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4 years ago

What is a decimal equivalent for 2/5 and 40/100

Mathematics

1 answer:

Simora [160]4 years ago

7 0

Answer:

For 2/5=0.4 For 40/100 is also 0.4

Step-by-step explanation:

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Quadratic Data

EastWind [94]

I think it’s quadratic

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3 years ago

What is the area of sector?

Shtirlitz [24]

Answer:

Step-by-step explanation:

since 90 degrees is 1/4 of a circle, the area of the sector is 5²pi/4 = 25/4pi = 19.63

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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.