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u ( φ , k ) = 0 φ d y 1 - k 2 sin 2 ( y )

The trigonometric sine of the inverse of this function is defined as the Jacobian elliptic sine of u with modulus k , and is denoted

s n ( u , k ) = sin ( φ ( u , k ) )

A special evaluation of [link] is known as the complete elliptic integral K = u ( π / 2 , k ) . It can be shown [link] that s n ( u ) and most of the other elliptic functions are periodic with periods 4 K if u is real. Because of this, K is also called the “quarter period". A plot of s n ( u , k ) for several values of the modulus k is shown in [link] .

Figure two is a graph titled elliptic sine of u, modulus k. The horizontal axis is labeled Independent Variable, u, and ranges in variable from 0 to 12 in increments of 2. The vertical axis is labeled sn(u, k) and ranges in value from -1 to 1 in increments of 0.5. There are four wavelike functions on this graph. The first, a blue curve, begins increasing from the origin and completes two peaks and nearly two troughs before it terminates with a positive slope at (12, -0.5). This curve is labeled k=0. The second, a green curve, is wavelike and begins positive in a similar fashion as the blue curve, but its trough is substantially wider, and it only completes one and a half peaks and one full trough before it terminates at (12, 1) when it is just about to reach the top of a peak. This curve is labeled k=0.9. The third, a red curve, has an even wider peak than the previous curve, completing one peak and half of one trough before it terminates traveling completely horizontal in a portion of the trough at (12, -1). This curve is labeled k=0.99. The fourth, a teal curve, is again an exaggeration of a sinusoidal function with an extremely wide peak, that does not even reach the horizontal apex of the trough when it terminates at (12, -1). This curve is labeled k=0.999.
Jacobian Elliptic Sine Function of u with Modulus k

For k=0, s n ( u , 0 ) = sin ( u ) . As k approaches 1, the s n ( u , k ) looks like a "fat" sine function. For k = 1 , s n ( u , 1 ) = tanh ( u ) and is not periodic (period becomes infinite).

The quarter period or complete elliptic integral K is a function of the modulus k and is illustrated in [link] .

Figure three is a graph titled complete elliptic integral, K_k. The horizontal axis is labeled modulus, k and ranges in value from 0 to 1 in increments of 1. The vertical axis is labeled complete elliptic integral, and ranges in value from 0 to 5 in increments of 0.5. There is one curve in this graph, and at its beginning at approximately (0, 1.5) is an arrow pointing at the place it begins, labeled π/2. The curve moves from left to right with a shallow slope at first, but the slope is slowly increasing across the page, and by the horizontal value 0.8 the graph has a sharp positive slope. At the horizontal value 1, the curve is completely vertical, and ends in the top-right corner of the graph, at (1, 5).
Complete Elliptic Integral as a function of the Modulus

For a modulus of zero, the quarter period is K = π / 2 and it does not increase much until k nears unity. It then increasesrapidly and goes to infinity as k goes to unity.

Another parameter that is used is the complementary modulus k ' defined by

k 2 + k ' 2 = 1

where both k and k ' are assumed real and between 0 and 1. The complete elliptic integral of the complementary modulus is denoted K ' .

In addition to the elliptic sine, other elliptic functions that are rather obvious generalizations are

c n ( u , k ) = c o s ( φ ( u , k ) )
s c ( u , k ) = t a n ( φ ( u , k ) )
c s ( u , k ) = c t n ( φ ( u , k ) )
n c ( u , k ) = s e c ( φ ( u , k ) )
n s ( u , k ) = c s c ( φ ( u , k ) )

There are six other elliptic functions that have no trigonometric counterparts [link] . One that is needed is

d n ( u , k ) = 1 - k 2 s n 2 ( u , k )

Many interesting properties of the elliptic functions exist [link] . They obey a large set of identities such as

s n 2 ( u , k ) + c n 2 ( u , k ) = 1

They have derivatives that are elliptic functions. For example,

d s n d u = c n d n

The elliptic functions are the solutions of a set of nonlinear differential equations of the form

x ' ' + a x ± b x 3 = 0

Some of the most important properties for the elliptic functions are as functions of a complex variable. For a purely imaginaryargument

s n ( j v , k ) = j s c ( v , k ' )
c n ( j v , k ) = n c ( v , k ' )

This indicates that the elliptic functions, in contrast to the circular and hyperbolic trigonometric functions, are periodic inboth the real and the imaginary part of the argument with periods related to K and K ' , respectively. They are the only class of functions that are “doubly periodic".

One particular value that the s n function takes on that is important in creating a rational function is

s n ( K + j K ' , k ) = 1 / k

The chebyshev rational function

The rational function G ( ω ) needed in [link] is sometimes called a Chebyshev rational function because of its equal-ripple properties.It can be defined in terms of two elliptic functions with moduli k and k 1 by

G ( ω ) = s n ( n s n - 1 ( ω , k ) , k 1 )

In terms of the intermediate complex variable φ , G ( ω ) and ω become

G ( ω ) = s n ( n φ , k 1 )
ω = s n ( φ , k )

It can be shown [link] that G ( ω ) is a real-valued rational function if the parameters k , k 1 , and n take on special values. Note the similarity of the definition of G ( ω ) to the definition of the Chebyshev polynomial C N ( ω ) . In this case, however, n is not necessarily an integerand is not the order of the filter. Requiring that G ( ω ) be a rational function requires an alignment of the imaginary periods [link] of the two elliptic functions in [link] , [link] . It also requires alignment of an integer multiple of the real periods. The integermultiplier is denoted by N and is the order of the resulting filter [link] . These two requirements are stated by the following very important relations:

Questions & Answers

can someone help me with some logarithmic and exponential equations.
Jeffrey Reply
sure. what is your question?
ninjadapaul
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
ninjadapaul
I don't understand what the A with approx sign and the boxed x mean
ninjadapaul
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
ninjadapaul
oops. ignore that.
ninjadapaul
so you not have an equal sign anywhere in the original equation?
ninjadapaul
Commplementary angles
Idrissa Reply
hello
Sherica
im all ears I need to learn
Sherica
right! what he said ⤴⤴⤴
Tamia
what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
Kevin Reply
a perfect square v²+2v+_
Dearan Reply
kkk nice
Abdirahman Reply
algebra 2 Inequalities:If equation 2 = 0 it is an open set?
Kim Reply
or infinite solutions?
Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Al
y=10×
Embra Reply
if |A| not equal to 0 and order of A is n prove that adj (adj A = |A|
Nancy Reply
rolling four fair dice and getting an even number an all four dice
ramon Reply
Kristine 2*2*2=8
Bridget Reply
Differences Between Laspeyres and Paasche Indices
Emedobi Reply
No. 7x -4y is simplified from 4x + (3y + 3x) -7y
Mary Reply
is it 3×y ?
Joan Reply
J, combine like terms 7x-4y
Bridget Reply
im not good at math so would this help me
Rachael Reply
yes
Asali
I'm not good at math so would you help me
Samantha
what is the problem that i will help you to self with?
Asali
how do you translate this in Algebraic Expressions
linda Reply
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
Crystal Reply
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
Chris Reply
what's the easiest and fastest way to the synthesize AgNP?
Damian Reply
China
Cied
types of nano material
abeetha Reply
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
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Yasmin
what is the function of carbon nanotubes?
Cesar
what is nanomaterials​ and their applications of sensors.
Ramkumar Reply
what is nano technology
Sravani Reply
what is system testing?
AMJAD
preparation of nanomaterial
Victor Reply
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
Himanshu Reply
good afternoon madam
AMJAD
what is system testing
AMJAD
what is the application of nanotechnology?
Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
Azam
anybody can imagine what will be happen after 100 years from now in nano tech world
Prasenjit
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
Azam
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
Prasenjit
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
Damian
silver nanoparticles could handle the job?
Damian
not now but maybe in future only AgNP maybe any other nanomaterials
Azam
can nanotechnology change the direction of the face of the world
Prasenjit Reply
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
Ali Reply
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
bamidele Reply
how did you get the value of 2000N.What calculations are needed to arrive at it
Smarajit Reply
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Source:  OpenStax, Digital signal processing and digital filter design (draft). OpenStax CNX. Nov 17, 2012 Download for free at http://cnx.org/content/col10598/1.6
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