Nuomegamath
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A maths blog for Advanced Computing, a blog on Computer Algebra, Mathematics, and Systems Engineering.
Nuomegamath
1w ago
That is,
And so,
By (1),
See also The Artistry of Mastery.
Exercise-1 Completing the ..read more
Nuomegamath
1w ago
Introducing Feynman’s Integral Method
Michael Xue
Vroom Laboratory
http://www.vroomlab.com
Indiana Section MAA Spring 2024 MEETING
Marian University – Indianapolis
April 5-6, 2024
“The first principle is that you must not fool yourself and you are the easiest person to fool.”
— Richard Feynman
Feynman’s Integral Method
Leibniz : “”
Feynman: “To evaluate hard differentiate under the integral sign!”
Two algorithms implementing Feynman’s Integral method
Algorithm #1 Reducing
Algorithm #2 Generating from a known definite integral
Example-1
Evaluate
Example-2
Evaluate
Example-3
Evalua ..read more
Nuomegamath
2w ago
Feynman’s trick, also known as parameter differentiation under the integral sign, is a powerful mathematical technique used for simplifying complex integrals. By introducing an auxiliary parameter into the integral and then differentiating with respect to that parameter, Feynman’s method often transforms difficult integrals into more manageable forms. It is a versatile tool in physics and mathematics, making the evaluation of certain integrals straightforward where traditional methods might falter.
Introducing Feynman’s Integral Method
Michael Xue
Vroom Laboratory
http://www.vroomlab.com
Indi ..read more
Nuomegamath
2w ago
In Deriving the Extraordinary Euler Sum , we derived one of Euler’s most celebrated results:
Now, we aim to provide a rigorous proof of this statement.
First, we expressed the partial sum of the left-hand side as follows:
This splits the partial sum into two parts: one involving the squares of even numbers and the other involving the squares of odd numbers. Simplifying the even part gives us:
Rearranging terms on one side, we obtain:
Since converges to (see My Shot at Harmonic Series)
converges to
to prove (1), it suffices to demonstrate that
or equivalently,
Let and we have ..read more
Nuomegamath
1M ago
Prove:
For any positive number , let
Since is a monotonic increasing function:
we have
That is,
or
By the definition of :
we also have
Combining (1) and (2) gives
This implies that
In other words,
See also The Sandwich Theorem for Functions 2 ..read more
Nuomegamath
1M ago
Evaluate
This integral is known as the Dirichlet Integral, named in honor of the esteemed German mathematician Peter Dirichlet. Due to the absence of an elementary antiderivative for the integrand, its evaluation by applying the Newton-Leibniz rule renders an impasse. However, the Feynman’s integral technique offers a solution.
The even nature of the function implies that
Let’s consider
and define
We can differentiate with respect to
Hence, we find
Integrating with respect to from to
gives
Since
and
,
we arrive at
It follows that by (*):
Show that
From the inequality
and ..read more
Nuomegamath
1M ago
For given functions and
The given condition gives
and
It means
Since we have
That is,
Or,
And so,
See also Sandwich Theorems and Their Proofs ..read more
Nuomegamath
1M ago
Show that
This integral is renowned in mathematics as the Gaussian integral. Its evaluation poses a challenge due to the absence of an elementary antiderivative expressed in standard functions. Conventionally, one method involves “squaring the integral” and subsequently interpreting the resulting double integral in polar coordinates. However, an alternative approach, which we present here, employs Feynman’s integral technique.
The even nature of the function implies that
Let’s consider
and define
We can differentiate with respect to
Given a differentiable function on with derivat ..read more
Nuomegamath
1M ago
Given is continuous on and satisfies the following conditions:
[1]
[2]
[3] has continuous derivative on
Prove:
The given premises
is a continuous function on
ensures the existence of the definite integral and the antiderivative of
Denoting the antiderivative of as we obtain
We also deduce from [3] that
is continuous on
Combining [3], [1] and (1),
is continuous on
Additionally, as per [3],
is continuous on
Together, (4) and (5) give
is continuous on
Consequently, exists as well.
Now let’s examine , where . We have
is the antiderivative of
This theorem serves as the ..read more