Problems

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Solution:

1. If we can show that \(0 < a - \sqrt{a^2-b} < 1\) then we will be done, since raising a number in \([0,1)\) to a positive integer power will always remain in the same interval. Clearly \(\sqrt{a^2-b} < \sqrt{a^2} = a\) so we have \(a-\sqrt{a^2-b} > 0\) We also have that \(\sqrt{a^2-b} > \sqrt{a^2-(2a-1)} = (a-1)\). Therefore \(a - \sqrt{a^2-b} < a - (a-1) = 1\) as required.
2. If we can show that \(\l a + \sqrt{a^2-b} \r^n + \l a - \sqrt{a^2-b} \r^n = N -r + s\) is an integer we will be done, since the only integer value \(-r+s\) can be is \(0\). This is easy to see, since \begin{align*} \l a + \sqrt{a^2-b} \r^n + \l a - \sqrt{a^2-b} \r^n &= \sum_{k=0}^n \binom{n}{k}a^{n-k}(\sqrt{a^2-b})^k +\sum_{k=0}^n \binom{n}{k}a^{n-k}(-\sqrt{a^2-b})^k \\ &= \sum_{k=0}^n \binom{n}{k}a^{n-k}\l (\sqrt{a^2-b})^k +(-\sqrt{a^2-b})^k \r \\ \end{align*} But every term where \(k\) is odd in this sum is \(0\) (since they cancel) and ever term where \(k\) is even in this sum is an integer. Therefore the sum is an integer and we're done.
3. \begin{align*} -r^2+rN &= -r(r-N) \\ &= s(r-N) \\ &=- \l a - \sqrt{a^2-b} \r^n \l a + \sqrt{a^2-b} \r^n \\ &= -\l \l a - \sqrt{a^2-b} \r \l a + \sqrt{a^2-b} \r\r^n \\ &= - \l a^2 - a^2+b\r^n \\ &= b^n \end{align*} Therefore \(r^2-rN + b^n = 0\)

Looking at \(\left(8+3\sqrt{7}\right)^{n}\) we have \(a = 8, b = 1\) (since \(8^2 - 1 = 9 \cdot 7\). So we can apply the result of the previous question to see that: \(\left(8+3\sqrt{7}\right)^{n}\) differs from an integer by \(\left(8-3\sqrt{7}\right)^{n}\). \begin{align*} 8-3\sqrt{7} &= \frac{1}{8+3\sqrt{7}} \\ &\approx \frac{1}{8 + 8} \\ &\approx 2^{-4} \end{align*} Therefore it differs by approximation \((2^{-4})^n = 2^{-4n}\)

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Solution: Let \({\displaystyle I_{m,n}=\int\cos^{m}x\sin nx\,\mathrm{d}x,}\) Then \begin{align*} && I_{m,n} &= \int\cos^{m}x\sin nx\,\mathrm{d}x \\ &&&= \left [ -\frac1n \cos^m x \cos n x \right] - \frac{m}{n} \int \sin^{m-1} x \cos x \cos n x \d x \\ &&&= \left [ -\frac1n \cos^m x \cos n x \right] + \frac{m}{n} \int \sin^{m-1} x (\cos (n-1)x -\sin x \sin nx) \d x\\ &&&= \left [ -\frac1n \cos^m x \cos n x \right] + \frac{m}{n} \int \sin^{m-1} x \cos (n-1)x \d x-\frac{m}{n} I_{m,n} \\ &&&= \left [ -\frac1n \cos^m x \cos n x \right] + \frac{m}{n} I_{m-1,n-1} -\frac{m}{n} I_{m,n} \\ \Rightarrow && nI_{m,n} &= -\cos^m x \cos n x + mI_{m-1,n-1} -mI_{m,n}\\ \Rightarrow && (m+n)I_{m,n} &= -\cos^m x \cos n x + mI_{m-1,n-1} \end{align*}

1. Note that \(I_{2m,0} = 0\) (the integrand is 0) and \(I_{0, 2m} = 0\) (symmetry for our limits). \(\displaystyle \left [-\cos^m x \cos n x \right]_0^\pi = \l - (-1)^m (-1)^n \r - \l -1 \r = 1 - (-1)^{m+n} = 0\) since \(m+n\) is even. Therefore all reductions are \(I_{m,n} = \frac{I_{m-1,n-1}}{m+n}\) terminating at \(0\), so all values are zero
2. \begin{align*} \int_{0}^{\frac{\pi}{2}}\sin^{2}x\sin3x\,\mathrm{d}x &= \int_{0}^{\frac{\pi}{2}}(1-\cos^2x)\sin3x\,\mathrm{d}x \\ &= I_{0,3} - I_{2,3} \\ &= \frac13 - \frac15 \l \left [-\cos^2 x \cos 3 x \right]_0^{\pi/2} + 2 \cdot I_{1,2} \r \\ &= \frac13 - \frac15 \l 1 + \frac23 \l \left [-\cos x \cos 2 x \right]_0^{\pi/2} + 1\cdot I_{0,1} \r \r \\ &= \frac13 - \frac15 -\frac2{15} - \frac2{15} \\ &= \frac{5}{15} - \frac{3}{15} - \frac{4}{15} \\ &= -\frac2{15} \end{align*}