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4 years ago

I need help w this question

Mathematics

1 answer:

ollegr [7]4 years ago

3 0

Answer:

360 degrees

Step-by-step explanation:

Remember that triangles have a sum of 180 degrees, and that quadrilaterals have 360 degrees.

You can either use that part, or actually solve for it.

The equation for the sum of the interior angles is:

$sum=180(n-2)$ [which "n" is the number of sides]

So, if we add in the number of sides to the equation, it changes to:

$180(4-2)=180(2)$ , which equals to 360 degrees.

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Find two linearly independent power series solutions about the point x0 = 0 of

aksik [14]

Assume a solution of the form

$y=\displaystyle\sum_{n\ge0}a_nx^n$

with derivatives

$y'=\displaystyle\sum_{n\ge0}(n+1)a_{n+1}x^n$

$y''=\displaystyle\sum_{n\ge0}(n+2)(n+1)a_{n+2}x^n$

Substituting into the ODE, which appears to be

$(x^2-4)y''+3xy'+y=0,$

gives

$\displaystyle\sum_{n\ge0}\bigg((n+2)(n+1)a_{n+2}x^{n+2}-4(n+2)(n+1)a_{n+2}x^n+3(n+1)a_{n+1}x^{n+1}+a_nx^n\bigg)=0$

$\displaystyle\sum_{n\ge2}n(n-1)a_nx^n-4\sum_{n\ge0}(n+2)(n+1)a_{n+2}x^n+3\sum_{n\ge1}na_nx^n+\sum_{n\g0}a_nx^n=0$

$(a_0-8a_2)+(4a_1-24a_3)x+\displaystyle\sum_{n\ge2}\bigg[(n+1)^2a_n-4(n+2)(n+1)a_{n+2}\bigg]x^n=0$

which gives the recurrence for the coefficients $a_n$ ,

$\begin{cases}a_0=a_0\\a_1=a_1\\4(n+2)a_{n+2}=(n+1)a_n&\text{for }n\ge0\end{cases}$

There's dependency between coefficients that are 2 indices apart, so we consider 2 cases.

If $n=2k$ , where $k\ge0$ is an integer, then

$k=0\implies n=0\implies a_0=a_0$

$k=1\implies n=2\implies a_2=\dfrac1{4\cdot2}a_0=\dfrac2{4\cdot2^2}a_0=\dfrac{2!}{2^4}a_0$

$k=2\implies n=4\implies a_4=\dfrac3{4\cdot4}a_2=\dfrac3{4^2\cdot4\cdot2}a_0=\dfrac{4!}{2^8(2!)^2}a_0$

$k=3\implies n=6\implies a_6=\dfrac5{4\cdot6}a_4=\dfrac{5\cdot3}{4^3\cdot6\cdot4\cdot2}a_0=\dfrac{6!}{2^{12}(3!)^2}a_0$

and so on, with the general pattern

$a_{2k}=\dfrac{(2k)!}{2^{4k}(k!)^2}a_0$

If $n=2k+1$ , then

$k=0\implies n=1\implies a_1=a_1$

$k=1\implies n=3\implies a_3=\dfrac2{4\cdot3}a_1=\dfrac{2^2}{2^2\cdot3\cdot2}a_1=\dfrac1{(3!)^2}a_1$

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$k=3\implies n=7\implies a_7=\dfrac6{4\cdot7}a_5=\dfrac{6\cdot4\cdot2}{4^3\cdot7\cdot5\cdot3}a_1=\dfrac{(3!)^2}{7!}a_1$

and so on, with

$a_{2k+1}=\dfrac{(k!)^2}{(2k+1)!}a_1$

Then the two independent solutions to the ODE are

$\boxed{y_1(x)=\displaystyle a_0\sum_{k\ge0}\frac{(2k)!}{2^{4k}(k!)^2}x^{2k}}$

and

$\boxed{y_2(x)=\displaystyle a_1\sum_{k\ge0}\frac{(k!)^2}{(2k+1)!}x^{2k+1}}$

By the ratio test, both series converge for $|x|$ , which also can be deduced from the fact that $x=\pm2$ are singular points for this ODE.

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