In general, the area of a sector bounded by a curve ##C## defined by polar radius ##\rho(\theta)##, such that ##C = \rho([\theta_1,\theta_2])## , is ## A= \frac{1}{2} \int_{\theta_1}^{\theta_2} \rho^2(\theta) \ d\theta ## . In your case, ##\rho(\theta)## is constant and equal to ##r## since...
It seems that it is just an application of the rule of composition of limits (if ##f## nice means that it has a finite limit in ##c##).
The rule states that if ##h(x) \to b ## as ## x\to a##, and if ##g(x) \to \ell ## as ##x \to b##, then ##g(h(x)) \to \ell ## as ##x\to a##.
If you add ##g##...
The way I've learned about the cross product is this:
In a euclidean space ##E## of dimension 3 fitted with a direct orthonormal basis ##{\cal B}## (in your case ##E = \mathbb{R}^3## fitted with the canonical basis), the vector product of ##\vec u## and ##\vec v## is the unique vector written...
Wow, this looks awful !
Let's write ## F(u,z) = \frac{u^{1+\epsilon} (1-z)^\epsilon }{1-uz} \frac{1}{1+au} ##, where ##a = s/t##.
I make the assumption that ##a,\epsilon >0##, I hope its OK for what you need.
The function ##F## is continuous on ##V = [0,1]^2 - \{1,1\} = [0,1]\times [0,1[...
Your remark about the nature of open sets in ##\mathbb{R}^2## makes the proof even simpler :
Assuming that ##U = \cup_{i \in I}\ U_i \times V_i##, ##U_i## and ##V_i## all open in ##\mathbb{R}##, then for any pair ##(x,y) \in U##, there exists ##i\in I## such that ##y## belongs to ##V_i##...
Hmmm, this looks suspicious :-)
But since I don't know a great deal about topology I'm not going to insist. However, with minimal knowledge as developped in post #3, can I conclude ?
Thank you, but why do you assume that ##U## has the form of a cartesian product of open sets ?
And also, I doubt that all open sets in ##\mathbb{R}^2## have the form ##\cup \ U_i\times V_i##, ##U_i,V_i## open sets in ##\mathbb{R}## ( I am not convinced that ##B(0,1)## be described in such a...
Think I've got more on this :
For any point ##a = (x_0,y_0) \in U## there is an open ball ##B(a,r)## included in ##U##.
For any point ##u = (x,y) ## in the ball, its symetric point by ##a##, called ##v= (2x_0 -x, 2y_0 -y )## is also in the ball, and by convexity of the ball, the lign joining...
I was trying to understand a proof in multivariable calculus (about fundamental pde's), and at some point it said that if ##U## is a convex open set of ##\mathbb{R}^2##, then it's second projection is an open set in ##\mathbb{R}##, without any further explanation. Could you explain please ?
Let's look at what happens on \int_a^b \frac{dx}{c- x}, and you'll do the necessary substitutions later.
If ## c \notin [a,b]##, then you are integrating a continuous function on ##[a,b]##. In this case, there is no discussion needed, and the theory says your integral is equal to ##F(b) -...
It comes from composition laws for polynomial approximations:
if two functions ##f## and ##g## have a polynomial approximation in 0 at order ##n##, say ## P## and ## Q## respectively, then if ##f(0) =0##, ## g \circ f ## has a polynomial approximation at order ## n##, which is the truncature at...
It looks difficult !
I don't know the answer, but have you tried this: assuming that you've shown integrability of your double integral on ##\mathbb{R}_+\times \mathbb{R}##, would a polar change of variable transform this complicated double integral to a single, parameter dependent integral ...
No, it is not true that all continuous functions are integrable (##f = 1## is continuous and not integrable on ##\mathbb{R}_+##). However, it is true that continuous functions on a segment are integrable (even piecewise continuous functions).
This comes from the fact that for every piecewise...
If you are starting on series, I think it is more likely you get questions regarding convergence-divergence rather than computing the sum of converging series which often require knowledge on power series and Fourier series. Maybe you can be asked to evaluate the sum of geometric and telescopic...
From the demo I read :
Set
##\phi(t) = f(x+th,y+k) - f(x+th,y) ##
and
## \psi(t) = f(x+h, y+tk) - f(x,y+tk) ##
Since these two functions are ##{\cal C}^1([0,1])##, the mean value theorem says there exists ##\alpha,\beta\in[0,1]## such that
##\phi(1) - \phi(0)= \phi'(\alpha) ## and ##...
This is THE problem in your demo. If it was possible to interchange the limits, the hypothesis of continuity for ##f_{xy}## and ##f_{yx}## would be useless.
In fact, you can lower the hypothesis to ##f\in{\cal C}^1(U)##, ##f_{xy}## and ##f_{yx}## exist and are continuous.
I've just read the demo, it is more subtle than I thought.
Try to find two distrinct single variable functions ##\phi## and ##\psi## such that
## f(x+h,y+k)-f(x,y+k) - f(x+h,y)...
It needs to be open because it makes legal such expressions as ##f(x+h,y+h)##, which you use a lot for the demo. That means you are sure that there is an open ball centered in ##(x,y)## that is included in ##U##, and it guarantees that their exist ##h## and ##k## small enough so that...