Derivation of the equation of Ellipse from its definition

    When we do the Kepler problem, we will obtain the result of a general conic section equation in polar form. But I got myself stuck looking at the equation of ellipse. I couldn't remember any damn thing I learned in my school, except the definition of ellipse. 

   But I know in mathematics, every thing can be derived from the fundamental postulates or axioms. So, I just proceed to build the equation starting from the definition. 

   The whole of the conic section is defined based on the concept of eccentricity. The definition I remember is that, Ellipse is the conic section with eccentricity value less than 1. And eccentricity is the ratio of the distance between focus to any point P to the perpendicular distance between point P and the directrix.

    To define a focus point, first we draw an axis and place the focus point. Directrix is the perpendicular line drawn anywhere to that axis except at the focus point.

   The condition is that you need to join all the points which obeys the rule that their eccentricity should be less than one and it is a constant. For each constant value, you will get different ellipse.    

   You don't even need the definition of axis to draw the ellipse, but it is just for convenience. All you need is just a point called focus and a line called directrix and follow the condition.
   

   From the definition of ellipse, we can see that there is two possible points which will obey the condition at the axis [which passes through the focus point and perpendicular to directrix]. Let me call those points $ P_1 and P_2 $ and the focus is F.  And also consider a general point P on the ellipse. 

    Giving separate symbols for each point helps to get clear understanding.  
   
    So far, we even don't know whether the ellipse would look symmetrical or not, or does have equal foci or etc. Yet, we haven't proved anything else.  

   The simple diagram we can draw with our so far concepts is,

From our definition, we know that the eccentricity value is constant for all the points of a given ellipse. Using the fact that   $P, P_1, P_2$ points are in the same ellipse, they all should have the eccentricity value - lets say "e".

    It gives $$ \frac{FP}{PM} = e and \frac{FP_1}{P_1M_1} = e and also, \frac{FP_2}{P_2M_1} = e $$

Let us say, $P_2P_1 = y $ 
   Using some elementary algebra, we can write $ \frac{FP_2}{P_2M_1} = e $ as, $$ FP_2 = e [P_2M_1] = e [P_2F + FP_1+P_1M_1]$$
   But $ FP_1 and P_1M_1$ are related by $$\frac{FP_1}{P_1M_1} = e $$ We can choose in either way.. I choose to write $P_1M_1$ in terms of $FP_1$ 
   So, $$ P_1M_1 = \frac{FP_1}{e}$$
Substituting this value in $FP_2$ we get,
  $$ FP_2 = e [FP_2 + FP_1 + \frac{FP_1}{e}]$$ 
$$ FP_2 = e \frac{[ e FP_2 + FP_1 (e+1)]}{e} $$
$$ FP_2 = e FP_2 + FP_1 (e+1) $$
$$ FP_2 = e (FP_2+FP_1) + FP_1$$ 
$FP_2 + FP_1 = P_2P_1 = y $
$$ FP_2 = e y + FP_1 $$ add $FP_1$ on both sides $$ FP_2 + FP_1 = e y + 2 FP_1 $$ and finally we get, $$ y = ey + 2FP_1$$
or simply, $$ FP_1 = \frac{y}{2} (1 - e) $$

That is all we want. If we draw our coordinate system at $\frac{y}{2}$ distance away from the focus on the axis, then focus would be situated exactly at $ \frac{y}{2}e $ distance away from the origin.

For simplicity, we take y = 2a so that y/2 = a. Thus, if the coordinate system is drawn at "ae" distance from the focus, the extreme points would have "2a" distance between them. And they are equidistant from the origin. 

Since we find out now that, both extreme points are symmetrical with each other with respect to the origin, we can conclude that based on symmetry arguments,
    Whatever the graph of ellipse drawn on the right side, would be similar if we choose the focus at the left side exactly at "-ae" distance from the origin at the axis of the ellipse. The same things would repeat. 

    Thus we conclude that, the graph of the ellipse has symmetry with respect to the y-axis - as shown in the figure below  

To find the distance of Directrix from the origin,

$$ OM_1 = OP_1 + P_1M_1 $$
where $OP_1 = a $ and $P_1M_1 = \frac{FP_1}{e} $ and $FP_1 = a - ae $
Gives, $$ OM_1 = a + \frac{a}{e} - a $$ 

Thus, we finally get the distance of directrix from origin as "a/e".

We are almost over finding the every relations along the axis of the ellipse. Now, we can go to the perpendicular to the axis i.e. the maximum height on the y - axis from the axis of the ellipse.

Since it deals with y - direction, let us consider the point $P_3$ which is the positive y - intercept of the ellipse. 

Again applying the eccentricity condition we get, $$ \frac{FP_3}{PM_3} = e $$
Drawing FT perpendicular to $P_3M_3$ helps us to use Pythagoras theorem, so that we can write everything in terms of known quantities. 
     
From Pythagoras theorem,
$$ {F_1P_3}^2 = {OP_3}^2 + {OF_1}^2 $$

Let us say, $ OP_3 = b $ for the sake of simplicity 
and we know $OF_1 = focal length = ae $ which gives 

So, $$ {F_1P_3}^2 = b^2 + (ae)^2 $$

From the geometry, it is understood that $TM_3 = F_1M_1$ and $ P_3M_3 = P_3T + TM_3$

$$ P_3M_3 = OM_1 = distance of the directrix from origin = \frac{a}{e}$$

And also $$ {F_1P_3}^2 = e^2 {P_3M_3}^2 = e^2 \frac{a^2}{e^2} = a^2 $$

Equating the two results of $ F_1P_3 $,
$$ b^2 + (ae)^2 = a^2 $$ Hence, 
$$b^2 = a^2 (1 - e^2) $$

Everything is in position to find the general equation. For the general equation, take a general point and apply the same old rules. 

$$ {FP}^2 = e^2 (PM)^2 $$ Let us take the co-ordinates of P as (x,y). 

Using the same Pythagoras theorem, distance between two points in a Cartesian plane can be determined using the formula,
If the points are $ A(x_1, y_1) and B(x_2, y_2)$ then distance between them $$ AB = \sqrt {(x_2 - x_1)^2 + (y_2 - y_1)^2} $$

Here we know the coordinates of all points i.e. F(ae,0), P(x,y), M(a/e, y). 
Using that , the equation $ {FP}^2 = e^2 {PM}^2 $ becomes, 
$$ (x - ae)^2 + y^2 = e^2 [ (x - \frac{a}{e})^2 + (y - y)^2] $$
$$ (x - ae)^2 + y^2 = (xe - a)^2 $$
$$ x^2 + (ae)^2 -2xae + y^2 = (xe)^2 + a^2 -2xae $$
$$ x^2 - (xe)^2 +y^2 = a^2 - (ae)^2 $$
$$ (1-e^2) x^2 + y^2 = a^2 (1- e^2) $$
Diving the equation by $ a^2 (1- e^2) $ We get,

$$\frac{x^2}{a^2} + \frac{y^2}{a^2(1 - e^2)} = 1 $$
But we know b^2 = a^2 (1 - e^2)

Hence, we derived the general equation of Ellipse in Cartesian co-ordinates as,

$$ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 $$ where a > b

Properties (measured from origin):
Focal length = ae
Distance of the directrix = a/e
$ b^2 = a^2 (1 - e^2) $
Length of the major axis = 2a
Length of the minor axis = 2b  

Infinite Series Solution for Differential Equations

Infinite Series Solutions is the One of the most helpful technique often used in solving differential Equations. They are called special functions and the method is known as "Frobenius method". It is an improvisation of the common "power series method". 

    Some of the famous solutions are known as Legendre function, Bessel function, Hermite functions and etc. These are just the name given for the solutions of different differential equations where else the method is quite general.  

   Let us say, for example, we encounter a differential equation like, $$ {d^2}y/d{x^2} - 2x dy/dx + 2ny = 0 $$ which is called the Hermite differential equation. You may need to solve this equation, if you try to solve Schrodinger equation for Simple Harmonic Oscillator. 

   To proceed,  first we are going to assume, there exists really a solution, for this differential equation which can be expressed in terms of power series solution. 

   Then, we will insert that solution in our differential equation and impose the necessary conditions to be the solution and finally will get our modified output solution. 
Note: Remember, here we just seek a solution that fulfills our conditions, not anything more.   

   The general input solution is given by, $$ y(x) = \sum_{m=0}^{\infty} C_m x^{m+r} ........ where C_0 \neq 0 $$ Substituting this on our differential equation, our differential equation gets modified into $$ \sum_{m=0}^\infty \left[ (m+r)(m+r-1) C_m x^{m+r-2} + 2 [ n - (m+r)] C_m x^{m+r} \right] = 0 $$ This should be equal to zer0, which can be obtained only if and only if all the coefficients are zero. 

   Equating to zero, the coefficients of,
$$ (1) x^{r-2}, \ldots... r(r-1) = 0 \rightarrow r=0 \; or \; r=1 \; since \; C_0 \neq 0 $$ $$ (2) x^{r-1}, \ldots... (r+1)rC_1 = 0 .... \; if \;\; r=0, \; C_1\neq 0 ; and \; if \;\; r=1, \; then \; C_1=0 $$ $$ (3) x^{m+r}, \\~\\ for \; r=0 \dots......... \frac{C_{m+2}}{C_m} = \frac{-2(n-m)} {(m+1)(m+2)} \\~\\ for \; r=1 \dots...........\frac {C_{m+2}}{C_m} = \frac{-2[n-m-1]}{(m+2)(m+3)}$$

  Considering r=0, the general expression for even coefficients expressed as, 
$$ C_{2s} = \frac{(-1)^s 2^s n (n-2).....(n-2s+2)}{(2s!)} C_0 $$
and the odd coefficients are expressed as,
$$ C_{2s+1} = \frac{ (-1)^s 2^s (n-1) (n-3).....(n-2s+1)}{(2s+1)!} C_1$$
The general solution is therefore,

$$ y(x) = C_0 \left[ 1 + \sum_{s=1}^\infty \frac{ (-1)^s 2^s n (n-2)....(n-2s+2)}{(2s!)} x^{2s}\right] + \\~\\ C_1x \left[ 1 + \sum_{s=1}^\infty \frac{(-1)^s 2^s (n-1)(n-3)....(n-2s+1)}{(2s+1)!} x^{2s+1} \right] $$   

For r=1, if we proceed like the same, we will get a solution exactly similar to the second series in the general solution. Since its property already inherent in the general solution, we don't need to put much attention on that. 

   That's it. We arrived to our general solution. Depending on the   

nature of the problem, we can make the series to converge or stop by appropriately choosing the values of $ \; C_0 \;and \;C_1$ 

   There is a conventional way of choosing the constants, which gives the series of many function known as Hermite polynomials. It should be dealt separately on  how the Hermite polynomials are derived from the general solution. 

   As far as now, the general solution is just an example of how to get an output from a differential equation. To understand this more, we need to separately solve more differential equations in different physical situations.  

How to install Android apps on your smartphone using your Computer's Internet Connection

   
    When you don't have any internet account balance in your smartphone, but wanted to download and install Android apps [say for example from google play store] from the Internet connection of your PC or laptop, the best solution I found is Mobogenie.  

[ Of course, if you have WiFi adapter in your laptop, you can create a WiFi hotspot and can use Internet connection to your smartphone. But, if you are using broadband connection and don't have WiFi adapters, you can use this way. ]

   I tried many other ways like "Reverse tethering" - where you can use your computer's Internet connection e.g. Broadband connection to surf on your smartphone. 

   Using "Reverse tethering", you can surf, download and install all things from your computer's Internet connection. 

    But I found, Reverse tethering needs a little technical things like Rooting your Android phone, super user permissions and etc. Really, there are quite many ways to Reverse tether. 

   Here, I want to share the simplest way only for dowloading and installing apps on your smartphone from your computer's Internet connection [ Note: You cannot surf using this method ].

The steps you need to follow is, 

First, go to the Mobogenie website by searching it on the Internet, download the mobogenie for PC and install it in your PC or laptop.

After you installed mobogenie on your computer, whenever you connect a new android device in your computer and open mobogenie, it automatically connects with your android phone. 

    It also installs the mobogenie app in your android phone. 

    Here is the very important thing, Mobogenie connects with your android phone only if and only, if you have enabled "USB debugging in your android phone - which you can find it in developer options in settings menu.

   I made some photos of each step, so you can confirm with those in case you have some problems. 

To enable USB debugging mode, Go to the Settings - Developer options

Go to Developer Options and Click the USB Debugging options which asks you for the confirmation like below

Click "Ok" and you are done with usb debugging. Now, whenever you open Mobogenie when your computer is connected with your android phone, it automatically installs and does the necessary thing in your android. 

You can see the mobogenie logo in your android phone once it is done with installation

      Now, with this installed mobogenie app, you can at any time download and install any Android apps to your smartphone via your PC's Internet connection. 

      You can also manage all your contacts, can download videos, games and etc. 

      The best advantage of using this is, you don't need to run out of your house to get internet signal. Instead you can get High speed Internet from your PC itself in your home. 

Lorentz Transformation Equations from the fundamental Postulates of Special theory of Relativity

The Two fundamental Postulates are,

               

           1. The laws of Physics are same in all inertial reference frames.

           2. The speed of light is always a constant independent of the motion of the observer. 

      As a student When I heard these postulates for the first time, I really never thought that there would be so much in it to understand. But the implications are quite significant to explain the Nature. Let us see, how it changes the universe...
      The most important term is that "the speed of light". We know the velocity follow Galilean addition rules withing one frame to another frame of reference. 

And we know how to make this Galilean transformation between two inertial reference frames.     
      Now, with surprise the postulate is given as the speed of light is constant in all reference frames, which means we cannot apply our previous Galilean transformations. It should be changed!!
     
      But we are already using these transformation and getting best results. That means we can be sure that these laws can never proved to be completely wrong. 

     All it can be done is some alteration without affecting the known phenomena. For the mathematics, let us take a case..  

     As usual we take two inertial reference frames S and S' with relative speed "v".  We want to keep the structure of Galilean equations at speed lower than c. 
     So, we assume that there is an extra factor comes into play with Galilean equations, which has the special property that it gets the value "1" when v<<c. Let us assume the factor is $ \gamma $

Rewriting the known Galilean transformation with the new factor "gamma".. $$ x' = \gamma (x - vt)..............eq.(1) $$ For convenience we always used to take the motion to be only in x-direction. And it will never going to change anything in y or z- direction. It is always y'=y and z'=z. 

    Inverse transformation similarly given as, $$ x = \gamma (x'+vt)..............eq.(2) $$ So far, we never used any information from the postulates. 
    Now from the second postulate, for a light wave speed is always the constant - c. Let us imagine that a light wave is allowed to propagate in the positive x direction, when the both reference frames starts from the common origin. 

[ Why I am taking the case of Light wave?

We assume the Universe is same for everything else whether it is a light or a particle. As the postulate defined at light speed, if we could find the value of the gamma factor at light speed, it will eventually obey for all velocities. Intrinsically, we believe that the Universe won't have difference structure at different velocities. 

Simply, it is finding the function by applying the boundary conditions.]   

    For the light wave, the equations become, x' = ct' and x = ct 
[We now thrown away the concept of space and time being flat and universal to everyone]    
Applying in eq.(1) and (2).. $$ ct' = \gamma (ct-vt)...............eq.(3)$$ 
and $$ ct = \gamma (ct'+vt')................eq.(4)$$  eq.(3) becomes $$ t' = \gamma \frac{(c-v)}{c}t $$ and eq.(4) becomes $$ t = \gamma \frac{(c+v)}{c} t' $$
   Combining the (4) with (3) we can cancel the time variables..
$$ t = \gamma \frac {(c+v)}{c} ( \gamma \frac{(c-v)}{c}t ) $$
$$ t = \gamma^2 \frac{(c^2 - v^2)}{c^2} t $$
Cancelling the 't' on both sides..
$$ \gamma^2 = \frac{1}{1 - (v^2/c^2)}$$ Thus we finally get the value of "gamma" as $$ \gamma = \frac {1}{\sqrt{1-(v^2/c^2)}}$$

    Remember, we took a special case at light speed and got the value for gamma - $\gamma$ factor. Now, we need to substitute this in the normal case i.e. eq.(1) and (2) which is given a new name - Lorentz Transformation equations. $$ x' = \frac{x - vt}{\sqrt{1 - (v^2/c^2)}} $$ and the inverse transformation is $$ x = \frac{x' + vt'}{\sqrt{1 - (v^2/c^2)}} $$

    To derive the general transformation equation for time, substitute eq.(2) in eq.(1) $$ x' = \gamma ((\gamma (x' + vt')) - vt) \\~\\ x' = \gamma^2 (x' + vt') - \gamma vt $$ Taking t in one side and the remaining in the other gives..$$ \gamma vt = (\gamma^2 - 1) x' + \gamma^2 vt' $$ Finally we get t in terms of x' and t'.. $$ t = \frac{\gamma^2 x' (v^2/c^2)} {\gamma v} + \frac{\gamma^2 vt'}{\gamma v} \\~\\ t = \gamma (t' + (x'v/c^2) = \frac{t' + (x'v/c^2)}{\sqrt{1 - (v^2/c^2)}} $$

Similarly, the inverse tranformation can be obtained as, $$ t' = \frac {t - (xv/c^2)}{\sqrt{1 - (v^2/c^2)}} $$

I think.. we finished getting the most basic Lorentz transformation equations from one inertial frame of reference to other which moves at a relative speed "v". 

From these transformation equations we may able to understand all those amazing, significant physical nature of Space and Time.  

Mathematics Problem:1 - Think Out of the Box

      I often used to remember a mathematical problem [from IIT entrance test papers], which took me a lot of days to solve in my school level. 

     The problem is to find the value of the following definite integral, $$ I = 5050 \frac{\int_0^1 (1-x^{50})^{100} \,dx }{\int_0^1(1-x^{50})^{101}\,dx}$$

    Giving some time before going to the solution, gives you enough insight into the problem.

     Of course I will give you the answer but not solution because the answer is what surprised more than anything else. The value of above integral is - 5051. 

     I took this problem just because of this answer because,
here 5050 * {Integral} = 5051
Simply the integral is so amazing that it just adds the amount by one number when multiplied!!! 

For the solution,

     Most of the time, we tend to solve the integral using all our pre-loaded formulas in our mind. I myself tend to solve this using many possible methods including series of expansion, integral by parts, even tried find in a numerical way, etc. 

     We always used to stick only with the given terms in the problem, but that won't help here in anyway

     Problem is all about a simple trick, which you can understand by just looking at the numbers i.e. 100 and 101

     To solve it, we should simplify the integral but it is possible only when they both have the same powers. 
So, you need to change the power of the integral but need to keep the integral unchanged. Perfect choice is multiply and divide by same value.
   Again it won't help with the integral since it would change power both in the numerator and the denominator. 

Note:You need to multiply only in the numerator.  

      The choice would be the number - "1" 
Yet since we also need $x^{50}$ here the best choice would be $$ 1- x^{50}+x^{50}$$
      Thus we never changed any value but multiplied a term only in the numerator to change the power. 
     The rest of the problem is just simple mathematics as follows,

$$ I = 5050 \left(\frac{\int_0^1 (1-x^{50})^{100} \times (1-x^{50}+x^{50}) \,dx }{ \int_0^1(1-x^{50})^{101}\,dx }\right)$$ 
which gives,
$$ I = 5050 \left( \frac{ \int_0^1 (1-x^{50})^{101}\,dx }{ \int_0^1 (1-x^{50})^{101}\,dx } + \frac {\int_0^1 (1-x^{50})^{100}x^{50} \,dx }{ \int_0^1 (1-x^{50})^{101} \,dx}\right) \\~\\ I = 5050 \left( 1 + \frac{\int_0^1 x^{50}\,dx}{\int_0^1 (1-x^{50})^{101}\,dx}\right)$$
The integral on applying the limits simplifies into...
$$ I = 5050 \left( 1 + \frac{\int_0^1(1-x^{50}) x^{50}\,dx}{\int_0^1(1-x^{50})^{101}\,dx}\right)$$ 

We will do the separately the integral in the denominator using "integration by parts method"..
$$ I_0 =\int_0^1(1-x^{50})^{101}\,dx = [x(1-x^{50})^{101}]|_0^1 - \int_0^1 (x) (101) (-50x^{49}) (1-x^{50})^{100} \,dx \\~\\ I_0 = 0 + (5050) \int_0^1(1-x^{50})^{100}x^{50} /,dx $$

Thus we got.. $$ I_0 = (5050) \int_0^1 (1-x^{50})^{100} x^{50} \,dx $$
substituting this in "I" .. we arrive at..
$$ I = 5050 \left( 1 + \frac{1}{5050} \frac {\int_0^1 (1-x^{50})^{100} x^{50} \,dx}{\int_0^1 (1-x^{50})^{100} x^{50} \,dx}\right)$$
$$ \rightarrow                         I =  5050 \left( 1 + \frac{1}{5050} \right) $$
Thus, we finally arrived at our amazing solution

                              I = 5050 + 1 = 5051

Now, we can go and look at the beauty of the solution!!!

New solution to Einstein's puzzle - "Who owns the Zebra?" or "Who owns the fish?"

I was just solving, the so called Einstein puzzle. I don't know whether it is Einstein's or not but it is a good puzzle. 

As given in Wikipedia, it was called "Who owns the Zebra? - Puzzle".

The puzzle was given as follows, 

1. There are five houses.
2. The Englishman lives in the red house.
3. The Spaniard owns the dog.
4. Coffee is drunk in the green house.
5. The Ukrainian drinks tea.
6. The green house is immediately to the right of the ivory house.
7. The Old Gold smoker owns snails.
8. Kools are smoked in the yellow house.
9. Milk is drunk in the middle house.
10. The Norwegian lives in the first house.
11. The man who smokes Chesterfields lives in the house next to the man with the fox.
12. Kools are smoked in the house next to the house where the horse is kept.
13. The Lucky Strike smoker drinks orange juice.
14. The Japanese smokes Parliaments.
15. The Norwegian lives next to the blue house.

Now, who drinks water? Who owns the zebra?
In the interest of clarity, it must be added that each of the five houses is painted a different color, and their inhabitants are of different national extractions, own different pets, drink different beverages and smoke different brands of American cigarets [sic]. One other thing: in statement 6, right means your right.
                                     — Life International, December 17, 1962

[Give some time to this puzzle and its previous solution and then scroll down to the end of this post for my new solution]

I don't know exactly, but I tried this for some 2 days. When I was travelling back to my home, I made some "hit and trial" between two choices to solve the puzzle. 
Surprisingly I got the answer with the given specifications. 
With too much joy, after reaching my home I just googled to check the answer.

Here is where I got the amazement. 

My answer is Different!!!

But it looks like it follows each and every rule. I was sure, my answer satisfies all the given conditions. 

When I checked the solutions, I found out the turning point. 

In the 11th, 12th and the 15th statement, it was used the word "next".
In the solution, "next" was used in the sense "to the right or the left". 

But I have been concluding that the fifth house and the first house could be thought as next to each other.

And my conclusion is not completely wrong because it was said in the first statement itself, there are only five houses considered.
So, I counted as the first house as the next house from the fifth one. 
Moreover it can be situated in a circular way.

Thus I end up with a new answer where the Norwegian owns the Zebra and drinks the water. 

Definitely, I believe it is a new solution. 

Away from this, there are other views such as, 
There needn't be Zebra and water. The solution can be some other things since it was never said that Zebra should be there in one of the houses. And there are some discussion on the words used in the puzzle and the way it was presented, which you can find it in the wikipedia.   

If those discussions are valid based on the words, then the above conclusion is also perfectly valid since it was never said that the houses should not be in a circular order. 


Thus I can justify my solution. 

The solution is 

For complete explanation - click Who owns the zebra solution

Inertial Frame of Reference

     This abstract started with the fundamental concept called "the frame of reference". 
     We know that the laws of physics are not same for all observers from the simple fact that a moving observer would explain things in a different way from a relative stationary observer (unless and until they both move relatively at constant speed). 
     Try it for yourself by explaining the motion of objects around you i.e. for example one from the train and one from the ground.   

Since physics is the most compact mathematical tool we can afford.. We don't want to create new and separate laws for moving observers and stationary observers. 
     And so, Galilean transformation rules are invented for the transformation of physical quantities from one reference frame to another. 

      Galilean transformation rules are just basic arithmetic thing. Let us consider two reference frame relatively moving at constant speed i.e. S and S' moving with relative speed "v" where else the movement considered only to be in the x-direction. 
     *Don't bother about the direction of motion. We can always chose the co-ordinate system such a way that x-axis lies in the direction of motion. 

     If a position is to be specified in the universe in "S- frame" the components are (x,y,z,) at time - t. Similarly the components measured in " S' -frame " are (x',y',z') at time - t. In Galilean time, time was considered to be the same for everyone. 

    If S' is thought to be moving in the positive x-direction relative to S then the components observed by two frames are related by,
                           x' = x -vt
                           y'= y
                           z'= z
the reverse is of course x = x' + v t ; y = y'; z = z'

    Of course there should be primed coordinates on the right [because we are relating the measurements made from one frame to other.] 
i.e.  x = x' + v t'
but we assumed t = t' and so the problem solved. 

   You may ask why I am concerned about just positions?
After all, positions and their change with respect to time are just needed to explain any motion from Newton's laws of motion.

   Now, Let us check how Newton's laws are behaving in these reference frames. In S - frame of reference
$$ \vec{F} = m \frac{d^2x}{dt^2} $$ Applying the same law in S' - frame, it gives $$ \vec{F'} = m \frac{d^2x'}{dt^2}$$ 
To compare the S' frame with S , substituting the value of x' in terms of co-ordinates observed in S-frame..$$ \vec{F'} = m \frac{d^2(x-vt)}{dt^2}$$ which gives $$ \vec{F'} = \frac {d^2x}{dt^2} - \frac{d^2(vt)}{dt^2}$$ "vt" term cancels on double integration $$ \rightarrow                    \vec{F'} = m \frac{d^2x'}{dt^2} = \vec{F} = \frac{d^2x'}{dt^2}$$

   As we can see that, both the observers obtain the same Newtonian force and will explain any phenomena exactly in the same way.  They can't even know, which one is moving from the observed motion. 

   Newton's laws alone cannot determine exactly either which one is moving when they move at relatively constant speed. 

Thus it was realized that reference frames moving with relatively constant speed has special meaning than the usual ones and they are called "inertial reference frames". 

Computer freezes with White lines on monitor screen

The most recent problem appeared on my computer is that....
      
Problem: 
       When I work for a while in my desktop PC, it automatically freezes down with white horizontal and vertical lines on the desktop monitor screen. All the hardware components such as mouse, keyboard or any input-output connections stopped working suddenly.
      The only thing you can do is just unplugging the PC..

Reason: 
      When I surfed on the internet, I learned that the common reasons would be such as,
     Loose connection of your VGA port - where you will connect your monitor cord (or) 
     Graphics card problem which you can find it in the motherboard
      But if your computer don't show any lines when you boot up in other operating system then the definite problem would be with your windows which may be corrupted. 

Solution:
     Try to reattach monitor cord firmly (use the same with graphics card). Do all the basic solution stuff  Free-of-cost-solutions-for-computers


Boot in safe mode and if your windows is corrupted - back up all your data. Insert your windows CD and try to repair the windows. 

Again if the problem continues, the only way is remove your windows OS and reinstall it using the CD. 

That is what I did finally for my computer!! Thank God I didn't have any important data for backup..