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Simplicity is the ultimate sophistication.” — Leonardo da Vinci
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Wednesday, 25 February 2015

A thought that rules you


A thought that rules you
 

We don't tend to think about how lucky we are if our thoughts are crowded with worries. Please don't stop reading, seeing this as another boring advice someone across the world gives you while you have many people around you to flood you with advices; this is going to be different. Before I proceed with this article, let me tell you that I am not an expert in subject areas like human psychology or social psychology or on any other related fields. So what I say now are just stuff I think are true and it does not go beyond that.

In this article, I am going to talk about thoughts which rule you through your daily course of life with or without your knowledge. As we cross our adolescence age, we tend to comprehend more information compared to our early life. I believe its because we retain more information and consider more thoughts at the same time. Have you ever realized how you could write down what was dictated without the slightest thought of actually writing it? What I mean is, when we note down something that's being dictated, our hand moves automatically, writing down the words. Have you ever realized how you could walk around without even thinking about walking?

This one answer depicts the solution to all our problems. I don't believe in giving meaningless advices that had been repeated over a million times in different places and different languages. If you had seen my other posts, you would realize that I love solving problems down to its finest detail rather than giving an abstract over view of the solution. The same way I am going to give you a fine detailed solution which states the root cause of all your problems. I believe its true; it doesn't mean that the scientific society must accept it (I have no professional knowledge in psychology). When I talk about a thought that rules you, I mean to say that certain thoughts exist within your head with or without your knowledge, which organizes your behavior at the most fundamental level. Now I guess my example of writing a dictated text in a piece of paper makes sense to you. Yes I believe its the "thought" that listens to the word that's being dictated, find its spelling, recollects how to write it down and finally writes it down without you recognizing all these details. Labeling this attribute as "thought" might not be correct; but in this article, that's how I am going to call it.

Now let me give you another example: "our planet earth is a part of a tiny solar system, among billion others, together which constitutes our galaxy". Did you think about galaxies a second before you read this? probably no, right? You might have had other thoughts in your mind before you read this sentence. And is what I said now new to you? Of course not! But why didn't you have that piece of thought in your mind when you actually knew it? Its because, we don't think about everything we know or have known before (or have forgotten it). Then how did my sentence make you recollect them in not less than a fraction of a second? I am not a scientist dealing with brain, so my answer would be "that's how our brain works!". My explanation is, once we have a thought similar or in someway associated with another thought or a peace of memory in our brain, then that piece of thought gets recollected automatically without us showing any effort. Once the new thought comes to your mind, either you accept it or reject it. when you reject it, then your thread of thoughts on that particular matter is broken. but if you persist thinking about it, newer thoughts associated with the previous thought starts popping up in your head.            

Until now, I guess I had talked about two kinds of thoughts; one that immediately induces an action and the other which just remains passive in our head (without inducing any action directly, unless we respond to it). There are parts of it you can control, and there are parts of it you cannot control. Lets consider the second example first, where you didn't have the slightest idea what a galaxy was the moment before you read about it. Now, being aware of the sentence I had written down above got you reminded about galaxies; do you think its ever possible for you to read that sentence and not be reminded about galaxies. I had used the word galaxy extensively, until now. Could you stop thinking about it the moment I ask you to do so? possibly not! So you need to become aware of what you cannot control, so that you could get yourself to control the parts of your life you can control. Every human on this world has the power to induce change. The most challenging part of it is, each of us have different means by which we can induce a change practically. The first example I had given you makes it clear that you don't consider each detail of the work you do, instead let your brain takeover your actions. So this is again something that your brain does without your knowledge. Can we give other examples similar to this? The example of walking and writing were positive ones, now let me give you a negative example from my life, before I had changed; realizing all this.

Almost every day, I plan a study schedule for the evening; but when I come home, I throw my backpack, and sit before the computer or read my novel. I just tend to think, "let me do it tomorrow. I am a bit too tired today". I keep postponing my work until it accumulates to a big bundle. In the morning, before going to my college, I feel that my mode is opt for studying and nothing in the world can stop me from doing so, when I get back home I find that I was wrong! not for a day or two, but for months! And this thought made me feel more miserable and made me avoid studying even more intensively. I would always have some work to do; but though I could do those works, I won't be able to bring myself to do my college work. Some other time in the same year, I start studying studiously (like a nerd) and score well. Why am I not able to both at the same time? and what's making it impossible for me to switch from one to another? During the time of the year when I do my project work, I don't maintain the continuity of my studies and hence loose my temporary habits (or I forget the "thoughts" that had helped me to keep my studying phase) that aid me in my academical studies.

Reading right before exams seem completely impossible for me. I won't be able to get myself seated before my study table. This is again because of that "thought" or the habit that got erased once I had reduced paying heed to my academical studies. These "thoughts" are usually impossible to appose and change in a short period of time. For stuffs that are good, then these thoughts need not be changed. But for stuff that are bad or harmful or depressing, we would desperately need a change. Why do you think happiness lies in possessing something you don't have? The explanation to this dates back to the days you were young. Young children (below the age of three), usually will want to hold new stuff in their hand, feel it, smell it and taste it. Older children (about age five, six and seven) will be curious to handle stuff they see their parents handle (like the car's driver's seat. Its the best example, children would want to drive a vehicle even if they are aware that they won't be allowed to do so). But unfortunately, children are not given everything they ask; it could either be an expensive toy or the aeroplane that flies overhead! This intention (to posses something they are not given)  gets cultivated and intensifies as they grow older. Initially, their intention to handle fancy objects were induced by their curiosity. But as they get older, they tend to associate possessing something they don't have with "victory" or "happiness". Later, as he grows more older, his desire to own stuff he doesn't have increases drastically.

So hence, most people "want everything". I don't have any objection to your views, but I just want to say that the source of happiness is yourself, and not something outside. Its because, it was you who associated happiness with materialistic gain when you were young.  


copyright (c) 2015 K Sreram. You may not distribute this article without the concerned permission from K Sreram.  
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Sunday, 22 February 2015

Proof the vector dot product and cross product are distributive

Proof that vector dot product is distributive

We may write a vector product $\vec{a}.\vec{b}$ as $ |a||b|cos\theta$, by definition. For expressing an n-dimensional Euclidean space we may use the summation notation $\vec{a} = \sum \limits_{i=0}^n a_i \hat k$. The magnitude of a vector quantity $|\vec{a}|$ can be expressed as $|\vec{a}|^2 = \sum\limits_{i=0}^n {a_i}^n$. The proof of this expression directly follows from the Pythagoras theorem which states that, the magnitude of the line joining the ends of the line segments that emerges mutually perpendicularly from a point in space is equal to the square root of the sum of the squares of the magnitude of the other two line segments. Here, two mutually perpendicular lines are taken; the same way, while extending this to three dimensional space in which we introduce another axis perpendicular to the plane in on which the right angled triangle lies. Lets extend another line across this third axis from the point we initially considered. The hypotenuses of the triangle, which was computed using the Pythagoras theorem, is defined to be perpendicular to the newly drawn line. This is because, the third axis is perpendicular to the plain on which we constructed our first triangle.

Because the hypotenuses lies on the plain, it becomes perpendicular to the third line. Now consider this new line and the hypotenuses to lie on another plane which cuts through the three dimensional coordinate system irregularly. We now have two perpendicular lines in this new plain which we again root the sum of the squares of their magnitudes to get the magnitude of another line which is the hypotenuse of this system of lines. Thereby we can say that the newly formed hypotenuse is the magnitude of the line joining the origin and the point $(x,y,z)$ in three dimensional space, if the point we initially considered is the origin of this three dimensional co-ordinate system. Now we may extend this to higher dimensions by introducing another axis which we define to be perpendicular to all the other three axis. Like our example of extending a two dimensional vector analysis to three dimensions by just considering the magnitude of a single line formed by the two axis; introducing another axis, does not complicate our process of generalizing our expression of vector magnitude to another dimension if we only consider the system of the third axis and the single line plotted previously.

Now, for extending this to higher dimensions, we may do the same. Lets just consider the three-dimensional magnitude of an n-dimensional line (by just neglecting the higher dimensions); its practically impossible to imagine another dimension. But we a find the magnitude of a four-spatial dimensional line by the expression $l_4 = \sqrt{{l_3}^2+{k_4}^2}$ (where, $l_3 = \sqrt{{k_1}^2+{k_2}^2+{k_3}^2}$ ). We know that the fourth line is perpendicular to the first three mutually perpendicular lines, but it's not necessary to imagine it practically to compute its magnitude in the 4D co-ordinate system. We have the magnitude $k_4$ and the line $l_3$ so they form a plane that irregularly cuts the 4D co-ordinate system, but again, we have a plain to which we can freely apply the Pythagoras theorem.

So now lets proceed to prove the distributive property of  vector dot product. To prove it, we have to show that $\vec{c}.(\vec{a}+\vec{b}) = |\vec{c}||\vec{a}+\vec{b}|cos{\theta}_{a+b,c} = |\vec{c}|     |\vec{a}|cos{\theta}_{a,c}+ |\vec{c}||\vec{b}|cos{\theta}_{b,c}$. To proceed with this proof, we have to show that $|\vec{a}+\vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + 2|\vec{a}||\vec{b}|cos{\theta}_{a,b}$ and then show that $\vec{a}.\vec{b} = \sum\limits_{i=0}^n a_ib_i$.

The parallelogram law of vector addition can be proved in the following way (proof for the parallelogram law of vector addition),


   Let, $OA$ be the vector component $\vec{a}$ and let $OC$ be the vector component $\vec{b}$ and let $OB$ be the vector component $\vec{a}+\vec{b}$. Now lets express the length $OB$ in terms of $|\vec{a}|$ and $|\vec{b}|$ and the angle $\angle{BCD}$.  $\angle{BCD} = \angle{AOC}$ as the above diagram is a parallelogram. We can say that $OA = CB = |\vec{a}|$ and so we get, $CD = CB.cos\angle{BCD} = |\vec{a}|.cos\angle{BCD}$. Now, as we have $CD$  and $OC$ we can write $OD = OC+CD$ or, $OD = |\vec{b}| + |\vec{a}|.cos\angle{BCD}$. Now we can express $BD$ as, $BD = |\vec{a}|.sin\angle{BCD}$.

$${OB}^2 = |\vec{a}+\vec{b}|^2 = {BD}^2 + {OD}^2 $$
So we can write $$|\vec{a}+\vec{b}|^2 =  (|\vec{a}|.sin\angle{BCD})^2 + (|\vec{b}| + |\vec{a}|.cos\angle{BCD})^2 $$.

After simplification we get, $$|\vec{a}+\vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + 2|\vec{a}||\vec{b}|cos{\theta}_{a,b}$$

proof that $\vec{a}.\vec{b} = \sum\limits_{i=0}^n a_ib_i$:

Now, lets proceed with proving the distributive property of vector dot product. We first need to show that,  $$\vec{a}.\vec{b} = \sum\limits_{i=0}^n a_ib_i$$. From the parallelogram law of vector addition we have $$cos\theta_{a,b} = \frac {|\vec{a}+\vec{b}|^2 - |\vec{a}|^2 - |\vec{b}|^2}{2|\vec{a}||\vec{b}|}$$. This can be rewritten in its summation form as,

$$cos\theta_{a,b} =
\frac {
              |\sum \limits_{i=0}^n a_i \hat k +
                                         \sum \limits_{i=0}^n b_i \hat k|^2 -
                                                               |\sum \limits_{i=0}^n a_i \hat k|^2 -
                                                                           |\sum \limits_{i=0}^n b_i \hat k|^2
         }
        {
                  2|\sum \limits_{i=0}^n a_i \hat k|
                   |\sum \limits_{i=0}^n b_i \hat k|
         }
$$ Or,

$$cos\theta_{a,b} =
\frac {
              |\sum \limits_{i=0}^n (a_i+b_i) \hat k|^2-
                                                               |\sum \limits_{i=0}^n a_i \hat k|^2 -
                                                                           |\sum \limits_{i=0}^n b_i \hat k|^2
         }
        {
                  2|\sum \limits_{i=0}^n a_i \hat k|
                   |\sum \limits_{i=0}^n b_i \hat k|
         }
$$ Or, 
$$cos\theta_{a,b} =
\frac {
              \sum \limits_{i=0}^n (a_i+b_i)^2-
                                                               \sum \limits_{i=0}^n {a_i }^2 -
                                                                           \sum \limits_{i=0}^n {b_i}^2
         }
        {
                  2\sqrt{\sum \limits_{i=0}^n {a_i}^2
                   \sum \limits_{i=0}^n {b_i}^2}
         }
$$ We know that, 

$$\vec{a}.\vec{b} = |\vec{a}||\vec{b}|cos\theta_{a,b}$$ Or, 

$$\vec{a}.\vec{b} = (\frac {
              \sum \limits_{i=0}^n (a_i+b_i)^2-
                                                               \sum \limits_{i=0}^n {a_i }^2 -
                                                                           \sum \limits_{i=0}^n {b_i}^2
         }
        {
                  2\sqrt{\sum \limits_{i=0}^n {a_i}^2
                   \sum \limits_{i=0}^n {b_i}^2}
         }
)                 \sqrt{ \sum \limits_{i=0}^n {a_i}^2                    \sum \limits_{i=0}^n {b_i}^2} $$ Or, 
$$ \vec{a}.\vec{b} = \frac{\sum \limits_{i=0}^n ({a_i}^2 + {b_i}^2 + 2{a_i}{b_i})-                                                               \sum \limits_{i=0}^n {a_i }^2 -
                                                                           \sum \limits_{i=0}^n {b_i}^2}{2} $$ Or,
$$\vec{a}.\vec{b} = \sum\limits_{i=0}^n a_ib_i$$

Proof that $(\vec{a} + \vec{b}) .\vec{c} = \vec{a}.\vec{c} + \vec{b}.\vec{c}$:

On substituting the result $\vec{x}.\vec{y} = \sum\limits_{i=0}^n x_iy_i$ in the expression we get,

$$
(\vec{a} + \vec{b}) .\vec{c} =
              \sum\limits_{i=0}^n (a_i+b_i)c_i =
                              \sum\limits_{i=0}^n (a_ic_i+b_ic_i) =
                                          \sum\limits_{i=0}^n (a_ic_i) + \sum\limits_{i=0}^n (b_ic_i)
 $$.
Now consider the RHS of the equation $(\vec{a} + \vec{b}) .\vec{c} = \vec{a}.\vec{c} + \vec{b}.\vec{c}$$; again on substituting the same result in it, we get:
$$\vec{a}.\vec{c} + \vec{b}.\vec{c} = \sum\limits_{i=0}^n (a_ic_i) + \sum\limits_{i=0}^n (b_ic_i)$$, which is same as the LHS. Hence we have proved that LHS = RHS or,
$(\vec{a} + \vec{b}) .\vec{c} = \vec{a}.\vec{c} + \vec{b}.\vec{c}$ 

Now lets show that $\vec{c}\times(\vec{a}+\vec{b}) = (\vec{c}\times \vec{a} + \vec{c}\times \vec{b})$. By proving this, we would be proving that the cross product of the sum of two vector quantities is distributive in nature. 

A vector cross product is defined as, $\vec{a}\times\vec{b} = |\vec{a}||\vec{b}|sin\theta_{a,b}\hat n$. This expression, like the expression for the vector dot product is considered to be the basic definition we introduce.The unit vector $\hat n$ shows the direction perpendicular to the plane which contains the lines $\vec{a}$ and $\vec{b}$. We know that 
$\vec{a}.\vec{b} = |\vec{a}||\vec{b}|cos\theta_{a,b}$, so we may convert it to $|\vec{a}||\vec{b}|sin\theta_{a,b}$ by squaring the dot product expression shown above and subtracting it from $|\vec{a}||\vec{b}|$. I.e., 

$$(|\vec{a}||\vec{b}|)^2 - (\vec{a}.\vec{b})^2 = (|\vec{a}||\vec{b}|)^2 -  (|\vec{a}|^2|\vec{b}|^2cos^2\theta_{a,b})^2$$ which is same as,

$$ (|\vec{a}||\vec{b}|)^2 - (\vec{a}.\vec{b})^2  = |\vec{a}|^2|\vec{b}|^2sin^2\theta_{a,b}$$

or we may write,

$$[(|\vec{a}||\vec{b}|)^2 - (\vec{a}.\vec{b})^2]\hat n  = |\vec{a}|^2|\vec{b}|^2sin^2\theta_{a,b}\hat n$$. Or, 

$$\vec{a}\times\vec{b}  = |\vec{a}||\vec{b}|sin\theta_{a,b}\hat n =\sqrt{ (|\vec{a}||\vec{b}|)^2 -  (|\vec{a}||\vec{b}|cos\theta_{a,b})^2}\hat n$$.

From the parallelogram law of vector addition, we know that $cos\theta_{a,b} = \frac {|\vec{a}+\vec{b}|^2 - |\vec{a}|^2 - |\vec{b}|^2}{2|\vec{a}||\vec{b}|}$ so, $sin\theta_{a,b} = \sqrt(1- [\frac {|\vec{a}+\vec{b}|^2 - |\vec{a}|^2 - |\vec{b}|^2}{2|\vec{a}||\vec{b}|}]^2)$. Now, we may write this as,
$$sin\theta_{a,b} = \sqrt{
\frac { 4(\sum \limits_{i=0}^n {a_i}^2
                  \sum \limits_{i=0}^n {b_i}^2) -
      (\sum \limits_{i=0}^n (a_i+b_i)^2-
         \sum \limits_{i=0}^n {a_i }^2-
                \sum \limits_{i=0}^n {b_i}^2)^2
         }
        {
                  4(\sum \limits_{i=0}^n {a_i}^2
                   \sum \limits_{i=0}^n {b_i}^2)
         }
   }
$$ Or,
$$|\vec a||\vec b|sin\theta_{a,b} =
\frac { \sqrt{
                4(\sum \limits_{i=0}^n {a_i}^2
                  \sum \limits_{i=0}^n {b_i}^2) -
      (\sum \limits_{i=0}^n (a_i+b_i)^2-
         \sum \limits_{i=0}^n {a_i }^2-
                \sum \limits_{i=0}^n {b_i}^2)^2
               }
         }
        {
                 2
         }

$$Or,
$$|\vec a||\vec b|sin\theta_{a,b} =
\frac { \sqrt{
                4(\sum \limits_{i=0}^n {a_i}^2
                  \sum \limits_{i=0}^n {b_i}^2) -
      (2\sum \limits_{i=0}^n (a_ib_i))^2
               }
         }
        {
                2
        }
$$ Or,
$$|\vec a||\vec b|sin\theta_{a,b} =
 \sqrt{
          (\sum \limits_{i=0}^n {a_i}^2
                  \sum \limits_{i=0}^n {b_i}^2) -
      (\sum \limits_{i=0}^n (a_ib_i))^2
               }
$$Or,

$$|\vec a||\vec b|sin\theta_{a,b} =
 \sqrt{
          (\sum \limits_{i=0}^n\sum \limits_{j=0}^n {a_i}^2{b_j}^2) -
      (\sum \limits_{i=0}^n\sum \limits_{j=0}^n (a_ib_i)(a_jb_j))
               }
$$ 

On simplification, we may write this as,

$$|\vec a||\vec b|sin\theta_{a,b} =
 \sqrt{
         (-\sum \limits_{i=0}^n\sum \limits_{j=0}^n a_ib_j)^2, i\not=j
               }
$$ Which implies,

$$|\vec a||\vec b|sin\theta_{a,b} = \sum \limits_{i=0}^n\sum \limits_{j=0}^n a_ib_j, i\not=j $$ Now, multiplying both sides with $\hat n$ we get, $$\vec{a}\times\vec{b} = (\sum \limits_{i=0}^n\sum \limits_{j=0}^n a_ib_j) \hat n, i\not=j $$







copyright (c) 2015 K Sreram. You may not distribute this article without the concerned permission from K Sreram.  

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