What is the relationship between power series and the sum function of a power series?
The examples in this article are quoted from: 80_1 Power series operations, item-by-item integration and derivation [Teacher Xiaoyuan] Advanced Mathematics, Postgraduate Entrance Examination Mathematics
Find the power series ∑ n = 1 ∞ 1 nxn \sum\limits_{n=1}^{\infty}\frac{1}{n}x^nn=1∑∞n1xThe sum function (1) of n
finds the radius of convergence. Since it is a series with no missing terms, you can uselim n → ∞ ∣ an + 1 an ∣ = ρ \lim\limits_{n\rightarrow\infty}|\frac{a_ {n+1}}{a_n}|=\rhon→∞lim∣anan+1∣=ρ , if it is a missing term series, onlylim n → ∞ ∣ un + 1 ( x ) un ( x ) ∣ = ρ ∣ ϕ ( x ) ∣ < 1 \lim\limits_{n\rightarrow\infty}| \frac{u_{n+1}(x)}{u_n(x)}|=\rho|\phi(x)|\lt 1n→∞lim∣un(x)un+1(x)∣=ρ∣ϕ(x)∣<1 , of course the latter can also be used if there is no lack of term series.
ρ = lim n → ∞ ∣ an + 1 an ∣ = lim n → ∞ ∣ 1 n + 1 1 n ∣ = 1 \rho=\lim\limits_{n\rightarrow\infty}|\frac{a_{n +1}}{a_n}|=\lim\limits_{n\rightarrow\infty}|\frac{\frac{1}{n+1}}{\frac{1}{n}}|=1r=n→∞lim∣anan+1∣=n→∞lim∣n1n+11∣=1
(2) Determine the convergence and divergence at the endpoints
Whenx = − 1 x=-1x=− 1 ,∑ n = 1 ∞ ( − 1 ) n 1 n \sum\limits_{n=1}^{\infty}(-1)^n\frac{1}{n}n=1∑∞(−1)nn1， u n = 1 n → 0 u_n=\frac{1}{n}\rightarrow0 un=n1→0且un = 1 n u_n=\frac{1}{n}un=n1Decreasing, the series converges (determined by Leibniz’s theorem of staggered series)
when x = 1 x=1x=1st ,∑ n = 1 ∞ 1 n \sum\limits_{n=1}^{\infty}\frac{1}{n}n=1∑∞n1， p = 1 p=1 p=1 , the series diverges (identified by p series)
(3) In summary, the convergence domain of this series is[ − 1 , 1 ) [-1,1)[−1,1 )
(4) Find the sum function of the power series in the convergence domain (In the convergence region, the power series is equal to its sum function, and beyond the convergence region, the two are not equal.）
s ( x ) = ∑ n = 1 ∞ 1 n x n = x + 1 2 x 2 + 1 3 x 3 + ⋯ + 1 n x n + ⋯ s(x)=\sum\limits_{n=1}^{\infty}\frac{1}{n}x^n=x+\frac{1}{2}x^2+\frac{1}{3}x^3+\cdots+\frac{1}{n}x^n+\cdots s(x)=n=1∑∞n1xn=x+21x2+31x3+⋯+n1xn+⋯
逐项求导
s ′ ( x ) = ( ∑ n = 1 ∞ 1 n x n ) ′ = 1 + x + x 2 + ⋯ + 1 n x n − 1 + ⋯ = 1 1 − x s'(x)=\big(\sum\limits_{n=1}^{\infty}\frac{1}{n}x^n\big)'=1+x+x^2+\cdots+\frac{1}{n}x^{n-1}+\cdots=\frac{1}{1-x} s′(x)=(n=1∑∞n1xn)′=1+x+x2+⋯+n1xn−1+⋯=1−x1
Integrate the left and right ends simultaneously (term-by-term integration on the right side)
s ( x ) = s ( 0 ) + ∫ 0 xs ′ ( t ) dt = 0 + ∫ 0 x 1 1 − tdt = − ln ( 1 − x ) s (x)=s(0)+\int_0^xs'(t)dt=0+\int_0^x\frac{1}{1-t}dt=-\ln(1-x)s(x)=s(0)+∫0xs′(t)dt=0+∫0x1−t1dt=−ln(1−
Why does the above formula of x ) also contain s ( 0 ) s(0)s(0)?
∫ 0 x s ′ ( t ) d t = s ( x ) ∣ 0 x = s ( x ) − s ( 0 ) s ( x ) = s ( 0 ) + ∫ 0 x s ′ ( t ) d t \int_0^xs'(t)dt=s(x)|_0^x=s(x)-s(0)\\ ~\\ s(x)=s(0)+\int_0^xs'(t)dt ∫0xs′(t)dt=s(x)∣0x=s(x)−s(0) s(x)=s(0)+∫0xs′ (t)dt
The sum function of the power series on the final convergence region is:
s ( x ) = − ln ( 1 − x ) , x ∈ [ − 1 , 1 ) s(x)=-\ln(1 -x),x\in[-1,1)s(x)=−ln(1−x)，x∈[−1,1 )
Why do we go around in circles first deriving (or integrating) the series and then integrating (or deriving) it?
I mainly want to use geometric series because its sum function is easy to find, and the purpose of term-by-term derivation and integration is to transform the given power series into a geometric series, and then use the geometric series to find the given power. sum function of series
Let’s see in the image what is the relationship between power series and the sum function of power series?
The image of the power series in the figure below is a green curve, which is not a real image because nnn is infinity, the author here isnnn only got 9, just for illustration. The red curve in the figure below is the image of the power series and function. We can find thatIn the convergence region, the power series is equal to its sum function. Beyond the convergence region, the two are not equal.
