PWNED!!

Dont ask me what the title of the post means. It is a direct plagiarism-it was the name of an event at Shaastra'08(IIT-M).And If my memory is strong, this was the event which tested the capacity of the participants to make things complex.And after a really looooooooooooooooooooooooooong(the oooooooo takes "au" sound) time, I have come up with an idea to compute '2+2'.So here goes the computation.

Let,

y=2+2

y=2(1+1)

y=2*2

Taking natural logarithms on both sides,

logy=log(2*2) ------------------(1)

We know that,

log(m*n)=logm+logn

Hence eqn (1) becomes

logy=log2+log2

logy=2log2 ------------------(2)

We know that,

log(m^n)=nlogm ------------------(3)

Applying the converse of eqn (3) in (2), we have

logy=log(2^2)

Now the above eqn becomes

y=2^2

If f(x)=a^x; So here [a=2;x=2]

We know that the Maclaurin series expansion of 'a^x' is

------(4)

f(o)=2^0=1 ;

f'(x)=(ln2)2^x; f'(o)=ln2;

f''(x)=[(ln2)^2]2^x; f''(0)=(ln2)^2;

..............................................................................................

...............................................................................................

Hence substituting the values of f(0),f'(0) and f''(0) in eqn(4),

f(x)=1+xln2+(x^2)(ln2)^2/2!+(x^3)(ln2)^3/3!+..............

f(x)=1+xln2+(xln2)^2/2!+(xln2)^3/3!+.......................

Therefore,

2^2=1+1.386294361+0.960906027+0.444032869

+0.153890065+0.042667386+0.009858259452+.............

=3.997648967

On approximation and rounding off,

2^2=4 .

Hence y=2+2=4.