Question

check this out:
[x+x+x+...(x times)]=x^2
differentiate both sides with respect to x
[1+1+1+............+1]=2x
x=2x
1=2
???
can anyone explain to me what the **** is this? check this out:
[x+x+x+...(x times)]=x^2
differentiate both sides with respect to x
[1+1+1+............+1]=2x
x=2x
1=2
???
can anyone explain to me what the **** is this? @Mathematics

Answer 1

can you show me how to write x out x times?

Answer 2

continue to add x for x times

Answer 3

?...show me how to do that if \[x=\pi\]

Answer 4

\[\Pi +\Pi +\Pi +\](\[\Pi -\]3)*\[\Pi\]

Answer 5

[1+1+1+............+1]=2x

that doesnt lead to this

x=2x

Answer 6

and why is that??????????

Answer 7

lol

Answer 8

becasue when you do real math, you use real math rules instead if made up ones ....

Answer 9

'lol' dosnt explain anything

Answer 10

thought george was loling :)

Answer 11

what you wrote is nonsense george. you want x+x+x +...+x but then you don't write pi like that

Answer 12

chicken, whale have u completely lost it?

Answer 13

x^2 mean x*x and there is no finite way to write that with [1+1+1+...+1]

any more than you can write infinity as [1+1+1+....+1]

Answer 14

i completely lost it along time ago, this is just the aftermath ;)

Answer 15

you are prettymuch before'math'

Answer 16

to equate x^2 as [x+x+x+.... (xtimes)] has no mathematical meaning to it in order to take a derivative of it

Answer 17

no actually there is nothing wrong in equating [x+x+x+...(x times)] to x^2

Answer 18

becasue we want to invent our own postulates and propositions?

Answer 19

apparently non of u are as good as me at math

Answer 20

apparently not :) but then what would that matter since you have no answers to your question to begin with lol

Answer 21

which postulate ?

Answer 22

the one where you claime that [x+x+x+ ...(xtimes)] = x^2 , or do you not recall your own statements?

Answer 23

postulate: A thing suggested or assumed as true as the basis for reasoning, discussion, or belief.

Answer 24

i have my answers i just wanted to test the knowledge of the world
(which i know now is not much)

Answer 25

then long live socrates ...

Answer 26

i know what a postulate is

Answer 27

you do now lol

Answer 28

clearly what you have written above george is wrong since it leads to a contradiction

Answer 29

long live 0.5*x^2

Answer 30

indeed, if [x+x+x+ ...(xtimes)] = x^2 were to mean anything; then to take the derivative of the whole would amount to: x+x = 2x

Answer 31

on taking derivative lhs=x and rhs=2x

Answer 32

becasue .... you wave your hands and it is so?

Answer 33

and zarkon im asking u whats wrong?

Answer 34

the derivative of the lhs \(\ne\) x

Answer 35

I already pointed that out...and so has amistre

Answer 36

why?????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????

Answer 37

because the left hand side means nothing.

Answer 38

thats bcoz u dont know math

Answer 39

chicken = beached whale.... how is this possible????????????????????? lol

Answer 40

its true simply because you postulate it

Answer 41

A Mr. is again coming with his favorite chicken and whale

Answer 42

its just as valid as yours ;)

Answer 43

my what??????

Answer 44

\[x^2=x\cdot x=x+x+\cdots+x +(x-\lfloor x\rfloor)\cdot x\]

where there are\[\lfloor x\rfloor, x's\] added together.

differentiate this :)

Answer 45

then anyone could postulate stuff( )

Answer 46

anyone is :)

where is your proof that [x+x+x+...(xtimes)] = x^2, other then your repeated declaration of it as a fact?

Answer 47

no i dont need a proof since it is a 'postulate'

Answer 48

zarkon u r very funny
just kidding

Answer 49

ok, if that is indeed your postulate; then prove that the derivative of the left side is "x" then. Or do you want to just make up a whole string of postulates

Answer 50

thats already proved. any loser knows that 1 added x times is x

Answer 51

thats not a proof the the left side derives to x.

Answer 52

since you say its not?

Answer 53

read that wrong lol

Answer 54

your the one making the claim, not me.

Answer 55

you claim the left side derives to x, wheres your proof?

Answer 56

each term is x. and derivative of each x is 1

Answer 57

ok...so far you are wrong...but continue ;)

Answer 58

where are you guys from?

Answer 59

which way do you wanna define it?

[x+x+x+....] or [x+x+x+....(xtimes)] ?

Answer 60

[x+x+x+...(x times)]

Answer 61

[x+x+x+...] is infinite

[x+x+x+...(xtimes)] is finite

they dont mean the same thing

so which definition do you claim you wanna ues?

Answer 62

its staring at ya

Answer 63

since you wanna use [x+x+x+...(xtimes)] then take its derivative

Answer 64

[1+1+1+...(x 1s)]

Answer 65

=x

Answer 66

xtimes is product rule .... try again

Answer 67

1*derivative of x +x*derivative of 1

Answer 68

=1*1+x*0

Answer 69

=1

Answer 70

lets try this:

x^2 = x*x

x = (1+n), where n = the number of times needed to satisfy x

x^2 = (1+n)*(1+n)

x^2 = 1 + 2n + n^2

the derivative of each side is then:

2x = 2 + 2n and since x=1+n

2(1+n) = 2+2n

2+2n = 2+2n

Answer 71

your definition of [x+x+x+...(xtimes)] is meaningless since it doesnt convey any meanigful information to use

Answer 72

and in that it is meaningless, to ascribe to it any derivative, much less "x" as its derivative is just as pointless