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2 years ago

The set {1,2,3,4,5,6...} is called the set of

Mathematics

1 answer:

Anit [1.1K]2 years ago

4 0

1,2,3,4,5,6... is the set of u

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Please help i'll give brainliest

Aliun [14]

Answer:

-01/0/0.8/1.2/1.6

Step-by-step explanation:

3 0

2 years ago

y′′ −y = 0, x0 = 0 Seek power series solutions of the given differential equation about the given point x 0; find the recurrence

sukhopar [10]

Let

$\displaystyle y(x) = \sum_{n=0}^\infty a_nx^n = a_0 + a_1x + a_2x^2 + \cdots$

Differentiating twice gives

$\displaystyle y'(x) = \sum_{n=1}^\infty na_nx^{n-1} = \sum_{n=0}^\infty (n+1) a_{n+1} x^n = a_1 + 2a_2x + 3a_3x^2 + \cdots$

$\displaystyle y''(x) = \sum_{n=2}^\infty n (n-1) a_nx^{n-2} = \sum_{n=0}^\infty (n+2) (n+1) a_{n+2} x^n$

When x = 0, we observe that y(0) = a₀ and y'(0) = a₁ can act as initial conditions.

Substitute these into the given differential equation:

$\displaystyle \sum_{n=0}^\infty (n+2)(n+1) a_{n+2} x^n - \sum_{n=0}^\infty a_nx^n = 0$

$\displaystyle \sum_{n=0}^\infty \bigg((n+2)(n+1) a_{n+2} - a_n\bigg) x^n = 0$

Then the coefficients in the power series solution are governed by the recurrence relation,

$\begin{cases}a_0 = y(0) \\ a_1 = y'(0) \\\\ a_{n+2} = \dfrac{a_n}{(n+2)(n+1)} & \text{for }n\ge0\end{cases}$

Since the n-th coefficient depends on the (n - 2)-th coefficient, we split n into two cases.

• If n is even, then n = 2k for some integer k ≥ 0. Then

$k=0 \implies n=0 \implies a_0 = a_0$

$k=1 \implies n=2 \implies a_2 = \dfrac{a_0}{2\cdot1}$

$k=2 \implies n=4 \implies a_4 = \dfrac{a_2}{4\cdot3} = \dfrac{a_0}{4\cdot3\cdot2\cdot1}$

$k=3 \implies n=6 \implies a_6 = \dfrac{a_4}{6\cdot5} = \dfrac{a_0}{6\cdot5\cdot4\cdot3\cdot2\cdot1}$

It should be easy enough to see that

$a_{n=2k} = \dfrac{a_0}{(2k)!}$

• If n is odd, then n = 2k + 1 for some k ≥ 0. Then

$k = 0 \implies n=1 \implies a_1 = a_1$

$k = 1 \implies n=3 \implies a_3 = \dfrac{a_1}{3\cdot2}$

$k = 2 \implies n=5 \implies a_5 = \dfrac{a_3}{5\cdot4} = \dfrac{a_1}{5\cdot4\cdot3\cdot2}$

$k=3 \implies n=7 \implies a_7=\dfrac{a_5}{7\cdot6} = \dfrac{a_1}{7\cdot6\cdot5\cdot4\cdot3\cdot2}$

so that

$a_{n=2k+1} = \dfrac{a_1}{(2k+1)!}$

So, the overall series solution is

$\displaystyle y(x) = \sum_{n=0}^\infty a_nx^n = \sum_{k=0}^\infty \left(a_{2k}x^{2k} + a_{2k+1}x^{2k+1}\right)$

$\boxed{\displaystyle y(x) = a_0 \sum_{k=0}^\infty \frac{x^{2k}}{(2k)!} + a_1 \sum_{k=0}^\infty \frac{x^{2k+1}}{(2k+1)!}}$

4 0

2 years ago

Consider a Triangle ABC like the one below. Suppose that C = 98, A = 74, and b = 11 (figure is not drawn to scale.) solve the tr

Degger [83]

Answer:

$A=73.8\°$

$B=8.2\°$

$c=76.3\ units$

Step-by-step explanation:

step 1

Find the measure of side c

Applying the law of cosines

$c^{2}= a^{2}+b^{2}-2(a)(b)cos(C)$

substitute the given values

$c^{2}= 74^{2}+11^{2}-2(74)(11)cos(98\°)$

$c^{2}=5,823.5738$

$c=76.3\ units$

step 2

Find the measure of angle A

Applying the law of sine

$\frac{a}{sin(A)}=\frac{c}{sin(C)}$

substitute the given values

$\frac{74}{sin(A)}=\frac{76.3}{sin(98\°)}$

$sin(A)=(74)sin(98\°)/76.3$

$A=arcsin((74)sin(98\°)/76.3)$

$A=73.8\°$

step 3

Find the measure of angle B

we know that

The sum of the internal angles of a triangle must be equal to 180 degrees

$A+B+C=180\°$

substitute the given values

$73.8\°+B+98\°=180\°$

$171.8\°+B=180\°$

$B=180\°-171.8\°=8.2\°$

5 0

3 years ago

Given that a=4, b=4 and c=-2 work out b(squared) -2ac

Nimfa-mama [501]

Answer:

Step-by-step explanation:

substitute the given values of a, b and c into the expression:

b² - 2ac = 4² - (2 x 4 x -2)

= 16 - (-16)

= 16 + 16

= 32

3 0

2 years ago

Solve this equation and explain the steps 9-2x= 7x

Hunter-Best [27]

Answer:

x = 1

Step-by-step explanation:

Step 1: Write equation

9 - 2x = 7x

Step 2: Solve for x

Add 2x on both sides: 9 = 9x

Divide both sides by 9: x = 1

Step 3: Check

Plug in x to verify it's a solution.

9 - 2(1) = 7(1)

9 - 2 = 7

7 = 7

3 0

3 years ago