Recent content by fred_91

<Moderator's note: Moved from a homework forum.> 1. Homework Statement I am wondering if it is the same to say : Function f is infinitely many times continuously differentiable AND Function f is infinitely many times differentiable And if it is not the same, which one defines: C^infinity...

Homework Statement I have E as a material parameter. It is a modulus of Elasticity It is given as: 109/E = 80/psi I want to write E as a value with units psi. Homework EquationsThe Attempt at a Solution Is this correct? 1/E = 80*109/psi So psi/E = 80*109 So E = 1.25*10^7 psi

Homework Statement Solve the differential equation: dy/dx = 2/(x+e^y) Homework EquationsThe Attempt at a Solution I tried to use the substitution v=x+e^y, but I didn't get very far: v’=1+e^y y’ v’-1=(v-x)y' y’ = (v’-1)/(v-x) (v’-1)/(v-x) (x+v-x)=2 V (v’-1)/(v-x)=2 vv’-v=2(v-x) vv’-3v=-2x...

Thank you. I hope it looks clearer now. Any help would be very much appreciated.

Homework Statement I have the following code (please see below). I want it to return [2,4,6,8,10] which is the function evaluated at the given list [1,2,3,4,5]. I want to do this using the process command. My problem is: print A returns [None, None, None, None, None] But I want it to return...

I was asking just to make sure :) It implies that the guy in the middle must also approach 0.

This implies that it tends to zero...?

As b \rightarrow a, we have that the upper bound tends to zero... ?

We need to find a bound for |b-x| for x between a and b So, \left|\int_{a}^{b} \int_{0}^{a} (f(a) - f(y)) (x-y) dy (b-x) dx\right| \leq \int_{a}^{b} \left|\int_{0}^{a} (f(a) - f(y)) (x-y) dy\right| |b-x| dx \leq M(b-a) Is that right? (Thank you by the way, very much appreciated.)

\left|\frac{1}{2}f(a)(2xa + a^2) - \int_{0}^{a} f(y)(x-y)dy \right| \leq \left| \frac{1}{2}f(a)(2xa + a^2)\right| + \left|\int_{0}^{a}f(y)(x-y)dy \right| \le \frac{1}{2}|f(a)|(2a^2 + a^2) + B(a+b) (with a>0) What then happens with the original inequality with the double integral...

Oh, I see now why f is bounded. thank you! I've just checked and we have that a and b are positive. I should have stated that in the problem. Also, b \ge a. So, can I say that |x-y|\le |x|+|y| \le a+b ?

Thank you, yes you're right. The f(b) should be f(y) in post numbers 3 and 4. We have fixed this in the following posts.

We have: |x|\le max(|a|,|b|). For the second term: We can say that f is continuous and bounded on [0,a] since we can take B=max(|f(a)|,f(0)| so that |f(y)|\le B\,\,\,\, \forall\,\, y\in[0,a] I think |x-y| must be bounded by b. is that right? It's the biggest difference it can take...

Thank you. I see, so I have: |\frac{1}{2}f(a)(2xa+a^2)-\int_0^a f(y)(x-y)dy|\le M Using the triangle inequality, we have: |\frac{1}{2}f(a)(2xa+a^2)|+|\int_0^a f(y)(x-y)dy|\le M The first term is bounded by \frac{1}{2}f(a)(2ba+a^2) So, we can write: |\frac{1}{2}f(a)(2ab+a^2)|+\int_0^a...

Thank you very much. To show that the inner integral is bounded, I first simplified |\int_0^a (f(a)-f(b))(x-y)dy | since f(a) is independent of x. So I split the integral into 2 parts: \int_0^a f(a)(x-y)dy-\int_0^a f(b)(x-y)dy Evaluating the first part: \int_0^a...