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# Tutor profile: Joseph V.

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Joseph V.
Mathematician
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## Questions

### Subject:Linear Algebra

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Question:

Find the volume of the tetrahedron embedded into three dimensional Euclidean space with vertices $$(1,4,2),(2,1,5),(3,4,7),(1,3,3)$$

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Joseph V.

The volume $$\text{Vol}(A)$$ of the tetrahedron $$A$$ with vertices $$\mathbf{a},\mathbf{b},\mathbf{c},\mathbf{d}$$ can be computed with the following formula: $(\text{Vol}(A)=\frac{1}{6}\cdot|\big{(}(\mathbf{b}-\mathbf{a})\cdot(\mathbf{c}-\mathbf{a})\big{)}\times(\mathbf{d}-\mathbf{a})| =\frac{1}{6}\cdot|\text{Det}(\mathbf{b}-\mathbf{a},\mathbf{c}-\mathbf{a},\mathbf{d}-\mathbf{a}).|$) Observe that since $$\text{Det}(D)=\text{Det}(D^{T})$$ whenever $$D$$ is a square matrix, the determinant does not depend on whether we consider $$\mathbf{b}-\mathbf{a},\mathbf{c}-\mathbf{a},\mathbf{d}-\mathbf{a}$$ to be row vectors or column vectors. Set $$\mathbf{a}=(1,4,2),\mathbf{b}=(2,1,5),\mathbf{c}=(3,4,7),\mathbf{d}=(1,3,3)$$. Then $(\mathbf{b}-\mathbf{a}=(2,1,5)-(1,4,2)=(1,-3,3)$) $(\mathbf{c}-\mathbf{a}=(3,4,7)-(1,4,2)=(2,0,5),$) and $(\mathbf{d}-\mathbf{a}=(1,3,3)-(1,4,2)=(0,-1,1).$) Now, $(\text{Det}(\mathbf{b}-\mathbf{a},\mathbf{c}-\mathbf{a},\mathbf{d}-\mathbf{a})=\begin{vmatrix} 1 & -3 & 3 \\ 2 & 0 & 5 \\ 0 & -1 & 1 \end{vmatrix} =0\cdot\begin{vmatrix} -3 & 3 \\ 0 & 5 \end{vmatrix}- (-1)\cdot\begin{vmatrix} 1 & 3 \\ 2 & 5 \end{vmatrix} +1\cdot\begin{vmatrix} 1 & -3 \\ 2 & 0 \end{vmatrix} =\begin{vmatrix} 1 & 3 \\ 2 & 5 \end{vmatrix} +\begin{vmatrix} 1 & -3 \\ 2 & 0 \end{vmatrix} =5-6+6=5.$) Therefore, $(\text{Vol}(A)=\frac{5}{6}.$)

### Subject:Discrete Math

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Question:

Prove that $(\sum_{k=1}^{n}k=\frac{n(n+1)}{2}$) whenever $n$ is a natural number with $$n\geq 1.$$

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Joseph V.

We shall prove this result by induction since induction is the most straightforward proof strategy for these kinds of problems. Let $$P(n)$$ denote the statement $(\sum_{k=1}^{n}k=\frac{n(n+1)}{2}.$) In order to prove $$P(n)$$ for all natural numbers $$n\geq 1$$, we need to prove both the base case $$P(1)$$ and the induction step which states that $$P(n)\rightarrow P(n+1)$$. For the base case, if $$n=1$$, then $(\sum_{k=1}^{n}k=1$) and $(\frac{n(n+1)}{2}=\frac{1(1+1)}{2}=\frac{2}{2}=1=\sum_{k=1}^{n}k,$) so we conclude that $$P(1)$$ holds. For the induction step, suppose that $$n\geq 1$$, and assume that $$P(n)$$ holds. Then $(\sum_{k=1}^{n+1}k=n+1+\sum_{k=1}^{n}k=n+1+\frac{n(n+1)}{2})$) $(=\frac{2(n+1)+n(n+1)}{2}=\frac{(n+2)(n+1)}{2}=\frac{(n+1)(n+2)}{2},$) so we conclude that $$P(n+1)$$ holds as well.

### Subject:Calculus

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Question:

Show that there exists a positive real number $$x$$ where $$x^{x}=x+1$$?

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Joseph V.

The intermediate value theorem states that if $$f:[a,b]\rightarrow(-\infty,\infty)$$ is a continuous function and $$f(a)\geq 0,f(b)\leq 0$$ or $$f(a)\leq 0,f(b)\geq 0$$, then there exists some $$c\in[a,b]$$ where $$f(c)=0$$. If we define $(f:[1,2]\rightarrow x^{x}-(x+1)$), then $(f(1)=1^{1}-(1+1)=1-2=-1,$) but $(f(2)=2^{2}-(2+1)=4-3=1.$) Therefore, by the intermediate value theorem, we conclude that there is some $$x$$ with $$1\leq x\leq 2$$, but where $$x^{x}-(x+1)=f(x)=0.$$ Therefore, $$x^{x}=x+1.$$

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