Find the open interval and the radius of absolute convergence for the following power series: (+1)^4t. (Note that (3r + 1)^n = 3^(r + 4) * 5 the above power series.) Use the Maclaurin power series for sin(x) (given below) to find the power series centered at the following derivative. Express the power series in sigma notation form. sin(1?) {which, by the way, equals 2r cos(r)} sin(1) (2n + 1) (absolutely converges for all real x) Find the first several terms (up to 1^n) for the solution of the following first-order linear differential equation centered at the ordinary point (that is, the solution will be of the form X = a0e^t). Express the coefficients exactly and not as decimal approximations. Do not attempt to find a generalized formula for each function of n. (s = 2) 2y = 0 Find the “guaranteed” minimum radius of absolute convergence, R_min, for the power series solution centered at x = -1 of the following second-order linear homogeneous differential equation. Do not attempt to find the power series solution. 3x^2 + 10x + 5y” + 76x – 48y’ + x^2 – 21 – y = 0 (Note that the five zeros of 15x^3 + 10x^2 + 76x – 48 are -4, 2, 3, 1+i, and 1-i.)
The Correct Answer and Explanation is:
1. Radius and Interval of Absolute Convergence
We are given a power series involving (3r+1)n(3r + 1)^n(3r+1)n, stated as: (3r+1)n=3r+4⋅5(3r + 1)^n = 3^{r + 4} \cdot 5(3r+1)n=3r+4⋅5
This expression appears unclear, but let’s proceed by interpreting the main task: find the open interval and radius of convergence for a general power series.
Assume we have a power series of the form: ∑n=0∞an(x−c)n\sum_{n=0}^{\infty} a_n (x – c)^nn=0∑∞an(x−c)n
To find the radius of convergence RRR, we use the Ratio Test: limn→∞∣an+1an∣=L⇒R=1L\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = L \Rightarrow R = \frac{1}{L}n→∞limanan+1=L⇒R=L1
But the precise coefficients ana_nan are not given clearly, so we cannot compute RRR directly here. Instead, for the second part, we’ll use known convergence properties.
2. Maclaurin Series for sin(x)\sin(x)sin(x)
The Maclaurin series for sin(x)\sin(x)sin(x) is: sin(x)=∑n=0∞(−1)n(2n+1)!x2n+1\sin(x) = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} x^{2n+1}sin(x)=n=0∑∞(2n+1)!(−1)nx2n+1
This series converges absolutely for all real x, so the radius of convergence is infinite, or R=∞R = \inftyR=∞.
3. Power Series Expansion of a Function Related to sin(1)\sin(1)sin(1)
If given that: sin(1?)=2rcos(r)\sin(1?) = 2r \cos(r)sin(1?)=2rcos(r)
That seems symbolic or possibly typographical, but for finding a power series derivative related to sin(x)\sin(x)sin(x), we differentiate the series: ddx[∑n=0∞(−1)n(2n+1)!x2n+1]=∑n=0∞(−1)n(2n+1)(2n+1)!x2n=∑n=0∞(−1)n(2n)!x2n\frac{d}{dx} \left[ \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} x^{2n+1} \right] = \sum_{n=0}^{\infty} \frac{(-1)^n (2n+1)}{(2n+1)!} x^{2n} = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!} x^{2n}dxd[n=0∑∞(2n+1)!(−1)nx2n+1]=n=0∑∞(2n+1)!(−1)n(2n+1)x2n=n=0∑∞(2n)!(−1)nx2n
Which is the Maclaurin series for cos(x)\cos(x)cos(x).
4. First Several Terms of a Solution for 2y=02y = 02y=0
The equation 2y=02y = 02y=0 implies: y=0y = 0y=0
So the solution is the trivial zero function. If the solution is of the form y=a0ety = a_0 e^ty=a0et, then y=0y = 0y=0 implies a0=0a_0 = 0a0=0. So all terms vanish.
5. Radius of Convergence for the DE Centered at x=−1x = -1x=−1
We are given a differential equation: (3×2+10x+5)y′′+(76x−48)y′+(x2−21)y=0(3x^2 + 10x + 5)y” + (76x – 48)y’ + (x^2 – 21)y = 0(3×2+10x+5)y′′+(76x−48)y′+(x2−21)y=0
To find the minimum radius of convergence, we locate the distance from the center x=−1x = -1x=−1 to the nearest singularity, which is a root of the leading coefficient: 3×2+10x+53x^2 + 10x + 53×2+10x+5
We are told that the roots of the full expression 15×3+10×2+76x−4815x^3 + 10x^2 + 76x – 4815×3+10×2+76x−48 include: −4, 2, 3, 1+i, 1−i-4,\ 2,\ 3,\ 1+i,\ 1-i−4, 2, 3, 1+i, 1−i
Compute distances from x=−1x = -1x=−1:
- ∣−1−(−4)∣=3|-1 – (-4)| = 3∣−1−(−4)∣=3
- ∣−1−2∣=3|-1 – 2| = 3∣−1−2∣=3
- ∣−1−3∣=4|-1 – 3| = 4∣−1−3∣=4
- ∣−1−(1+i)∣=(1+1)2+12=5|-1 – (1+i)| = \sqrt{(1+1)^2 + 1^2} = \sqrt{5}∣−1−(1+i)∣=(1+1)2+12=5
- ∣−1−(1−i)∣=(1+1)2+12=5|-1 – (1 – i)| = \sqrt{(1+1)^2 + 1^2} = \sqrt{5}∣−1−(1−i)∣=(1+1)2+12=5
The minimum distance is R=3R = 3R=3
✅ Final Answers
- Maclaurin series for sin(x)\sin(x)sin(x):
∑n=0∞(−1)n(2n+1)!x2n+1\displaystyle \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!} x^{2n+1}n=0∑∞(2n+1)!(−1)nx2n+1, radius of convergence: R=∞R = \inftyR=∞ - Solution to 2y=02y = 02y=0:
y=0y = 0y=0, with coefficients all zero - Minimum radius of convergence centered at x=−1x = -1x=−1:
Rmin=3R_{\text{min}} = 3Rmin=3
✍️ Explanation
To analyze power series, we apply convergence techniques and standard expansions. The Maclaurin series for sin(x)\sin(x)sin(x) is widely known and converges absolutely for all real xxx, meaning its radius of convergence is infinite. Differentiating term-by-term helps derive related functions like cos(x)\cos(x)cos(x), whose series expansion is also absolutely convergent.
For the equation 2y=02y = 02y=0, the function yyy must be identically zero. This makes all coefficients in any power series expansion zero, giving a trivial solution. No further series expansion is required because the function is constant zero.
For a more complex differential equation with polynomial coefficients, the solution’s convergence behavior near a point is determined by how close that point is to a singularity of the equation. Singularities occur where the coefficient of the highest derivative vanishes, causing the equation to become undefined or unstable. In this case, finding the roots of the leading coefficient identifies those points.
Since the equation is centered at x=−1x = -1x=−1, the guaranteed radius of convergence is the shortest distance from −1-1−1 to any singularity. Using known roots, we compute distances and select the minimum. This guarantees that the series will converge in an open disk centered at −1-1−1 with a radius of 3. Beyond that radius, the series may diverge due to approaching a singular point. This method avoids solving the full differential equation but still gives essential information about the behavior of the solution’s power series.
