### Wednesday, July 27, 2005

## Not a bad thing to remember

*"If you are in good health, not in a lot of debt, and have at least one passion, there is no real reason you should be depressed"*

Remember this when you think you are depressed.

### Monday, July 18, 2005

## School math, brain power, age and everything

The sum of the consecutive numbers starting from 1 to any number N is N*(N+1)/2. I was reading the proof of this in some blog or article. It went like this:

1 + 2 + 3 + .... N

Rearranging so that the higher numbers are alongside the lower numbers, we get

1 + N + 2 + N-1 + 3 + N-2 + ...

Now since the number of terms will differ if N is either odd or even, we consider each case separately.

(1+N) + (2+N-1) + (3+N-2) + .... (N/2 + N/2 + 1) This has now N/2 groupings

The above expression can be viewed as N/2 occurrences of the the term (N+1)

So the value of he expression = (N/2) * (N+1) = N*(N+1)/2

then we group together all terms except the middle term which is (N+1)/2 .

(1+N) + (2+N-1) + (3+N-2) + ....+ (N+1)/2

The above expression can be viewed as (N-1)/2 groupings of the term (N+1) plus the middle term (N+1)/2

The value of the expression = (N+1) * (N-1)/2 + (N+1)/2

= (N+1)/2 [N - 1 +1] = N*(N+1)/2

So we have proved that the sum of consecutive numbers from 1 to N is N*(N+1)/2, regardless of whether N is even or odd.

This led me to think - the result is regardless of whether N is even or odd, so shouldn't there way to calculate this expression without bringing 2 separate cases (when N is even and when is odd) into the picture?

So lets see if we can calculate the value, without applying any strategies that will force us to consider the two separate cases.

Let S denote the sum of the consecutive numbers from 1 to N.

S = 1 + 2 + 3 + .... N

Lets write the term on the right hand side backwards, counting from the highest term to the lowest.

S = N + N-1 + N-2 + N-3 + .... + N-(N-2) + N-(N-1)

Group all the Ns together

S = N + N + N +N .... -1 -2 -3 ....- N-2 - N-1

S = N + N + N +N .... - (1 + 2 + 3 ....+ N-2 + N-1)

Look at the grouped term furthest to the right. Its actually the sum of N consecutive terms minus the last term N! So we can write it as:

S = N*N - (S - N)

S = N*N - S + N

2S = N*N + N

2S = N*(N+1)

S = N*(N+1)/2

So there we have it. We have proved that the sum of consecutive numbers from 1 to N is N*(N+1)/2 whether N is even or odd.

Now, lets get to what I REALLY wanted to say. The fact is, when I was in school, I would have never applied the reasoning like I applied it here. I wouldn't go searching for a proof just because I knew it should exist. Strangely, it seems that if I tackle my math and physics problems in school now, I feel I would be more successful at it. Thats because I feel my reasoning is much more mature, plus the fact that I view it as an enjoyable, leisurely activity. I am somewhat puzzled at how people think that as they grow older, they do not have the mental capabilities of solving problems that they had when they were in school. To me, I believe that you can actually increase your capacity to solve problems once you mature. This strengthens my belief in the theory that if we keep exercising our brain, our capacity to think clearly increases. Its not a question of age. Age might play a part much much later, (if ones about 80 or so maybe, when your brain retires, like everybody else ;)) but if anybody says they cant solve problems as good as they did before, they are just not making it a point to exercise their brain enough.

So go and find ways to test and exercise your brain. If you think you are in a job that doesnt give you enough opportunities to do that, make time to solve some puzzles, or draw something or take some music or art classes - anything that makes you think.

1 + 2 + 3 + .... N

Rearranging so that the higher numbers are alongside the lower numbers, we get

1 + N + 2 + N-1 + 3 + N-2 + ...

Now since the number of terms will differ if N is either odd or even, we consider each case separately.

**If N is even:**(1+N) + (2+N-1) + (3+N-2) + .... (N/2 + N/2 + 1) This has now N/2 groupings

The above expression can be viewed as N/2 occurrences of the the term (N+1)

So the value of he expression = (N/2) * (N+1) = N*(N+1)/2

**If N is odd:**then we group together all terms except the middle term which is (N+1)/2 .

(1+N) + (2+N-1) + (3+N-2) + ....+ (N+1)/2

The above expression can be viewed as (N-1)/2 groupings of the term (N+1) plus the middle term (N+1)/2

The value of the expression = (N+1) * (N-1)/2 + (N+1)/2

= (N+1)/2 [N - 1 +1] = N*(N+1)/2

So we have proved that the sum of consecutive numbers from 1 to N is N*(N+1)/2, regardless of whether N is even or odd.

This led me to think - the result is regardless of whether N is even or odd, so shouldn't there way to calculate this expression without bringing 2 separate cases (when N is even and when is odd) into the picture?

So lets see if we can calculate the value, without applying any strategies that will force us to consider the two separate cases.

Let S denote the sum of the consecutive numbers from 1 to N.

S = 1 + 2 + 3 + .... N

Lets write the term on the right hand side backwards, counting from the highest term to the lowest.

S = N + N-1 + N-2 + N-3 + .... + N-(N-2) + N-(N-1)

Group all the Ns together

S = N + N + N +N .... -1 -2 -3 ....- N-2 - N-1

S = N + N + N +N .... - (1 + 2 + 3 ....+ N-2 + N-1)

Look at the grouped term furthest to the right. Its actually the sum of N consecutive terms minus the last term N! So we can write it as:

S = N*N - (S - N)

S = N*N - S + N

2S = N*N + N

2S = N*(N+1)

S = N*(N+1)/2

So there we have it. We have proved that the sum of consecutive numbers from 1 to N is N*(N+1)/2 whether N is even or odd.

Now, lets get to what I REALLY wanted to say. The fact is, when I was in school, I would have never applied the reasoning like I applied it here. I wouldn't go searching for a proof just because I knew it should exist. Strangely, it seems that if I tackle my math and physics problems in school now, I feel I would be more successful at it. Thats because I feel my reasoning is much more mature, plus the fact that I view it as an enjoyable, leisurely activity. I am somewhat puzzled at how people think that as they grow older, they do not have the mental capabilities of solving problems that they had when they were in school. To me, I believe that you can actually increase your capacity to solve problems once you mature. This strengthens my belief in the theory that if we keep exercising our brain, our capacity to think clearly increases. Its not a question of age. Age might play a part much much later, (if ones about 80 or so maybe, when your brain retires, like everybody else ;)) but if anybody says they cant solve problems as good as they did before, they are just not making it a point to exercise their brain enough.

So go and find ways to test and exercise your brain. If you think you are in a job that doesnt give you enough opportunities to do that, make time to solve some puzzles, or draw something or take some music or art classes - anything that makes you think.

### Thursday, July 14, 2005

## Mind-boggling concept of special relativity!

Uncle Bob, one of my favorite software authors, is also a science enthusiast. He writes about how special relativity says that one is always traveling at a velocity of the speed of light (C). This is startling at first. If one is sitting at one place, is he traveling at the velocity of light? The idea is to consider space and time as different dimensions. So one's velocity in the universe is one's velocity through space and one's velocity through time. Through an equation called the Lorentz transformation, we can arrive at an equation which is like this(for the derivation, see the link to the article above):

C = velocity of light

T = coefficient of velocity through the time dimension

so CT = velocity through the time dimension

V = velocity through the space dimension

Now since the velocity of light is constant, this means that the vector sum of our velocity through space and velocity through time is also constant. This brings us to a mind-boggling conclusion. If I am sitting still (my V = 0), then to a stationary observer near me, I am traveling through the time dimension at the velocity of light. If I get up and walk (my V > 0) then to the stationary observer beside me, my velocity through the time dimension decreases slightly below the velocity of light. To me however, even when I get up and walk, my velocity through space is zero relative to myself. So as far as I am concerned, my velocity through time is still the speed of light.

Lets take this further. Lets say my stationary observer sees me moving at the speed of light. Then to him, my velocity through time is zero. This means if he records my time at the start and the finish, he will see that no time has passed! So no matter how much I travel through space, he will always record my start and end times as the same! Again, relative to myself I am at a zero velocity through space. If I see my start and end times on my watch, they will be different!

So then what happens if, according to my poor startled stationary observer I start moving at a speed faster than light through space? Then according to him, I will start my journey now and end it a few minutes earlier! Meaning I would have traveled backwards through time! And of course, if you ask me, everything is still perfectly normal since I am at a velocity of zero through space.

Mind-boggling, isn't it??

**C^2 = (CT)^2 + V^2**C = velocity of light

T = coefficient of velocity through the time dimension

so CT = velocity through the time dimension

V = velocity through the space dimension

Now since the velocity of light is constant, this means that the vector sum of our velocity through space and velocity through time is also constant. This brings us to a mind-boggling conclusion. If I am sitting still (my V = 0), then to a stationary observer near me, I am traveling through the time dimension at the velocity of light. If I get up and walk (my V > 0) then to the stationary observer beside me, my velocity through the time dimension decreases slightly below the velocity of light. To me however, even when I get up and walk, my velocity through space is zero relative to myself. So as far as I am concerned, my velocity through time is still the speed of light.

Lets take this further. Lets say my stationary observer sees me moving at the speed of light. Then to him, my velocity through time is zero. This means if he records my time at the start and the finish, he will see that no time has passed! So no matter how much I travel through space, he will always record my start and end times as the same! Again, relative to myself I am at a zero velocity through space. If I see my start and end times on my watch, they will be different!

So then what happens if, according to my poor startled stationary observer I start moving at a speed faster than light through space? Then according to him, I will start my journey now and end it a few minutes earlier! Meaning I would have traveled backwards through time! And of course, if you ask me, everything is still perfectly normal since I am at a velocity of zero through space.

Mind-boggling, isn't it??

### Thursday, July 07, 2005

## On that note...

Was listening to a song this morning that I thought would share with the world. Its amazing how sometimes when you are a little down in the dumps, the appropriate message reaches you bang on target in some way or the other. The one particular stanza in between in the song that seemed sweet goes like this:

Translated as:

*Tujhe pyaar bhi mile path mein to samajhna yahi,*

Ye hai raaste ki chhaon, saathi teri manzil nahin,

Tera to apna aur hai sapna

Apne hi sapnon ki manzil se na reh jaana tu...

Ye hai raaste ki chhaon, saathi teri manzil nahin,

Tera to apna aur hai sapna

Apne hi sapnon ki manzil se na reh jaana tu...

Translated as:

*Even if you find love in the journey of life,*

realize that it is the intermittent shade, not your destination.

You have your own dreams,

do not stray away from the destination of your dreams.realize that it is the intermittent shade, not your destination.

You have your own dreams,

do not stray away from the destination of your dreams.

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