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设f(x)=∑x^n/n^2,证明f(x)+f(1-x)+lnxln(1-x)=∑1/n^2

2022-07-06 23:18:00 【深海里的鱼(・ω＆lt;)*】

题目

设 f ( x ) = ∑ n = 1 ∞ x n n 2 ，证明： f ( x ) + f ( 1 − x ) + ln ⁡ x ln ⁡ ( 1 − x ) = ∑ n = 1 ∞ 1 n 2 \text{设}f\left( x \right) =\sum_{n=1}^{\infty}{\frac{x^n}{n^2}}\text{，证明：}f\left( x \right) +f\left( 1-x \right) +\ln x\ln \left( 1-x \right) =\sum_{n=1}^{\infty}{\frac{1}{n^2}} 设f(x)=n=1∑∞​n2xn​，证明：f(x)+f(1−x)+lnxln(1−x)=n=1∑∞​n21​

解答

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​=x1​n=1∑∞​nxn​=x1​∫0x​n=1∑∞​tn−1dt=x1​∫0x​n=0∑∞​tndt=x1​∫0x​1−t1​dt=−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
令 g ( x ) = f ( x ) + f ( 1 − x ) + ln ⁡ x ln ⁡ ( 1 − x )    x ∈ ( 0 , 1 ) \text{令}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) 令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 ∵∫0x​g′(u)du=g(x)−t→0+lim​g(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+lim​g(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+lim​g(t)=t→0+lim​f(t)+t→0+lim​f(1−t)+t→0+lim​lnt⋅ln(1−t)
其中 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{其中}\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 其中t→0+lim​f(t)=0, t→0+lim​f(1−t)=n=1∑∞​n21​, t→0+lim​lnt⋅ln(1−t)=t→0+lim​tlnt=t→0+lim​t1​lnt​=t→0+lim​−t21​t1​​=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+lim​g(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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版权声明
本文为[深海里的鱼(・ω＆lt;)*]所创，转载请带上原文链接，感谢
https://blog.csdn.net/qq_50710984/article/details/125629549

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