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marissa [1.9K]
3 years ago
11

Determine the amplitude or period as requested. Period of y = - 1/3 * sin 2x

Mathematics
1 answer:
Sedbober [7]3 years ago
3 0

Answer:

Period = \pi

Step-by-step explanation:

Given

y = -\frac{1}{3} * \sin(2x)

Required

The period

A sine function is represented as:

y =A \sin(Bx + C) + D

Where

Period = \frac{2\pi}{B}

By comparing:

y =A \sin(Bx + C) + D and y = -\frac{1}{3} * \sin(2x)

B = 2

So, we have:

Period = \frac{2\pi}{2}

Period = \pi

Hence, the period of the function is \pi

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Solve using Fourier series.
Olin [163]
With 2L=\pi, the Fourier series expansion of f(x) is

\displaystyle f(x)\sim\frac{a_0}2+\sum_{n\ge1}a_n\cos\dfrac{n\pi x}L+\sum_{n\ge1}b_n\sin\dfrac{n\pi x}L
\displaystyle f(x)\sim\frac{a_0}2+\sum_{n\ge1}a_n\cos2nx+\sum_{n\ge1}b_n\sin2nx

where the coefficients are obtained by computing

\displaystyle a_0=\frac1L\int_0^{2L}f(x)\,\mathrm dx
\displaystyle a_0=\frac2\pi\int_0^\pi f(x)\,\mathrm dx

\displaystyle a_n=\frac1L\int_0^{2L}f(x)\cos\dfrac{n\pi x}L\,\mathrm dx
\displaystyle a_n=\frac2\pi\int_0^\pi f(x)\cos2nx\,\mathrm dx

\displaystyle b_n=\frac1L\int_0^{2L}f(x)\sin\dfrac{n\pi x}L\,\mathrm dx
\displaystyle b_n=\frac2\pi\int_0^\pi f(x)\sin2nx\,\mathrm dx

You should end up with

a_0=0
a_n=0
(both due to the fact that f(x) is odd)
b_n=\dfrac1{3n}\left(2-\cos\dfrac{2n\pi}3-\cos\dfrac{4n\pi}3\right)

Now the problem is that this expansion does not match the given one. As a matter of fact, since f(x) is odd, there is no cosine series. So I'm starting to think this question is missing some initial details.

One possibility is that you're actually supposed to use the even extension of f(x), which is to say we're actually considering the function

\varphi(x)=\begin{cases}\frac\pi3&\text{for }|x|\le\frac\pi3\\0&\text{for }\frac\pi3

and enforcing a period of 2L=2\pi. Now, you should find that

\varphi(x)\sim\dfrac2{\sqrt3}\left(\cos x-\dfrac{\cos5x}5+\dfrac{\cos7x}7-\dfrac{\cos11x}{11}+\cdots\right)

The value of the sum can then be verified by choosing x=0, which gives

\varphi(0)=\dfrac\pi3=\dfrac2{\sqrt3}\left(1-\dfrac15+\dfrac17-\dfrac1{11}+\cdots\right)
\implies\dfrac\pi{2\sqrt3}=1-\dfrac15+\dfrac17-\dfrac1{11}+\cdots

as required.
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3 years ago
Can u help it simple geometry
Paul [167]

Answer:

yes

Step-by-step explanation:

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Korolek [52]
5.12-1.30= your answer
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3 years ago
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Should we simplify or solve for m...?

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What is -11+6x+25-3x
evablogger [386]

Answer:

3 x + 14

Step-by-step explanation:

Simplify the following:

6 x - 3 x - 11 + 25

Grouping like terms, 6 x - 3 x - 11 + 25 = (6 x - 3 x) + (-11 + 25):

(6 x - 3 x) + (-11 + 25)

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Answer:  3 x + 14

8 0
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