Soal dari Monthly
Beberapa minggu yang lalu, saya sempat pengen mengirim jawaban ke rubrik American Mathematical Monthly tapi pas saya cek deadline nya sudah lewat, jadi ya saya post disini saja.
Solution
Define the sequence \{a_n\}_{n\geq 0} as
a_n := (-1)^n (a^n+b^n)
where a and b are the roots of x^2+x+\frac{1}{2}=0.
Note that \alpha:=-a and \beta:=-b are the roots of quadratic equation x^2-x+\frac{1}{2}=0, and so a_n can also be written as a_n= (-a)^n + (-b)^n = \alpha^n+\beta^n. We can directly check that a_0=2, and by Viete's Formula we have a_1=\alpha+\beta=1 and a_2=\alpha^2 + \beta^2= (\alpha+\beta)^2-2\alpha \beta= 1-1=0.
Also by multiplying the equation with \alpha^n we have \alpha^{n+2} = \alpha^{n+1}-\frac{\alpha^n}{2}, similarly \beta^{n+2} = \beta^{n+1}-\frac{\beta^n}{2} and hence
\begin{align*} \alpha^{n+2} + \beta^{n+2} = \alpha^{n+1} + \beta^{n+1} - \frac{\alpha^n + \beta^n}{2} \end{align*}
Therefore we have the following recursive formula for the sequence a_n = (-1)^n (a^n + b^n) = \alpha^n + \beta^n
a_0= 2\quad a_1=1 \qquad a_{n+2} = a_{n+1} - \frac{a_n}{2} \quad n\geq 0
Since a_2 = \alpha^2+\beta^2 = 0, we have for j \geq 0 a_{4j+2} = (\alpha^2)^{2j+1} + (\beta^2)^{2j+1} = (\alpha^2+\beta^2)(\alpha^{2j-1} + \cdots + \alpha^{2j-1})=0
Thus by the recursive formula
a_{4j+3} = -\frac{a_{4j+1}}{2} \qquad \text{and}\qquad a_{4j+4} = a_{4j+3} \qquad \text{for $j\geq 0$}
This gives us the formula of a_{4k+1} which is
a_{4k+1} = a_{4k} - \frac{a_{4k-1}}{2} = \frac{a_{4k-1}}{2} = -\frac{a_{4(k-1)+1}}{4} = \frac{a_{4(k-2)+1}}{4^2}=\cdots = \frac{(-1)^{k}}{4^{k}} a_1 = \frac{(-1)^{k}}{4^{k}}
From a_{4k+2}=0 and a_{4k+2} = a_{4k+1} - \frac{a_{4k}}{2} we have a_{4k} =2 a_{4k+1} =2\cdot \frac{(-1)^k}{4^k}
And using a_{4k+3} = -\frac{a_{4k+1}}{2} we have
a_{4k+3} = \frac{(-1)^{k+1} }{2\cdot 4^k} = 2\cdot \frac{(-1)^{k+1}}{ 4^{k+1}}
So the formula for a_n is
a_{4k} = 2\cdot \frac{(-1)^k}{4^k} \qquad a_{4k+1}=\frac{(-1)^{k}}{4^{k}} \qquad a_{4k+2}=0 \qquad a_{4k+3}=2\cdot \frac{(-1)^{k+1}}{ 4^{k+1}}
For |x| \leq 1, we have for any n \geq 1
\left| a_n x^{n+1} \right| \leq |a_{n}|
While
\sum_{n=1}^{\infty} |a_n| = \sum_{j=1}^{\infty} \left(\frac{2}{4^j} + \frac{1}{4^j} + \frac{2}{4^{j+1}} \right) = \frac{7}{6}
Therefore by the Weierstrass M-Test, \sum_{n=1}^{\infty} a_n x^{n+1} converges uniformly on [-1,1], let say it converges to a function f(x). We have
\begin{align*} f(x)=\sum_{n=1}^{\infty}a_n x^{n+1} &= \sum_{n=1}^{\infty} a_{n+2} x^{n+3} + a_2 x^3 + a_1 x^2 \\ &= \sum_{n=1}^{\infty} a_{n+1}x^{n+3} - \sum_{n=1}^{\infty} \frac{a_{n}}{2}x^{n+3} + a_2 x^3 + a_1 x^2 \\ &= xf(x) -\frac{x^2f(x)}{2} + x^2 \end{align*}
Thus
f(x)=\frac{2x^2}{x^2-2x+2}
By uniform convergence we have
\int_0^1 f(x) \, dx = \int_0^1 \sum_{n=1}^{\infty}a_n x^{n+1} \, dx = \sum_{n=1}^{\infty} \int_0^1 a_n x^{n+1} \, dx = \sum_{n=1}^{\infty} \frac{a_n}{n+2}
Therefore the sum equals to
\int_0^1 \frac{2x^2}{(x^2-2x+2)}\, dx = \int_0^1 1 + \frac{4x-4}{x^2-2x+2} \, dx = 2x + 2\ln (x^2-2x+2) \bigg|_0^1 = 2-2\ln 2
11876 Let a and b be the roots of x^2+x+\frac{1}{2}=0. Find\sum_{n=1}^{\infty} \frac{(-1)^n (a^n+b^n)}{n+2}
Solution
Define the sequence \{a_n\}_{n\geq 0} as
a_n := (-1)^n (a^n+b^n)
where a and b are the roots of x^2+x+\frac{1}{2}=0.
Note that \alpha:=-a and \beta:=-b are the roots of quadratic equation x^2-x+\frac{1}{2}=0, and so a_n can also be written as a_n= (-a)^n + (-b)^n = \alpha^n+\beta^n. We can directly check that a_0=2, and by Viete's Formula we have a_1=\alpha+\beta=1 and a_2=\alpha^2 + \beta^2= (\alpha+\beta)^2-2\alpha \beta= 1-1=0.
Also by multiplying the equation with \alpha^n we have \alpha^{n+2} = \alpha^{n+1}-\frac{\alpha^n}{2}, similarly \beta^{n+2} = \beta^{n+1}-\frac{\beta^n}{2} and hence
\begin{align*} \alpha^{n+2} + \beta^{n+2} = \alpha^{n+1} + \beta^{n+1} - \frac{\alpha^n + \beta^n}{2} \end{align*}
Therefore we have the following recursive formula for the sequence a_n = (-1)^n (a^n + b^n) = \alpha^n + \beta^n
a_0= 2\quad a_1=1 \qquad a_{n+2} = a_{n+1} - \frac{a_n}{2} \quad n\geq 0
Since a_2 = \alpha^2+\beta^2 = 0, we have for j \geq 0 a_{4j+2} = (\alpha^2)^{2j+1} + (\beta^2)^{2j+1} = (\alpha^2+\beta^2)(\alpha^{2j-1} + \cdots + \alpha^{2j-1})=0
Thus by the recursive formula
a_{4j+3} = -\frac{a_{4j+1}}{2} \qquad \text{and}\qquad a_{4j+4} = a_{4j+3} \qquad \text{for $j\geq 0$}
This gives us the formula of a_{4k+1} which is
a_{4k+1} = a_{4k} - \frac{a_{4k-1}}{2} = \frac{a_{4k-1}}{2} = -\frac{a_{4(k-1)+1}}{4} = \frac{a_{4(k-2)+1}}{4^2}=\cdots = \frac{(-1)^{k}}{4^{k}} a_1 = \frac{(-1)^{k}}{4^{k}}
From a_{4k+2}=0 and a_{4k+2} = a_{4k+1} - \frac{a_{4k}}{2} we have a_{4k} =2 a_{4k+1} =2\cdot \frac{(-1)^k}{4^k}
And using a_{4k+3} = -\frac{a_{4k+1}}{2} we have
a_{4k+3} = \frac{(-1)^{k+1} }{2\cdot 4^k} = 2\cdot \frac{(-1)^{k+1}}{ 4^{k+1}}
So the formula for a_n is
a_{4k} = 2\cdot \frac{(-1)^k}{4^k} \qquad a_{4k+1}=\frac{(-1)^{k}}{4^{k}} \qquad a_{4k+2}=0 \qquad a_{4k+3}=2\cdot \frac{(-1)^{k+1}}{ 4^{k+1}}
For |x| \leq 1, we have for any n \geq 1
\left| a_n x^{n+1} \right| \leq |a_{n}|
While
\sum_{n=1}^{\infty} |a_n| = \sum_{j=1}^{\infty} \left(\frac{2}{4^j} + \frac{1}{4^j} + \frac{2}{4^{j+1}} \right) = \frac{7}{6}
Therefore by the Weierstrass M-Test, \sum_{n=1}^{\infty} a_n x^{n+1} converges uniformly on [-1,1], let say it converges to a function f(x). We have
\begin{align*} f(x)=\sum_{n=1}^{\infty}a_n x^{n+1} &= \sum_{n=1}^{\infty} a_{n+2} x^{n+3} + a_2 x^3 + a_1 x^2 \\ &= \sum_{n=1}^{\infty} a_{n+1}x^{n+3} - \sum_{n=1}^{\infty} \frac{a_{n}}{2}x^{n+3} + a_2 x^3 + a_1 x^2 \\ &= xf(x) -\frac{x^2f(x)}{2} + x^2 \end{align*}
Thus
f(x)=\frac{2x^2}{x^2-2x+2}
By uniform convergence we have
\int_0^1 f(x) \, dx = \int_0^1 \sum_{n=1}^{\infty}a_n x^{n+1} \, dx = \sum_{n=1}^{\infty} \int_0^1 a_n x^{n+1} \, dx = \sum_{n=1}^{\infty} \frac{a_n}{n+2}
Therefore the sum equals to
\int_0^1 \frac{2x^2}{(x^2-2x+2)}\, dx = \int_0^1 1 + \frac{4x-4}{x^2-2x+2} \, dx = 2x + 2\ln (x^2-2x+2) \bigg|_0^1 = 2-2\ln 2
Post a Comment: