tHIS Rajni Kanth's கணக்கு. பதிவியலில் இடம் பெறட்டும்.

Post by Devanand Sadasivam.

இடுக்கண் வருங்கால் நகுக

Post by Devanand Sadasivam.

Arnold in TN [Watch Sun (Sn) Network]

Post by Devanand Sadasivam.

Teacher's Days ka Doodle

Post by Devanand Sadasivam.

General Locus for differentiation

== The problem of differential equations ==
The problem of differential equations is - "cyclic redemption" with the consideration that there weren't any z axis. Or it that when there is x,y axis it is always that z axis is implicit with unit 1.

It is known that point lies on the circle (x,y) represented by (rCosθ,rSinθ).

By that we can say
x=rCosθ
y=rSinθ

In nature, we cannot find anything perfect as circle or sphere, but it is almost to it. Having that we can say,

x=rCosθ+Δx
y=rSinθ+Δy

x/y=rCosθ+Δx/rSinθ+Δy

To avoid abnormalities consider x/y=1 as z=1 so its equality is maintained. See, NP₀ = 1. Sinθ/θ = 1 when θ tends to 0.

rCosθ+Δx/rSinθ+Δy=1

rCosθ+Δx/rSinθ+Δy=1

rCosθ+Δx=rSinθ+Δy

Δy=rCosθ+Δx-rSinθ

Δy/Δx=[rCosθ/Δx]+[1]-[rSinθ/Δx]

Δy/Δx=[(rCosθ-rSinθ)/Δx]+1

Δy/Δx=[rCosθ+Δx-rSinθ/Δx]

[Δy=rCosθ+Δx-rSinθ]

'''When you say dy then it is that,'''

dy=y+Δy

dy=y+rCosθ+Δx-rSinθ [as x=rCosθ, y=rSinθ So,] [or logical plane z=1]

dy=y+x+Δx-y

dy=x+Δx

as dx=x+Δx so

[dy=dx]

=== The General Equality Locus of differentiation is ===
dy=dx when plane-z=1, if it is not (z!=1) this equality is accords. So it is not dy/dx gives some ratio rather, make it dy=dx leverage with z-plane analysis like 1,NP_r series.

This will avoid cyclic redemption problem of differentiation that it is ignoring that there is a z-axis

~~~~

Ref:

https://en.wikipedia.org/wiki/Group_structure_and_the_axiom_of_choice
http://en.wikipedia.org/wiki/Talk:Derivative#The_problem_of_differential_equations

S.Anand TAC [Proprietor, CEO, Founder, Director]

Post by Devanand Sadasivam.

[Draft]VB neither or a PERL..... Both are same on its PACT level....... Or anything on & around SO[Draft]

<>
Lets harvest our grooves by memory ref or so by SUB'sub'o' Namo Naraya||

Advanced Ping Concepts

Everyone know of the command 'ping' which helps to see whether internet connection is available in a Computer (of any sort - [i.e., laptop, desktop, server]). And prior to check the internet connection whether internet gateway is up (i.e., either firewall or the gateway device). In general Routers acts as gateway device. Either it is per-configured with least parameters or can be configured on a very low level on several networking algorithms with wide set of options and configurability For ex:- Cisco Routers.

This command is useful to check whether this kind of ROUTER is up [or] to check the availability of any specialized device which has a valid IP in order to access (like NAS, SAN, Firewall, configured UNIX boxes which are meant for some purpose like LDAP server, DNS server, Version control systems like CVS, SVN etc., DHCP server, Clustered Application Servers like Weblogic, Websphere, JBoss etc., Well Configured Web Servers - Apache and so on )

And simply to say this command helps to check the network availability of a system in a local area network (LAN) and to see the given host is reachable in a network of any size. When you ping the host which is not in shortest path as per the Router's algorithm, some packets might gets lost, because time-to-live (TTL) makes the pact on the time on the ICMP messages anticipated from the host. This command is commonly available in all O.S [MS-DOS, Windows, UNIX, Linux etc.,]. But other than -t option which helps to ping infinite ICMP packets, there are many options in 'ping' which can be intelligently used.

Note: ICMP stands for Internet Controlled Message Protocol

Here is the ping options taken from MS-DOS:-

-a option gives the host-name of the system you ping

Lets Assess one by one Slowly:-

பங்குனி வெயிலில் பட்ட பகலில் எனை பார்த்து சிரிக்கும் நிலா

How-to find the area of the Triangle - ∆

Can be found by the below Square Matrix. Determinant value of the Matrix yields the surface of the triangle ∆. And it can be denoted by the symbol delta as well - δ

|x1 y1 1|
|x2 y2 1| . 1/2
|x3 y3 1|



Let AL, BM, CN be the ordinates of the points A, B and C respectively.

Now  ∆ ABC = trapezium ABML + trap. ALNC - trap BMNC
                      = 1/2(AL+BM)ML+1/2(AL+CN)LN-1/2(BM+CN)MN
=>              ∆ = 1/2 (y1+y2)(x1-x2)+1/2(y1+y3)(x3-x1)-1/2(y2+y3)(x3-x2)
                      = 1/2[x1y2-x2y1+x3y1-x1y3+x2y3-x3y2]
∴              ∆   = 1/2[x1(y2-y3)+x2(y3-y1)+x3(y1-y2)]
i.e.,           ∆   = 1/2Σx1(y2-y3) numerically.

Note:
        |x1 y1 1|
1. ∆=|x2 y2 1| . 1/2
        |x3 y3 1|
2. The area obtained by the formula is positive or negative according as the vertices are taken round the triangle anti-clockwise or clockwise
3. If the area of triangle formed by the three points is zero, then the points are collinear,
                                                             |x1 y1 1|
i.e., if ∆=0, then 1/2.Σx1(y2-y3) = 0 or |x2 y2 1| =0
                                                             |x3 y3 1|
is the condition for collinearity of three points.



Source: Polytechnic Mathematics-2, S.Janakiraman, J.J Publications

The Matrix - Spin One Mathematical Mutation Algorithm