What is the classification for this polynomial? u^2 v^3 w
1 answer:
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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.
Answer:
<u>Algebraically</u>
0.5x-7= √(-5x+29)
(0.5x-7)² = -5x+29
(0.5x)² - 2*0.5x*7 + 7² = -5x+29
0.25x² - 7x + 49 + 5x - 29 = 0
0.25x² - 2x + 20 = 0
Its discriminant is:
b² - 4*a*c
(-2)² - 4*0.25*20
-16 < 0
Therefore, the equation has no solutions.
<u>Graphically</u>
In the picture attached, the plots of the functions are shown. We can see that they don't intersect each other, and that is why there is no solution.
