Investigating an Unexpected Error: What's Wrong?

[jaumzaum]

What is wrong??

Hi,

I want to know where is the mistake in the statement below (I really don't know what is wrong, but it is obviously wrong)

\frac{\partial ^{2}x}{\partial t^{2}} = \frac{\partial ^{2}x}{\partial x^{2}} (\frac{\partial x}{\partial t} )^{2}

But
\frac{\partial ^{2}x}{\partial x^{2}} = \frac{\partial \frac{\partial x}{\partial x}}{\partial x} = \frac{\partial 1}{\partial x} = 0

This way:
\frac{\partial ^{2}x}{\partial t^{2}} = 0

Thanks,
John

 

[voko]

After so many posts, you surely know that you should post what is given and what you are supposed to find.

Really.

 

[1MileCrash]

Hi,

I want to know where is the mistake in the statement below (I really don't know what is wrong, but it is obviously wrong)

\frac{\partial ^{2}x}{\partial t^{2}} = \frac{\partial ^{2}x}{\partial x^{2}} (\frac{\partial x}{\partial t} )^{2}

That first statement doesn't look true.

 

[Mark44]

What does the following simplify to?
$$ \frac{\partial x}{\partial x}$$

 

[jaumzaum]

After so many posts, you surely know that you should post what is given and what you are supposed to find.

Really.

Sorry.

Imagine a car traveling in the x axis. It's initial position is zero and its initial velocity ∂x/∂t is zero too. Although it has a variable acceleration a.

a = \frac{\partial ^{2}x}{\partial t^{2}}

I've just multiplied by
\frac{\partial x^{2}}{\partial x^{2}}
to get
\frac{\partial ^{2}x}{\partial x^{2}} (\frac{\partial x}{\partial t} )^{2}

In my conception I just multiplied by 1 and got an absurd result. "a" is not always zero (I haven't even mentioned the function a(t))

 

[voko]

First of all, it is far from clear why you are using partial derivatives. You have functions of just one argument - the time - so ordinary derivatives would work fine.

Second, $a = \frac {d^2x} {dt^2} \ne (\frac {dx} {dt})^2 = v^2$.

 

[1MileCrash]

Sorry.

Imagine a car traveling in the x axis. It's initial position is zero and its initial velocity ∂x/∂t is zero too. Although it has a variable acceleration a.

a = \frac{\partial ^{2}x}{\partial t^{2}}

I've just multiplied by
\frac{\partial x^{2}}{\partial x^{2}}
to get
\frac{\partial ^{2}x}{\partial x^{2}} (\frac{\partial x}{\partial t} )^{2}

In my conception I just multiplied by 1 and got an absurd result. "a" is not always zero (I haven't even mentioned the function a(t))

Yikes, I think you're having a problem with notation vs. concepts.

First of all,

\frac{\partial ^{2}x}{\partial x^{2}} (\frac{\partial x}{\partial t} )^{2}

is in no way, shape, or form, the same as

\frac{\partial ^{2}x}{\partial t^{2}}

You did multiply by 1, but your mistake is thinking that the Leibniz notation is just like a simple fraction involving numbers being multiplied because it looks the same. It is a fraction, but ∂ is not a number. ∂/∂ is not 1, it's not anything, it's like saying +/+ or (grapefruit)/(I think Michael Shannon is a talented actor.)-edited out-

I proved that it couldn't be true, but misread the equation myself in doing so. :/

 

[jaumzaum]

Yikes, I think you're having a problem with notation vs. concepts.

First of all,

\frac{\partial ^{2}x}{\partial x^{2}} (\frac{\partial x}{\partial t} )^{2}

is in no way, shape, or form, the same as

\frac{\partial ^{2}x}{\partial t^{2}}

You did multiply by 1, but your mistake is thinking that the Leibniz notation is just like a simple fraction involving numbers being multiplied because it looks the same. It is a fraction, but ∂ is not a number. ∂/∂ is not 1, it's not anything, it's like saying +/+ or (grapefruit)/(I think Michael Shannon is a talented actor.)


-edited out-

I proved that it couldn't be true, but misread the equation myself in doing so. :/

But why can I multiply dx/dt by dθ/dθ to get (dx/dθ)(dθ/dt) = w (dx/dθ).
I've just multipied by one too, and only rearranged the terms, as I did in the initial derivative. What's the difference between them? Why in the first one I cannot do this and in the second one I can?

 

[1MileCrash]

Ok... this may be challenging to explain. What you've done is used symbols that look algebraic, who's values are not algebraic but defined, then did algebra with these symbols, assuming their values would be preserved.

Put it this way, some people write arcsin(x) as sin^-1(x). But, that doesn't mean I can do algebra to both sides and get a true statement, because the symbol "sin^-1(x)" gets its value from definition.

Here:

Hi,

I want to know where is the mistake in the statement below (I really don't know what is wrong, but it is obviously wrong)

\frac{\partial ^{2}x}{\partial t^{2}} = \frac{\partial ^{2}x}{\partial x^{2}} (\frac{\partial x}{\partial t} )^{2}

See,

\frac{\partial ^{2}x}{\partial x^{2}}

Is 0.

This symbol, as a whole, means the acceleration of x with respect to itself. It will always be changing at a constant rate relative to itself (namely, 1) so you have made the claim:


\frac{\partial ^{2}x}{\partial t^{2}} = 0(\frac{\partial x}{\partial t} )^{2}

\frac{\partial ^{2}x}{\partial t^{2}} = 0

That's the claim, period. The fact that if you leave 0 as you wrote it and do algebra to get something that looks true means absolutely nothing because our "zero" is zero through conceptual definition and not algebraically so (IE, the numerator is not 0.)

So it's like "changing your mind" mid way through.

"This is zero because of what we say the symbols mean, not algebraically"
to
"Do normal algebra with those symbols"
back to
"Now let's return to what the symbols mean again and observe the result."

Which will of course, end up as nonsense.

 

[Mark44]

\frac{\partial ^{2}x}{\partial x^{2}}

Is 0.

Yes. That's where I was going when I asked this question.

What does the following simplify to?
$$ \frac{\partial x}{\partial x}$$

This symbol, as a whole, means the acceleration of x with respect to itself.

Minor quibble. It means the second derivative of x with respect to itself. Acceleration usually means the second derivative of position with respect to time.

It will always be changing at a constant rate relative to itself (namely, 1) so you have made the claim:


\frac{\partial ^{2}x}{\partial t^{2}} = 0(\frac{\partial x}{\partial t} )^{2}

\frac{\partial ^{2}x}{\partial t^{2}} = 0

That's the claim, period. The fact that if you leave 0 as you wrote it and do algebra to get something that looks true means absolutely nothing because our "zero" is zero through conceptual definition and not algebraically so (IE, the numerator is not 0.)

So it's like "changing your mind" mid way through.

"This is zero because of what we say the symbols mean, not algebraically"
to
"Do normal algebra with those symbols"
back to
"Now let's return to what the symbols mean again and observe the result."

Which will of course, end up as nonsense.

 

[1MileCrash]

Yes. That's where I was going when I asked this question.

Minor quibble. It means the second derivative of x with respect to itself. Acceleration usually means the second derivative of position with respect to time.

Well, of course I agree, but I wanted to stay in the OP's terms.

 

[voko]

In fact, the symbols were not used correctly even by the 17th century's standard. Mr Leibniz did a marvelous job to ensure that $\frac {d} {dt} \frac {d} {dt}$ naturally ends up being $\frac {d^2} {dt^2}$, not $(\frac {d \text{something}} {dt})^2$, simply because it has two d's and none of the 'something'.

 

[1MileCrash]

I think
\frac{\partial ^{2}x}{\partial t^{2}} = \frac{\partial ^{2}x}{\partial x^{2}} (\frac{\partial x}{\partial t} )^{2}

was thought to be true because if you treat the RHS purely algebraically, you get


\frac{\partial ^{2}x\partial ^{2}x^{2}}{\partial ^{2} x^{2} \partial ^{2} t ^{2}}

to which "cancelations" give

\frac{\partial ^{2}x}{ (\partial t)^{2}}


Not due to an error involving

$\frac {d} {dt} \frac {d} {dt}$ naturally ends up being $\frac {d^2} {dt^2}$, not $(\frac {d \text{something}} {dt})^2$

Unless I'm missing it somewhere. :)