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
\[\int_0^1 \frac{2x^2}{(x^2-2x+2)}\, dx = \int_0^1 1 + \frac{4x-4}{x^2-2x+2} \, dx = 2x + 2\ln (x^2-2x+2) \bigg|_0^1 = 2-2\ln 2 \]
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