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# Tutor profile: Yannic V.

Yannic V.
Ph.D. in mathematics

## Questions

### Subject:Discrete Math

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Question:

Find a formula for the sum $\sum_{k=0}^n k \binom{n}{k},$ by means of the binomial theorem. What can be deduced about the average number of elements in a random subset of $\{1,2, \hdots, n\}$?

Yannic V.

Let $x,y \in \mathbb{R}$. Remember that the binomial theorem sates that $(x+y)^n = \sum_{k=0}^n \binom{n}{k}x^{k}y^{n-k},$ for all $n \in \mathbb{N}$. In particular, for $y=1$ we have $(x+1)^n=\sum_{k=0}^n \binom{n}{k}x^k.$ Considering $x$ as a variable, we differentiate both sides of the above equation to obtain $n(x+1)^{n-1}= \sum_{k=0}^n k\binom{n}{k}x^{k-1}.$ It remains to plug in $x=1$: $\sum_{k=0}^n k \binom{n}{k} = n2^{n-1}.$ The average number of elements in a random subset of $\{1,2, \hdots, n\}$ is thus $\frac{\sum_{k=0}^n k \binom{n}{k}}{\sum_{k=0}^n \binom{n}{k}} = \frac{n 2^{n-1}}{2^n}=\frac{n}{2}.$ This is not surprising, since the number of subsets of size $k$ is exactly the same as the number of subsets of size $n-k$, for any $k$.

### Subject:Calculus

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Question:

Solve the integral $\displaystyle \int 3 \sec(2x-1)\tan(2x-1) \,dx$.

Yannic V.

As the arguments of both functions in the integral are not single variables, we can start to consider a change of variable in order to just have one variable in the secant and tangent function. This can be made using the change of variable \begin{equation}\label{cambio1} \left\lbrace\begin{array}{l} y = 2x-1 \\ dy = 2 \,dx \end{array}\right. \end{equation} Notice that the derivative $dy$ is easy to obtain from the expression of the integral: $dx$ can be expressed in terms of $dy$ as \begin{equation}\label{eq1} dx = \frac{dy}{2}. \end{equation} Substituting the change of variable \eqref{cambio1} in the integral, we obtain \begin{align} \int 3 \sec(2x-1)\tan(2x-1)\, dx &= \int 3 \sec(y)\tan(y) \,\frac{dy}{2} \tag{\footnotesize by \eqref{cambio1} and \eqref{eq1}}\\ &= \frac{3}{2} \int \sec(y)\tan(y) \,dy. \label{da1} \end{align} We have now reduced our problem to a more simple integral. We want to integrate the expression $\sec(y)\tan(y)$, which is the derivative of\footnote{This comes from the following calculation: $(\sec(y))'= \left(\frac{1}{\cos(y)} \right)' = \frac{(0\cos(y))-1(-\sin(y))}{\cos(y)^2}= \frac{\sin(y)}{\cos(y)^2} = \frac{1}{\cos(y)}\frac{\sin(y)}{\cos(y)}=\sec(y)\tan(y).$ } the function $\sec(y)$. It is then appropriate to consider the following change of variable: \begin{equation}\label{cambio2} \left\lbrace\begin{array}{l} u = \sec(y)\\ du = \sec(y)\tan(y) \,dy \end{array}\right. \end{equation} We can now replace \eqref{cambio2} into the expression \eqref{da1}. Remark that \eqref{da1} just contain the derivative part of\footnote{This is not a problem. When doing a change of variable, the most important part is that the expression corresponding to the derivative of the change of basis appears in the integral (up to a scalar). } \eqref{cambio2}. Thus, we have: \begin{align} \int 3 \sec(2x-1)\tan(2x-1)\, dx &= \frac{3}{2} \int \sec(y)\tan(y) \,dy. \tag{\footnotesize{by \eqref{da1}}}\\ &= \frac{3}{2} \int du \tag{\footnotesize{by \eqref{cambio2}}}\\ &= \frac{3}{2}\, u + C, \label{da2} \end{align} where $C$ is a constant. Finally, we just have to substitute $u$ by its value in terms of $y$ (given by \eqref{cambio2}), and replace $y$ by its value in terms of $x$ (given by \eqref{cambio1}): \begin{align} \int 3 \sec(2x-1)\tan(2x-1)\, dx &= \frac{3}{2}\, u + C \tag{\footnotesize{by \eqref{da2}}}\\ &= \frac{3}{2}\sec(y) + C \tag{\footnotesize{substitute $u=\sec(y)$}}\\ &= \frac{3}{2}\sec(2x-1) + C. \tag{\footnotesize{substitute $y=2x-1$}} \end{align}

### Subject:Algebra

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Question:

If $G$ is a finite group, show that the number of elements in $G$ of order greater than $2$ must be even. In particular, prove that any group of even order must contain an element of order 2.

Yannic V.

We will solve the question by a counting argument as follows. Let $X$ be the set of all elements of G with order greater than two. If $X = \emptyset$ then $|X|=0$ and we’re done. Suppose now that $X \neq \emptyset$. Since the orders of $g$ and $g^{-1}$ coincides for any $g \in G$, we have that $g \in X \Longleftrightarrow g^{-1} \in X.$ Moreover, if $g \in X$ then $g^2\neq e_G$. From here, $g \neq g^{-1}$. It follows that $X$ can be written as the disjoint union of two element sets of the form $\{g, g^{-1}\}$ {x,x−1}, and hence that $|X|$ is even. Suppose that $|G|$ is even. Since $e_G$ has order $1$, $e_G \notin X$. It follows that $G \setminus X \neq \emptyset$. So $0 < |G \setminus X|=|G|-|X|$. Since $|G|$ and $|X|$ are both even, it follows that $|G\setminus X|$ is a nonzero even integer, i.e. is at least 2. Thus, there is an $z \in G \setminus X$ such that $z \neq e_G$ . Since $X$ consists of all elements in $G$ of order greater than $2$, it must be the case that $|z|=2$.

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