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valina [46]
3 years ago
12

A lawyer drives from her home, located 3 miles east and 5 miles north of the town courthouse, to her office, located 6 miles wes

t and 7 miles south of the courthouse.
Find the distance between the lawyer's home and her office.
miles.
The distance between her home and her office is
(Type an integer or a decimal.)
Mathematics
1 answer:
vekshin13 years ago
5 0
This is more of a physics question. Basically you have to separate all the given values into x- and y- coordinates. You are given the home (3 miles East)(5 miles up), so the 3 miles would be on the x axis and the 6 miles on the 6 axis

Given that, you can determine the total displacement of the house from the town using Pythagorean’s theorem. You then do the same with the office and then add your 2 resultant answers together.

The home will be square root 34 miles away from the town, and the office will be square root 85 miles. Adding these together, the answer would be

15,05 miles
Meaning that the lawyers home is 15,05 miles from her office (home to office would be in a south western direction)
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Answer:

The Taylor series for sin^2(3 x) = - \sum_{n=1}^{\infty} \frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}, the first three non-zero terms are 9x^{2} -27x^{4}+\frac{162}{5}x^{6} and the interval of convergence is ( -\infty, \infty )

Step-by-step explanation:

<u>These are the steps to find the Taylor series for the function</u> sin^2(3 x)

  1. Use the trigonometric identity:

sin^{2}(x)=\frac{1}{2}*(1-cos(2x))\\ sin^{2}(3x)=\frac{1}{2}*(1-cos(2(3x)))\\ sin^{2}(3x)=\frac{1}{2}*(1-cos(6x))

   2. The Taylor series of cos(x)

cos(y) = \sum_{n=0}^{\infty}\frac{-1^{n}y^{2n}}{(2n)!}

Substituting y=6x we have:

cos(6x) = \sum_{n=0}^{\infty}\frac{-1^{n}6^{2n}x^{2n}}{(2n)!}

   3. Find the Taylor series for sin^2(3x)

sin^{2}(3x)=\frac{1}{2}*(1-cos(6x)) (1)

cos(6x) = \sum_{n=0}^{\infty}\frac{-1^{n}6^{2n}x^{2n}}{(2n)!} (2)

Substituting (2) in (1) we have:

\frac{1}{2} (1-\sum_{n=0}^{\infty}\frac{-1^{n}6^{2n}x^{2n}}{(2n)!})\\ \frac{1}{2}-\frac{1}{2} \sum_{n=0}^{\infty}\frac{-1^{n}6^{2n}x^{2n}}{(2n)!}

Bring the factor \frac{1}{2} inside the sum

\frac{6^{2n}}{2}=9^{n}2^{2n-1} \\ (-1^{n})(9^{n})=(-9^{n} )

\frac{1}{2}-\sum_{n=0}^{\infty}\frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}

Extract the term for n=0 from the sum:

\frac{1}{2}-\sum_{n=0}^{0}\frac{-9^{0}2^{2*0-1}x^{2*0}}{(2*0)!}-\sum_{n=1}^{\infty}\frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}\\ \frac{1}{2} -\frac{1}{2} -\sum_{n=1}^{\infty}\frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}\\ 0-\sum_{n=1}^{\infty}\frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}\\ sin^{2}(3x)=-\sum_{n=1}^{\infty}\frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}

<u>To find the first three non-zero terms you need to replace n=3 into the sum</u>

sin^{2}(3x)=\sum_{n=1}^{\infty}\frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}\\ \sum_{n=1}^{3}\frac{-9^{3}2^{2*3-1}x^{2*3}}{(2*3)!} = 9x^{2} -27x^{4}+\frac{162}{5}x^{6}

<u>To find the interval on which the series converges you need to use the Ratio Test that says</u>

For the power series centered at x=a

P(x)=C_{0}+C_{1}(x-a)+C_{2}(x-a)^{2}+...+ C_{n}(x-a)^{n}+...,

suppose that \lim_{n \to \infty} |\frac{C_{n}}{C_{n+1}}| = R.. Then

  • If R=\infty, the the series converges for all x
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  • If R=0, the the series converges only for x=a

So we need to evaluate this limit:

\lim_{n \to \infty} |\frac{\frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}}{\frac{-9^{n+1}2^{2*(n+1)-1}x^{2*(n+1)}}{(2*(2n+1))!}} |

Simplifying we have:

\lim_{n \to \infty} |-\frac{(n+1)(2n+1)}{18x^{2} } |

Next we need to evaluate the limit

\lim_{n \to \infty} |-\frac{(n+1)(2n+1)}{18x^{2} } |\\ \frac{1}{18x^{2} } \lim_{n \to \infty} |-(n+1)(2n+1)}|}

-(n+1)(2n+1) is negative when n -> ∞. Therefore |-(n+1)(2n+1)}|=2n^{2}+3n+1

You can use this infinity property \lim_{x \to \infty} (ax^{n}+...+bx+c) = \infty when a>0 and n is even. So

\lim_{n \to \infty} |-\frac{(n+1)(2n+1)}{18x^{2} } | \\ \frac{1}{18x^{2}} \lim_{n \to \infty} 2n^{2}+3n+1=\infty

Because this limit is ∞ the radius of converge is ∞ and the interval of converge is ( -\infty, \infty ).

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