Good luck on your journey. I think you're making a very positive choice. I think a lot of people in your position would be too scared to change their life path because it's kind of drilled into our heads that we should finish school and get settled down asap. I found out myself that life isn't...
I think in terms of increasing retention and also preventing burnout, it makes the most sense to take just one of the mathematics courses.
The material will be condensed but if that's all you're focusing on then you'll have plenty of time to keep up. Also, since you'll be spending a significant...
It sounds like you're not completely sure what is meant by the median or how to find the distribution of it. I want to explain those concepts to you so you can try to tackle the problem.
You have a sample of 5 random variables all distributed uniformly over (0,1). So we have X_1, X_2, X_3...
Homework Statement
Prove or give a counter example of the following statement:
If f: [a,b] \to [c,d] is linear and g:[c,d] \to \mathbb{R} is Riemann integrable then g \circ f is Riemann integrable
Homework Equations
The Attempt at a Solution
I'm going to attempt to prove the statement...
Hey there's actually a way you can do it without too much thinking:
y = x^r \\
y' = rx^{r-1} \\
y'' = r(r-1)x^{r-2} \\
\implies xy'' - y' = x[r(r-1)x^{r-2}] - rx^{r-1} = x^{r-1}r(r-1 -1) = x^{r-1}r(r-2)
Which equals zero for all x when r = 0 or r = 2. Sorry I didn't notice this before. A bit...
You kind of have to think a bit outside the box to figure out which values of r are "different".
If r = 0 then y is a constant so all if its derivatives are 0
If r = 2 then its first derivative is a function of x but its second derivative is a constant
if r < 0 or r >2 then all the derivatives...
What you should do is plug in some concrete examples. I'll give you two and see if you can work out the rest:
y = x^1 \\
y' = 1 \\
y'' = 0 \\
\implies xy'' - y' = x(0) - 1 \neq 0
So you can see it doesn't work for r = 1. Now let's see why it does work for r = 2
y = x^2 \\
y' = 2x \\
y'' = 2...
Yep you've got it. Think of the two critical regions as events. If the test statistic falls in the first region OR the second region we reject the null hypothesis. Since The two regions are disjoint, we add the probabilities using the usual probability rules (P(A or B) = P(A) + P(B) if A and B...
Imagine they didn't say anything about vectors and just gave you the equation "2*a + 1.5*b + 1*c", would you know what it describes? You're multiplying the price of something by the number of those things you sold. What does that mean in real life?
As for the vectors:
Without all this...
One thing I noticed right away is that you're confusing the unknown population means with the sample means that can be calculated by doing the survey. The symbol "mu" always refers to the unknown population means while the symbol x with a bar over it represents the mean you get from your sample...
Actually a relatively simple function works lol. Sorry for all this confusion:
h(x) = xf(x)
There exists a c in (0, 1) such that
0 = cf'(c) + f(c)
by Rolle's Theorem.
Hey sorry I edited my previous post. What I wrote doesn't work because the two c's might be different. The c that works for the equation you posted might be a different c from the one I used for h(x).
When trying to define an h(x) we want it to be 0 at the end points so we can apply Rolle's...
edit
Sorry the post I made contained an error. I'm not quite sure how to fix it yet but basically what I think you need to do is find a function h(x) and define it in terms of some linear combination of f(x), f(0), f(1) so that it equals zero at the end points and then apply Rolle's Theorem.
When x is between -1 and 2 one of the functions is lower than the other. When you plot them can you see which it is? That is the lower bound. And the upper bound is the function that is higher.
The upper and lower bound varies with x. That's why you can't just use constants...you need to write...
You can see easily what the region is by plotting your boundary functions here:
http://www.mathsisfun.com/data/function-grapher.php
When x is between -1 and 2, what is the upper and lower bound of y? You can get it just by looking at the graph
I think you just need to use the fact that Ax = b
From what you wrote: x_{k+1} = x_k + B^{-1}\left(b - Ax_k\right) subsituting Ax = b we get x_{k+1} = x_k + B^{-1}\left(Ax - Ax_k\right) \\ x_{k+1} = x_k + B^{-1}A\left(x - x_k\right) \\ x_{k+1} - x = x_k - x+ B^{-1}A\left(x - x_k\right) \\...
I think your reduced form is off.
I get
1 0 -1
0 a-2 0
0 1 -1
If you want to find the solutions then you need to figure out what this matrix is telling you. Start with rows 1 and 3. What do they tell you about the relationship between x, y and z? If you can figure out...
Homework Statement
-A bus is moving along an infinite route and has stops at n = 0, 1, 2, 3, ......
-The bus can hold up to B people
-The number of people waiting at stop n, Yn, is distributed Poisson(10) and is independent from the number waiting at all the other stops.
-At any given stop each...
Thanks guys. I hadn't considered the symmetry of the problem. Does this look alright:
E[X_1 + X_2]\\
= \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} (x_1 + x_2) f_{X_1,X_2}(x_1, x_2) dx_1 dx_2 \\
= \int_{-\infty}^{\infty}\int_{-\infty}^{x_2} (x_1 +...
Homework Statement
Have 2 iid random variables following the distribution f(x) = \frac{\lambda}{2}e^{-\lambda |x|}, x \in\mathbb{R}
I'm asked to solve for E[X_1 + X_2 | X_1 < X_2]
Homework Equations
The Attempt at a Solution
So what I'm trying to do is create a new random variable Z =...
If 2 < x < 4 then that's the same as saying 2 < x and x < 4 You can manipulate inequalities the same way you do equalities so this means \frac{1}{x} < \frac{1}{2} and \frac{1}{4} < \frac{1}{x} We only need the 1/2 term since we're looking for an upper bound on 1/x. The 1/3 comes from the fact...
Thanks a lot, you've really helped me get my head around the issues with this problem. I'm going to work on the below and hopefully get an answer. Cheers :)
The big property I remember is that they're continuous. But the derivative itself doesn't necessarily have to be continuous.
edit: Yeah it's g(x) that is differentiable. f(x) is the derivative
We haven't touched antiderivatives yet. So far we've only done the definition of derivative and associated laws (sum, product, quotient, chain rule) and theorems related to the mean value theorem (Rolle's Theorem, local maximums have a derivative equal to 0, mean value theorem, if f'(x) = 0...
If \lim_{x\to 0} f(x) = L then \lim_{x\to 0} f(x)*x = \lim_{x\to 0} f(x)*\lim_{x\to 0} x = L*0 = 0
So if f is continuous at x=0 then \lim_{x\to 0} f(x)*x = \lim_{x\to 0} f(x)*\lim_{x\to 0} x = f(0)*0 = 0
So we can conclude that f isn't continuous at 0. And more generally \lim_{x\to 0} f(x) does...
Thanks that's helpful. So if I can define a function c(x), is the following true:
If \lim_{x\to 0^+} c(x) = 0 then \lim_{x\to 0^+} x f(c(x)) = \lim_{x\to 0^+} x f(x)
And if so, does c(x) have to have any properties like being continuous or injective?
Homework Statement
Let f: R -> R be a function such that \lim_{z\to 0^+} zf(z) \gt 0 Prove that there is no function g(x) such that g'(x) = f(x) for all x in R.
Homework Equations
Supposed to use the mean value theorem. If f(x) is continuous on [a,b] and differentiable on (a,b) then...