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ludmilkaskok [199]
3 years ago
5

Can I get help with finding the Fourier cosine series of F(x) = x - x^2

Mathematics
1 answer:
trapecia [35]3 years ago
4 0
Assuming you want the cosine series expansion over an arbitrary symmetric interval [-L,L], L\neq0, the cosine series is given by

f_C(x)=\dfrac{a_0}2+\displaystyle\sum_{n\ge1}a_n\cos nx

You have

a_0=\displaystyle\frac1L\int_{-L}^Lf(x)\,\mathrm dx
a_0=\dfrac1L\left(\dfrac{x^2}2-\dfrac{x^3}3\right)\bigg|_{x=-L}^{x=L}
a_0=\dfrac1L\left(\left(\dfrac{L^2}2-\dfrac{L^3}3\right)-\left(\dfrac{(-L)^2}2-\dfrac{(-L)^3}3\right)\right)
a_0=-\dfrac{2L^2}3

a_n=\displaystyle\frac1L\int_{-L}^Lf(x)\cos nx\,\mathrm dx

Two successive rounds of integration by parts (I leave the details to you) gives an antiderivative of

\displaystyle\int(x-x^2)\cos nx\,\mathrm dx=\frac{(1-2x)\cos nx}{n^2}-\dfrac{(2+n^2x-n^2x^2)\sin nx}{n^3}

and so

a_n=-\dfrac{4L\cos nL}{n^2}+\dfrac{(4-2n^2L^2)\sin nL}{n^3}

So the cosine series for f(x) periodic over an interval [-L,L] is

f_C(x)=-\dfrac{L^2}3+\displaystyle\sum_{n\ge1}\left(-\dfrac{4L\cos nL}{n^2L}+\dfrac{(4-2n^2L^2)\sin nL}{n^3L}\right)\cos nx
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Under the last-in, first-out (LIFO) inventory valuation method, a price index for inventory must be established for tax purposes
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Step-by-step explanation:

First of all we would have to perform the following table:

                                             

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Therefore, using laspeyres index number, we calculate the following:

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<u>ANSWER:</u>

The midpoint of AB is M(-5,1). The coordinates of B are (-6, 7)

<u>SOLUTION: </u>

Given, the midpoint of AB is M(-5,1).  

The coordinates of A are (-4,-5),  

We need to find the coordinates of B.

We know that, mid-point formula for two points A(x_{1}, y_{1}) and B (x_{1}, y_{2}) is given by

M\left(x_{3}, y_{3}\right)=\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)

Here, in our problem, \mathrm{x}_{3}=-5, \mathrm{y}_{3}=1, \mathrm{x}_{1}=-4 \text { and } \mathrm{y}_{1}=-5

Now, on substituting values in midpoint formula, we get

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On comparing, with the formula,

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