Homework Statement
As given in the title. The law of cosines, the law of sines, or any other aspect of trigonometry may be used. Ultimately, I need to show that when two triangles have two pairs of proportional sides and the included angles congruent, that they are similar - that is, the...
That's ok. I'm satisfied with what I've got so far... I see what you're saying. There are obviously a few ways to do it (I can think of a couple of others right now which involve approaching it from other angles, like countability), but the professor gave us a hint and I based my work off of...
Dick, I saw your hint when I had just finished posting my solution. Thanks! It looks like I had the same general idea as you hinted at, but I'm not sure if what I did is precisely the same.
Ok, here is the solution I came up with. Any comments are appreciated.
By the given theorem, a bounded function f is integrable on [a,b] if and only if \forall \epsilon > 0 \exists some partition P_{\epsilon} of [a,b] such that U(f,P_{\epsilon})-L(f,P_{\epsilon}) < \epsilon.
Clearly...
Homework Statement
Let f(x)= { 1 if x=\frac{1}{n} for some n\in the natural numbers,
or 0 otherwise}
Prove f is integrable on [0,1], and evaluate the integral.
Homework Equations
This is using Riemann Integrability. I know that the method of providing the solution is supposed to be by...
Looks like or similar to a more concise version of what I did, but what are you saying when you say M:=sup(t|0<t<x||f(x)|) ? I'm unsure what t|0<t<x||f(x)| means.
I think there are a few problems... It looks like you are trying to dosomethingvery similar to what I did, so I like it :). That being said, here are the issues I see.
First, I'm not sure how you conclude that the Mean Value Theorem requires there to be a point \xi\in (0,x_0) with...
I think you are correct. The point we would have to look at if we were to work it out with epsilon and delta would be 0, that is, prove thatthe resultingfunction would not be continuous at 0. As I said, I was working on that pretty late and was fuzzy about that portion. The point is that there...
Pere's solution seems much simpler than mine, but for the curious I've checked it with my Analysis professor who agrees with it.
However, I made one error that requires correcting. I said to choose any c0 where |f(c0)-0| < 1. I made two blunders: first, I confused f(c0) with c0, and second...
I'm not sure if you realize this, and I apologize if I'm just telling you something you already know... it looked like in an earlier post on this problem you treated wlog as a function of some kind, perhaps a logarithm, when you wrote wlog(x)<y. If I'm mistaken, most sincere apologies...
Not quite. I'd start by making it simpler by multiplying out the polynomials. Then you'd have:
(x^{3}-3x^{2})^{\frac{1}{3}}
Then, use the chain rule:
If f(x)=g(h(x)), then f ' (x) = g ' (f(x))*f ' (x).
For example, look at f(x) = (sin(x))2.
So if g(x) = x2, and h(x) = sin(x)...
It's very late so perhaps I'm not thinking this through, but try this...
First, let me give a general description of what I'm trying to say. Take your calculator, computer, whatever, and graph f(x)=x2. Now evaluate the derivative at some point close to 0, say, .01. At .01, the derivative is...
I never said that |f'(c0)|=|f(0)|. I said:
|f'(c0)|>0=|f(0)|
This is a shorthand/symbolic/whatever-the-proper-term-is-way of saying that |f'(c0)|>0, and that in turn 0=f(0). In other words, |f'(c0)|>f(0).
That being said, wackikat you're right I misread the thing, or misthought it...