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3 years ago

I'll mark brainliest!

Mathematics

1 answer:

GuDViN [60]3 years ago

4 0

Answer:

im answering so he can have it :D

Step-by-step explanation:

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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

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

Evaluating both sides at x for x ∈ {1/8, 3/8, 5/8, 7/8} and solving the system of equations yields the dependent solution

$\begin{cases}{S_4}' = \dfrac{\pi^2}{256} \\\\ {S_1}' - {S_3}' - {S_5}' + {S_7}' = \dfrac{\pi^2}{8\sqrt 2}\end{cases}$

It turns out that

${S_1}' - {S_3}' - {S_5}' + {S_7}' = S_1 - S_3 - S_5 + S_7$

so we're done, and the sum's value is $\boxed{\dfrac{\pi^2}{8\sqrt2}}$ .

6 0

2 years ago

(Add 34 and 6) , Times 3 as an expression

Colt1911 [192]

(34 + 6 ) * 3

there would be no equal sign since it is an expression

3 0

3 years ago

Find the volume in cubic millimeters.

forsale [732]

Answer:

Option B is correct.

volume in cubic millimeters is, 8000 cubic mm

Step-by-step explanation:

Volume of a cube is found by multiplying the length of any edge by itself twice.

It is given by:

$V = a^3$ cubic units,......[1] where V represents the volume of a cube and a represents the side of a length.

From the given figure;

Length of a side(a) = 2 cm

Use conversion, to convert cm into mm;

1 cm = 10 mm

then;

2cm =20 mm

⇒length of a side(a) = 20 mm

Substitute the value of a = 20 mm in [1];

$V = (20)^3$ cubic mm

Simplify:

$V = 8000 mm^3$

Therefore, the volume in cubic millimeters is, 8000 cubic mm

8 0

3 years ago

Read 2 more answers

Which number is equivalent to 3 exponent 4 over 3 exponent 2

choli [55]

Answer:

9 or B

Step-by-step explanation:

3^4=81

3^2=9

81/9=9

3 0

3 years ago

Read 2 more answers

State the degree and leading coefficient of each polynomial in one variable.if it is not a polynomial in one variable explain wh

valina [46]

Answer:

2. degree 3; leading coefficient 8

4. degree 6; leading coefficient 7

6. not a polynomial

Step-by-step explanation:

The "leading coefficient" is the coefficient of the highest-degree term in the sum of terms that makes up a polynomial.

These expressions all have one variable, so the number of variables in not an issue in any case. All of the exponents are positive integers, so that is not an issue in any case. However, the variable appears in the denominator in the expression of problem 6, so that sum is not a polynomial.

2. In order to put this into the form we recognize as a polynomial, the expression must be "simplified' by performing the multiplication of the two factors:

= 8x³ -4x² +6x -3

The leading coefficient is the coefficient of the highest-degree term, which is the product of the highest-degree terms of the factors. That product is ...

(2x)(4x²) = 8x³

so the leading coefficient is 8. The variable is to the 3rd power, so the degree is 3.

You don't actually have to do the rest of the multiplication in order to find the required answer.

4. The expression is already written as a sum, so we only need to find the term of highest degree. That is the last one: 7y^6. Its degree is 6 and its leading coefficient is 7.

6. Variables are not allowed in the denominator of a polynomial. This expression is not a polynomial.

_____

<em>Comment on degree</em>

The degree of a term is the exponent of the variable. If there is more than one variable, the degree of the term is the sum of their exponents. For example, the polynomial ...

x² +xy +y²

has three terms, each of degree 2.

If this example were one of your problems, it would be rejected as "not a polynomial in one variable," since two variables are involved.

7 0

2 years ago