The post is made by Mohit Khandelwal .
WE KNOW THAT d/dx of x2 is 2x.
now consider this
x2 = x+x+x+x+x+x+x+x+x+x+x+............(x times)
if we now differentiate it with respect to x then we get
d/dx(x2) = 1+1+1+1+1+1+1+1+1+1+1+1+...........(x times)
= x
x = 2x
1= 2
how is it possible??
hey! MOHIT,
there is certainly some fallacy in the argument...but i dont know what......
d(F(x))/dx represents the slope of the tangent to F(x) at pt. x.......if we go by your argument then:
x=1+1+1+1+1+1..........(x times)
differentiating w.r.t x on both sides we get
1 = 0+0+0+0+0+0+0....
and we also know that slope of the tangent at any x on the fnxn F(x)=x is 1
so there is certainly some fallacy which we can not detect
Hey Mohit! i think the mistake you are doing is that you are changing a continous function into a non-continous function.
Actually X2 is continous but your equivalent function x+x+x+x+x..... is only defined for integral values and not for all real numbers. What will you write
(.5)2 . So due to this discontinuity might be your fallacy formed.
You can also consider -12 = -1+-1+-1+-1......(-1 times) no meaning
If you find anything wrong then please do reply.
Consider one more fact
X2 = x + x + x + x + ...... (x times)
Now each x = 1 + 1 + 1 + 1 + .... +(x times)
Thus we have x2 = 1+1+1+1+1+1+......(x2 times)
Diff. wrt to x we have
2x = 0 or x = 0
Thus for any polynomial function we just have slope 0
i.e all polynomial functions are parallel to X - axis
OH! GOD have mercy on Maths.........;-)
Rajat just consider that this is a diff equation so we cant put "X"=1
X2=X+X+X+X+X+.......................X times
diff wrt X we get and please dont put X=1
2X=1+1+1+.............X times
and rajat please tell me about that in X2 we just add X for X times how can you add 1 for X2 times if we can replace X by 1 than also we can add 1 to X times and if X=1 then X2=12=1 how can you add 1 to X2 times
2X=X
and think that it is actually a strange fact according to my knowledge and please Rajat answer soon
Oh! sorry Aman if mine explanation hurted you but it is going to hurt your heart more that i couldn't understand any of your objection clearly.
Still i will try to explain my point again :
As we see if we write X square as x times x then how will you express .5 as this because i dont think one can write .5 times .5 ...:-(
Now for compatible transmission of our function we need X to be a positive integer and all positive integers can be written as sum of 1 ..... By induction :-)
So we can write X square as x times x where each x can be expressed as sum of 1 x times. So X2 = 1 + 1 + 1 + 1 + 1 +...... (X square times)
eg 32 = 3 + 3 + 3 = (1+1+1) + (1+1+1) + (1+1+1)
And thus 2x = 0 or x = 0.
And from next time dont even try to use *** because that hurts me.
Hey Rajat sorry friend if i hurt you but i think that you just put the value of X=1 hey i m sorry again
x2 is different function &
x+X+X+X+X+X+X+X+X+X+X+X+X+X+------(x times ) is different function both have diff domain
you can't put x=-1 in x+x+x+x+x+x+x---(x times)
the proof is obviously wrong.
f(x)=x2
the expansion of f(x) as given is clearly a discontinuous function.(you cannot write say.5 ,.5 times) so the funcion can also not be differentiated.
well i will tell you what my sir told me when i asked the same 2 months back in my class
he came the other way round
x=1+1+1+1........(x times) (1)
since x is variable ......it is compulsory that you have to multiply the below way
x2=x+x+x+x+x............(x2 times)
when you multiply a variable on both sides ...you need to multpily tht variable on both the sides ....including the no.of terms too (dont ask me why )
now differentiate w.r.t. x
2x = 1+1+1+1+1......... (2x times ) =2x (2)
such calculations must be done because .....x is unknown ....and variable
if x was not a variable ........what ever you gave in the question wud have been applicable
if you are not satisfied .....chck the eqns 1 and 2 by putting x values
for e.g. x=3
=> 3=1+1+1(3 times )
where as 2x is 6 = 1+1+1+1...(2x times i.e.6 times)
now dont ask me about the other eqn x2=x+x+x....... i am helpless over there 
...even i was not completely satisfied .
Hello,
This is an oft repeated argument.The fallacy is in the last step.If x = 2x,then clearly one cannot conclude 1=2 as it amounts to division by zero on both sides,which is meaningless.
Sathej
I agree with you, Sathej,because it is just like,
if you differentiate for eg:,the foll eqn.,
2x2=3x+1,
4x=3,
x=3/4. Now if we substitute this value of x in the original eqn, we get,
2 * (9/16)=(9/4)+1,
(9/8)=(13/4), which is not true........
in the same way the eqn 2x=x,
we can write as x(2-1)=0
x=0, or 1=0.
Obviously 1=0 is not possible therefore the slope is 0 at the pt (0,0).
tell if there are any mistakes.......
The post is made by jenish shivshaktiwala .
hi mohit
i have ur answer.
you ask that :x2= x+x+x+x+......(x times) o.k.
1st reply
x2= [x+x+x+x+........(x/2times)]+[x+x+x+x+...(x/2times)]
x2= x*x/2 + x*x/2
2x= 2x/2 + 2x/2
2x=x+x
2x=2x
2nd reply
x2=[x+x+x+...(x/3 times)]+[x+x+x+...(x/3times)]+[x+x+x+..(x/3times)]
x2=x*x/3 + x*x/3 + x*x/3
2x=2x/3 + 2x/3 + 2x/3
2x=2x
ok
NOW I PROVE THE CHAPTER OF COMPLEX NUMBER PARTIALLY FAILED
____________________________________________________________________________________
(1) i = root minus one
(2) i square= - one (here i stands for iota)
(3) i cube = - iota
(4) i power 4 = 1
now 3rd equation ,,, i cube = - iota
i cube + i =0
i(i square + 1)= 0
i = 0 and i square = - one
so,these are the 2 values of iota .....(1)
now 4th equation ....i power 4 - 1 = 0
( i power2-1)(ipower2+1)= 0
i power2 -1 = 0 and i power 2 = -1
i power 2 =1 and i power2 = -1
i = + - one and i = + - root minus one
so thus are 4 values of iota
so ,the total 5 values of iota are o,+1,-1,,-i,,+i
and if we put the values of iota in their general form then the equation becomes
(1) if we put i=o then x + i y = x + o y = x
(2)if we put i = + - one x + i y = x+ y and x-y
(3)if we put i = + - i then x + i y = x + i y and x - i y
if conclusion is x + i y = x - i y is only possible if y =0
thus all are straight line equations so complex no chapter is not totally correct
