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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 :

(1)
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
QED
(2)
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
QED
(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.
QED
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 :

5 comments:

Anonymous 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!
spidy

RaSh 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!

Anonymous 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
or
(a+b) = a + b + 0+...+ a + b
or
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
spidy

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