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# Tutor profile: Enzo A.

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Enzo A.
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## Questions

### Subject:Linear Algebra

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Question:

Let's suppose $$J$$ is a $$n \times n$$ matrix such that every entry is equal to $$1$$. Which numbers are their eigenvalues ?

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Enzo A.

First, it is not hard to check the following equality $( J^2 = nJ$) Therefore, $$J^2 - nJ = 0$$. The important fact, is that we've computed the minimal polynomial of $$J$$ ! In fact, since $$J$$ is not the identity matrix (or $$\lambda I$$ for some $$\lambda$$) its minimal polynomial must have degree greater or equal than 2. On the other hand, taking $(p(x) = x^2 - nx = x(x - n),$) we have found a monic polynomial of degree 2 such that $$p(J)= 0$$. Then, such $$p(x)$$ is the minumal polynomial of $$J$$. The eigenvalues of $$J$$ are the roots of such polynomial. In other words, its eigenvalues are $$0$$ and $$n$$.

### Subject:Number Theory

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Question:

What is last digit of $$3^{2018}$$ ?

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Enzo A.

My first thought when I used to face this kind of problem was: "How the heck I am supposed to find the last digit of such a big number ?! ". Fun fact: you don't need to calculate that number and then check the last digit. Since we are interested in the last digit, the question could be paraphrased as: which number $$x \in \left \{ 0, \ldots, 9\right \}$$ satisfy the following equation? $( 3^{2018} \equiv x \ \text{mod} \ 10$) If we start computing the first powers of $$3$$, we'll realize that $$3^4 \equiv 81 \equiv 1 \ \text{mod} \ 10$$. Therefore, powering the previous equality, we obtain $(3^4 \equiv 1 \ \text{mod } 10 \implies 3^{2016} \equiv \left (3^4 \right )^{504} \equiv 1^{504} \equiv 1 \ \text{mod }10$) Finally, multiplying by $$3^2$$, we obtain $( 3^{2018} \equiv 3^2 \equiv 9 \ \text{mod }10$) Therefore, the last digit of $$3^{2018}$$ is $$9$$.

### Subject:Calculus

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Question:

Calculate the following integral $(\displaystyle{\int_{0}^{\frac{\pi}{2}}\dfrac{\sin ^{2018}(x)}{\sin^{2018}(x) + \cos^{2018}(x)}\mathrm{d} x }$)

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Enzo A.

At first, it seems borderline impossible to calculate such an integral. It even has a $$2018$$ as exponent ! Well, let's denote $( I = \displaystyle{\int_{0}^{\frac{\pi}{2}}\dfrac{\sin ^{2018}(x)}{\sin^{2018}(x) + \cos^{2018}(x)}\mathrm{d} x }$) In other words, we are looking for the value of $$I$$. We are integrating from $$0$$ to $$\pi /2$$. What if we integrate in the opposite direction ? and what do we mean by a statement like that? Well, if you define $$y = \frac{\pi}{2} - x$$ and you look at how $$y$$ changes as $$x$$ moves from $$\pi/2$$ to $$0$$, it is basically to move in the opposite direction. In this case, using this change of variable, we'll obtain $$\mathrm{d}x = -\mathrm{d}y$$ and hence $( I = \displaystyle{\int_{\frac{\pi}{2}}^{0}\dfrac{\sin ^{2018}\left (\frac{\pi}{2} - y\right )}{\sin^{2018}\left (\frac{\pi}{2} - y \right ) + \cos^{2018}\left (\frac{\pi}{2} - y\right)}(-\mathrm{d} y) }$) Using the following well-known trigonometric identities ( \begin{align*} \sin \left (\frac{\pi}{2} - y\right ) &= \cos (y) \\ cos\left ( \frac{\pi}{2} - y \right ) &= \sin(y)\end{align*}) Therefore, \begin{align*} I &= \displaystyle{\int_{\frac{\pi}{2}}^{0}\dfrac{\sin ^{2018}\left (\frac{\pi}{2} - y\right )}{\sin^{2018}\left (\frac{\pi}{2} - y \right ) + \cos^{2018}\left (\frac{\pi}{2} - y\right)}(-\mathrm{d} y) } \\ \\ &= \displaystyle{\int_{\frac{\pi}{2}}^{0}\dfrac{\cos ^{2018}(y)}{\cos^{2018}\left (y \right ) + \sin^{2018}\left ( y\right)}(-\mathrm{d} y) } \\ \\ &= \displaystyle{\int_{0}^{\frac{\pi}{2}}\dfrac{\cos ^{2018}(y)}{\cos^{2018}\left (y \right ) + \sin^{2018}\left ( y\right)}\mathrm{d} y } \end{align*} Adding $$I$$ and $$I$$ (using the two equalities we have obtained for $$I$$), we have \begin{align*} I + I &= \left ( \displaystyle{\int_{0}^{\frac{\pi}{2}}\dfrac{\sin ^{2018}(x)}{\sin^{2018}(x) + \cos^{2018}(x)}\mathrm{d} x } \right ) + \left ( \displaystyle{\int_{0}^{\frac{\pi}{2}}\dfrac{\cos ^{2018}(y)}{\cos^{2018}\left (y \right ) + \sin^{2018}\left ( y\right)}\mathrm{d} y } \right ) \\ \\ &= \displaystyle{\int_{0}^{\frac{\pi}{2}}\dfrac{\sin ^{2018}(x)+ \cos^{2018}(x)}{\sin^{2018}(x) + \cos^{2018}(x)}\mathrm{d} x } \\ \\ &= \displaystyle{\int_{0}^{\frac{\pi}{2}}1 \mathrm{d} x = \dfrac{\pi}{2}}\end{align*} Finally $( 2 I = \dfrac{\pi}{2 } \implies I = \dfrac{\pi}{4}$)

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