Express 1/(1 − x)2 as a power series by differentiating the equation below. what is the radius of convergence? 1 1 − x = 1 + x + x2 + x3 + = [infinity] xn n = 0 |x| < 1

S₂=1/(1-x)² = (1/x)∑ n*xⁿ

the power series converges also for |r|<1

Step-by-step explanation:

we know that the infinite geometric series

S= ∑ rⁿ  , from n=0 to n=∞

S= ∑ rⁿ = 1/(1-r)

then differentiating S with respect to r

dS/dr=d[1/(1-r)] = 1/(1-r)²

and

dS/dr=d(∑ rⁿ)/dr =∑ d(rⁿ)/dr  = ∑ n*rⁿ⁻¹ = (1/r)∑ n*rⁿ

then

S₂= (1/r)∑ n*rⁿ =  1/(1-r)²

thus

r²*S₂- r*S₂ = ∑ n*rⁿ⁺¹-∑ n*rⁿ = ∑ (n+1-1)*rⁿ⁺¹ - ∑ n*rⁿ  = (∑ (n+1)*rⁿ⁺¹ - ∑ n*rⁿ )- ∑rⁿ⁺¹ = (n+1)*rⁿ⁺¹ - (1-rⁿ⁺²)/(1-r)

r²*S₂- r*S₂ = r*S₂*(r-1) = [(n+1)*rⁿ⁺¹] - (1-rⁿ⁺²)/(1-r)

S₂ = (1-rⁿ⁺²)/[r*(1-r)²] - [(n+1)*rⁿ⁺¹ ]/[r*(1-r)]

thus |r|<1  in the first term to converge and for the second term for n→∞  , S₂= 1/(1-r)² → (n+1)*rⁿ⁺¹=n*rⁿ= 1 → ln r= -(ln n) /n

then knowing that lim (ln x)= ln (lim x)

lim (ln r)=  -lim [(ln n) /n ] = - lim ( 1/n) = 0 = ln r → r=1

therefore for the second term |r|< 1

then S₂ converges for |r|< 1

a) $ 9.50

b) 17

c) y-intercept (0, 12). this is the initial amount of money that joe had prior to playing any game.

d) x-intercept ( 24, 0). this is the number of games that joe will play before he depletes the initial amount of money that he had.

e) slope is -0.5. joe's balance will reduce by 0.5 for every game that he plays.

f)

domain; [0,24]

range; [0,12]

g) x + 2y = 24

step-by-step explanation:

a)

if joe plays 5 games, this will imply that x = 5. substitute x = 5 in the equation and solve for y;

y = -.50x + 12

y = -0.50(5) + 12

y = 9.50

b)

if joe leaves the arcade with $3.50, this will mean that y = 3.50. plug in y = 3.50 in the given equation and solve for x;

y = -.50x + 12

3.50 = -0.50x + 12

0.50x = 12 - 3.50

0.5x = 8.5

x = 17

c)

the function given is   y = -.50x + 12.

the y-intercept is the point where the graph of this function will intersect the y-axis. at this point, the value of x is usually 0. plug in x = 0 in the given equation and determine y;

y = -0.50(0) + 12

y = 12

the y-intercept is thus (0, 12). this is the initial amount of money that joe had prior to playing any game.

d)

the x-intercept refers to the point where the graph of a function intersects with the x-axis. at this point, the value of y is 0. substitute y = 0 in the given equation and solve for x;

0 =   -.50x + 12

0.5x = 12

x = 24

x-intercept ( 24, 0). this is the number of games that joe will play before he depletes the initial amount of money that he had.

e)

the slope of the given equation is -0.5. the equation has been given in slope-intercept form. thus the slope is simply the coefficient of x.

joe's balance will reduce by 0.5 for every game that he plays. this is the interpretation of the slope.

f)

domain is the set of x values for which the function is real and defined. the domain will thus be; [0,24]. this will restrict the balance to only assume positive values.

on the other hand, the range is the set of y values for which a function is real and defined. the range in our case will be [0,12]. joe can have between $0 and $12 balance.

g)

the standard form of the equation of a straight line is;

ax + by = c

where a, b, and c are integers.

our equation is given as;

y = -.50x + 12

re-writing the equation we have;

0.5x + y = 12

divide all through by 0.5;

x + 2y = 24