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muminat
4 years ago
8

Determine the first three nonzero terms of the Taylor polynomial approximations for the given initial value problem.

ttps://tex.z-dn.net/?f=y%27%20%3D%209%20sin%28y%29%2B2%20e%5E%7Bx%7D%20" id="TexFormula1" title="y' = 9 sin(y)+2 e^{x} " alt="y' = 9 sin(y)+2 e^{x} " align="absmiddle" class="latex-formula"> and y(0)=0

Mathematics
1 answer:
pychu [463]4 years ago
6 0
y=\displaystyle\sum_{n\ge0}a_nx^n
\sin x=\displaystyle\sum_{n\ge0}\frac{(-1)^n}{(2n+1)!}x^{2n+1}
\implies\sin y=\displaystyle\sum_{n\ge0}\frac{(-1)^n}{(2n+1)!}\left(\sum_{m\ge0}a_mx^m\right)^{2n+1}

Note that with y(0)=0, the first sum reduces immediately (after taking out the first term in the sum) to a_0, which implies that a_0=0.

So we have

\displaystyle\sum_{n\ge1}na_nx^{n-1}=9\sum_{n\ge0}\frac{(-1)^n}{(2n+1)!}\left(\sum_{m\ge1}a_mx^m\right)^{2n+1}+2\sum_{n\ge0}\frac1{n!}x^n
a_1+2a_2x+3a_3x^2+4a_4x^3+\cdots=9\mathbf S+2\left(1+x+\frac12x^2+\frac16x^3+\cdots\right)

where we can expand \mathbf S as

\mathbf S=\displaystyle\sum_{n\ge0}\frac{(-1)^n}{(2n+1)!}(a_1x+a_2x^2+a_3x^3+\cdots)^{2n+1}
=a_1x+a_2x^2+a_3x^3+\cdots
\,\,-\dfrac1{3!}(a_1x+a_2x^2+a_3x^3+\cdots)^3+\cdots
=a_1x+a_2x^2+\left(a_3-\dfrac{{a_1}^3}{3!}\right)x^3+\cdots

Dropping all terms with order greater than 2 (because this is enough to generate 3 nonzero terms) reduces the equation to

a_1+2a_2x+3a_3x^2=9\left(a_1x+a_2x^2\right)x^3\right)+2\left(1+x+\dfrac12x^2\right)
a_1+2a_2x+3a_3x^2=2+\left(9a_1+2\right)x+\left(9a_2+1\right)x^2
\implies\begin{cases}a_1=2\\2a_2=9a_1+2\\3a_3=9a_2+1\end{cases}\implies a_1=2,a_2=10,a_3=\dfrac{91}3

so that the Taylor polynomial approximating the solution y is

y(x)\approx2x+10x^2+\dfrac{91}3x^3

Just to demonstrate that the result is reasonable, I've attached a plot of an approximate solution with higher accuracy (blue) and the one we found (orange) over the interval -\dfrac\pi6.

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Zinaida [17]

Answer:

6

Step-by-step explanation:

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4 0
3 years ago
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On december 31, there were 40 units remaining in ending inventory. these 40 units consisted of 5 from january, 6 from february,
worty [1.4K]

Answer:

$5,675

Step-by-step explanation:

Answer:

Step-by-step explanation:

Calculation for the cost of the ending inventory using specific identification method.

Using this formula

Cost of ending inventory =

(January units ×January cost) +( February units ×February cost) + (May units × May cost) + (September units ×September cost) + (November units × November cost)

Cost of ending inventory =

January 5 units ×$116=$580

February 6 units ×$127=$762

May 10 units ×$139=$1,390

September 4 units ×$147=$588

November 15 units×$157=$2,355

Total =$5,675

Therefore the ending inventory using the specific identification method will be $5,675

5 0
3 years ago
Give the equation of a line that goes through the point ( − 21 , 2 ) and is perpendicular to the line 7 x − 4 y = − 12 . Give yo
nlexa [21]

Given:

Equation of line 7x-4y=-12.

To find:

The equation of line  that goes through the point ( − 21 , 2 ) and is perpendicular to the given line.

Solution:

The given equation of line can be written as

7x-4y+12=0

Slope of line is

\text{Slope}=-\dfrac{\text{Coefficient of x}}{\text{Coefficient of y}}

m_1=-\dfrac{7}{(-4)}

m_1=\dfrac{7}{4}

Product of slopes of two perpendicular lines is -1. So, slope of perpendicular line is

m_1m_2=-1

m_2=-\dfrac{1}{m_1}

m_2=-\dfrac{4}{7}           [\because m_1=\dfrac{7}{4}]

Now, the slope of perpendicular line is m_2=\dfrac{4}{7} and it goes through (-21,2). So, the equation of line is

y-y_1=m_2(x-x_1)

y-2=-\dfrac{4}{7}(x-(-21))

y-2=-\dfrac{4}{7}x-\dfrac{4}{7}(21)

y-2=-\dfrac{4}{7}x-12

y=-\dfrac{4}{7}x-12+2

y=-\dfrac{4}{7}x-10

Therefore, the required equation in slope intercept form is y=-\dfrac{4}{7}x-10.

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4 years ago
Find the sum of the measures of the interior angles of a 21-gon.
aleksandrvk [35]
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7 0
3 years ago
Please help!!!!!
jasenka [17]

Answer:

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