MATH1251 Questions HELP (4 Viewers)

InteGrand · Oct 7, 2016

1008 said:
These as well, please, if anyone could solve these that would be great. I'll be posting more questions up as exams are roughly in a month...

$\\\text{(1) Two examples involving time explicitly: }\\\text{(a) Seasonal variation: Solve }\\ \frac{\mathrm{d} y}{\mathrm{d} t}=k(1+\text{a} \cos(2\pi t))y.\\\text{(b) It is estimated that the population growth rate of a certain developing country will fall linearly from }\\\text{2\% per year to 1.5\% per year over the next decade. The present population is 10 million. What is the}\\\text{ estimated population in 10 years time? }$

$\\\text{(2) The differential equation } \frac{\mathrm{d} y}{\mathrm{d} x} =\lambda \frac{y}{x}\text{ is employed as a simple model for comparative growth.}\\\text{(a) Solve.}\\\text{(b) For the fiddler crab, the weights of the large claw, y grams, }\\\text{and the body (without claw), x grams, were found to be }$

$\\\text{Show that the above model appears to be reasonable in this example. Determine approximately }\lambda \text{ and the}\\\text{constant of integration.}$

$\noindent (1) (a) Just separate variables and integrate, so $\ln y = k \left(t + \frac{a}{2\pi}\sin (2\pi t)\right) + c \Rightarrow y = A \exp \left(k \left(t + \frac{a}{2\pi} \sin (2\pi t)\right)\right)$, where $A$ is an arbitrary constant.$

$\noindent (b) The growth rate is $k(t) = 0.02 - \frac{0.005}{10}t$, for $0\leq t \leq 10$, and for this time period we have $\frac{\mathrm{d}y}{\mathrm{d}t} = k(t)y$, where $y(t)$ is the population at time $t$. So $y(10) = y(0) \exp \left( \int _{0}^{10} k(t)\, \mathrm{d}t\right)$. Since we are given $y(0)$ and have the expression for $k(t)$ (which is easy to integrate), we can compute $y(10)$.$

InteGrand · Oct 7, 2016

1008 said:
These as well, please, if anyone could solve these that would be great. I'll be posting more questions up as exams are roughly in a month...

$\\\text{(1) Two examples involving time explicitly: }\\\text{(a) Seasonal variation: Solve }\\ \frac{\mathrm{d} y}{\mathrm{d} t}=k(1+\text{a} \cos(2\pi t))y.\\\text{(b) It is estimated that the population growth rate of a certain developing country will fall linearly from }\\\text{2\% per year to 1.5\% per year over the next decade. The present population is 10 million. What is the}\\\text{ estimated population in 10 years time? }$

$\\\text{(2) The differential equation } \frac{\mathrm{d} y}{\mathrm{d} x} =\lambda \frac{y}{x}\text{ is employed as a simple model for comparative growth.}\\\text{(a) Solve.}\\\text{(b) For the fiddler crab, the weights of the large claw, y grams, }\\\text{and the body (without claw), x grams, were found to be }$

$\\\text{Show that the above model appears to be reasonable in this example. Determine approximately }\lambda \text{ and the}\\\text{constant of integration.}$

$\noindent (2) (a) Separate variables again and we obtain $\ln y = \lambda \ln x + c$. So $y = Ax^{\lambda}$, where $A$ is an arbitrary constant.$

$\noindent (b) Notice that $\ln y$ is linearly related to $\ln x$. So we could plot the data using a \textsl{log}-\textsl{log plot}. You could use a computer to get the log-log plot (which will give a straight line approximately, if the model is a good fit). So to show the model is reasonable, check if the log-log plot yields a roughly linear plot (try fitting a line to the logs of the data and check the $R^2$ value for example. It should be near $1$ for a good fit). You can look at the value of $\lambda$ and $c$ from the slope and intercept that the computer gives for the line. If you wanted to do it by hand and calculator, you could calculate the logs of the data and fit it to a line $Y = \lambda X + c$ (where $Y = \ln y$, $X = \ln x$) by solving the normal equations $A^T A \bold{x} = A^T \bold{y}$, where as usual $\bold{x} = \begin{pmatrix} \lambda \\ c\end{pmatrix}$ say and so on.$

1008 · Oct 8, 2016

InteGrand said:
$\noindent (2) (a) Separate variables again and we obtain $\ln y = \lambda \ln x + c$. So $y = Ax^{\lambda}$, where $A$ is an arbitrary constant.$

$\noindent (b) Notice that $\ln y$ is linearly related to $\ln x$. So we could plot the data using a \textsl{log}-\textsl{log plot}. You could use a computer to get the log-log plot (which will give a straight line approximately, if the model is a good fit). So to show the model is reasonable, check if the log-log plot yields a roughly linear plot (try fitting a line to the logs of the data and check the $R^2$ value for example. It should be near $1$ for a good fit). You can look at the value of $\lambda$ and $c$ from the slope and intercept that the computer gives for the line. If you wanted to do it by hand and calculator, you could calculate the logs of the data and fit it to a line $Y = \lambda X + c$ (where $Y = \ln y$, $X = \ln x$) by solving the normal equations $A^T A \bold{x} = A^T \bold{y}$, where as usual $\bold{x} = \begin{pmatrix} \lambda \\ c\end{pmatrix}$ say and so on.$

Thanks! I've got a few more...

EDIT: Figured out two, I've deleted those from here

InteGrand · Oct 8, 2016

1008 said:
Thanks! I've got a few more...

EDIT: Figured out two, I've deleted those from here

$\noindent (1) The motion is $\ddot{x} = - \omega^{2} x$, where $\omega^{2} = \frac{\pi\cdot 40^{2}g}{m}$. Hence the period is $T = \frac{2\pi}{\omega} = 2\pi \times \sqrt{\frac{m}{\pi \cdot 40^{2}g}}$. We are told the period is $2.5$ seconds, so equate the expression for $T$ to 2.5 and solve for $m$ to get the answer (it'll be in kg).$

$\noindent (2) In general for an overdamped system, this is true. An overdamped harmonic oscillator system is $m\ddot{x} + c\dot{x} + k = 0$, where the discriminant $\Delta = c^{2} - 4mk$ is positive (the $m$ is the mass of the spring, the $c$ is a damping coefficient, and $k$ is the spring constant. You can find out more about this here:$

http://hyperphysics.phy-astr.gsu.edu/hbase/oscda.html#c1 .)

$\noindent Since the discriminant is positive, there are two real roots $\lambda_{1} = \frac{-c + \sqrt{\Delta}}{2m}$ and $\lambda_{2} = \frac{-c-\sqrt{\Delta}}{2m}$, so the solution is of the form $x(t) = A e^{\lambda_1 t} + B e^{\lambda_{2}t}$. Hence $\dot{x}(t) = A\lambda_1 e^{\lambda_1 t} + B\lambda_{2} e^{\lambda_2 t}$. So if $x(0) = 0$, we have $A + B = 0 \Rightarrow A = -B$. So $x(t) = A \left(e^{\lambda_1 t} - e^{\lambda_2 t}\right)$. So if $x(t) = 0$ for some $t \neq 0$, we have $e^{\lambda_1 t}= e^{\lambda_2 t}$ for some $t \neq 0$ (note $A \neq 0$ since the initial velocity is non-zero. If $A$ were $0$, the initial velocity would be $0$. In fact, the motion would be identically at rest.). But this is clearly impossible, since $\lambda_{1} \neq \lambda_{2}$. Hence $x$ can't equal $0$ again.$

$\noindent (3) Just solve the ODE and you should obtain exponential parts of the solution that decay to $0$, and also a sinusoidal part that doesn't ever decay to $0$. This sinusoidal part is the steady state oscillations of this driven harmonic oscillator.$

1008 · Oct 8, 2016

InteGrand said:
$\noindent (1) The motion is $\ddot{x} = - \omega^{2} x$, where $\omega^{2} = \frac{\pi\cdot 40^{2}g}{m}$. Hence the period is $T = \frac{2\pi}{\omega} = 2\pi \times \frac{m}{\pi \cdot 40^{2}g} = \frac{2m}{40^{2}g}$. We are told the period is $2.5$ seconds, so $\frac{2m}{40^{2}g} = 2.5$. Solve for $m$ here to get the answer (it'll be in kg).$

$\noindent (2) In general for an overdamped system, this is true. An overdamped harmonic oscillator system is $m\ddot{x} + c\dot{x} + k = 0$, where the discriminant $\Delta = c^{2} - 4mk$ is positive (the $m$ is the mass of the spring, the $c$ is a damping coefficient, and $k$ is the spring constant. You can find out more about this here:$

http://hyperphysics.phy-astr.gsu.edu/hbase/oscda.html#c1 .)

$\noindent Since the discriminant is positive, there are two real roots $\lambda_{1} = \frac{-c + \sqrt{\Delta}}{2m}$ and $\lambda_{2} = \frac{-c-\sqrt{\Delta}}{2m}$, so the solution is of the form $x(t) = A e^{\lambda_1 t} + B e^{\lambda_{2}t}$. Hence $\dot{x}(t) = A\lambda_1 e^{\lambda_1 t} + B\lambda_{2} e^{\lambda_2 t}$. So if $x(0) = 0$, we have $A + B = 0 \Rightarrow A = -B$. So $x(t) = A \left(e^{\lambda_1 t} - e^{\lambda_2 t}\right)$. So if $x(t) = 0$ for some $t \neq 0$, we have $e^{\lambda_1 t}= e^{\lambda_2 t}$ for some $t \neq 0$ (note $A \neq 0$ since the initial velocity is non-zero. If $A$ were $0$, the initial velocity would be $0$. In fact, the motion would be identically at rest.). But this is clearly impossible, since $\lambda_{1} \neq \lambda_{2}$. Hence $x$ can't equal $0$ again.$

$\noindent (3) Just solve the ODE and you should obtain exponential parts of the solution that decay to $0$, and also a sinusoidal part that doesn't ever decay to $0$. This sinusoidal part is the steady state oscillations of this driven harmonic oscillator.$

Thanks for that, I'll figure (3) out now too... Got another one, from series this time, I'm completely stuck:

Also, is this correct?

InteGrand · Oct 8, 2016

1008 said:
Thanks for that, I'll figure (3) out now too... Got another one, from series this time, I'm completely stuck:

a) Actually follows immediately from the p-series test (here, it's summing 1/n^p, where p = 1/2 < 1 ==> divergence).

$\noindent To show it the way they want, we can use induction. Note that $s_{2} = 1 + \frac{1}{\sqrt{2}} > \sqrt{2}$ (since the LHS is greater than $1.5$, since $\frac{1}{\sqrt{2}} > \frac{1}{2}$, so $LHS^{2} > 1.5^{2} = 2.25 > 2 \Rightarrow LHS > \sqrt{2}$). Now, suppose $s_{n} = 1 + \frac{1}{\sqrt{2}} + \cdots + \frac{1}{\sqrt{n}} > \sqrt{n}$ for some particular integer $n \geq 2$. Then $s_{n+1} = s_{n} + \frac{1}{\sqrt{n+1}} > \sqrt{n} + \frac{1}{\sqrt{n+1}}$ (induction assumption). This is greater than $\sqrt{n+1}$, since $\sqrt{n}\sqrt{n+1} + 1 > n+1$. Why is this latter inequality true? Because $\sqrt{n}\sqrt{n+1} > \sqrt{n}\sqrt{n} = n \Rightarrow \sqrt{n}\sqrt{n+1} + 1 > n + 1$. Thus $s_{n+1} > \sqrt{n+1}$, and the result is thus true by induction.$

InteGrand · Oct 8, 2016

Part b) is a famous elementary proof of the Harmonic series' divergence:

S = 1 + (1/2 + 1/3) + (1/4 + 1/5 + 1/6 + 1/7) + (1/8 + 1/9 + 1/10 + 1/11 + 1/12 + 1/13 + 1/14 + 1/15) + ...

> 1 + (1/4 + 1/4) + (1/8 + 1/8 + 1/8 + 1/8) + (1/16 + 1/16 + 1/16 + 1/16 + 1/16 + 1/16 + 1/16 + 1/16) + ...

= 1 + 1/2 + 1/2 + 1/2 + ...

= ∞,

so the Harmonic series diverges.

Edit: Just realised the Q. wanted a different bracketing to the one I did here. But it's the same idea, and you should be able to do it their way if you understand what was done here.

leehuan · Oct 8, 2016

I think I did a) like this

$\begin{align*}1+\frac{1}{\sqrt2}+\frac{1}{\sqrt3}+\dots+\frac{1}{\sqrt{n}}&\ge \underbrace{\frac{1}{\sqrt n}+\frac{1}{\sqrt n}+\frac{1}{\sqrt n}+\dots+\frac{1}{\sqrt n}}_{n\text{-terms}}\\ &= \frac{n}{\sqrt n}\\ &= \sqrt{n}\end{align*}$

leehuan · Oct 8, 2016

Paradoxica · Oct 8, 2016

Alternatively, using the simple fact that

$\frac{1}{2\sqrt{n}} > \frac{1}{\sqrt{n+1} + \sqrt{n}} \equiv \sqrt{n+1} - \sqrt{n}$

The sum is reduced to a telescopic inequality, and it is trivial to obtain:

$$\noindent$ \sum_{k=1}^n \frac{1}{2\sqrt{n}} > \sqrt{n+1} - \sqrt{n} + \sqrt{n} - \sqrt{n-1} + \cdots + \sqrt{3} - \sqrt{2} + \sqrt{2} - \sqrt{1} \\ \sum_{k=1}^n \frac{1}{\sqrt{n}} > 2\sqrt{n+1} - 2$

And so the infinite sum clearly diverges.

Paradoxica · Oct 8, 2016

leehuan said:
$\text{By writing the expanded form of the series I was able to achieve a correct result.}\\ (x+1)e^x=\sum_{k=0}^\infty \frac{(k+1)x^k}{k!}$

$\text{What's the problem with this again}\\ \begin{align*}(x+1)\sum_{k=0}^\infty \frac{x^k}{k!}&= x \sum_{k=0}^\infty \frac{x^k}{k!}+\sum_{k=0}^\infty \frac{x^k}{k!} \\&=\sum_{k=0}^\infty \frac{x^k(x+1)}{k!}\end{align*}$

Both sides are the same.

1008 · Oct 8, 2016

Thanks IG, leehuan and Paradoxica:

leehuan, how'd you get rid of the equality in this one?:

leehuan said:
I think I did a) like this

$\begin{align*}1+\frac{1}{\sqrt2}+\frac{1}{\sqrt3}+\dots+\frac{1}{\sqrt{n}}&\ge \underbrace{\frac{1}{\sqrt n}+\frac{1}{\sqrt n}+\frac{1}{\sqrt n}+\dots+\frac{1}{\sqrt n}}_{n\text{-terms}}\\ &= \frac{n}{\sqrt n}\\ &= \sqrt{n}\end{align*}$

Also, is this correct?

leehuan · Oct 8, 2016

Paradoxica said:
Both sides are the same.

Oh lol I see now, whoops.

1008 said:
Thanks IG, leehuan and Paradoxica:

leehuan, how'd you get rid of the equality in this one?:

Also, is this correct?

Yeah your /img is right. We did that one in my tutorial. Just compare.

Technically they aren't equal and I should've just used > and not \ge

InteGrand · Oct 8, 2016

leehuan said:
$\text{By writing the expanded form of the series I was able to achieve a correct result.}\\ (x+1)e^x=\sum_{k=0}^\infty \frac{(k+1)x^k}{k!}$

$\text{What's the problem with this again}\\ \begin{align*}(x+1)\sum_{k=0}^\infty \frac{x^k}{k!}&= x \sum_{k=0}^\infty \frac{x^k}{k!}+\sum_{k=0}^\infty \frac{x^k}{k!} \\&=\sum_{k=0}^\infty \frac{x^k(x+1)}{k!}\end{align*}$

They are the same.

Edit: said above.

Paradoxica · Oct 8, 2016

leehuan said:
Oh lol I see now, whoops.

Yeah your /img is right. We did that one in my tutorial. Just compare.

they aren't equal and I should've just used > and not \ge

if at least one term in the inequality sum isn't equal, then it's unequal. FTFY.

note the same holds true for integrals, if two functions are equal at a finite number of points, then provides one is larger than the other over the domain of integration, the integrals are not equal.

1008 · Oct 8, 2016

Thanks guys.

How'd you do this one?

I'm guessing it's really simple comparison like the one I posted above, just can think of the proper one...

InteGrand · Oct 8, 2016

1008 said:
Thanks IG, leehuan and Paradoxica:

leehuan, how'd you get rid of the equality in this one?:

Also, is this correct?

$\noindent To use the comparison test, we'd need the summands to have non-negative terms. We can get around this by showing the desired sum converges absolutely though, because then the series we'd consider would have positive terms. I.e. note $c_n := |a_n b_n| \leq \frac{a_n ^2 + b_n^2}{2}$ for all $n$. The LHS is always non-negative, and the RHS has convergent sum due to the assumptions in question, so $\sum c_{n}$ converges by the comparison test. Hence $\sum a_n b_n$ converges absolutely (and so converges).$

leehuan · Oct 8, 2016

InteGrand said:
$\noindent To use the comparison test, we'd need the summands to have non-negative terms. We can get around this by showing the desired sum converges absolutely though, because then the series we'd consider would have positive terms. I.e. note $c_n := |a_n b_n| \leq \frac{a_n ^2 + b_n^2}{2}$ for all $n$. The LHS is always non-negative, and the RHS has convergent sum due to the assumptions in question, so $\sum c_{n}$ converges by the comparison test. Hence $\sum a_n b_n$ converges absolutely (and so converges).$

Oh yeah, in the question book they said it was safe to assume that the sequences are non-negative. I think he just forgot to mention that.

1008 · Oct 8, 2016

leehuan said:
Oh yeah, in the question book they said it was safe to assume that the sequences are non-negative. I think he just forgot to mention that.

I don't think it says it's safe to assume that for this question.... Here's the proper screenshot:

Paradoxica · Oct 8, 2016

InteGrand said:
$\noindent Again we show absolute convergence, which implies the convergence. Let $c_{n} := |a_n|$. Note that $a_1 ^2 + a_2^2 + \cdots + a_n^2 = \left(c_1 ^2 + c_2^2 + \cdots + c_n^2\right)\left(\underbrace{1 + 1 + \cdots + 1}_{n\text{ times}}\right) \geq c_1 + c_2 + \cdots + c_n$ by the Cauchy-Schwarz Inequality. I.e. $c_1 ^2 + c_2^2 + \cdots + c_n^2 \geq \frac{c_1}{n} + \frac{c_2}{n} + \cdots + \frac{c_n}{n}$. The LHS has convergent sum, and the RHS is a sequence of partial sums that is increasing and bounded above (by the convergent sum), so it has a limit, i.e. $\sum _n \frac{c_n}{n}$ converges. This implies the desired sum converges absolutely, so converges.$

are you sure the n is independent of the a_n? I read the n as depending on the index.

MATH1251 Questions HELP (4 Viewers)

Well-Known Member

Well-Known Member

Active Member

Well-Known Member

Active Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

-insert title here-

-insert title here-

Active Member

Well-Known Member

Well-Known Member

-insert title here-

Active Member

Well-Known Member

Well-Known Member

Active Member

-insert title here-

Users Who Are Viewing This Thread (Users: 0, Guests: 4)