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Mathematics Maths Logic?, hope you guys can make sense out of it

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TSlin00b
post Jul 10 2009, 11:58 AM, updated 17y ago

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one of my friend pointed this out. admittedly it confuses the heck out of me.

2 * 2 = 2+2;
2 * 3 = 2+2+2;
2 * 4 = 2+2+2+2;
...
x * x = x+x+x+....+x+x (x number of times)

x * x =x^2
therefore,
x^2=x+x+x+x+...+x+x (x number of times)
differentiate,
d/dx(x^2)=2x
d/dx(x+x+x+...+x+x)=1+1+1+1+...+1+1+1 (x number of times)
which equals to x

therefore,
2x=x,
2=1
rclxub.gif rclxub.gif rclxub.gif

This post has been edited by lin00b: Jul 10 2009, 01:01 PM
TSlin00b
post Jul 10 2009, 12:54 PM

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QUOTE(exsea @ Jul 10 2009, 12:15 PM)
i think ur logic is flawed
*
please point out the flaw


Added on July 10, 2009, 12:55 pm
QUOTE(gengenrinni @ Jul 10 2009, 12:06 PM)
rclxub.gif  mana question??
*
no question (ok, maybe i mis-named the topic). sorry
just a solution?

This post has been edited by lin00b: Jul 10 2009, 12:55 PM
TSlin00b
post Jul 10 2009, 03:18 PM

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I'm following right up to
2x=(1+1+1+...+1)(x+x+x+...+x)

then its rclxub.gif again
TSlin00b
post Jul 15 2009, 06:03 PM

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QUOTE(darkwall @ Jul 14 2009, 11:26 PM)
why not like this??

2 * 2 = 2+2;
2 * 3 = 2+2+2;
2 * 4 = 2+2+2+2;
...
x * y = x+x+x+....+x+x (y number of times)
brows.gif  brows.gif
*
sorry then, lets make it this way:
2 * 2 = 2+2
3 * 3 = 3+3+3
4 * 4 = 4+4+4+4
x * x =x+x+x+...+x (x number of times)


Added on July 15, 2009, 6:08 pm
QUOTE(loongchai @ Jul 14 2009, 06:12 PM)
d/dx(x^2)=2x
d/dx(x+x+x+...+x+x)=1+1+1+1+...+1+1+1 (2x number of times)
which equals to 2x

therefore,
2x=2x,
2=2
*
how does d[x+x+x+..+x (x times)]/dx = 1+1+1+...+1 (2x times)?

This post has been edited by lin00b: Jul 15 2009, 06:08 PM
TSlin00b
post Jul 15 2009, 11:44 PM

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QUOTE(loongchai @ Jul 15 2009, 11:37 PM)
d/dx(x^2)=2x
d/dx(x+x+x+...+x+x)=1+1+1+1+...+1+1+1 (2x number of times)

2x because of d/dx(x^2)=2x

d/dx(x^2) = d/dx(x+x+x+...+x+x)

and

2x = 1+1+1+1+...+1+1+1 (2x number of times)

rclxub.gif
*
d/dx(x) = 1
d/dx(x+x) = 1+1
d/dx(x+x+x)=1+1+1
...
d/dx(x+x+x+...+x [x number of times])=1+1+1+...+1 [x number of times]

no?
TSlin00b
post Jul 30 2009, 01:28 PM

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QUOTE(MattL @ Jul 16 2009, 08:42 PM)
Wah this discussion still going on arr... See the wiki site on the previous page la... While it's true that
x^2=x+x+x+x+...+x+x (x number of times)
and
d/dx(x^2)=2x,

the mistake is
d/dx(x+x+x+...+x+x)=1+1+1+1+...+1+1+1 (x number of times)
because x here is not held constant.
*
true, the explanation is there, but how many actually understand the explanation?

and whats the problem with x not being held constant? is any mathematical laws being broken by letting x remain a variable?


Added on July 30, 2009, 1:39 pm
QUOTE(Aurora @ Jul 28 2009, 11:44 PM)
d/dx (x+x+...x+x) is in fact flawed. Remember that x increase at the magnitude of power of 2;
Let say x = 10
d/dx(x^2) = x+x+...x+x (10 times, with the value of x = 10)
Let say x = 11
d/dx(x^2) = x+x+...x+x (11 times, with the value of x = 11)
Why I say it is flawed, because the "x number of times" is not reflected in the equation, but only the value of x.

The true definition of d/dx is rate of change. For the equation y=x^2, the derivative (d/dx=2x) simply means the rate of change for the function when x is at certain value.

Let x = 10;
d/dx (x^2) = 2x = 20;

By rate of change, it means that any changes to the value, it is effectively scaled with that magnitude.

Say x = 10.001; the change (dx) = 0.001;
Effectively, the result should be = 10 + 0.001*20 = 10.02
Double check 10.001^2 = 10.020001
let x = 10,
2x = 1+1+1+... (10 times)
20 = 10 sweat.gif

iirc, small changes is only used when there is 2 or more variable in a function? (could be wrong though)

This post has been edited by lin00b: Jul 30 2009, 01:39 PM

 

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