Find a power series whose sum is:
(a) 5 / (1-3x)
(b) 9 / (1-4x)
(c) 6 / (2-4x)
suppose I have
1 + x + x^2 + x^3 + ...
this is a geometric series with
a = 1 and r = x
S∞ = a/(1-r)
= 1/(1 - x)
so in a) you have
5/(1-3x)
this would match the terms
5 + 5(3x) + 5(3x)^2 + 5(3x)^3 + ...
You did not state whether you have to give the answer in Sigma notation, but I am pretty sure you could change it into that notation, if you study series at this level.
for c), simplify
6/(2-4x) to 3/(1 - 2x) and proceed as above.
MY ANSWERS:
(a) ax = 5(3x)^(x-1)
(b) ax = 4(4x)^(x-1)
(c) ax = 3(2x)^(x-1)
I'm confused because all of those series have a divergent sum, and it looks like they're each supposed to be convergent.
To find a power series whose sum is given by a given rational function, we can make use of the geometric series formula.
The geometric series formula states that for any real number r, the following series is equal to 1 / (1 - r):
1 + r + r^2 + r^3 + ...
Now, let's use this formula to find the power series for each given rational function:
(a) For the given function 5 / (1 - 3x):
Step 1: Rewrite the function in terms of a geometric series. We know that 1 / (1 - r) = 1 + r + r^2 + r^3 + ...
Therefore, we can rewrite 5 / (1 - 3x) as:
5 * (1 / (1 - (-3x)))
Step 2: Recognize the form of the geometric series. In this case, r = -3x.
Step 3: Apply the geometric series formula. The power series for 5 / (1 - 3x) is:
5 * (1 + (-3x) + (-3x)^2 + (-3x)^3 + ...)
Simplifying further, we have:
5 * (1 - 3x + 9x^2 - 27x^3 + ...)
So the power series for 5 / (1 - 3x) is 5 * (1 - 3x + 9x^2 - 27x^3 + ...).
(b) For the given function 9 / (1 - 4x):
Step 1: Rewrite the function in terms of a geometric series. We know that 1 / (1 - r) = 1 + r + r^2 + r^3 + ...
Therefore, we can rewrite 9 / (1 - 4x) as:
9 * (1 / (1 - (-4x)))
Step 2: Recognize the form of the geometric series. In this case, r = -4x.
Step 3: Apply the geometric series formula. The power series for 9 / (1 - 4x) is:
9 * (1 + (-4x) + (-4x)^2 + (-4x)^3 + ...)
Simplifying further, we have:
9 * (1 - 4x + 16x^2 - 64x^3 + ...)
So the power series for 9 / (1 - 4x) is 9 * (1 - 4x + 16x^2 - 64x^3 + ...).
(c) For the given function 6 / (2 - 4x):
Step 1: Rewrite the function in terms of a geometric series. We know that 1 / (1 - r) = 1 + r + r^2 + r^3 + ...
Therefore, we can rewrite 6 / (2 - 4x) as:
6 * (1 / (1 - (-2x)))
Step 2: Recognize the form of the geometric series. In this case, r = -2x.
Step 3: Apply the geometric series formula. The power series for 6 / (2 - 4x) is:
6 * (1 + (-2x) + (-2x)^2 + (-2x)^3 + ...)
Simplifying further, we have:
6 * (1 - 2x + 4x^2 - 8x^3 + ...)
So the power series for 6 / (2 - 4x) is 6 * (1 - 2x + 4x^2 - 8x^3 + ...).
Note: The power series representation of a rational function may only be valid within the interval of convergence, where the series converges.