#1

I tried finding that last thread about a similar subject. My teacher showed this problem and I can't figure it out.

Bear in mind:

2^2= 2+2

3^2= 3+3+3

4^2= 4+4+4+4

etc.

Therefore, we can say:

n^2= n+n+n+n+n... n times, right?

Now, we take the derivative of that:

2n= 1+1+1+1+1... n times.

2n=n.

Therefore, 2=1.

I'm sure there's a problem here, but I can't find it.

Bear in mind:

2^2= 2+2

3^2= 3+3+3

4^2= 4+4+4+4

etc.

Therefore, we can say:

n^2= n+n+n+n+n... n times, right?

Now, we take the derivative of that:

2n= 1+1+1+1+1... n times.

2n=n.

Therefore, 2=1.

I'm sure there's a problem here, but I can't find it.

#2

maths is for teh g33kz

#3

yes 2=1, whatever you say...

#4

There was a thread on this a couple of days ago...

#5

nm

#6

n^2= n.n right?

#7

no, 2n=n+n, so where n is one, 2=2

if you meant n^2,

then, where n is one,

n^2=1 (n+, n times)

so n is still 1

if you meant n^2,

then, where n is one,

n^2=1 (n+, n times)

so n is still 1

#8

i havent done this in 2 years or so, but if u take the derivitive of a number (say 5), it goes away doesnt it?

if it does (i dont really remember), then the deritive of n^2= n+n+n+n+n would not be 2n=1+1+1+1+1....

if it does (i dont really remember), then the deritive of n^2= n+n+n+n+n would not be 2n=1+1+1+1+1....

#9

think about it, 2 does't equal 1

theres your problem

theres your problem

#10

No, 2n =/= n, unless it is zero.

#11

n is a constant not a variable hence

0=0

EDIT: Basicly n is just a number and you can't differentiate with respect to a constant (i think ?), and if you differentiated wrt a variable say x you'd get 0.

If n was a variable

n ^2=n.n

diff wrt n

using product rule

(Vdu+Udv)

2n=n.1+n.1

2n=2n

0=0

EDIT: Basicly n is just a number and you can't differentiate with respect to a constant (i think ?), and if you differentiated wrt a variable say x you'd get 0.

If n was a variable

n ^2=n.n

diff wrt n

using product rule

(Vdu+Udv)

2n=n.1+n.1

2n=2n

#12

TS = owned like Mark Hamill's movie career was after Star Wars VI

#13

i havent done this in 2 years or so, but if u take the derivitive of a number (say 5), it goes away doesnt it?

if it does (i dont really remember), then the deritive of n^2= n+n+n+n+n would not be 2n=1+1+1+1+1....

n is a variable; that's why the derivative of n^2 is 2n. This is also why the derivative of n is 1.

If n was a variable

n ^2=n.n

diff wrt n

using product rule

(Vdu+Udv)

2n=n.1+n.1

2n=2n

I understand it this way, but n.n is the same as n+n+n... n times.

*Last edited by tylerishot at May 25, 2008,*

#14

think about it, 2 does't equal 1

theres your problem

This.

#15

what are you smoking?

#16

that is epically wrong on so many levels. 2n =/= n and 2 certainly never equals 1!! u sir phail!! but no hard feelings, at least u didnt divide by a certain digit.......

#17

n is a variable; that's why the derivative of n^2 is 2n. This is also why the derivative of n is 1.

its not a very applicable variable it seems. because if u plug in 5^2=5+5+5+5+5

and took the derivitve, the 5's on the right hand side would not be equal to 1

#18

No, 2n =/= n, unless it is zero.

Exactly

the graph is y = 2^2

2^2 = 4 = constant

The gradient of this graph = 0

The 1st derivative of this graph also must = 0

2n = n = 0

Its just a stupid trick

#19

its not a very applicable variable it seems. because if u plug in 5^2=5+5+5+5+5

and took the derivitve, the 5's on the right hand side would not be equal to 1

If you take the derivative of 5^2=5+5+5+5+5, it's 0=0. Just so you know.

#20

the only one of these that works is:

Say 0.9 recurring is y.

so y x 10, ie. 10y = 9.9 recurring

10y - y = 9y = 9.9 recurring - 0.9 recurring = 9

so 9y = 9

divide by 9

y=1

therefore, 0.9 recurring is equal to 1

Say 0.9 recurring is y.

so y x 10, ie. 10y = 9.9 recurring

10y - y = 9y = 9.9 recurring - 0.9 recurring = 9

so 9y = 9

divide by 9

y=1

therefore, 0.9 recurring is equal to 1

#21

If you take the derivative of 5^2=5+5+5+5+5, it's 0=0. Just so you know.

well there ya go :-D

#22

the only one of these that works is:

Say 0.9 recurring is y.

so y x 10, ie. 10y = 9.9 recurring

10y - y = 9y = 9.9 recurring - 0.9 recurring = 9

so 9y = 9

divide by 9

y=1

therefore, 0.9 recurring is equal to 1

10y isn't 9.99999 recurring.

It's 9.99999....9990.

Just so you know.

EDIT: And because I know you won't take my word for it, look at it this way:

Say y=0.9.

Therefore, 2y=1.8

3y=2.7

4y=3.6

etc.

if y=0.9999...

2y=1.999999....98

3y=2.999999...97

etc.

It's a stupid problem.

*Last edited by tylerishot at May 25, 2008,*

#23

So you're saying if I have two apples, I really only have one?

Mind blooooowwwwing.....

Mind blooooowwwwing.....

#24

I tried finding that last thread about a similar subject. My teacher showed this problem and I can't figure it out.

Bear in mind:2^2= 2+2

3^2= 3+3+3

4^2= 4+4+4+4

etc.

Therefore, we can say:

n^2= n+n+n+n+n... n times, right?

Now, we take the derivative of that:

2n= 1+1+1+1+1... n times.

2n=n.

Therefore, 2=1.

I'm sure there's a problem here, but I can't find it.

you're wrong already. 2^2= 2x2

#25

The Sky Is Falling!!!!

#26

you're wrong already. 2^2= 2x2

yes, but if you try it in the pattern he had it in, it works. 2^2=4 2+2=4. 4=4.

10y isn't 9.99999 recurring.

It's 9.99999....9990.

Just so you know.

EDIT: And because I know you won't take my word for it, look at it this way:

Say y=0.9.

Therefore, 2y=1.8

3y=2.7

4y=3.6

etc.

if y=0.9999...

2y=1.999999....98

3y=2.999999...97

etc.

It's a stupid problem.

Technically, you can't divide or multiply irrational numbers, because they have no end. They're meaningless.

*Last edited by that_1_dude24 at May 25, 2008,*

#27

ya the best argument may be that n cant be applied as a variable

#28

10y isn't 9.99999 recurring.

It's 9.99999....9990.

Just so you know.

EDIT: And because I know you won't take my word for it, look at it this way:

Say y=0.9.

Therefore, 2y=1.8

3y=2.7

4y=3.6

etc.

if y=0.9999...

2y=1.999999....98

3y=2.999999...97

etc.

It's a stupid problem.

I think you're missing the point of recurring numbers.

you're thinking in finite terms, your logic only works with a large but not infinite number of .9s. It is mathematically accepted that 0.9 recurring is equal to 1.

#29

Just look at it this way, 2, does not equal 1.Period. There you go, problem solved.

#30

Therefore, we can say:

n^2= n+n+n+n+n... n times, right?

Now, we take the derivative of that:

2n= 1+1+1+1+1...n times.

2n=n.

Therefore, 2=1.

I'm sure there's a problem here, but I can't find it.

when you say n times, that is a constant

derivative of (2n) = derivative of a (sum of a sequence of terms)

0 = 0

#31

the equation fails at http://upload.wikimedia.org/math/2/e/1/2e182ba47dd0ff925c5b7a4acf0522a2.png

you need a continuous function if you want to be able to differentiate it

you need a continuous function if you want to be able to differentiate it

#32

he said that not too many posts before (that 0=0)

#33

to put it another way what you just did was

n^2 = n+n+... n times

3^2 != 3+3+3

EPIC FAIL

n^2 = n+n+... n times

3^2 != 3+3+3

*** 3**EPIC FAIL