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Ahmed Mehedi
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Does $$\int_{t=0}^{\infty}f(t)dt=\int_{t=0}^{\infty}g(t)dt$$ imply $$f(t)=g(t)$$ ?
How could that possibly be true?Ahmed Mehedi said:Does $$\int_{t=0}^{\infty}f(t)dt=\int_{t=0}^{\infty}g(t)dt$$ imply $$f(t)=g(t)$$ ?
etotheipi said:I would suspect that if ##\int_{a}^{b}f(t)dt=\int_{a}^{b}g(t)dt## for all possible ##a, b##, then you would be able to say ##f(t) = g(t)##. But that's quite different to having fixed limits.
etotheipi said:I would suspect that if ##\int_{a}^{b}f(t)dt=\int_{a}^{b}g(t)dt## for all possible ##a, b##, then you would be able to say ##f(t) = g(t)##. But that's quite different to having fixed limits.
PeroK said:Are you really asking that if:
$$\int_{t=0}^{\infty}f(t)dt= 0$$
Then ##f(t) = 0## everywhere?
Ahmed Mehedi said:I am very sorry that I can not make my message clear. To be specific my question goes as follows:
Let us consider the following Lagrangian in the context of an optimization problem:
$$L=B\int_{t=0}^{\infty}e^{-\beta t}\frac{c(t)^{1-\theta}}{1-\theta}dt+\lambda \left[ k(0)+\int_{t=0}^{\infty} e^{-R(t)}e^{(n+g)t}w(t)dt - \int_{t=0}^{\infty} e^{-R(t)}e^{(n+g)t}c(t)dt \right]$$
After taking the first partial derivative of the above Lagrangian with respect to c(t) (and setting it to zero as first order optimality condition) it is written that
$$Be^{-\beta t} c(t)^{-\theta}-\lambda e^{-R(t)}e^{(n+g)t}=0$$
It seems that they take the first partial derivative of the Lagrangian with respect to c(t) and simply ignore the dt from both sides.
How the second line follows from the first line? They said it can be proved using calculus of variation. But, how can I prove it?
dRic2 said:That's a functional derivative, not a partial derivative. It's an other story.
If you want a quick recipe, a functional derivative like ##\frac {\delta L} {\delta g}## where ##L =\int h[g]## can be evaluated by trowing away the integral and calculating ##\frac {\partial h[g]} {\partial g}##. Ok don't hate me for this comment.
etotheipi said:LOL this question took a turn...
I don't know if I can help from here on, I've only ever skimmed through the first chapter of a calculus of variations textbook.
Sorry, I'm not very familiar with functional derivatives. I just worked out some tricks to evaluate them in case of need. All I know comes from pag 54-56 from Lanczos' book on analytical mechanics and from 5 pages of notes by a professor of mine.Ahmed Mehedi said:Can you provide me any good quick read regarding functional derivatives?
Adesh said:Given
$$\int_{0}^{\infty} f(t) dt = \int_{0}^{\infty} g(t) dt$$ the most we can get is :
$$
\int_{0}^{\infty} f(t) dt - \int_{0}^{\infty} g(t) dt = 0 $$
$$\lim_{x\to \infty} \int_{0}^{x} \left[f(t) -g(t) \right] dt = 0 $$
$$\lim_{x \to \infty} \frac{d}{dx} \int_{0}^{x} \left[f(t) - g(t) \right] = 0$$
$$\lim_{x \to \infty} f(x) - g(x) = 0$$
That is to say, functions ##f## and ##g## converge to each other as they approach to infinity.
Adesh said:Given
$$\int_{0}^{\infty} f(t) dt = \int_{0}^{\infty} g(t) dt$$ the most we can get is :
$$
\int_{0}^{\infty} f(t) dt - \int_{0}^{\infty} g(t) dt = 0 $$
$$\lim_{x\to \infty} \int_{0}^{x} \left[f(t) -g(t) \right] dt = 0 $$
$$\lim_{x \to \infty} \frac{d}{dx} \int_{0}^{x} \left[f(t) - g(t) \right] = 0$$
$$\lim_{x \to \infty} f(x) - g(x) = 0$$
That is to say, functions ##f## and ##g## converge to each other as they approach to infinity.
I assumed the functions to be continuous.Math_QED said:This is false. Consider ##f=0## and $$g(x) = \begin{cases}1 \quad x \in \mathbb{N} \\ 0 \quad x \notin \mathbb{N}\end{cases}$$
Then $$\int_0^\infty f = 0 = \int_0^\infty g$$ yet $$\lim_{x \to \infty}[ f(x)-g(x)] = -\lim_{x \to \infty} g(x) $$ does not exist.
The flaw happens when you introduce the derivative.
Adesh said:I assumed the functions to be continuous.
It's still false. There are continuous functions that do not converge to ##0## as ##t \rightarrow \infty## yet the integral exists.Adesh said:I assumed the functions to be continuous.
We can do the switching when they are monotone.Math_QED said:Even then you must justify switching the limits involved, which you didn't. I'm pretty sure the statement is even false for continuous functions.
Adesh said:We can do the switching when they are monotone.
I didn’t say they converge to zero, I said they get closer and closer to each other as ##x## goes to ##\infty##.PeroK said:There are continuous functions that do not converge to 0 as t→∞ yet the integral exists.
Which is false. The difference of the functions need not converge to ##0##.Adesh said:I didn’t say they converge to zero, I said they get closer and closer to each other as ##x## goes to ##\infty##.
In this case they are coming out be zero, .PeroK said:Which is false. The difference of the functions need not converge to ##0##.
Adesh said:@Infrared Did you study from Rudin by yourself? I mean self-study or was your Univeristy good enough from where you can learned it? (Because as far as universities that I know, you cannot learn subjects which require time, like analysis, because teachers are not very interested in teaching).
People whom I know (in real life) learned by themselves, teachers only graded them.Math_QED said:so I'm not sure where your claim comes from.
Adesh said:People whom I know (in real life) learned by themselves, teachers only graded them.
Then what does it mean when so many people say “Rudin is not good for self-study” ?Math_QED said:My opinion is that most of mathematics is self-study. The professor is just there to motivate the concepts and give a first exposure, or to ask questions to.
Adesh said:Then what does it mean when so many people say “Rudin is not good for self-study” ?
The equality of integrals means that two integrals have the same value, while equality of integrands means that the functions being integrated are equal. In other words, equality of integrals relates to the result, while equality of integrands relates to the process.
Yes, it is possible for two integrals with different integrands to have the same value. This is because the value of an integral is dependent on the limits of integration and not just the integrand itself.
To determine if two integrands are equal, you can use algebraic manipulation or substitution to simplify the expressions and see if they are identical. You can also graph the two functions to visually compare them.
No, equality of integrands is not necessary for equality of integrals. As long as the limits of integration and the function being integrated are the same, the integrals will have the same value.
Understanding the difference between equality of integrals and equality of integrands is important in calculus and other areas of mathematics. It can also be useful in physics and engineering, where integrals are used to solve problems related to motion, work, and other physical quantities.