A Nifty Proof - Science, Math and Philosophy Forum

Hi Everyone:

2 squared = 2+2

3 squared = 3+3+3

4 squared is equal to 4+4+4+4

so X squared is equal to X+X+X+...+X (X amount of times)

If we take the derivatie of each side we get

2X = 1+1+1+...+1 (X amount of times) and 1+1+1+...+1 (X amount of times) is equal to X. So we now have 2X = X

Divide each side by X and we get 2 = 1

Best wishes,
Mason

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07-03-2017 , 07:20 AM

jeccross

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Quote:

Originally Posted by Mason Malmuth

If we take the derivatie of each side we get

This is meaningless, you haven't defined a function.

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07-03-2017 , 07:31 AM

lastcardcharlie

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Originally Posted by jeccross

This is meaningless, you haven't defined a function.

Let f : N -> N, x -> x^2, and let g : N -> N, x -> x + ... + x (x times), where N is the set of natural numbers.

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07-03-2017 , 07:33 AM

ToothSayer

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Quote:

Originally Posted by jeccross

This is meaningless, you haven't defined a function.

The right side is a function

sum [1=>X] (X)

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07-03-2017 , 09:41 AM

masque de Z

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You cannot take a derivative when the other side is integer number of x in order for the summation to make sense.

It's like taking the derivative of n, an integer variable!

The summation presented makes sense only in integers. The derivative requires a limit.

Last edited by Zeno; 08-16-2017 at 11:19 AM. Reason: Off topic comment

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07-03-2017 , 12:30 PM

gangip

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+1 That's not how derivatives work

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07-03-2017 , 01:40 PM

plaaynde

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Originally Posted by masque de Z

PS: Now how about doing the patriotic thing on the 4th of July inviting back BruceZ as probability forum moderator again and hopefully getting our friend back where he offered so much?

Supported!

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07-03-2017 , 01:49 PM

Aaron W.

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Quote:

Originally Posted by Mason Malmuth

so X squared is equal to X+X+X+...+X (X amount of times)

If we take the derivatie of each side we get

2X = 1+1+1+...+1 (X amount of times)

For forgot to take a derivative.

2X = 1+1+1+...+1 (1 amount of times)

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07-03-2017 , 05:12 PM

lastcardcharlie

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Quote:

Originally Posted by Mason Malmuth

so X squared is equal to X+X+X+...+X (X amount of times)

The trick is that the function X squared is being treated as the sum of X copies of the identity function:

the function X -> X^2 is equal to the function id+id+id+...+id (X times)

but the right hand function is not well-defined because X varies.

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07-03-2017 , 07:01 PM

#10

PairTheBoard

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You forgot to divide by zero.

Form the difference quotient. Let h be small.

([(x+h)+...+(x+h)] (x+h) times - [x+...+x] x times ) / h =

[(x+...+x) x+h times + (h+...h) x+h times - (x+...x) x times ] / h =

[(x+...+x) x times + (x+...+x) h times + (h+...+h) x times + (h+...+h) h times - (x+...x) x times ] / h =

[hx + xh + (h+...+h)h times] / h =

[ 2hx + (h+...h)h times] / h =

2x + h

which goes to 2x as h-->0

PairTheBoard

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07-03-2017 , 08:35 PM

#11

Mason Malmuth

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Quote:

Originally Posted by PairTheBoard

There's no division by zero.

Best wishes,
Mason

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07-03-2017 , 09:15 PM

#12

ecriture d'adulte

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Whats the point of this thread? Are you genuinely confused about how functions you've defined for zero measure won't have the same shorthand high school derivative as functions that vaguely look the same but are defined almost everywhere? This obv isn't a math/science forum, but this seems like trolling.

Last edited by ecriture d'adulte; 07-03-2017 at 09:20 PM.

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07-03-2017 , 10:12 PM

#13

PairTheBoard

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Quote:

Originally Posted by Mason Malmuth

There's no division by zero.

Best wishes,
Mason

Precisely.

Then there's the product rule:

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

(x'+...+x') x times + (x+...+x) x' times =

(1+...+1) x times + (x+...+x) 1 time =

x + x =

2x

PairTheBoard

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07-04-2017 , 12:22 AM

#14

TomCowley

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Maybe this is how derivatives wrecked the economy.

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07-04-2017 , 07:51 AM

#15

lastcardcharlie

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I have often wondered whether there is a purely algebraic theory of differentiation, without any mention of limits.

Functions R -> R from a ring to itself can be added and multiplied. If a function [R -> R] -> [R -> R] satisfies the chain and product rules then voila, an abstract theory of differentiation!

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07-04-2017 , 11:40 AM

#16

Aaron W.

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Quote:

Originally Posted by lastcardcharlie

https://en.wikipedia.org/wiki/Formal_derivative

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07-04-2017 , 12:06 PM

#17

lastcardcharlie

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Nice one.

(FWIW, what prompted my thought was teaching differentiation a lot to 17-year olds. Some lip service is paid to limits in the textbook, but that's not examined on, and I ignore it entirely. Rightly or wrongly, all they're doing is getting fluent with the chain and product rules, which forms a major part of their exam.)

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07-04-2017 , 12:48 PM

#18

Aaron W.

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Originally Posted by lastcardcharlie

The ongoing problem of why it is that people don't like or understand math. We're still trapped in a world where computational ability is emphasized to the detriment of conceptual fluency.

The best example of this is long division. A lot of adults can do it, but very few of them can actually explain it in conceptual terms. (What are you *actually* doing?)

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07-12-2017 , 11:22 PM

#19

:::grimReaper:::

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The derivative of the right is not "1+1+1+...+1 (X amount of times) ", because while the derivative is linear, i.e.

d/dx sigma(1 to n) x = sigma(1 to n) d/dx x = sigma(1 to n) 1 = n

that's not the same as:

d/dx sigma(1 to x) x

You can't move d/dx passed the sigma.

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07-13-2017 , 04:18 PM

#20

saw7988

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Originally Posted by Aaron W.

A lot of adults can do long division?

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07-13-2017 , 06:37 PM

#21

CORed

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Quote:

Originally Posted by Mason Malmuth

There's no division by zero.

Best wishes,
Mason

There is in a lot of "proofs" that 1 = 0, 2 = 1, etc.

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07-14-2017 , 03:58 AM

#22

masque de Z

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1+2+3+4+...= 1+3+5+...+2*(1+2+3+4+...)=> 1+3+5+...+1+2+3+4+...=0 but of course...

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07-14-2017 , 08:35 AM

#23

PairTheBoard

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Lift sanctions on Fake Math. Free the Insurgent Numbers.

PairTheBoard

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07-16-2017 , 04:34 AM

#24

Mason Malmuth

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Quote:

Originally Posted by CORed

There is in a lot of "proofs" that 1 = 0, 2 = 1, etc.

Hi CORed:

Yes. But not in this one.

Best wishes,
Mason

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08-15-2017 , 11:23 AM

#25

Gcm1998

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When you say "X amount of times"(OP) you are using X as a defined number and not a variable. You can only sum X, X amount of times if X is a real, defined number.

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