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3 years ago

Whats is 4 hundreds 13 tens 5 ones in standard form

Mathematics

1 answer:

mars1129 [50]3 years ago

6 0

<span>4 hundreds 13 tens 5 ones = 400+13*10+5
= 400+130+5
= 535</span>

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What is the solution to this system of equations?

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Answer:

Step-by-step explanation:

y=2x²+2x+1.... i

y=2x+3......... ii

Equating i and ii,

2x²+2x+1=2x+3

2x²+1-3=0

2x²-2=0

x²-1=0

x=1 or -1.

From ii, if x=1,y=5

If x=-1, y=1.

The points are (-1,1) and (1,5)

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<img src="https://tex.z-dn.net/?f=%5Clarge%20%5Crm%20%5Csum%20%5Climits_%7Bn%20%3D%200%7D%5E%20%5Cinfty%20%20%20%5Cfrac%7B%28%20

Fynjy0 [20]

The sum we want is

$\displaystyle \sum_{n=0}^\infty \frac{(-1)^{T_n}}{(2n+1)^2} = 1 - \frac1{3^2} - \frac1{5^2} + \frac1{7^2} + \cdots$

where $T_n=\frac{n(n+1)}2$ is the n-th triangular number, with a repeating sign pattern (+, -, -, +). We can rewrite this sum as

$\displaystyle \sum_{k=0}^\infty \left(\frac1{(8k+1)^2} - \frac1{(8k+3)^2} - \frac1{(8k+7)^2} + \frac1{(8k+7)^2}\right)$

For convenience, I'll use the abbreviations

$S_m = \displaystyle \sum_{k=0}^\infty \frac1{(8k+m)^2}$

${S_m}' = \displaystyle \sum_{k=0}^\infty \frac{(-1)^k}{(8k+m)^2}$

for m ∈ {1, 2, 3, …, 7}, as well as the well-known series

$\displaystyle \sum_{k=1}^\infty \frac{(-1)^k}{k^2} = -\frac{\pi^2}{12}$

We want to find $S_1-S_3-S_5+S_7$ .

Consider the periodic function $f(x) = \left(x-\frac12\right)^2$ on the interval [0, 1], which has the Fourier expansion

$f(x) = \frac1{12} + \frac1{\pi^2} \sum_{n=1}^\infty \frac{\cos(2\pi nx)}{n^2}$

That is, since f(x) is even,

$f(x) = a_0 + \displaystyle \sum_{n=1}^\infty a_n \cos(2\pi nx)$

where

$a_0 = \displaystyle \int_0^1 f(x) \, dx = \frac1{12}$

$a_n = \displaystyle 2 \int_0^1 f(x) \cos(2\pi nx) \, dx = \frac1{n^2\pi^2}$

(See attached for a plot of f(x) along with its Fourier expansion up to order n = 10.)

Expand the Fourier series to get sums resembling the $S'$ -s :

$\displaystyle f(x) = \frac1{12} + \frac1{\pi^2} \left(\sum_{k=0}^\infty \frac{\cos(2\pi(8k+1) x)}{(8k+1)^2} + \sum_{k=0}^\infty \frac{\cos(2\pi(8k+2) x)}{(8k+2)^2} + \cdots \right. \\ \,\,\,\, \left. + \sum_{k=0}^\infty \frac{\cos(2\pi(8k+7) x)}{(8k+7)^2} + \sum_{k=1}^\infty \frac{\cos(2\pi(8k) x)}{(8k)^2}\right)$

which reduces to the identity

$\pi^2\left(\left(x-\dfrac12\right)^2-\dfrac{21}{256}\right) = \\\\ \cos(2\pi x) {S_1}' + \cos(4\pi x) {S_2}' + \cos(6\pi x) {S_3}' + \cos(8\pi x) {S_4}' \\\\ \,\,\,\, + \cos(10\pi x) {S_5}' + \cos(12\pi x) {S_6}' + \cos(14\pi x) {S_7}'$