Prove this conjecture
1 answer:
Answer:
See explanation
Step-by-step explanation:
Consider the natural number
If is the natural number, then
are next two natural numbers, so numbers
and
are three consecutive numbers.
Given three consecutive natural numbers, one of them is always even number and one of them is always multiple of 3, so the product
is divisible by 6.
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Answer:
The Fourier series of f(x) converges to 3 at the points x= π+2kπ, where k is an integer.
Step-by-step explanation:
First, recall that the function f(x) is extended 2π periodic to the whole real line, in order to obtain a valid Fourier expansion. Remember that a Fourier series is formed by a sines and cosines, which are 2π-periodic.
So, the 2π-periodic expansion of f(x) is discontinuous at the points π+2kπ, in particular π and -π. Check the attached figure to a better understanding.
Now, the Dirichlet theorem on the convergence of a Fourier series tells us that the series converges to the function at the points of continuity, and at points of discontinuity the sum of the series is
.
Here we understand the notation and
as
and
.
In this particular case
.
For the limit , with
recall that our function is 2π-periodic, so the values of f near π, with x>π are the same when x is near -π and x>-π. Again, check the attached figure. So,
.
Thus,
.
Note: In the attached figure we only have drawn three repetitions of the 2π-periodic extension of , recall that the extension is <em>ad infinitum</em>. Also, the points drawn in the dotted lines are the sum of the series at the points of discontinuity.
