Enable contrast version

Tutor profile: Christopher T.

Inactive
Christopher T.
Tutor with 14 years experience at the college level.
Tutor Satisfaction Guarantee

Questions

Subject: Discrete Math

TutorMe
Question:

Use the Principle of Mathematical induction to show that $$\displaystyle\sum_{j=1}^nj\binom{n}{j} = n2^{n-1}$$ for all $$n\geq 1$$.

Inactive
Christopher T.
Answer:

$$\textit{Proof}$$: $$\textbf{Base Case } (n=1)$$: We see that $(\sum_{j=1}^1j\binom{1}{k} = 1\cdot \binom{1}{1} = 1 = 1\cdot 2^0 = 1\cdot 2^{1-1}.$) Therefore, the base case is true. $$\textbf{Inductive Hypothesis}$$: Assume for any $$k\geq 1$$ that $$\displaystyle\sum_{j=1}^k j\binom{k}{j} = k2^{k-1}$$. $$\textbf{Inductive Step } (n=k+1)$$: We seek to show that $(\sum_{j=1}^{k+1}j\binom{k+1}{j} = (k+1)2^k.$) We first note that $(\sum_{j=1}^{k+1}j\binom{k+1}{j} = \sum_{j=1}^{k}j\binom{k+1}{j} + (k+1)\binom{k+1}{k+1} = \sum_{j=1}^{k}j\binom{k+1}{j} + k+1$) Next, recall Pascal's identity $(\binom{n+1}{k} = \binom{n}{k}+\binom{n}{k-1}$) Thus, $(\begin{aligned} \sum_{j=1}^{k}j\binom{k+1}{j} + k+1 &= \sum_{j=1}^{k}j\left(\binom{k}{j}+\binom{k}{j-1}\right) + k+1\\ &= \sum_{j=1}^{k}j\binom{k}{j} + \sum_{j=1}^{k}j\binom{k}{j-1} + k+1 \\ &= \sum_{j=1}^{k}j\binom{k}{j} + \sum_{j=1}^{k}(j-1+1)\binom{k}{j-1} + k+1 \\ &= \sum_{j=1}^{k}j\binom{k}{j} + {\color{red}\sum_{j=1}^{k}(j-1)\binom{k}{j-1}} + {\color{red}k} + {\color{blue}\sum_{j=1}^k \binom{k}{j-1}} +{\color{blue}1}\end{aligned}\tag{1}$) Let's focus on the sum $(\sum_{j=1}^k (j-1)\binom{k}{j-1}.$) Expanding it out, we see that $(\sum_{j=1}^k (j-1)\binom{k}{j-1} = 0\cdot\binom{k}{0} + 1\cdot \binom{k}{1}+2\cdot \binom{k}{2}+\cdots + (k-1)\binom{k}{k-1}.$) Now note that $$k = k\cdot 1 = k\cdot \dbinom{k}{k}$$. Therefore, $(\begin{aligned}{\color{red}\sum_{j=1}^{k}(j-1)\binom{k}{j-1}}+{\color{red}k} &= 0\cdot\binom{k}{0} + 1\cdot \binom{k}{1}+2\cdot \binom{k}{2}+\cdots + (k-1)\binom{k}{k-1} + k\binom{k}{k} \\ &= 1\cdot \binom{k}{1}+2\cdot \binom{k}{2}+\cdots + (k-1)\binom{k}{k-1} + k\binom{k}{k}\\ &= {\color{red}\sum_{j=1}^k j\binom{k}{j}}.\end{aligned}\tag{2}$) Now let us consider the sum $(\sum_{j=1}^k\binom{k}{j-1}.$) Expanding it out, we see that $(\sum_{j=1}^k\binom{k}{j-1} = \binom{k}{0} + \binom{k}{1}+\binom{k}{2} + \cdots + \binom{k}{k-1}.$) Now note that $$1=\binom{k}{k}$$. Therefore, $(\begin{aligned}{\color{blue}\sum_{j=1}^k\binom{k}{j-1}}+{\color{blue}1} &= \binom{k}{0} + \binom{k}{1}+\binom{k}{2} + \cdots + \binom{k}{k-1} + \binom{k}{k} \\ &= {\color{blue}\sum_{j=0}^k\binom{k}{j}}\end{aligned}\tag{3}$) Substituting the results (2) and (3) into the last equation in (1) gives us $(\begin{aligned} \sum_{j=1}^{k}j\binom{k}{j} + {\color{red}\sum_{j=1}^{k}(j-1)\binom{k}{j-1}} + {\color{red}k} + {\color{blue}\sum_{j=1}^k \binom{k}{j-1}} +{\color{blue}1} &= \sum_{j=1}^kj\binom{k}{j} + {\color{red}\sum_{j=1}^kj\binom{k}{j}} + {\color{blue}\sum_{j=0}^k\binom{k}{j}}\\ &= 2\sum_{j=1}^kj\binom{k}{j} + \sum_{j=0}^k\binom{k}{j} \\ &= 2\cdot k2^{k-1}+ 2^k\quad\left(\text{by IH and $\sum_{i=0}^m\binom{m}{i}=2^m$}\right)\\ &= k2^k+2^k\\ &= (k+1)2^k.\end{aligned}$) This completes the inductive step. Therefore, $$\displaystyle\sum_{j=1}^nj\binom{n}{j} = n2^{n-1}$$ for all $$n\geq 1$$.$$\hspace{.25in}\blacksquare$$

Subject: Statistics

TutorMe
Question:

A survey of 15 large U.S. cities finds that the average commute time one way is 25.4 minutes. A chamber of commerce executive feels that the commute time in his city is less and wants to publicize this. He randomly selects 25 commuters and finds the average is 22.1 minutes with a standard deviation of 5.3 minutes. At $$\alpha=0.1$$, is he correct?

Inactive
Christopher T.
Answer:

We outline all the steps in the hypothesis test. We apply a $$t$$-test for means here. $$\textbf{Step 1}$$: We identify our hypotheses as follows: $$\hspace{.25in}\begin{aligned}[t]H_0: \mu &= 25.4\\ H_1: \mu &< 25.4\quad (\text{claim})\end{aligned}$$. This is a left-tailed test. $$\textbf{Step 2}$$: Find the critical value(s). Since $$\alpha=0.1$$, we find that our critical value is $$CV=-1.318$$ (note that we use the $$t$$-distribution table with $$\alpha=0.1$$ in one-tail and $$25-1=24$$ degrees of freedom to find $$t_{0.1}$$ but take the CV to be negative since it's a left-tail test); any test value that is less than $$-1.318$$ will lie in the rejection region for our test and would force us to reject $$H_0$$. $$\textbf{Step 3}$$: Find the test value. For the $$t$$-test, the test value is given by $$t=\dfrac{\overline{X}-\mu}{s/\sqrt{n}}$$. In the problem, we're given $$\overline{X}=22.1$$, $$s=5.3$$, $$n=25$$, and $$\mu=25.4$$ (from the null hypothesis). Thus, our test value is $$t=\dfrac{22.1-25.4}{5.3/\sqrt{25}}\approx -3.113$$. $$\textbf{Step 4}$$: Decide whether to reject or not reject $$H_0$$. Since our test value is smaller than the critical value ($$-3.113<-1.318$$), we reject $$H_0$$. $$\textbf{Step 5}$$: Summarize the results. There is sufficient evidence to support the claim that the average commute time in his city is less than $$25.4$$ minutes.

Subject: Differential Equations

TutorMe
Question:

Find the solution to the initial value problem $(\mathbf{x}^{\prime} = \begin{bmatrix}2 & \frac{3}{2}\\ -\frac{3}{2} & -1\end{bmatrix}\mathbf{x},\qquad \mathbf{x}(0)=\begin{bmatrix}3\\-2\end{bmatrix}.$)

Inactive
Christopher T.
Answer:

We first find the eigenvalues and eigenvectors of $$A$$. Since $$A=\begin{bmatrix}2 & \frac{3}{2}\\ -\frac{3}{2} & -1\end{bmatrix}$$, we find that $(p(\lambda) = \lambda^2-(\mathrm{tr} A)\lambda + \det A = \lambda^2-\lambda+\tfrac{1}{4}.$) Thus, $$p(\lambda)= 0 \implies \lambda^2-\lambda+\tfrac{1}{4}= 0 \implies (\lambda-\tfrac{1}{2})^2=0\implies \lambda=\tfrac{1}{2}$$ with multiplicity two. Since $$A$$ isn't a diagonal matrix, $$A$$ isn't diagonalizable. We now find the eigenvector associated with $$\lambda=\tfrac{1}{2}$$. If $$\lambda=\tfrac{1}{2}$$, then $$(A-\lambda I)\mathbf{v} = \mathbf{0}\implies \begin{bmatrix}\frac{3}{2} & \frac{3}{2}\\ -\frac{3}{2} & -\frac{3}{2}\end{bmatrix}\begin{bmatrix}v_1\\v_2\end{bmatrix} = \begin{bmatrix}0\\0\end{bmatrix}\implies \left\{\begin{aligned} \tfrac{3}{2}v_1+\tfrac{3}{2}v_2 &= 0,\\ -\tfrac{3}{2}v_1-\tfrac{3}{2}v_2 &= 0.\end{aligned}\right.$$ This system is redundant (since second equation is negative of the first), so we have $$\tfrac{3}{2}v_1+\tfrac{3}{2}v_2=0$$. Letting $$v_2$$ be free, we have that $$v_1=-v_2$$. Hence, $(\mathbf{v} = \begin{bmatrix}v_1\\v_2\end{bmatrix} = \begin{bmatrix}-v_2\\v_2\end{bmatrix} = v_2\begin{bmatrix}-1\\1\end{bmatrix}\implies \mathbf{v} = \begin{bmatrix}-1\\1\end{bmatrix}\quad (v_2=1).$) We now seek to find a vector $$\mathbf{w}$$ such that $((A-\lambda I)\mathbf{w}=\mathbf{v} \implies \begin{bmatrix}\frac{3}{2} & \frac{3}{2}\\ -\frac{3}{2} & -\frac{3}{2}\end{bmatrix}\begin{bmatrix}w_1\\w_2\end{bmatrix} = \begin{bmatrix}-1\\1\end{bmatrix}\implies \left\{\begin{aligned} \tfrac{3}{2}w_1+\tfrac{3}{2}w_2 &= -1,\\ -\tfrac{3}{2}w_1-\tfrac{3}{2}w_2 &= 1.\end{aligned}\right.$) This system is redundant (since second equation is negative of the first), so we have $$\tfrac{3}{2}w_1+\tfrac{3}{2}w_2=-1$$. Letting $$w_2$$ be free, we have that $$\tfrac{3}{2}w_1=-1-\tfrac{3}{2}w_2\implies w_1=-\tfrac{2}{3}-w_2$$. Hence, $(\mathbf{w} = \begin{bmatrix}w_1\\w_2\end{bmatrix} = \begin{bmatrix}-\frac{2}{3}-w_2\\w_2\end{bmatrix} = \begin{bmatrix}-\frac{2}{3}\\0\end{bmatrix}+w_2\begin{bmatrix}-1\\1\end{bmatrix}\implies \mathbf{w} = \begin{bmatrix}-\frac{2}{3}\\0\end{bmatrix}\quad (w_2=0).$) Thus, $$\mathbf{x}(t) = c_1\mathbf{v}e^{\lambda t}+c_2(t\mathbf{v}+\mathbf{w})e^{\lambda t}\implies \mathbf{x}(t) = c_1\begin{bmatrix}-1\\1\end{bmatrix}e^{\frac{t}{2}}+c_2\left(t\begin{bmatrix}-1\\1\end{bmatrix}+\begin{bmatrix}-\frac{2}{3}\\0\end{bmatrix}\right)e^{\frac{t}{2}}$$. Applying the initial condition $$\mathbf{x}(0)=\begin{bmatrix}3\\-2\end{bmatrix}$$, we have $(c_1\begin{bmatrix}-1\\1\end{bmatrix}+c_2\begin{bmatrix}-\frac{2}{3}\\0\end{bmatrix} = \begin{bmatrix}3\\-2\end{bmatrix}\implies \left\{\begin{aligned} -c_1-\tfrac{2}{3}c_2 &= 3\\ c_1 &= -2\end{aligned}\right.\implies c_1=-2,c_2=-\tfrac{3}{2}.$) Therefore, $$\boxed{\mathbf{x}(t) = -2\begin{bmatrix}-1\\1\end{bmatrix}e^{\frac{t}{2}}-\tfrac{3}{2}\left(t\begin{bmatrix}-1\\1\end{bmatrix}+\begin{bmatrix}-\frac{2}{3}\\0\end{bmatrix}\right)e^{\frac{t}{2}}}$$.

Contact tutor

Send a message explaining your
needs and Christopher will reply soon.
Contact Christopher

Request lesson

Ready now? Request a lesson.
Start Lesson

FAQs

What is a lesson?
A lesson is virtual lesson space on our platform where you and a tutor can communicate. You'll have the option to communicate using video/audio as well as text chat. You can also upload documents, edit papers in real time and use our cutting-edge virtual whiteboard.
How do I begin a lesson?
If the tutor is currently online, you can click the "Start Lesson" button above. If they are offline, you can always send them a message to schedule a lesson.
Who are TutorMe tutors?
Many of our tutors are current college students or recent graduates of top-tier universities like MIT, Harvard and USC. TutorMe has thousands of top-quality tutors available to work with you.
BEST IN CLASS SINCE 2015
TutorMe homepage
Made in California by Zovio
© 2013 - 2021 TutorMe, LLC
High Contrast Mode
On
Off