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3 years ago

Plsplsplsplsplsplspls help!!!!!!!!

Mathematics

1 answer:

True [87]3 years ago

5 0

Answer:

1. My estamite, close to -1

2. 1.5

3. Remote interior angles

4. 84

5. 60 degrees

Step-by-step explanation:

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8)Four added to twice a number yields. Find the number.

Murrr4er [49]

Answer:

X=-2

Step-by-step explanation:

Let no =X

So 2x+4=0

2x=-4

X=-4/2

X=-2

So the answer is -2

5 0

3 years ago

18. The sum of 3 integers is 137. The 2nd number is 4 more than two times the 1st

hichkok12 [17]

Answer:

The second number is four more than, two times the first number. The third ... Start by writing an expression for each of the three numbers. They are all formed from the first number, so let's call the first number x . Let the first number be x. The second number is 2x+4. The third number is 3x−5. We are told .

Step-by-step explanation:

8 0

4 years ago

In a week, 20 hens laid 60 eggs. What is the unit rate for eggs per hen?

Serggg [28]

That’s a fat 100 girls

5 0

3 years ago

192 = 3w^2 solve using square root

exis [7]

Answer:

W=8

Step-by-step explanation:

First you divide both sides by 3 then you square root both sides getting 8

192/3=3w^2/3

64=w^2

w=8

6 0

3 years ago

Taylor Series Questions!

riadik2000 [5.3K]

5.
$f(x)=\sin x\implies f(\pi)=0$
$f'(x)=\cos x\implies f'(\pi)=-1$
$f''(x)=-\sin x\implies f''(\pi)=0$
$f'''(x)=-\cos x\implies f'''(\pi)=1$

Clearly, each even-order derivative will vanish, and the terms that remain will alternate in sign, so the Taylor series is given by

$f(x)=-(x-\pi)+\dfrac{(x-\pi)^3}{3!}-\dfrac{(x-\pi)^5}{5!}+\cdots$
$f(x)=\displaystyle\sum_{n\ge0}\frac{(-1)^{n-1}(x-\pi)^{2n+1}}{(2n+1)!}$

Your answer is off by a sign - the source of this error is the fact that you used the series expansion centered at $x=0$ , not $x=\pi$ , and so the sign on each derivative at $x=\pi$ is opposite of what it should be. I'm sure you can figure out the radius of convergence from here.

- - -

6. Note that this is already a polynomial, so the Taylor series will strongly resemble this and will consist of a finite number of terms. You can get the series by evaluating the derivatives at the given point, or you can simply rewrite the polynomial in $x$ as a polynomial in $x-2$ .

$f(x)=x^6-x^4+2\implies f(2)=50$
$f'(x)=6x^5-4x^3\implies f'(2)=160$
$f''(x)=30x^4-12x^2\implies f''(2)=432$
$f'''(x)=120x^3-24x\implies f'''(2)=912$
$f^{(4)}(x)=360x^2-24\implies f^{(4)}(2)=1416$
$f^{(5)}(x)=720x\implies f^{(5)}(2)=1440$
$f^{(6)}(x)=720\implies f^{(6)}(2)=720$
$f^{(n\ge7)}(x)=0\implies f^{(n\ge7)}(2)=0$

$\implies f(x)=50+160(x-2)+216(x-2)^2+152(x-2)^3+59(x-2)^4+12(x-2)^5+(x-2)^6$

If you expand this, you will end up with $f(x)$ again, so the Taylor series must converge everywhere.

I'll outline the second method. The idea is to find coefficients so that the right hand side below matches the original polynomial:

$x^6-x^4+2=(x-2)^6+a_5(x-2)^5+a_4(x-2)^4+a_3(x-2)^3+a_2(x-2)^2+a_1(x-2)+a_0$

You would expand the right side, match up the coefficients for the same-power terms on the left, then solve the linear system that comes out of that. You would end up with the same result as with the standard derivative method, though perhaps more work than necessary.

- - -

7. It would help to write the square root as a rational power first:

$f(x)=\sqrt x=x^{1/2}\implies f(4)=2$
$f'(x)=\dfrac{(-1)^0}{2^1}x^{-1/2}\implies f'(4)=\dfrac1{2^2}$
$f''(x)=\dfrac{(-1)^1}{2^2}x^{-3/2}\implies f''(4)=-\dfrac1{2^5}$
$f'''(x)=\dfrac{(-1)^2(1\times3)}{2^3}x^{-5/2}\implies f'''(4)=\dfrac3{2^8}$
$f^{(4)}(x)=\dfrac{(-1)^3(1\times3\times5)}{2^4}x^{-7/2}\implies f^{(4)}(4)=-\dfrac{15}{2^{11}}$
$f^{(5)}(x)=\dfrac{(-1)^4(1\times3\times5\times7)}{2^5}x^{-9/2}\implies f^{(5)}(4)=\dfrac{105}{2^{14}}$

The pattern should be fairly easy to see.

$f(x)=2+\dfrac{x-4}{2^2}-\dfrac{(x-4)^2}{2^5\times2!}+\dfrac{3(x-4)^3}{2^8\times3!}-\dfrac{15(x-4)^4}{2^{11}\times4!}+\cdots$
$f(x)=2+\displaystyle\sum_{n\ge1}\dfrac{(-1)^n(-1\times1\times3\times5\times\cdots\times(2n-3)}{2^{3n-1}n!}(x-4)^n$

By the ratio test, the series converges if

$\displaystyle\lim_{n\to\infty}\left|\frac{\dfrac{(-1)^{n+1}(-1\times\cdots\times(2n-3)\times(2n-1))(x-4)^{n+1}}{2^{3n+2}(n+1)!}}{\dfrac{(-1)^n(-1\times\cdots\tiems(2n-3))(x-4)^n}{2^{3n-1}n!}}\right|$
$\implies\displaystyle\frac{|x-4|}8\lim_{n\to\infty}\frac{2n-1}{n+1}=\frac{|x-4|}4$
$\implies |x-4|$

so that the ROC is 4.

- - -

10. Without going into much detail, you should have as your Taylor polynomial

$\sin x\approx T_4(x)=\dfrac12+\dfrac{\sqrt3}2\left(x-\dfrac\pi6\right)-\dfrac14\left(x-\dfrac\pi6\right)^2-\dfrac1{4\sqrt3}\left(x-\dfrac\pi6\right)^3+\dfrac1{48}\left(x-\dfrac\pi6\right)^4$

Taylor's inequality then asserts that the error of approximation on the interval $0\le x\le\dfrac\pi3$ is given by

$|\sin x-T_4(x)|=|R_4(x)|\le\dfrac{M\left|x-\frac\pi6\right|^5}{5!}$

where $M$ satisfies $|f^{(5)}(x)|\le M$ on the interval.

We know that $(\sin x)^{(5)}=\cos x$ is bounded between -1 and 1, so we know $M=1$ will suffice. Over the given interval, we have $\left|x-\dfrac\pi6\right|\le\dfrac\pi6$ , so the remainder will be bounded above by

$|R_4(x)|\le\dfrac{1\times\left(\frac\pi6\right)^5}{5!}=\dfrac{\pi^5}{933120}\approx0.000328$

which is to say, over the interval $0\le x\le\dfrac\pi3$ , the fourth degree Taylor polynomial approximates the value of $\sin x$ near $x=\dfrac\pi6$ to within 0.000328.

7 0

4 years ago