What is wrong in this proof?

[amriths04]

let y=x^2,----------------eq1
so, y=x+x+...x times,------------eq2
differentiate eq1 and eq2 with respect to x,
2x=x,
there fore 2=1,

what is wrong in this proof?

 

[Element_115]

Re: time to think

"You forgot C."

Sorry, C is for integration and not differentiation, my mistake. It's been a while since
I've used Calculus.

 

[Phrzby Phil]

time to think

I don't think that the constant of integration C is involved in differentiation.

Your first function is a simple polynomial, but your second function is an open formula whose length depends on the value of the variable. Since you cannot write it as a closed formula, nor can you write it as a summation of a given finite number of terms, you cannot write the expression for its slope, so you cannot take that (non-existent) expression's limit, hence you cannot differentiate it.

 

[cedance]

time to think

1 argument, in the RHS of equ. 2, x is a constant not variable. diff(x) is 0 there. so 2x = 0 => x is 0.

and the other thing is, the equation 2 is not valid fox x < 0. equ 1 and 2 are strictly not equal.

cedance.

 

[svicent]

Re: time to think

The equality

y = x^2 = x+x+...x times

is valid only is x is an natural number. If x is 9.7 is not possible to add 9.7 + 9.7 + ..., 9.7 times. In the second case the function is not continuous and then is not differentiable.

 

[Phrzby Phil]

time to think

So amriths04, what is your answer, or your analyses of our answers??

 

[psmon]

time to think

y=X^2 (quadratic eqn)
y=X+X+... (linear eqn)
It's wrong to equate them!

 

[aersoy]

Re: time to think

dy/dx = d/dx (X^2) >>>>> dy/dx = 2 ; from eqn1;

y = x+x+.... x times >>>> y = X * X = X^2 am i wrong

:?:

dy/dx = d/dx (X^2)>>>>>dy/dx = 2; from eq2;

both equations and their dy/dx are equal

i hope this helps you

 

[nonlinear]

time to think

It's a mind blowing question!!!!!!

 

[amriths04]

Re: time to think

yes friends,

always in maths there are very many different ways to look over the same problem.

1)un-arguably that it is not a continuous function is right, so differentiation cannot be applied.

2)another nice point is that x^2=x+x+x+..x times is only valid for integers. this can act as a corollary of the 1st argument.

3)here x^2=x+x+x...xtimes, involves 2 x's which are not the same. so if this can be considered like that, partial differentiation can be applied to x times x which would give dx(1)+(1)(dx) which is 2.

i am not sure of my 3rd analysis. do you all agree with my points?

Added after 5 minutes:

your argument is correct phrzby phil. actually i had 2 different ways to solve this before i could post this question. so i wanted to know if there is some other angle to look at this and prove this wrong.

I don't think that the constant of integration C is involved in differentiation.

Your first function is a simple polynomial, but your second function is an open formula whose length depends on the value of the variable. Since you cannot write it as a closed formula, nor can you write it as a summation of a given finite number of terms, you cannot write the expression for its slope, so you cannot take that (non-existent) expression's limit, hence you cannot differentiate it.

 

[tzushky]

time to think

Here's more puzzles so you can see that only by not taking into account CORRECT mathematical reasoning you will always be able to demo anything you like, as absurd as it may seem...even that 1=2...

Yeahhhh..this is fun!

Suppose a=b . That implies a^2=ab ...

Now
a^2-b^2=ab-b^2.. which implies...

(a- b )(a+ b )=b(a- b )

Simplify both sides by (a- b ) and get:
a+b=b
Knowing that a=b ...we get a+a=a=2a ==> 1=2


Since we have now demonstrated that: 1=2 why not also 1+1=4?


Another example:
Say i=i^5 .
Apply log i = log i^5...adica ..log i= 5log i ==> 1=5

So...It's easy to prove anything..

 

[Phrzby Phil]

time to think

If you divide by zero.

 

[marko_benigno]

Re: time to think

let y=x^2,----------------eq1
so, y=x+x+...x times,------------eq2
differentiate eq1 and eq2 with respect to x,
2x=x,
there fore 2=1,

what is wrong in this proof?

these are the mistakes....
d/dx of x is 1, not x and you don't differentiate when comparing the absolute values of functions.

 

[pareshatlnmiit]

Re: time to think

what r u doing man,u r differenciatiin 1 with respet to 2 but i m not able to see variable other then x,so its just ridiculous to equate them