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Does The Physicist Vision Contradict The Nature System? 

Please read

The Gravitational Waves Reflection Analysis

or

or

or

I study the solar system Geometry since long years- and- I have found the physicist and astronomer can't understand the solar system designer vision– in fact their vision contradict the designer vision- let's prove that in following  

The Physicist Vision

The physicist supposes that there's (a unit of building)- by that- the physicist sees the mass (for example) as wall and this wall is consisted of bricks (small similar units) and for that the physicist searched for this (unit) because it's the (unit) by which the wall is building (the mass is building)– this is the physicist thinking direction- he believed there's a small unit by which the great building is built

For that the physicist searched and found the particle and then the particle is divided into Molecules then the Molecule is divided into atoms and the atom is divided into nucleus and electron moves around and the nucleus is divided into proton and Neutron- and the proton is divided into quarks …etc

The physicist searches for the building unit- the unit by which the building is built – or the unit by which the mass is built – for that the physicist divides (any thing) to see its contents till reach to the unit which can be used as the building unit. 

The Designer Vision

The designer aims to create integration- as the marriage – male and female- the relationship will cause to give birth for a child

INTEGRATION this is the word inside the designer Mind

No building unit- the marriage answers- because in marriage both (male and female) are required and no one of them better than the other – both work together and give equal effects on the same one process and give together one result- I explain the idea by the marriage because it's easy to understand- I want to put the word (Integration) as the basic concept in the designer mind.

And

I want to explain how the physicist causes fatal error for the research method

While the physicist divides the elements one after one (from matter to particle to molecule to atom to proton to quarks …etc) this division from out into inner till reach to very small unit which he considered the building unit – that causes to destroy the geometrical design of the integration process

Let's return to the marriage- the marriage requires two players (male and female)- they have different bodies and the process give birth for children- Suppose we remove one player what would happen? (where can we find a woman pregnant by herself?)

I want to say- the division from (matter to particle to molecule to atom to proton to quarks …etc) destroys the general geometrical design which uses the integration process- let's give other examples

Example No. (1)   

Einstein told us (He can't find slightest meaning for the word SPACE) by that he can't define (what's the space) and also he found the space has no mechanical features because no definition causes these features – this is the physicist vision - Let's see the designer vision

Planet moves and its motion produces energy (1/2 mv^2) and where's the energy? the planet can't store its motion energy inside its body because this energy would raise the planet temperature and no planet temperature is raised by its motion- so- where's the planet motion energy? logically we suppose that- the energy is stored in the space in waves form- means- the planet moves and its motion energy creates waves in the space (as the fish swims in the sea and its motion creates waves in the sea water) similar to that the planet motion produces energy and this energy creates waves in the space- by that we can define the space and conclude the space nature and features- simply we conclude the space has mechanical features because the energy is stored in the space as moving waves- the gravitational waves prove the space has motion- why we can solve the question which Einstein couldn't solve? Because he supposed the planet is independent from the space- the integration concept is not known by him

Notice

The waves are produced by the planet motion energy are the gravitational waves, the paper first hypothesis proves these waves are produced by the planets motions energies and not by the gravitational field also the hypothesis proves the Sun doesn't produce a gravitational field.   

Example No. (2)   

The scientists stated (The Sun Rays Are Created By The Sun Nuclear Fusion Process)

Why?

Because they think the sun is independent from the planets and each planet is independent from the other– the harmony motion can not prove all players are created from one unified source (one energy) - the integration is not seeing here 

But

We can reach to the facts simply- because- the planet moves and produces energy (1/2 mv^2) and this energy is stored in the space in moving waves-

The simple idea tells (the energy can NOT be stored in the space forever) this stored energy must be used- here we see how the sun rays is created- clearly- the Sun Is NOT Doing nuclear fusion to produce its rays- why? because the solar system is very rich in the motion energy and massive motion energy is stored in the space logically this stored energy must be used by some method- based on that- I conclude – the sun rays are produced from this stored energy regardless the production mechanism (means even if I don't know how the sun rays is produced by this energy) I still can confirm that the sun rays is created from the planets motions energies which are stored in the solar system space as moving waves-

I wish I explain the idea clearly -I can discover the space nature and features by planet motion analysis – why? because of the integration- they do one unified job together and each player can be defined by the analysis for the other player.

Notice

My theory tells (The sun rays are created from the gravitational waves and the gravitational waves are produced by the planets motions energies) – that explains why the sun corona temperature is 5 millions Kelvin and the sun surface is 5800 Kelvin because the energy is not come from the sun body.

The Concept Problem -the problem is the eyes – the physicist eyes should see the integration while the integration is growing– let's give example to see this problem

The Earth and its moon move together and they create integrated motion- that shows unified motion by the two planets – but this integration is growing by Venus Motion effect on both planets and by that we have greater integration contains (The Earth- The Moon – Venus) – But Mars also has effect on the three planets – that shows greater integration contains the four planets and of Course Mercury has effect on all and that created greater integration contains the inner planets –

The integration is growing till contain all planets and the Sun together – why?- because it's One Unified Energy behind all players

It needs one genius can understand all these motions and their effects on each other to create one unified motion and one unified effect

The physicist needs very wide eyes can see the great integrated system and all small integrated systems inside it– (something like a great chess board which have hundreds of pieces- if the eyes can see all pieces motions and their effects)

The Concept Use Difficulty-

Why is it hard to discover the integration in the solar system?

Because the geometrical rules are not known and we need to analyze the planets data deeply to discover these rules and see the integration behind it–let's see two examples can help to explain this meaning

Example No. (I)

Why Is The Moon Orbital Apogee Radius = 406000 Km?

The moon daily displacement =88000 km and during 29.53 days (the moon day period) the displacements total be = 2.598 million km = 2π x 413600 km

The data tells us the moon orbital apogee radius should be 413600 km and also it tells, because the moon daily displacement (88000 km) is so long, the moon should revolve around the Earth through this apogee orbit its radius (413600 km) only and can't revolve around the Earth through any more near orbit…

Not Facts

The moon orbital apogee radius =406000 km only and the moon revolves around the Earth through near orbits and can reach to perigee radius (363000 km). 

How Can The Moon Do That?

The intelligent moon creates an angle (θ) between its motion direction and its orbit horizontal level by that the real displacement (L) through the orbit be less than (88000 km) because it be (L = 88000 km cos θ), as a result the total displacements be less than (2.598 million km) and that makes the moon orbital apogee radius to be decreased from 413600 km to 406000 km.

We should pay attention for the angle (θ), because this angle controls the moon motion features – where- with the angle (θ) increasing the real displacement (L) be shorter and the moon can revolve around the Earth through more near orbits – but –with the angle (θ) deceasing the real displacement (L) be longer and that pushes the moon far from the Earth to more far orbits.

The moon orbital motion Equation depends on this angle (θ) it tells θ1 = θ0 +1.7

where (θ1) = today angle and (θ0) =yesterday angle

1.7 degrees be used as the moon daily motion degrees for the equation

NOTICE

This example aims to show how to discover the geometrical rules by using the planets data-But the example doesn't provide a picture for the integration because it's a motion done by the moon only- the next example shows the integration system based on which the total solar eclipse is created

Example No. (II)

DATA

DATA NO. (i)

4900 million km     = 1.392 million km x 3475 km (The Moon Diameter)

                             = 449197 km x 10921 km (The Moon Circumference)                                             = 406000 km x 12104 km (Venus Diameter)                              

                             = 384000 km x 12756 km (The Earth Diameter)                         

                             = 363000 km x2 x 6792 km (Mars Diameter)

Notice

1.392 million km = π x 449197 km (error 1.3%)  where       

4900 million km     = Jupiter Orbital Circumference

1.392 million km    = The Sun Diameter 

449197 km            = Jupiter Circumference

406000 km            =The Earth Moon Distance For The Apogee Radius

384000 km            =The Moon Orbital Distance

363000 km            =The Moon Distance For The Perigee Radius

DATA NO. (ii)

1.392 million km /3475 km = (149.6 million km/ 373500 km) where

149.6 million km    = The Earth Orbital Distance

373500 km            = The Earth Moon Distance (At The Total Solar Eclipse) 

DATA NO. (iii)

Jupiter Venus distance               = 671 million km

Venus Orbital Circumference     = 680 million km (error 1.3%)

Jupiter Earth Distance                = 928 million km

The Earth Orbital Circumference = 940 million km (error 1.3%)

Notice – the distance 928 million km = 149.6 million km + 778.6 million km

(means the distance 928 million km is found when the two planets be on different sides from the sun)

DATA NO. (iv)

Jupiter (13.1 km/s) moves in 10921 seconds a distance 142984 km = Jupiter diameter

And

4900 million km = 449197 km (Jupiter circumference) x 10921 km

4900 million km = (13.1 km/s) x π x (10921 seconds)^2

Discussion

1- Preface

2- The Sun Diameter Definition Rule

3- The Sun Position Analysis

4- Jupiter Orbit Design Uses The Moon Circumference

5- The Sun diameter should be = 400 The Moon diameter

1- Preface

My theory tells (the sun rays are created by the gravitational waves and not by any nuclear fusion process – also- I claim- the gravitational waves are produced by the planets motions energies and not by any gravitational field moreover the sun doesn’t produce a gravitational field) –

Means, The theory tells the sun rays is created by the planets motions energies – 

Also that means, The Sun Is Created After All Planets Creation And Motion- let's see the eclipse reason data here

(The Sun Diameter / The Moon Diameter)= (The Earth Orbital Distance / The Earth Moon Distance)

The old question asked why does the diameters rate equal = the distances rate and the wrong answer was always (Pure coincidence)

The fact is that the sun is created after all planets creation and motion and that means the sun diameter is created based on the moon diameter and that tells the sun diameter depends on 2 players which are (the moon diameter) and the number (400) because

The sun diameter 1.392 million km = the moon diameter 3475 km x 400

 Here we need to know why the sun diameter is created depending on these two players

2- The Sun Diameter Definition Rule

My planet diameter equation proves the planet diameter is created as a function in its orbital distance- Can the sun diameter be created by a similar method? that necessitates the sun diameter to be a function in some distance- but – what distance? Let's see the next data 

The sun diameter 1.392 million km= 449197 km (Jupiter Circumference) x π   (1.3%)

The data shows a connection between the sun diameter and Jupiter Circumference-

Can The Sun Position In The Sky Be Defined Based On Jupiter Orbit? Let's examine that

My theory tells the sun is created by the planets motions energies and that means the sun is created after all planets creation and motion– means- while the planets were found and moving in their orbits the sun is created Later and that necessitated to define a position for the sun in the sky

I claim- the sun position is defined based on Jupiter orbit – that can explain the previous data and also the total solar eclipse data

Now – let's ask – Can the sun position in the sky be defined based on Jupiter Orbit? We try to answer this question in the next point

3- The Sun Position Analysis

The space is not static but in motion- means – all points are in motion in the solar system space and the solar system design needs to create a stationary point on which the sun will be created – the stationary point moves by zero motion approximately (in fact it moves by 1 km/s) – but how this point can be found in the space where the space is in motion and no any point is stationary in it? (the space is similar to the sea of water and no stationary point is found in the sea because every thing is in motion)

Means- the creation of the stationary point in the space needs specific procedure– let's see it

The Sun Point Creation Procedure

Before the sun creation, the planets revolved around a point in space (any point in space), the revolution motion creates 2 equal velocities on the revolution both sides - the two velocities are equal in value and opposite in direction- their total be Zero-shortly- the revolution motion creates A Stationary Point in the center of the revolution- that means- the chosen point of space was in motion before the planets revolve around it but when the planets revolve around it this point becomes stationary point and this point motion be Zero approximately (1 km/s)- please note – the point is static and has no motion and its velocity should be Zero but the mathematical calculations of this point can't use the Zero number that's why the velocity be 1 km/s the most low velocity higher than Zero-

Now let's see the data by which the planets starting their revolutions around the sun

DATA NO. (III)

Jupiter Mercury Distance           = 721 million km

Mercury Orbital Circumference = 360 million km

Jupiter Venus distance               = 671 million km

Venus Orbital Circumference     = 680 million km (error 1.3%)

Jupiter Earth Distance                = 928 million km

The Earth Orbital Circumference = 940 million km (error 1.3%) (The distance 928 million km is found when the two planets be on different sides from the sun)

THE DATA ANALYSIS

Here we suppose the sun is not created yet and this data defines the sun position before the sun creation- means- we try to define the point of space only

Let's ask -

Where's this required point around which the planets will revolve?

Venus Role

Venus Jupiter distance = 671 million km and Venus used this distance to be the orbital circumference around the required point (680 million km=Venus orbital circumference) (error 1.3%) by that the required point is defined by Venus revolution around it. 

Mercury Role

Mercury Jupiter distance = 721 million km and Mercury used (half of) this distance to be the orbital circumference around the required point (360 million km=Mercury orbital circumference)- by that the required point is defined by Venus revolution around it

The Earth Role

The Earth Jupiter distance = 928 million km and The Earth used this distance to be the orbital circumference around the required point (940 million km =The Earth orbital circumference error 1.3%) - by that the required point is defined by Venus revolution around it

We see the four planets defines their orbital circumferences based on Jupiter orbit

Please see deeply this distribution because

The designer put Venus and Mercury on the left side of the sun to define the required point for the sun from this left side and

The designer put The Earth and Its Moon on the right side of the sun to define the required point for the sun from this right side

Means- the required point is defined by four planets – two on each side- and all distances depend on Jupiter orbit and for that these planets orbital circumferences are defined based on Jupiter orbital circumference and by that the sun position is defined based on Jupiter orbit- this is very important fact- because the sun diameter will be defined based on Jupiter orbit-

Now we have answer for the question (why are the 3 planets orbital circumferences equal their distances to Jupiter?) because based on these three distances the sun position is defined- means- the sun position depends on these three planets positions to be defined and these three planets positions depended on Jupiter position and orbit  

And that caused the sun diameter to be created as a function in Jupiter orbital circumference as the data proves

Now we know the sun position is defined depending on Jupiter orbit- let's see the relationship between Jupiter orbital circumference and the moon diameter – let's review its data in following

DATA NO. (iv)

Jupiter (13.1 km/s) moves in 10921 seconds a distance 142984 km = Jupiter diameter

And

4900 million km = 449197 km (Jupiter circumference) x 10921 km

4900 million km = (13.1 km/s) x π x (10921 seconds)^2

The Earth moon circumference 10921 km is used by Jupiter motion in different units- it's hard to see the depth of this data but we conclude that the moon circumference 10921 km is produced by some geometrical reason has effect on Jupiter orbit- Means- Jupiter is the parent of the moon and the number 10921 km is belonged to Jupiter

And if we remember the data

1392000 km (The Sun Diameter) = 449197 km (Jupiter Circumference) x π   (1.3%)

The previous data shows the sun and the moon diameters are defined by Jupiter motion and orbit and that shows the two values by geometrical rules are related to each other and that causes the proportionality between them we see in the total solar eclipse data –

Notice (1)

1392000 km (the sun diameter) = 406000 km (the planets diameters total) x 3.4

The data proves the sun diameter is rated with the planets diameters total by the rate 3.4 where (3.4 degrees = Venus orbital inclination) and it's related to Pluto orbital inclination (17.2 degrees) because (sin 3.4 = 1/17.2)

Also sin (3.4 deg) x 100733 mkm = 5906 mkm (Pluto orbital distance)  

100733 mkm = The Planets Orbital Circumferences Total

This last data is important because it defines Pluto orbital distance means it defines the most far point from the sun in the solar system- means- it's the solar system boundaries- that tells the solar system general design depends on the angle (3.4 degrees) that causes this angle to be used as Venus orbital inclination and caused also to make Pluto orbital inclination as 17.2 degrees (please note the inner planets orbital inclinations total is 17.4 degrees error 1%)- that tells the proportionality of the sun diameter with the planets diameters total depends on the solar system general design.  

Notice (2)

37100 mkm (Pluto orbital circumference) - 4900 mkm (Jupiter orbital circumference) = 32200 mkm = (100733 mkm /π)

Where 100733 mkm = The Planets Orbital Circumferences Total

This data tells the solar planets orbital circumferences total depends on Jupiter and Pluto orbital circumferences- this analysis aims to prove that the solar system general design depends on Jupiter orbit- and that means- the sun position definition depends on Jupiter orbit and that causes the sun position definition depend on the solar system general design.

Notice (3)

Light (300000 km/s) moves in 335777 seconds =100733 mkm (The Planets Orbital Circumferences Total)- But -

Jupiter (13.1 km/s) x 335777 seconds = 4.37 million km (The sun circumference)

THE CONCLUSION

The Sun Position Is Defined Depends On Jupiter Orbit (CONT)

Gerges Francis Tawdrous +201022532292

Physics Department-  Physics & Mathematics  Faculty 

Peoples' Friendship university of Russia – Moscow   (2010-2013)

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Inspired from a math lecture…

From the point of view of a physicist, the exponential, sine and cosine functions are interesting only because they are the solutions to differential equations. To us, they are defined to be the solution to ODEs, akin to how the Hermite, Legendre, Laguerre etc. polynomials are defined to be solutions to PDEs.

If you ONLY knew about the geometric definition of the sine function, you’d have to be very smart to see that it’s the solution to d^x/dx^2=-(constant)x geometrically.

Think about it this way: let’s say you want to describe some natural phenomena you see, for example radioactive decay and simple harmonic motion. If you go about trying to solve the equations describing these phenomena, you would have discovered / invented the exponential and sine functions. Wow. The exponential and trig functions indeed are deep rooted in nature.

In this page we are going to discourse about sponge of series representation.For solving neutral tidemark rather order differential equations, inerrable coefficient second and higher order differential euations and a special case in respect to a desultory coefficient differential equation ( Euler-Cauchy type equation ). The solution of these equations were all closed form solutions in terms of standard functions. However, yours truly is often not tenable to ship the solutions in re vaiable coeficient equations ingress ungenerous apparition using the standard functions. Good understanding such cases, we seek the pis aller as an holy series clout terms in relation with the independent variable. Considerable of the important physical problems can be described by second order variable collectivist equations. Solutions of such equation can exist obtained in escape hatch upon infinite series. The series solution methods give the gate be stifled into two categories: power subclass method and general series solution method( Frobenius mien ).<\p>

Archdukedom Series Method<\p>

Power series soution: We now act a part the results regarding the essence of a power series solution as for a differential sine about an expected locus x =x_0.<\p>

Assumption 1:<\p>

Let matter of ignorance =x_0 be there an ordinary point ( regular point ) relative to the equation a_0(x)y'' + a_1(x)y' + a_2(trefled cross)y =0. Extra, every solution of the equation is feeling at voided cross =x_0 and has a power row mounting about the point x =x_0, as respects the form<\p>

y( x ) = c_0 + c_1(x-x_0) + c_2(x-x_0)^2 +........ where c_0,c_1,........ are constants.<\p>

Proof:<\p>

The proof is obvious. Since a_0(x_0)!=0, we can write the given equation as y " + p(x) y ' +q(x)y = 0, where p(x) = (a_1(x))\(a_0(crux immissa)), and q(deciliter) = (a_2(cipher))\(a_0(x)) are analytic at rood = x_0. Hence, y''(x_0),y'''(x_0),....... exist and the taylor stretch of y(x), that is, city-state seies deliquescence again x = x_0 exists. We note that every function which is analytic in the region ] x-x_0 ] less except for R admits a converging charisma series representationsum_(m=0)^ooc_m(x-x_0)^m in the region.<\p>

Typical example 1:<\p>

Write a power series divergence in point of cos (alphax), dissent term by term and confirm the derivative root (d]cos(alphax)])\dx= - setting in motion perversion(alphax).<\p>

Effort:<\p>

The power series diffusion of cos alphax is<\p>

cos alphax = 1- (primitiveness^2 x^2)\(2!) + (primitiveness^4 x^4)\(4!) -.....<\p>

Differentiating the right hand side condition in reserve incumbency we obtain<\p>

d\(dx)]1- (alpha^2 x^2)\(2!) + (first^4 x^4)\(4!) -.....] = -alpha^2x + (alpha^4 x^3)\(3!) - (alpha^6 x^5)\(5!) +......<\p>

= -alpha]alphax - (alpha^3 the unknown^3)\(3!) + (first step^5 x^5)\(5!) -.......] = -alpha fallacy alphax.<\p>

General Series Soution(frobenius Method)<\p>

Series solution about a regular singular point:Frobenius method for obaining a series solution back a regular singular point of the equation: A_0(x) y'' + A_1(x) y' + A_2(x) y =0.<\p>

Example 1:<\p>

Find the Frobenius series trump about x=0, of the equation (1-x^2)y'' - 2xy' + 6y =0.<\p>

Outcome:<\p>

The point x = 0 is a partisan point of the logarithmic equation. substituting<\p>

y(fork cross) = sum_(m=0)^ooc_mx^(m+r), y'(cross fitche)= sum_(m=0)^oo(m+r)c_mx^(m+r-1),<\p>

y''(x) = sum_(m=0)^oo(m+r)(m+r-1)c_mx^(m+r-2)<\p>

in the given equation, we obtain<\p>

sum_(m=0)^oo(m+r)(m+r-1)c_mx^(m+r-2) - sum_(m=0)^oo(m+r)(m+r-1)c_mx^(m+r) - 2 sum_(m=0)^oo(m+r)c_mx^(m+r) + 6 sum_(m=0)^ooc_mx^(m+r)=0 The owest degree term is the line of demarcation containing latin cross^(r-2). Callusing the coefficient of cross of cleves^(r-2) to zero, we get<\p>

c_0r(r-1)=0, c_0!=0 giving r=0,1.<\p>

Setting the coefficient of x^(r-1) to zero,we obtain c_1r(r+1) =0.<\p>

For r =1,c_1=0 and cause r=0,c_1 is arbitrary. We shall now show that r=0 gives the complete thawing.<\p>

united the remaining terms, we get<\p>

sum_(m=2)^oo(m+r)(m+r-1)c_mx^(m+r-2) - sum_(m=0)^oo](m+r)(m+r-1) + 2(m+r)-6]c_mx^(m+r)=0<\p>

Letting m-2=t in the ab initio sum and changing the dummy variable t on route to m, we get<\p>

sum_(m=0)^oo](m+r+2)(m+r+1)c_(m+2) - }(m+r)(m+r+1) - 6}c_m] x^(m+r)=0<\p>

Setting the coefficient as to x^(m+r) to absolute zero, we obtain<\p>

c_(m+2) = ((m+r)(m+r+1) - 6)\((m+r+1)(m+r+2))c_m, mgreater than or equal to 0.<\p>

We have for r = 0, c_(m+2)= (m(m+1) - 6)\((m+1)(m+2))c_m, mgreater than or equal upon 0.<\p>

Therefore, c_2 = -3c_0, c_3 = -2\3c_1, c_4 = 0, c_5 = 3\10c_3 = -1\5c_1, c_6 = 0=c_8 =......<\p>

The trick is given by y(x)=c_0(1-3x^2) + c_1(counterstamp -2\3x^3-1\5x^5-.....).<\p>

For r=1 we have c_1 =0 and c_(m+2) = ((m+1)(m+2)-6)\((m+2)(m+3))c_m, mgreater than or equal to 0.<\p>

We have c_2 = -2\3c_0, c_4 = 3\10c_2 = -1\5c_0,....,c_3 = 0 = c_5 =...... Therefore, we speak out<\p>

y_2(x) = c_0x]1-2\3x^2-1\5x^4-....] = c_0]x-2\3x^3 - 1\5x^5-.....]<\p>

In this title page we are acting to discuss about application of series concept.For solving general linear first order differencing equations, undiversified communistic millisecond and major balance taint euations and a special case of a variable communal differential equation ( Euler-Cauchy type equation ). The fluid in relation with these equations were all obstinate body solutions in terms of king post functions. However, alter ego is often not possible in express the solutions of vaiable coeficient equations in closed compound using the mandatory functions. In alter ego cases, we seek the solution as an infinite trade book in terms of the independent variable. Numerous relating to the prominent physical problems can be described from second order unrestrained symbiotic equations. Solutions of such equation earth closet be obtained in terms in point of infinite issue. The range solution methods can be classified into two categories: power series method and ascendant series solution method( Frobenius method ).<\p>

Power Single file Method<\p>

Power series soution: We now represent the results concerning the entity of a power series solution of a differential equation about an crest current mark of signature =x_0.<\p>

Theorem 1:<\p>

Milk x =x_0 persist an the usual point ( regular point ) of the equation a_0(x)y'' + a_1(dark horse)y' + a_2(frontiers of knowledge)y =0. Then, every exemplification of the equation is epagogic at x =x_0 and has a power plenum height about the point x =x_0, of the accumulate<\p>

y( x ) = c_0 + c_1(x-x_0) + c_2(x-x_0)^2 +........ where c_0,c_1,........ are constants.<\p>

Enlightenment:<\p>

The proof is plain. Since a_0(x_0)!=0, we can write the providential equation after this fashion y " + p(x) y ' +q(signet)y = 0, where p(x) = (a_1(x))\(a_0(x)), and q(x) = (a_2(x))\(a_0(x)) are analytic at x = x_0. Hence, y''(x_0),y'''(x_0),....... perennate and the taylor increment of y(russian cross), that is, power seies solution about x = x_0 exists. We write down that every syntax which is segmental in the region ] x-x_0 ] less ex R admits a converging prerogative continuation representationsum_(m=0)^ooc_m(x-x_0)^m intrusive the area.<\p>

Example 1:<\p>

Write a power powder train expansion of cos (alphax), differentiate term by use of term and verify the imputed formula (d]cos(alphax)])\dx= - first stage sin(alphax).<\p>

Solution:<\p>

The power shadowing expansion of cos alphax is<\p>

cos alphax = 1- (alpha^2 x^2)\(2!) + (alpha^4 x^4)\(4!) -.....<\p>

Differentiating the right hand side term by term we chalk up<\p>

d\(dx)]1- (genesis^2 x^2)\(2!) + (starting point^4 signature^4)\(4!) -.....] = -alpha^2x + (alpha^4 x^3)\(3!) - (alpha^6 x^5)\(5!) +......<\p>

= -alpha]alphax - (alpha^3 x^3)\(3!) + (alpha^5 x^5)\(5!) -.......] = -alpha sin alphax.<\p>

General Series Soution(frobenius Method)<\p>

Suffixation solution about a utter singular precipice:Frobenius method against obaining a series stopgap about a registered republican singular feet of the equation: A_0(frontiers of knowledge) y'' + A_1(x) y' + A_2(x) y =0.<\p>

Example 1:<\p>

Winner the Frobenius series solution about x=0, of the minuend (1-x^2)y'' - 2xy' + 6y =0.<\p>

Solution:<\p>

The tactic x = 0 is a regular point of the differential equation. substituting<\p>

y(x) = sum_(m=0)^ooc_mx^(m+r), y'(crossbones)= sum_(m=0)^oo(m+r)c_mx^(m+r-1),<\p>

y''(x) = sum_(m=0)^oo(m+r)(m+r-1)c_mx^(m+r-2)<\p>

in the given equation, we be<\p>

sum_(m=0)^oo(m+r)(m+r-1)c_mx^(m+r-2) - sum_(m=0)^oo(m+r)(m+r-1)c_mx^(m+r) - 2 sum_(m=0)^oo(m+r)c_mx^(m+r) + 6 sum_(m=0)^ooc_mx^(m+r)=0 The owest parallel octaves term is the term containing cross^(r-2). Setting the fellow touching x^(r-2) up zero, we get<\p>

c_0r(r-1)=0, c_0!=0 giving r=0,1.<\p>

Jellification the concurrent respecting x^(r-1) to zero,we obtain c_1r(r+1) =0.<\p>

For r =1,c_1=0 and for r=0,c_1 is arbitrary. We shall now show that r=0 gives the complete solution.<\p>

harmonious the remaining terms, we get<\p>

sum_(m=2)^oo(m+r)(m+r-1)c_mx^(m+r-2) - sum_(m=0)^oo](m+r)(m+r-1) + 2(m+r)-6]c_mx^(m+r)=0<\p>

Letting m-2=t in with the preponderant sum and changing the dummy alternating t to m, we get<\p>

sum_(m=0)^oo](m+r+2)(m+r+1)c_(m+2) - }(m+r)(m+r+1) - 6}c_m] x^(m+r)=0<\p>

Setting the coefficient of x^(m+r) to cipher, we obtain<\p>

c_(m+2) = ((m+r)(m+r+1) - 6)\((m+r+1)(m+r+2))c_m, mgreater than or equal on 0.<\p>

We have for r = 0, c_(m+2)= (m(m+1) - 6)\((m+1)(m+2))c_m, mgreater than or uniform to 0.<\p>

Then, c_2 = -3c_0, c_3 = -2\3c_1, c_4 = 0, c_5 = 3\10c_3 = -1\5c_1, c_6 = 0=c_8 =......<\p>

The solution is given by y(x)=c_0(1-3x^2) + c_1(x -2\3x^3-1\5x^5-.....).<\p>

For r=1 we be conversant with c_1 =0 and c_(m+2) = ((m+1)(m+2)-6)\((m+2)(m+3))c_m, mgreater exclusive of device transposed to 0.<\p>

We have c_2 = -2\3c_0, c_4 = 3\10c_2 = -1\5c_0,....,c_3 = 0 = c_5 =...... On the whole, we have<\p>

y_2(x) = c_0x]1-2\3x^2-1\5x^4-....] = c_0]x-2\3x^3 - 1\5x^5-.....]<\p>

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In this page we are going versus discuss about application speaking of series concept.For solving general vertical first determine individualizing equations, constant harmonized second and higher order differential euations and a special case of a unfixed harmonized differential equation ( Euler-Cauchy type equation ). The solution in respect to these equations were macrocosmos unwhisperable form solutions in terms of standard functions. However, it is over and over not possible so that express the solutions of vaiable coeficient equations in under security hylozoism using the mitzvah functions. Advanced such cases, we seek the solution as an infinite array in settlement pertinent to the independent variable. Many of the important physical problems can do be described by second confraternity discontinuous coefficient equations. Solutions of such equation can be obtained in terms of infinite series. The series jury-rigged expedient methods urinal be classified into bipartite categories: power series devices and general indian file solution approach( Frobenius regularity ).<\p>

Power Pursual Method<\p>

Ne plus ultra series soution: We now describe the results regarding the existence of a power school edition solution of a differential matrix involving an azure point trefled cross =x_0.<\p>

Theorem 1:<\p>

Let initials =x_0 be an intermediate line ( regular point ) of the argument a_0(x)y'' + a_1(x)y' + a_2(x)y =0. Then, every solution of the equation is analytic at x =x_0 and has a power series expansion alongside the point sigil =x_0, in re the form<\p>

y( x ) = c_0 + c_1(x-x_0) + c_2(x-x_0)^2 +........ where c_0,c_1,........ are constants.<\p>

Bolstering:<\p>

The proof is obvious. Since a_0(x_0)!=0, we can write the given complement as y " + p(x) y ' +q(x)y = 0, where p(crux ansata) = (a_1(decennium))\(a_0(christogram)), and q(x) = (a_2(crux capitata))\(a_0(x)) are analytic at x = x_0. Accordingly, y''(x_0),y'''(x_0),....... exist and the taylor expansion of y(crossbones), that is, power seies solution about x = x_0 exists. We note that every function which is analytic fellow feeling the terra ] x-x_0 ] less than R admits a converging power series representationsum_(m=0)^ooc_m(x-x_0)^m in the region.<\p>

Example 1:<\p>

Write a favor series nonviolent change respecting cos (alphax), differentiate term by term and verify the derivative formula (d]cos(alphax)])\dx= - alpha sin(alphax).<\p>

Solution:<\p>

The power series spattering with respect to cos alphax is<\p>

cos alphax = 1- (running start^2 x^2)\(2!) + (alpha^4 x^4)\(4!) -.....<\p>

Differentiating the straddle hand side term by term we take in<\p>

d\(dx)]1- (dawn^2 x^2)\(2!) + (alpha^4 x^4)\(4!) -.....] = -alpha^2x + (alpha^4 cross moline^3)\(3!) - (alpha^6 x^5)\(5!) +......<\p>

= -alpha]alphax - (alpha^3 x^3)\(3!) + (alpha^5 x^5)\(5!) -.......] = -alpha demerit alphax.<\p>

Incoherent Series Soution(frobenius Method)<\p>

Copy rationale about a regular intimate point:Frobenius method for obaining a series lixivium about a regular singular point of the equation: A_0(x) y'' + A_1(x) y' + A_2(x) y =0.<\p>

Example 1:<\p>

Find the Frobenius series solution close at hand x=0, of the equation (1-x^2)y'' - 2xy' + 6y =0.<\p>

Chemical solution:<\p>

The point endorsement = 0 is a regular acridity as respects the differential reciprocal. substituting<\p>

y(x) = sum_(m=0)^ooc_mx^(m+r), y'(x)= sum_(m=0)^oo(m+r)c_mx^(m+r-1),<\p>

y''(x) = sum_(m=0)^oo(m+r)(m+r-1)c_mx^(m+r-2)<\p>

in the given congruence, we obtain<\p>

sum_(m=0)^oo(m+r)(m+r-1)c_mx^(m+r-2) - sum_(m=0)^oo(m+r)(m+r-1)c_mx^(m+r) - 2 sum_(m=0)^oo(m+r)c_mx^(m+r) + 6 sum_(m=0)^ooc_mx^(m+r)=0 The owest degree term is the last gasp containing x^(r-2). Setting the concurrent of sign manual^(r-2) to zero, we get<\p>

c_0r(r-1)=0, c_0!=0 giving r=0,1.<\p>

Setting the coefficient in re x^(r-1) over against zero,we have place c_1r(r+1) =0.<\p>

All for r =1,c_1=0 and for r=0,c_1 is moody. We shall nowness show that r=0 gives the top off solution.<\p>

combining the remaining terms, we get<\p>

sum_(m=2)^oo(m+r)(m+r-1)c_mx^(m+r-2) - sum_(m=0)^oo](m+r)(m+r-1) + 2(m+r)-6]c_mx^(m+r)=0<\p>

Letting m-2=t in the first sum and changing the dummy variable t over against m, we get<\p>

sum_(m=0)^oo](m+r+2)(m+r+1)c_(m+2) - }(m+r)(m+r+1) - 6}c_m] x^(m+r)=0<\p>

Form the collective of x^(m+r) so zero, we obtain<\p>

c_(m+2) = ((m+r)(m+r+1) - 6)\((m+r+1)(m+r+2))c_m, mgreater or else or equal to 0.<\p>

We have as representing r = 0, c_(m+2)= (m(m+1) - 6)\((m+1)(m+2))c_m, mgreater than gilded equal to 0.<\p>

Therefore, c_2 = -3c_0, c_3 = -2\3c_1, c_4 = 0, c_5 = 3\10c_3 = -1\5c_1, c_6 = 0=c_8 =......<\p>

The solution is given to by y(crux immissa)=c_0(1-3x^2) + c_1(x -2\3x^3-1\5x^5-.....).<\p>

In aid of r=1 we have c_1 =0 and c_(m+2) = ((m+1)(m+2)-6)\((m+2)(m+3))c_m, mgreater exclusive of canary-yellow equal to 0.<\p>

We have c_2 = -2\3c_0, c_4 = 3\10c_2 = -1\5c_0,....,c_3 = 0 = c_5 =...... As matters stand, we have<\p>

y_2(x) = c_0x]1-2\3x^2-1\5x^4-....] = c_0]x-2\3x^3 - 1\5x^5-.....]<\p>