当前位置:网站首页>Let f (x) = Σ x^n/n^2, prove that f (x) + F (1-x) + lnxln (1-x) = Σ 1/n^2
Let f (x) = Σ x^n/n^2, prove that f (x) + F (1-x) + lnxln (1-x) = Σ 1/n^2
2022-07-07 05:17:00 【Fish in the deep sea (・ ω& lt;)*】
subject
set up f ( x ) = ∑ n = 1 ∞ x n n 2 , prove : f ( x ) + f ( 1 − x ) + ln x ln ( 1 − x ) = ∑ n = 1 ∞ 1 n 2 \text{ set up }f\left( x \right) =\sum_{n=1}^{\infty}{\frac{x^n}{n^2}}\text{, prove :}f\left( x \right) +f\left( 1-x \right) +\ln x\ln \left( 1-x \right) =\sum_{n=1}^{\infty}{\frac{1}{n^2}} set up f(x)=n=1∑∞n2xn, prove :f(x)+f(1−x)+lnxln(1−x)=n=1∑∞n21
answer
f ′ ( x ) = ∑ n = 1 ∞ x n − 1 n = 1 x ∑ n = 1 ∞ x n n = 1 x ∫ 0 x ∑ n = 1 ∞ t n − 1 d t = 1 x ∫ 0 x ∑ n = 0 ∞ t n d t = 1 x ∫ 0 x 1 1 − t d t = − ln ( 1 − x ) x f'\left( x \right) =\sum_{n=1}^{\infty}{\frac{x^{n-1}}{n}}=\frac{1}{x}\sum_{n=1}^{\infty}{\frac{x^n}{n}}=\frac{1}{x}\int_0^x{\sum_{n=1}^{\infty}{t^{n-1}}dt}=\frac{1}{x}\int_0^x{\sum_{n=0}^{\infty}{t^n}dt}=\frac{1}{x}\int_0^x{\frac{1}{1-t}dt}=-\frac{\ln \left( 1-x \right)}{x} f′(x)=n=1∑∞nxn−1=x1n=1∑∞nxn=x1∫0xn=1∑∞tn−1dt=x1∫0xn=0∑∞tndt=x1∫0x1−t1dt=−xln(1−x)
f ′ ( x ) − f ′ ( 1 − x ) = − ln ( 1 − x ) x + ln x 1 − x f'\left( x \right) -f'\left( 1-x \right) =-\frac{\ln \left( 1-x \right)}{x}+\frac{\ln x}{1-x} f′(x)−f′(1−x)=−xln(1−x)+1−xlnx
∵ [ ln x ln ( 1 − x ) ] ′ = ln ( 1 − x ) x − ln x 1 − x \because \left[ \ln x\ln \left( 1-x \right) \right] '=\frac{\ln \left( 1-x \right)}{x}-\frac{\ln x}{1-x} ∵[lnxln(1−x)]′=xln(1−x)−1−xlnx
∴ f ′ ( x ) − f ′ ( 1 − x ) + [ ln x ln ( 1 − x ) ] ′ = 0 \therefore f'\left( x \right) -f'\left( 1-x \right) +\left[ \ln x\ln \left( 1-x \right) \right] '=0 ∴f′(x)−f′(1−x)+[lnxln(1−x)]′=0
Make g ( x ) = f ( x ) + f ( 1 − x ) + ln x ln ( 1 − x ) x ∈ ( 0 , 1 ) \text{ Make }g\left( x \right) =f\left( x \right) +f\left( 1-x \right) +\ln x\ln \left( 1-x \right) \ \ x\in \left( 0,1 \right) Make g(x)=f(x)+f(1−x)+lnxln(1−x) x∈(0,1)
∴ g ′ ( x ) = 0 \therefore g'\left( x \right) =0 ∴g′(x)=0
∵ ∫ 0 x g ′ ( u ) d u = g ( x ) − lim t → 0 + g ( t ) = 0 \because \int_0^x{g'\left( u \right)}du=g\left( x \right) -\underset{t\rightarrow 0^+}{\lim}g\left( t \right) =0 ∵∫0xg′(u)du=g(x)−t→0+limg(t)=0
∴ g ( x ) = lim t → 0 + g ( t ) \therefore g\left( x \right) =\underset{t\rightarrow 0^+}{\lim}g\left( t \right) ∴g(x)=t→0+limg(t)
∵ lim t → 0 + g ( t ) = lim t → 0 + f ( t ) + lim t → 0 + f ( 1 − t ) + lim t → 0 + ln t ⋅ ln ( 1 − t ) \because \underset{t\rightarrow 0^+}{\lim}g\left( t \right) =\underset{t\rightarrow 0^+}{\lim}f\left( t \right) +\underset{t\rightarrow 0^+}{\lim}f\left( 1-t \right) +\underset{t\rightarrow 0^+}{\lim}\ln t\cdot \ln \left( 1-t \right) ∵t→0+limg(t)=t→0+limf(t)+t→0+limf(1−t)+t→0+limlnt⋅ln(1−t)
among lim t → 0 + f ( t ) = 0 , lim t → 0 + f ( 1 − t ) = ∑ n = 1 ∞ 1 n 2 , lim t → 0 + ln t ⋅ ln ( 1 − t ) = lim t → 0 + t ln t = lim t → 0 + ln t 1 t = lim t → 0 + 1 t − 1 t 2 = 0 \text{ among }\underset{t\rightarrow 0^+}{\lim}f\left( t \right) =0,\ \underset{t\rightarrow 0^+}{\lim}f\left( 1-t \right) =\sum_{n=1}^{\infty}{\frac{1}{n^2}},\ \underset{t\rightarrow 0^+}{\lim}\ln t\cdot \ln \left( 1-t \right) =\underset{t\rightarrow 0^+}{\lim}t\ln t=\underset{t\rightarrow 0^+}{\lim}\frac{\ln t}{\frac{1}{t}}=\underset{t\rightarrow 0^+}{\lim}\frac{\frac{1}{t}}{-\frac{1}{t^2}}=0 among t→0+limf(t)=0, t→0+limf(1−t)=n=1∑∞n21, t→0+limlnt⋅ln(1−t)=t→0+limtlnt=t→0+limt1lnt=t→0+lim−t21t1=0
∴ lim t → 0 + g ( t ) = ∑ n = 1 ∞ 1 n 2 \therefore \underset{t\rightarrow 0^+}{\lim}g\left( t \right) =\sum_{n=1}^{\infty}{\frac{1}{n^2}} ∴t→0+limg(t)=n=1∑∞n21
∴ g ( x ) = ∑ n = 1 ∞ 1 n 2 \therefore g\left( x \right) =\sum_{n=1}^{\infty}{\frac{1}{n^2}} ∴g(x)=n=1∑∞n21
∴ f ( x ) + f ( 1 − x ) + ln x ln ( 1 − x ) = ∑ n = 1 ∞ 1 n 2 \therefore f\left( x \right) +f\left( 1-x \right) +\ln x\ln \left( 1-x \right) =\sum_{n=1}^{\infty}{\frac{1}{n^2}} ∴f(x)+f(1−x)+lnxln(1−x)=n=1∑∞n21
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