UTC)

When you define a function $\scriptstyle \gamma (h)$ dependent on the step size $\scriptstyle h$ such that $\scriptstyle \gamma (1)$ is the Euler gamma constant, can't you express $\scriptstyle \gamma (h)$ elementarily in terms of $\scriptstyle \gamma (1)$? Bo Jacoby (talk) 06:29, 16 May 2010 (UTC).

Depending on what you mean by step size, not necessarily. The way I interpreted it, for each specific rational step size you can find an expression with Euler's Gamma and logs. For irrational step sizes the expression requires advanced functions, which means that the general function is also non-elementary. -- Meni Rosenfeld (talk) 08:49, 16 May 2010 (UTC)

Tasdelen, the formulas you write don't make sense. Please see how we've written summation and integration formulas and write similarly, so we understand what you mean. -- Meni Rosenfeld (talk) 08:49, 16 May 2010 (UTC)

• Meni,here is my evaluations:

On an excel table ,with 65500 working lines,I choose step dx=1 and then x=1;1+dx;1+2*dx;1+3*dx;1+4*dx;...1+65500*dx

• SUM(dx/x)=11,6670287200934
• Ln (65500)=11,089805421623300
• (SUM-Ln)=0,577223298469860 .This confirm Euler.

On an excel table,I choose step dx=0,1 and then x=1;1+dx;1+2*dx;1+3*dx;1+4*dx;....1+65500*dx

• SUM(dx/x)=11,140654397331200
• Ln(65500)=11,089805421623300
• (SUM-Ln)=0,050848975707812 .This does not confirm Euler.

On an excel table,if I choose dx=0,000000...1 and then x=1;1+dx,1+2*dx;1+3*dx;......1+65500*dx

• SUM(dx/x)=11,089805421623300
• Ln(65500)=11,089805421623300
• (SUM-Ln)= 0. It is normal that this will not confirm Euler as the steps tend to Zero.Then why Euler constant? I may arrange a dx and find PI as constant.TASDELEN (talk) 21:13, 18 May 2010 (UTC)

• • Meni,if I choose dx=4,361917092296070, for 65500 working lines on excel, I have:
• x=(1; 1+1*dx; 1+2*dx; 1+3*dx;....(1+65500*dx=285702,2)) and then
• SUM(dx/x)=15,704297968863200
• Ln(285702,2)=12,562705315273400
• (SUM-ln)=3,141592653589790=Pi.......I don't see any gamma constant....TASDELEN (talk) 22:26, 19 May 2010 (UTC)

• • More constant depending of the choose of dx:

*dx--------------------(SUM-ln)------------explication

• 0,001-----------------0,000507602
• 0,01-------------------0,005015955
• 0,1---------------------0,050840136
• 1------------------------0,577223298-----Euler
• 2------------------------1,270370479
• 3,850601754--------2,718281828-----e
• 4,361917092--------3,141592654-----Pi
• 10-----------------------8,121177481
• 16,17709104--------13,87338721----Sum=2*ln
• (SUM-LN) value is not a constant.Then,what is the use of gamma,if this is not an "artistic math"?TASDELEN (talk) 21:34, 20 May 2010 (UTC)

• Any comment? TASDELEN (talk) 18:12, 21 May 2010 (UTC)