hobz

Suppose I have f(x)=e^x, how can it be proven that it is smooth?

Ah yes. Of course. Thank you both.

I don't think I have made my self clear.

The problem is that if multiplication is repeated addition, how do I add the units to make them square?

To my knowledge:

[math]

3 \cdot 5 = 5 + 5 + 5

[/math]

However

[math]

3 \, \mathrm{m} \cdot 5 \,\mathrm{m} = 15 \, \mathrm{m}^2 \neq 5 \,\mathrm{m} + 5 \,\mathrm{m} + 5 \,\mathrm{m}

[/math]

What is wrong?

In practice if you keep the gills wet, the fish will be able to survive above water, breathing air.

If a function is not injective, then it does not have an inverse function so the derivative cannot be found.

Surely an injective function can still has a derivative? You mean as a ratio?

I see. So with ordinary "high school" mathematics, the derivative as a ratio holds just fine. Which also would explain how the founding fathers (before the derivative as a limit) were able to use and deduce stuff.

Concerning differentials, when is 1/dy/dx not equal to dx/dy?

When can the derivative not be thought of as a ratio?

Excellent!

Is there a different meaning to hypothesis in maths?

Merged post follows:

Wikipedia http://en.wikipedia.org/wiki/Theorem equates a theorem to a statement proven true and says that the theorem consists of hypotheses and a conclusion which is a necessary consequence of the hypothesis.

Merged post follows:

Suggesting that the definitions are different depending on which topic they are used in.

I see. I had the impression that a hypothesis was more like "P" and a theorem was "If P then Q".

What is a theorem?

A side note. A statement is sort of a binary axiom?

What is the difference between a statement and a hypothesis?

this method can create a linear gradient ...

 

now with any help from linear algebra mathematicians, we can improve general methods, to iterate through this matrix, with more unique mapping ...

I don't think you can do it simpler than

for i=0:N
for j=0:M
...

which is the general method for iterating through a matrix.

I will try to dig deeper using the links you posted.

Thanks for a good discussion. I appreciate it.

So is [math]\frac{\partial}{\partial x}[/math] more "correct" than [math]\frac{d}{dx}[/math] which is more of a special case?

The notation looks advanced. I will have to look it up somewhere.

Is it fair to say, that in Leibniz' days, the [math]\frac{dy}{dx}[/math] was thought of as a ratio, until the more rigid approach with limits gave a definition where nilpotents and arguments as to [math](dx)^2[/math] could be discarded while [math]dx[/math] could not, were not needed?

The notation [math]\frac{dy}{dx}[/math] is a legacy that reminds us of the origin (a ratio) but really isn't. If division is defined, then [math]dy[/math] and [math]dx[/math] can be split up into each of their own limits and be multiplied on both sides etc. rendering algebraic operations valid in algebra that has +,-,*,/ defined (as I suppose was the case around Leibniz days).

If division is undefined, the notion of [math]\frac{dy}{dx}[/math] doesn't change; it is still the derivative or how much [math]dy[/math] changes when [math]dx[/math] changes, but it is no longer a ratio, and the derivative has to be defined in another way.

I mean, when introduced in high school, the classic secant to tangent method is introduced, and the limit is called the derivative. While I always thought that this presented the idea of the derivative in any space, it really presents a way of calculating the derivative in that particular space where +,-,*,/ are defined.

If we move into a vector field, the idea of measuring how the vector field changes is still perfectly valid. But to calculate the change is no longer a ratio issue, but the term [math]\frac{dy}{dx}[/math] lingers on, and becomes an operator in its entirety. [math]dx[/math] looses it meaning, the only thing that matters is the operator: [math]\frac{dy}{dx}[/math].

Can you verify or completely obliterate my thoughts the past days?

I don't know what do you mean when you say: "there is no division" ,but since

 

dy = f'(x)dx ,if you divide by dx ,then dy/dx = f'(x) = [math]\frac{dy}{dx}[/math].

 

Hence the ratio of dy to dx is equal to the derivative of y w.r.t x

I mean, when the notion of division is not defined. For instance in a vector space as ajb suggested.

I think we arrived at the special notation [math]\frac{dy}{dx}[/math] is not a ratio, but defined in terms of limits of a ratio (the normal derivative definition).

However, I have seen plenty of derivations of formulas in physics, where dy is treated as a "thing" that easily can be divided with dx. Or multiplying [math]\frac{dy}{dx} = f(x)[/math] by dx and integrate both sides is also frequently used.

I am having an extremely difficult time trying to understand this post.

I think you should reformulate your idea.

Take a look at this

I has little to do with guns though..

They will probably start moving right away. They will most certainly not move on the host machine before the move order is received with a 2 sec delay.

The game could also wait for the host to confirm that your troops have received new orders, and not allow the move order to be executed before the host has confirmed. In this case, there would be a 4 sec delay. But by that time you would probably have quit the game and found something else to do.

You'd be far better of defining pi as a constant at compile time. First of all you can get exact precision to the degree that 'double' (or long double if you prefer) offers.

Second, calculating a constant value every time the function is called, is just a waste of power. (Hopefully the compiler will fix this for you.)

So how would you go about defining differentiation where the is no division?

hobz ,how do you know that:

[math]

y + dy = (x + dx)^2

[/math]

I assume that by changing x I change y since y=f(x).

So by changing x by dx, I change y by dy.

My goal is then to find out how much the change in y (dy) is in response to a change in x (dx).

Merged post follows:

Try it yourself. Take the operator [math]\frac{\partial^{2}}{\partial x \partial x}[/math] and apply it to a product of functions.

I will attempt this a quiet evening.

You usually write [math]\frac{dy}{dx} = \frac{\partial y }{\partial x}[/math] in the situation you have described. But saying [math]dx = \partial x[/math] I do not understand.

 

You can define the notion derivatives or differential on vector spaces, algebras, Lie algebras etc... You don't need the notion of division to do this.

If I understand your notation correctly, then [math]\frac{\partial y}{\partial x}[/math] means the derivative. This a notation, and not a division; the notation cannot be separated into a [math]\partial y[/math] and [math]\partial x[/math]. It serves as an operator.

If so, then this operator relates the change in x [math]dx[/math] to the change in y [math]dy[/math]. Am I off?

It makes sense in systems where division is not defined that [math]\frac{\partial y}{\partial x}[/math] is NOT a division.

Is the notation [math]\frac{\partial y}{\partial x}[/math] still related to Taylor's theorem?

Thanks for the reply.

So we see that first order term is the derivative and the second order term contains term that measures the failure of a second order derivative [math]\frac{\partial^{2}}{\partial x^{2}}[/math] ("Laplacian") to be a derivation.

I don't understand this. Failure of a derivative to be a derivation?

So under a true diffeomorphism of the real line [math]\gamma : x \rightarrow y[/math] with the usual abuse [math]\gamma^{*}y = y(x)[/math] (now a function in terms of the original coordinate) we have

 

[math]dy = dx \frac{\partial y(x)}{\partial x}[/math].

I am unfamiliar with this notation. Does it mean that [math]\gamma[/math] maps x to y? (Can't figure out [math]\gamma^{*}y=y(x)[/math].)

Note we have not dived by [math]dx[/math]. The partial derivatives are defined in terms of limits, the differentials are formal.

That is to say that [math]dx[/math] is a differential where [math]\partial x[/math] is not? I thought they were identical execept the partials denoted a function of several variables being differentiated with respect to one while treating the other as constant.

You are almost right, but dividing by dx is problematic. I mean in general we will not have a clear notion of dividing.

Does that mean that the derivative is not dependent on division to exist in a given context in order to work? (This would mean that the notation is sort of abused). I cases where division is defined (a*b=c <=> b=c/a i guess) is it then problematic to divide?

[math]

dy = dx\frac{\partial y}{\partial x}

[/math]

Is this different from

[math]

dy = dx\frac{dy}{dx}

[/math]

when y is only dependent on x? (Related to a previous encounter http://www.scienceforums.net/forum/showthread.php?t=44865)

But surely there are no terms in the Taylor expansion that are not there when taking the derivative. I mean, the Taylor expansion must yield zero for the second term, right?

If the Taylor expansion is used on [math]y=u(x)\cdot v(x)[/math] would the term [math]\frac{dudv}{dx}[/math] also appear?

On another note. Is the differential notation not perfectly valid and equivalent to the difference quotient notation?

If

[math]

\frac{dy}{dx} = \lim_{\Delta x \rightarrow 0} \frac{f(x+\Delta x)-f(x)}{\Delta x}

[/math] then what is [math]\frac{dy}{dx}[/math] in the differential notation?

If:

[math]

y = x^2

[/math]

Changing x by the amount dx gives the change dy in y:

[math]

y + dy = (x + dx)^2

[/math]

[math]

y + dy = x^2 + 2x\cdot dx + (dx)^2

[/math]

[math]

dy = 2x\cdot dx + (dx)^2

[/math]

[math]

\frac{dy}{dx} = 2x + dx

[/math]

Suggesting that the rate of change in y with respect to x also depends on dx.

Now you have to argue that dx is negligible, which is handled by limits in the difference quotient way of writing this.

So is the differential notation inferior?

What I write looks identical to both http://en.wikipedia.org/wiki/Product_rule#Discovery_by_Leibniz and http://en.wikipedia.org/wiki/Product_rule#Using_non-standard_analysis.

In both cases the dudv is discarded. But what happens if I keep it?