(a) Use differentiation to find a power series representation for
f(x) = 1/(5 + x)^2
What is the radius of convergence, R?
(b) Use part (a) to find a power series for
f(x) = 1/(5 + x)^3
What is the radius of convergence, R?
(c) Use part (b) to find a power series for
f(x) = x^2 /(5 + x)^3
What is the radius of convergence, R?
(a) To find a power series representation for f(x) = 1/(5 + x)^2, we can use the formula for the derivative of a power series.
Let's start by finding the derivative of the function:
f'(x) = d/dx (1/(5 + x)^2)
= -2/(5 + x)^3
Now, we can write f'(x) as a power series:
f'(x) = -2/(5 + x)^3
= -2 * (5 + x)^(-3)
= -2 * (5^(-3)) * (1 + x/5)^(-3)
= -2 * (5^(-3)) * (1 - (-x/5))^(-3)
Using the binomial series formula (1 + u)^(-n) = 1 - nu + (n(n + 1)/2!)u^2 - (n(n + 1)(n + 2)/3!)u^3 + ... , with u = -x/5 and n = 3, we can express f'(x) as a power series:
f'(x) = -2 * (5^(-3)) * [1 - (3)(-x/5) + (3)(4)(-x/5)^2 - (3)(4)(5)(-x/5)^3 + ...]
Now, integrating f'(x) term by term, we can obtain the power series representation for f(x):
f(x) = ∫ f'(x) dx
= ∫ [-2 * (5^(-3)) * [1 - (3)(-x/5) + (3)(4)(-x/5)^2 - (3)(4)(5)(-x/5)^3 + ...] dx
= -2 * (5^(-3)) * [x + (3)(5)x^2/10 + (3)(4)(5^2)x^3/10^2 + (3)(4)(5)(5^2)x^4/10^3 + ...] + C
where C is the constant of integration.
The power series representation for f(x) is:
f(x) = -2 * (5^(-3)) * [x + (3)(5)x^2/10 + (3)(4)(5^2)x^3/10^2 + (3)(4)(5)(5^2)x^4/10^3 + ...] + C
The radius of convergence, R, can be found using the ratio test. Assuming the power series converges, we have:
lim┬(n→∞)|aₙ₊₁ / aₙ| = lim┬(n→∞)|[(3)(5)(-1)(-x/5) / [(n + 1)](x/5)| = 5|x|
Since the series will converge if the absolute value of 5|x| is less than 1, the radius of convergence is R = 1/5.
(b) Using the result from part (a), we can find a power series for f(x) = 1/(5 + x)^3 by differentiating f'(x):
f''(x) = d/dx (f'(x))
= d/dx (-2/(5 + x)^3)
= 6/(5 + x)^4
Now, we can represent f''(x) as a power series using the binomial series formula:
f''(x) = 6/(5 + x)^4
= 6 * (5^(-4)) * [1 - 4(-x/5) + (4)(5)(-x/5)^2 - (4)(5)(6)(-x/5)^3 + ...]
Integrating f''(x) term by term, we can obtain the power series representation for f(x):
f(x) = ∫ f''(x) dx
= ∫ [6 * (5^(-4)) * [1 - 4(-x/5) + (4)(5)(-x/5)^2 - (4)(5)(6)(-x/5)^3 + ...]] dx
= 6 * (5^(-4)) * [x + 4(5)(x^2)/10 + (4)(5)(6)(x^3)/10^2 + ...] + C
where C is the constant of integration.
The power series representation for f(x) is:
f(x) = 6 * (5^(-4)) * [x + 4(5)(x^2)/10 + (4)(5)(6)(x^3)/10^2 + ...] + C
The radius of convergence, R, can be found using the ratio test. Assuming the power series converges, we have:
lim┬(n→∞)|aₙ₊₁ / aₙ| = lim┬(n→∞)|6 * (5^(-4)) * [(n + 1)/(n + 1)](x/5)| = |x/5|
Since the series will converge if the absolute value of x/5 is less than 1, the radius of convergence is R = 5.
(c) Using the result from part (b), we can find a power series for f(x) = x^2 /(5 + x)^3 by multiplying f(x) by x^2:
f(x) = x^2 /(5 + x)^3
= x^2 * (6 * (5^(-4)) * [x + 4(5)(x^2)/10 + (4)(5)(6)(x^3)/10^2 + ...] + C)
Simplifying, we get:
f(x) = 6 * (5^(-4)) * [x^3 + 4(5)(x^4)/10 + (4)(5)(6)(x^5)/10^2 + ...] + C * x^2
The radius of convergence, R, remains the same as in part (b), which is R = 5.
(a) To find the power series representation for f(x) = 1/(5 + x)^2, we can use the formula for the geometric series:
1/(1 - u) = 1 + u + u^2 + u^3 + ...
In this case, u = -x/5. So we have:
1/(5 + x)^2 = (1/5^2)/(1 + (-x/5))^2
= (1/25)/(1 - (x/5))^2
Now, we can use the binomial series expansion to obtain the power series representation. The binomial series expansion is given by:
(1 + u)^n = 1 + n*u + (n(n-1)/(2!))*u^2 + (n(n-1)(n-2)/(3!))*u^3 + ...
In this case, u = -x/5 and n = 2. So we have:
(1 - (x/5))^(-2) = 1 + 2*(-x/5) + (2(2-1)/(2!))*(-x/5)^2 + ...
Simplifying this expression gives us:
(1 - (x/5))^(-2) = 1 - (2x/5) + (4x^2)/(25)
So the power series representation for f(x) = 1/(5+x)^2 is:
f(x) = (1/25) - (2/5)x + (4/25)x^2 - ...
Now let's find the radius of convergence, R. The radius of convergence can be found using the ratio test. The ratio test states that if the limit of |a_n+1/a_n| as n approaches infinity is L, then the series converges if L < 1 and diverges if L > 1.
In this case, a_n = (4/25)x^n and a_n+1 = (4/25)x^(n+1). So we have:
|r| = |(a_n+1/a_n)| = |((4/25)x^(n+1))/((4/25)x^n)|
= |(x/n)*((4/25)x^n)/(4/25)x^n|
= |x/n|
Taking the limit of |x/n| as n approaches infinity gives us:
lim(n->∞) |x/n| = 0
Since 0 < 1, the series converges for all x. Therefore, the radius of convergence, R, is infinity.
(b) To find the power series representation for f(x) = 1/(5 + x)^3, we can use the result from part (a) where we found the power series representation for f(x) = 1/(5+x)^2.
The power series representation for f(x) = 1/(5 + x)^2 is:
f(x) = (1/25) - (2/5)x + (4/25)x^2 - ...
To find the power series representation for f(x) = 1/(5 + x)^3, we can differentiate the power series representation for f(x) = 1/(5 + x)^2 term by term. This is allowed because we have established that the power series converges for all x.
Differentiating term by term, we get:
f'(x) = 0 - (2/5) + 2*(4/25)x - ...
Simplifying this gives us:
f'(x) = -(2/5) + (8/25)x - ...
Now we can see that the power series representation for f'(x) = 1/(5 + x)^3 is:
f'(x) = -(2/5) + (8/25)x - ...
The radius of convergence remains the same as in part (a), which is R = infinity.
(c) To find the power series representation for f(x) = x^2 /(5 + x)^3, we can use the result from part (b) where we found the power series representation for f(x) = 1/(5 + x)^3.
The power series representation for f(x) = 1/(5 + x)^3 is:
f(x) = -(2/5) + (8/25)x - ...
To find the power series representation for f(x) = x^2 /(5 + x)^3, we can multiply each term in the power series representation for f(x) = 1/(5 + x)^3 by x^2. This is allowed because we have established that the power series converges for all x.
Multiplying term by term, we get:
f(x) = -(2/5)x^2 + (8/25)x^3 - ...
So the power series representation for f(x) = x^2 /(5 + x)^3 is:
f(x) = -(2/5)x^2 + (8/25)x^3 - ...
The radius of convergence remains the same as in part (b), which is R = infinity.