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Thursday, June 01, 2006

DEAD ENDS : Part II : Burning Reality

Hi Again
As promised, here is PART II ...
Bewarethough !!! I'd certainly say this one is more wierd, more interesting but still a LIE which has been imposed wrongly upon reality :

Every1 here knows the pretty little Binomial thorem going as :

(a+b)n = an + n*b*an-1 + ... + n*bn-1*a + bn
Observing we find that leaving the first and last term, all other terms are a multiple of n.
For some wierd results, let us put n = 0 ;
LHS = (a+b)0 = 1
RHS = a0 + 0 ... + 0 + b0
= 1 + 0 ... 0 + 1
= 2
:. since LHS = RHS
2 = 1
Consider the integral :
I = ∫ 1/x ∂x
Intigrating by parts ;
I = ∫ 1*1/x ∂x
= x*(1/x) - ∫ x*(-1/x2) ∂x
= 1 + ∫ 1/x ∂x
= 1 + I
:. 0 = 1
or, 2 = 1
(3) *** CLASSIC ***
Have a look at the figure. O is the origin. C is the centre of the circle. P a variable point (x,y).
t = OP ; R = CP ; a = OC
The equation of the circle would be:
x2 + y2 -2*a*x + a2 - R2
Now, t2 = x2 + y2
Also putting P(x,y) in circles equation we get :
t2 - 2*a*x + R2 - a2 = 0
:. t2 = 2*a*x + R2 - a2
2*∂t/ ∂ x = 2*a
Now for the point with minimum distance , The above expression equals 0.
For that to be possible, a = 0, that is O and C are the same points !
Hence Whenever O is not the center of the circle, there is no point on the circle from which the distant of O is maximum or minimum.
There is still more to come. Before thinking about the faults, appreciate the beauty of these propositions, how decievingly they proove the unproovable !
This in not the end.
~ Twish ~

NOTE : Those having difficulty typing math visit this link for keyboard codes :


sparekh said...

great going twish. your the one who is most enthusiastic. not surprising considering your online all the time. your little '1=2'-s leave me surprised wondering what happened!

Gr81 said...

Well part two is better than part one - but all these ques arise because of one thing - assumptions or definitions of the functions that we use (refer to my post too!) Like your 1st ques here violates the definition of binomial theorem - It is valid for n being a natural no only (Thats why we were taught the other way when n is a rational no.)
So incase u can't find the mistakes in such problems - think from the most basic facts you know - that might help!

bravura said...

I have been asked to post an appreciation for this post. and looks like if i don't i ain't gonna be around for very long. Jaan bachane ke liye likh raha hun.
Some of these posts by twish hav been really good. they hav made me think and think and think and think and think and think and think and think and think ........... Huh lemme think folks.

bravura said...

on a more serious note these posts hav been really good. they make u wonder and wander.

sparekh said...

(a+b)^n = a^n + n.a^n-1.b + ...
+ n.a.b^n-1 + b^n

now the number of terms should always be n+1. but how do you know when to stop.
for example: if n=1
(a+b)^n = a^1 + 1.a^0.b
+ 1.(1-1).a^-1.b^2 +...
+ n.a.b^0 + b^1
(a+b) = a + b + 0+...+ a + b
1 = 2 !
so i think we should stop the sequence once we get a term to be zero. or procede till only n+1 terms.
that way even for n = 0 the expression boils down to an identity

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