Please remind me (it's been a really long time since I've dealt with this stuff) what ##f \in L_2[0,1]## means. One of my topology books used ##\mathcal l^2##, but this had to do with Hilbert spaces and sequences that were square-integrable.joshmccraney said:Problem Statement: Does there exist a function ##f \in L_2[0,1]## such that for all natural ##n##, we have $$\int_0^1 x^n f(x) \, dx = 1$$
Relevant Equations: Nothing comes to mind.
At first I was thinking about using the dirac delta function ##\delta(x-1)##, but then I recalled ##\delta \notin L_2[0,1]##. Any ideas? I'm thinking no such function exists.
A function ##f## is in ##L_2[0,1]## if ##\int_0^1 |f|^2\, dx## exists.Mark44 said:Please remind me (it's been a really long time since I've dealt with this stuff) what ##f \in L_2[0,1]## means.
The Cauchy-Schwartz inequality for this case would be $$\int_0^1| f x^n| \, dx \leq \int_0^1|f|^2\, dx\int_0^1|x^n|^2\, dx = \frac{1}{2n+1}\int_0^1|f|^2\, dx$$ and if we assume such a function ##f## exists, then we have $$1 \leq \frac{1}{2n+1}\int_0^1|f|^2\, dx.$$LCKurtz said:Have you thought about using the Cauchy-Schwartz inequality to disprove it?
But isn't it possible ##n\to\infty \implies f \to \infty##, and so balances? Like if ##f = n##, or would this imply ##f \notin L_2##?LCKurtz said:What happens if n is large?
No. You are using an indirect argument, supposing there such an ##f##. It doesn't get to change and it has a finite ##\|f\|##.joshmccraney said:But isn't it possible ##n\to\infty \implies f \to \infty##, and so balances? Like if ##f = n##, or would this imply ##f \notin L_2##?
An integral polynomial is a polynomial with coefficients that are integers. This means that all the terms in the polynomial have whole number coefficients. For example, 2x^3 + 5x^2 + 3x + 1 is an integral polynomial, while 2.5x^2 + 3x + 1/2 is not.
To find the integral of a polynomial, you can use the power rule of integration. This means that you raise the power of each term by 1 and then divide by the new power. For example, the integral of 3x^2 would be (3x^3)/3 = x^3. You can also use the linearity property of integration to solve for the integral of more complex polynomials.
The integral of a polynomial is used to calculate the area under the curve of the polynomial. This can be useful in many real-world applications, such as calculating the work done by a variable force or finding the average value of a function over a specific interval.
Yes, the Fundamental Theorem of Calculus states that the integral of a function can be used to find the derivative of that function. This means that if you know the integral of a polynomial, you can find the derivative of that polynomial.
Yes, there are a few special cases when finding the integral of a polynomial. One example is when the polynomial has a constant term. In this case, you need to add a constant of integration to your final answer. Another special case is when the polynomial has a term with a negative power. In this case, you need to use the power rule for integration with a negative power and then add a constant of integration.