Find the taylor series about 0 for each of the functions below. give the first three non-zero terms for each.

Algebra 2 Math question Grade 11

dlinn [17]

Answer: y=-3x+5

Step-by-step explanation:

6x+2y=10

Solve for y

2y=-6x+10

y=-3x+5

5 0

3 years ago

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Of a classroom filled with 20 students, 2 will be selected to stay after school and correct homework for extra credit. How

Tcecarenko [31]

Answer: 10

Step-by-step explanation:

20/2=10

pairs of two

3 0

3 years ago

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Solve these recurrence relations together with the initial conditions given. a) an= an−1+6an−2 for n ≥ 2, a0= 3, a1= 6 b) an= 7a

8_murik_8 [283]

Answer:

a) 3/5·((-2)^n + 4·3^n)
b) 3·2^n - 5^n
c) 3·2^n + 4^n
d) 4 - 3 n
e) 2 + 3·(-1)^n
f) (-3)^n·(3 - 2n)
g) ((-2 - √19)^n·(-6 + √19) + (-2 + √19)^n·(6 + √19))/√19

Step-by-step explanation:

These homogeneous recurrence relations of degree 2 have one of two solutions. Problems a, b, c, e, g have one solution; problems d and f have a slightly different solution. The solution method is similar, up to a point.

If there is a solution of the form $a[n]=r^n$ , then it will satisfy ...

$r^n=c_1\cdot r^{n-1}+c_2\cdot r^{n-2}$

Rearranging and dividing by $r^{n-2}$ , we get the quadratic ...

$r^2-c_1r-c_2=0$

The quadratic formula tells us values of r that satisfy this are ...

$r=\dfrac{c_1\pm\sqrt{c_1^2+4c_2}}{2}$

We can call these values of r by the names r₁ and r₂.

Then, for some coefficients p and q, the solution to the recurrence relation is ...

$a[n]=pr_1^n+qr_2^n$

We can find p and q by solving the initial condition equations:

$\left[\begin{array}{cc}1&1\\r_1&r_2\end{array}\right] \left[\begin{array}{c}p\\q\end{array}\right] =\left[\begin{array}{c}a[0]\\a[1]\end{array}\right]$

These have the solution ...

$p=\dfrac{a[0]r_2-a[1]}{r_2-r_1}\\\\q=\dfrac{a[1]-a[0]r_1}{r_2-r_1}$

_____

Using these formulas on the first recurrence relation, we get ...

a)

$c_1=1,\ c_2=6,\ a[0]=3,\ a[1]=6\\\\r_1=\dfrac{1+\sqrt{1^2+4\cdot 6}}{2}=3,\ r_2=\dfrac{1-\sqrt{1^2+4\cdot 6}}{2}=-2\\\\p=\dfrac{3(-2)-6}{-5}=\dfrac{12}{5},\ q=\dfrac{6-3(3)}{-5}=\dfrac{3}{5}\\\\a[n]=\dfrac{3}{5}(-2)^n+\dfrac{12}{5}3^n$

__

The rest of (b), (c), (e), (g) are solved in exactly the same way. A spreadsheet or graphing calculator can ease the process of finding the roots and coefficients for the given recurrence constants. (It's a matter of plugging in the numbers and doing the arithmetic.)

_____

For problems (d) and (f), the quadratic has one root with multiplicity 2. So, the formulas for p and q don't work and we must do something different. The generic solution in this case is ...

$a[n]=(p+qn)r^n$

The initial condition equations are now ...

$\left[\begin{array}{cc}1&0\\r&r\end{array}\right] \left[\begin{array}{c}p\\q\end{array}\right] =\left[\begin{array}{c}a[0]\\a[1]\end{array}\right]$

and the solutions for p and q are ...

$p=a[0]\\\\q=\dfrac{a[1]-a[0]r}{r}$

__

Using these formulas on problem (d), we get ...

d)

$c_1=2,\ c_2=-1,\ a[0]=4,\ a[1]=1\\\\r=\dfrac{2+\sqrt{2^2+4(-1)}}{2}=1\\\\p=4,\ q=\dfrac{1-4(1)}{1}=-3\\\\a[n]=4-3n$

__

And for problem (f), we get ...

f)

$c_1=-6,\ c_2=-9,\ a[0]=3,\ a[1]=-3\\\\r=\dfrac{-6+\sqrt{6^2+4(-9)}}{2}=-3\\\\p=3,\ q=\dfrac{-3-3(-3)}{-3}=-2\\\\a[n]=(3-2n)(-3)^n$

_____

<em>Comment on problem g</em>

Yes, the bases of the exponential terms are conjugate irrational numbers. When the terms are evaluated, they do resolve to rational numbers.

6 0

3 years ago

(6y + 3) minus (3y + 6) when y=7

never [62]

Answer:

y

Step-by-step explanation:

((((2•3y3) - 22y2) - 3y) - —) - 2

y

STEP

4

:

Rewriting the whole as an Equivalent Fraction

4.1 Subtracting a fraction from a whole

Rewrite the whole as a fraction using y as the denominator :

6y3 - 4y2 - 3y (6y3 - 4y2 - 3y) • y

6y3 - 4y2 - 3y = —————————————— = ————————————————————

1 y

Equivalent fraction : The fraction thus generated looks different but has the same value as the whole

Common denominator : The equivalent fraction and the other fraction involved in the calculation share the same denominator

STEP

5

:

Pulling out like terms

5.1 Pull out like factors :

6y3 - 4y2 - 3y = y • (6y2 - 4y - 3)

Trying to factor by splitting the middle term

5.2 Factoring 6y2 - 4y - 3

The first term is, 6y2 its coefficient is 6 .

The middle term is, -4y its coefficient is -4 .

The last term, "the constant", is -3

Step-1 : Multiply the coefficient of the first term by the constant 6 • -3 = -18

Step-2 : Find two factors of -18 whose sum equals the coefficient of the middle term, which is -4 .

-18 + 1 = -17

-9 + 2 = -7

-6 + 3 = -3

-3 + 6 = 3

-2 + 9 = 7

-1 + 18 = 17

Observation : No two such factors can be found !!

Conclusion : Trinomial can not be factored

Adding fractions that have a common denominator :

5.3 Adding up the two equivalent fractions

Add the two equivalent fractions which now have a common denominator

Combine the numerators together, put the sum or difference over the common denominator then reduce to lowest terms if possible:

y • (6y2-4y-3) • y - (6) 6y4 - 4y3 - 3y2 - 6

———————————————————————— = ———————————————————

y y

Equation at the end of step

5

:

(6y4 - 4y3 - 3y2 - 6)

————————————————————— - 2

y

STEP

6

:

Rewriting the whole as an Equivalent Fraction :

6.1 Subtracting a whole from a fraction

Rewrite the whole as a fraction using y as the denominator :

2 2 • y

2 = — = —————

1 y

Checking for a perfect cube :

6.2 6y4 - 4y3 - 3y2 - 6 is not a perfect cube

Trying to factor by pulling out :

6.3 Factoring: 6y4 - 4y3 - 3y2 - 6

Thoughtfully split the expression at hand into groups, each group having two terms :

Group 1: -3y2 - 6

Group 2: 6y4 - 4y3

Pull out from each group separately :

Group 1: (y2 + 2) • (-3)

Group 2: (3y - 2) • (2y3)

Bad news !! Factoring by pulling out fails :

The groups have no common factor and can not be added up to form a multiplication.

Polynomial Roots Calculator :

6.4 Find roots (zeroes) of : F(y) = 6y4 - 4y3 - 3y2 - 6

Polynomial Roots Calculator is a set of methods aimed at finding values of y for which F(y)=0

Rational Roots Test is one of the above mentioned tools. It would only find Rational Roots that is numbers y which can be expressed as the quotient of two integers

The Rational Root Theorem states that if a polynomial zeroes for a rational number P/Q then P is a factor of the Trailing Constant and Q is a factor of the Leading Coefficient

In this case, the Leading Coefficient is 6 and the Trailing Constant is -6.

The factor(s) are:

of the Leading Coefficient : 1,2 ,3 ,6

of the Trailing Constant : 1 ,2 ,3 ,6

Let us test ....

P Q P/Q F(P/Q) Divisor

-1 1 -1.00 1.00

-1 2 -0.50 -5.88

-1 3 -0.33 -6.11

-1 6 -0.17 -6.06

-2 1 -2.00 110.00

Note - For tidiness, printing of 13 checks which found no root was suppressed

Polynomial Roots Calculator found no rational roots

Adding fractions that have a common denominator :

6.5 Adding up the two equivalent fractions

(6y4-4y3-3y2-6) - (2 • y) 6y4 - 4y3 - 3y2 - 2y - 6

————————————————————————— = ————————————————————————

y y

Polynomial Roots Calculator :

6.6 Find roots (zeroes) of : F(y) = 6y4 - 4y3 - 3y2 - 2y - 6

See theory in step 6.4

In this case, the Leading Coefficient is 6 and the Trailing Constant is -6.

The factor(s) are:

of the Leading Coefficient : 1,2 ,3 ,6

of the Trailing Constant : 1 ,2 ,3 ,6

Let us test ....

P Q P/Q F(P/Q) Divisor

-1 1 -1.00 3.00

-1 2 -0.50 -4.88

-1 3 -0.33 -5.44

-1 6 -0.17 -5.73

-2 1 -2.00 114.00

Note - For tidiness, printing of 13 checks which found no root was suppressed

Polynomial Roots Calculator found no rational roots

Final result :

6y4 - 4y3 - 3y2 - 2y - 6

————————————————————————

y

4 0

2 years ago

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Help help math math math

Alborosie

Answer:

2/15

Step-by-step explanation:

3 0

2 years ago

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