Math Brain Teaser

[ArmenK]

Originally posted by: MacBaine
While it doesn't matter really... I think some of you are still misunderstanding it. Whatever number X represents, is how many X^2 terms you have on the right side of the equation. if X=5, then X^3 = 5(x^5)

you mean 5(x^2)

 

[AgentDib]

ok i'll try to make it more clear.

Where it works:

x^3 does equal x^2 + x^2 +... (x terms).
x^2 + x^2 + .... x^2 (x terms) is just saying add x^2 for every instance of x... or x * x ^2 or x^3.

Where it fails:

The problem in this is that it is taking the derivative of an equation that is only defined at a single point. Every X you choose is a new equation and obviously there is an undefined change of rate of a function identified with a single point for any X that you pick.

 

[gopunk]

Originally posted by: AgentDib
ok i'll try to make it more clear.

Where it works:

x^3 does equal x^2 + x^2 +... (x terms).
x^2 + x^2 + .... x^2 (x terms) is just saying add x^2 for every instance of x... or x * x ^2 or x^3.

Where it fails:

The problem in this is that it is taking the derivative of an equation that is only defined at a single point. Every X you choose is a new equation and obviously there is an undefined change of rate of a function identified with a single point for any X that you pick.

goddammit, i was just about post that the function is defined in terms of itself, but the internet like freaking crashed :|

 

[eakers]

when you take the derivative of x(x^2) you have to use the product rule

d/dx = 1*x^2 + 2X*x
= 3X^2

 

[GroundZero]

i think you are both correct...
but it doesn't matter if you are talking to the power of.
x can only = one if we are talking powers
edit> x can also = zero

 

[Pepsi90919]

Originally posted by: gopunk

Originally posted by: AgentDib
ok i'll try to make it more clear.

Where it works:

x^3 does equal x^2 + x^2 +... (x terms).
x^2 + x^2 + .... x^2 (x terms) is just saying add x^2 for every instance of x... or x * x ^2 or x^3.

Where it fails:

The problem in this is that it is taking the derivative of an equation that is only defined at a single point. Every X you choose is a new equation and obviously there is an undefined change of rate of a function identified with a single point for any X that you pick.

goddammit, i was just about post that the function is defined in terms of itself, but the internet like freaking crashed :|

sorry i had to unplug it for a sec to use the microwave

 

[eakers]

Originally posted by: eakers
when you take the derivative of x(x^2) you have to use the product rule

d/dx = 1*x^2 + 2X*x
= 3X^2

 

[Hammer]

Originally posted by: eakers
when you take the derivative of x(x^2) you have to use the product rule

d/dx = 1*x^2 + 2X*x
= 3X^2

 

[MacBaine]

Originally posted by: Pepsi90919

Originally posted by: gopunk

Originally posted by: AgentDib
ok i'll try to make it more clear.

Where it works:

x^3 does equal x^2 + x^2 +... (x terms).
x^2 + x^2 + .... x^2 (x terms) is just saying add x^2 for every instance of x... or x * x ^2 or x^3.

Where it fails:

The problem in this is that it is taking the derivative of an equation that is only defined at a single point. Every X you choose is a new equation and obviously there is an undefined change of rate of a function identified with a single point for any X that you pick.

goddammit, i was just about post that the function is defined in terms of itself, but the internet like freaking crashed :|

sorry i had to unplug it for a sec to use the microwave

LOL! If my sig weren't packed, that would be in there 😛

 

[MacBaine]

Yes, sorry.

 

No, this isn't any teaser. The logic behind Mears' argument is very similar to that of the motel visitors who had $30 to spend and somehow $3 dollars couldn't be accounted for when recalculated. It's simply confusing because of the magic of words.

Here's really what Mears forgot about derivatives: There's the chain rule for composite functions. He failed to apply the chain rule here. And according to his reasoning, even the derivative of x^2 expressed as x*x is x. It's an erroneous reasoning.

Here's exactly what I mean in notations:

x^3 = x*x^2 = x^2 + x^2 + . . . + x^2 (x times) (using your choice of expression)

The derivative accordingly would be

3x^2 = 2x^2 + x^2 = x *2x + 1*x^2 = (2x + 2x + . . . + 2x) (x times) + 1*x^2

You forgot the chain rule used to find the derivative of two functions expressed as f(x)g(x) (i.e., composite functions).

Basically, your failure to apply the chain rule caused the inconsistency. Try a real brain teaser this time. 😉

 

[Mears]

I didn't formulate that proof. The question came from a book of (*gasp*) math brain teasers, which is why I chose the title.

 

[Mears]

Here is a simpler one:

a = x [given]

a+a = a+x
2a = a+x
2a-2x = a+x-2x
2(a-x) = a+x-2x
2(a-x) = a-x
2 = 1 [So]

 

[jarsoffart]

Originally posted by: Mears
Here is a simpler one:

a = x [given]

a+a = a+x
2a = a+x
2a-2x = a+x-2x
2(a-x) = a+x-2x
2(a-x) = a-x
2 = 1 [So]

a=x
a-x=0

(a-x)/(a-x)=(a-x)/0=undefined

 

[her209]

Okay, I think I got the answer. x^2+x^2+...+x^2 where there are x x^2-terms is not a valid one.

From the same logic you can write,

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

Now if you derive both sides, you get, 1 = 0 which is not correct.

Now say, x is not an integer. How do you write for example, 1.5 terms? :Q

 

[TuxDave]

ahhh.. messed up... but yeah, the line x^2 + x^2 + x^2... (x times). The number of terms is a function of x, and so if you have to take the derivative in respect to x (basically saying, what a small change in x would do), you would not be able to just find the derivative of the individual terms and leave the total number of terms unchanged in the derivative. But I like.. .I'm writing this one down...hehe

 

[dighn]

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