The derivative of uv wrt x using st function (homework problem)

In summary, the individual is attempting to find the derivative of uv with respect to x using non standard analysis, hyperreals, and the standard part function st. They take u and v to be functions of x and have made certain assumptions in their solution. They also discuss the advantages of using non standard analysis in conjunction with standard analysis and the interpretation of the definite and indefinite integrals in both contexts.
  • #1
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TL;DR Summary: I attempt to find the derivative of uv with respect to x using non standard analysis, hyperreals, and the standard part function st; I take u to be a function of x, and I also take v to be a function of x.

Hello everyone!

I've been learning about non standard analysis concepts (shout-out to @Dale for clarifying a lot of concepts and taking time out of his day to help me) and thought I would try something super basic with what I learned to see if I'm actually understanding the concepts correctly, I worked out a problem, and my AI assistant objected to what I thought was a correctly worked out solution, so I wanted to double check here to see if my logic is correct, please feel free to critique my approach and show any misconceptions I may have.

My solution:

d(uv)/dx = st(((u+du)(v+dv) - uv)/dx) = st(((uv) + (u)(dv)+(v)(du)+(du)(dv)-(uv))/dx) = st((u)(dv/dx)+(v)(du/dx) + ((du)(dv)/dx)) = (u)(dv/dx)+(v)(du/dx)

Some underlying assumptions:

I assume that given some infinitesimal e and certain non zero real valued variables a, b, c, that du = ae, dv = be, and dx = ce.

I assume we can distribute st over a summation, i.e, I assume st(p1 + p2 + p3) = st(p1) + st(p2) + st(p3) where p1, p2, and p3 are parameters.

I assume that we can round the hyperreal in the analysis by assuming st((du)(dv)/dx) is zero.

Because I'm using hyperreals I'm pretty sure that means the infinitesimal e is not nilpotent.

If you feel you can shed some light, please feel free to reply, thank you!
 
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  • #2
Chenkel said:
TL;DR Summary: I attempt to find the derivative of uv with respect to x using non standard analysis, hyperreals, and the standard part function st; I take u to be a function of x, and I also take v to be a function of x.

My solution:

d(uv)/dx = st(((u+du)(v+dv) - uv)/dx) = st(((uv) + (u)(dv)+(v)(du)+(du)(dv)-(uv))/dx) = st((u)(dv/dx)+(v)(du/dx) + ((du)(dv)/dx)) = (u)(dv/dx)+(v)(du/dx)
I am not certain of the middle process but
[tex]\frac{d\ uv}{dx}=u\frac{dv}{dx}+v\frac{du}{dx}[/tex]
seems right to me. What is the motivation of your way ?
 
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  • #3
What does st( ... ) mean?
Chenkel said:
d(uv)/dx = st(((u+du)(v+dv) - uv)/dx) = ...

I assume that we can round the hyperreal in the analysis by assuming st((du)(dv)/dx) is zero.

In ordinary analysis, if x, y, and z are all approaching zero, the fraction ##\frac{xy} z## is indeterminate. To determine a value, one must use limits.
 
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  • #4
anuttarasammyak said:
I am not certain of the middle process but
[tex]\frac{d\ uv}{dx}=u\frac{dv}{dx}+v\frac{du}{dx}[/tex]
seems right to me. What is the motivation of your way ?
I noticed that standard calculus makes much more sense in conjunction with non standard analysis because non standard analysis rigorously defines differentials as infinitesimals, and also defines the derivative as the quotient of infinitesimals instead of the standard approach of defining the derivative as the limit of a quotient of real numbers.

One disadvantage of the standard way of defining the derivative is that we can't rigorously define differentials using infinitely small reals, because there's no such thing as an infinitely small real number; this results in sometimes not being able to analytically break down certain problems into their most basic parts.

Generally speaking because standard analysis uses the same notation as non standard analysis we can also conceptually switch between non standard analysis and standard analysis where it benefits us.

One more thing that makes more sense with non standard analysis is the interpretation of the definite integral and indefinite integral, because the integrand in non standard analysis adds more meaning to the mathematical interpretation.

An example of the previous paragraph is suppose we are integrating x^2 over the interval a to b, both in standard analysis and non standard analysis we use the same notation so we can write:$$\int_a^b x^2 \, dx$$The interpretation that standard analysis tries to convey is that dx is interpreted as an operator specifying which variable we are integrating wrt, and x^2 is interpreted as the function we're integrating, however in non standard analysis, it is safe to interpret dx as a multiplication factor that is part of the integrand and multiplies the function we're integrating, this has applications in many fields including physics, one example is easily rewriting the integrand to change variables in an intuitive way.
 
  • #5
Mark44 said:
What does st( ... ) mean?
It's the "standard part function" which takes a hyperreal and rounds to the nearest real finite number if I'm not mistaken.
 
  • #6
Mark44 said:
In ordinary analysis, if x, y, and z are all approaching zero, the fraction xyz is indeterminate. To determine a value, one must use limits.
I agree, in ordinary "standard" analysis one uses limits, however I'm trying to use non standard analysis.

Using your variable names, if x y and z are infinitesimal hyperreals I think we might be able to safely assume there is an infinitesimal e such that we can write x = ae, y = be, and z = ce, where a, b, and c are real numbers that exist in certain ratios to each other. With non nilpotent infinitesimals, there are infinitesimal hyperreals that are sometimes a real number times e^2, this is concerning to me because one might accidentally create an infinite hyperreal by taking certain quotients using different kinds of the hyperreal infinitesimals, and not be able to compute the standard part function.

In the bese scenario, this means that xy/z would end up being (xe)(ye)/ze which is xye, which is still an infinitesimal, the finite real value for a function that is a real number + an infinitesimal should be the real number if I'm not mistaken.

The concern I have is that an infinite hyperreal could sneak into a standard part function, and be considered arbitrarily close to zero, when it's actually arbitrarily close to infinity.
 
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  • #7
Chenkel said:
My solution:

d(uv)/dx = st(((u+du)(v+dv) - uv)/dx) = st(((uv) + (u)(dv)+(v)(du)+(du)(dv)-(uv))/dx) = st((u)(dv/dx)+(v)(du/dx) + ((du)(dv)/dx)) = (u)(dv/dx)+(v)(du/dx)
Usually we derive it as follows.
[tex]\frac{d\ uv}{dx}=\lim_{\triangle x \rightarrow 0}\frac{(u+\triangle u)(v+ \triangle v)-uv}{\triangle x}=\lim_{\triangle x \rightarrow 0}[u\frac{\triangle v}{\triangle x}+v\frac{\triangle u}{\triangle x}+\frac{\triangle u \triangle v}{\triangle x}][/tex]
[tex]\lim_{\triangle x \rightarrow 0}u\frac{\triangle v}{\triangle x}=u\frac{dv}{dx}[/tex]
[tex]\lim_{\triangle x \rightarrow 0}v\frac{\triangle u}{\triangle x}=v\frac{du}{dx}[/tex]
[tex]\lim_{\triangle x \rightarrow 0}\frac{\triangle u \triangle v}{\triangle x}=0[/tex]
 
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1. What is the formula for finding the derivative of uv with respect to x using the st function?

The formula for finding the derivative of uv with respect to x using the st function is d/dx(uv) = u(dv/dx) + v(du/dx), where u and v are functions of x.

2. How do you apply the st function to find the derivative of uv with respect to x?

To apply the st function, you simply plug in the functions u and v into the formula d/dx(uv) = u(dv/dx) + v(du/dx) and then take the derivatives of u and v with respect to x.

3. What does the st function stand for in this context?

In this context, the st function stands for the product rule of derivatives, which is used to find the derivative of a product of two functions.

4. Can you provide an example of using the st function to find the derivative of uv with respect to x?

Sure, let's say we have the functions u(x) = 2x and v(x) = x^2. To find the derivative of uv with respect to x, we would use the formula d/dx(uv) = u(dv/dx) + v(du/dx) and plug in the values for u and v. This would give us d/dx(2x * x^2) = 2x(2x) + x^2(2) = 4x^2 + 2x^2 = 6x^2.

5. Are there any special cases to consider when using the st function to find the derivative of uv with respect to x?

Yes, there are a few special cases to consider. For example, if one of the functions u or v is a constant, its derivative will be 0 and can be ignored in the formula. Also, if one of the functions is a power of x, you can use the power rule instead of the st function. Additionally, if the functions are more complex, you may need to use the chain rule in addition to the st function.

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