11:20
@soupless If a matrix is normal, then it's inverse is also normal. This follows from the fact that (A^T)^(-1) = (A^(-1))^T
@00GB It should be able to help : but if you let me know the question (1) I may be able to answer it (2) If not I may be able to frame your question so that you can post it on MSE with no problems.
@Bhavay I was left thoroughly confused after long periods of thinking about this question, but then I realized something that I should have, miles earlier.
@Bhavay We have f(x,y) , where $x = r \cos \theta$ and $y = r \sin theta$. Now, the point is that f is NOT a function or r : it's a function of some things which are functions of r.
So what does df/dr even mean here? It doesn't even have a meaning.
Usually, when we talk about the partial derivative, we ask for the partial derivative of a function with respect to one of its arguments. For example, if we have f(a,b,c), then we can talk about df/da,df/db/df/dc. But , if secretly, a,b,c ended up depending on x, we can't talk about df/dx , for example, because x isn't even a parameter of f.
So the question itself, is : what does df/dr mean when f=f(x,y), where x,y are functions of r?
There is only ONE way in which you can give this any meaning : rewrite the function $f$ so that it's a function of r and some other arguments.
So for example, when we simplify f(x,y) = x^2+y^2 where x=r cos theta and y = r sin theta, then we know that $f(r,\theta) = r^2$ so NOW we can talk about df/dr and df/d theta.
The other way of giving meaning, which is the chain rule, is actually a workaround in case you can't make f into an explicit function of r. In that case, you just abstractly know that f is a function of r, but don't have an expression beyond what is given in x and y. In THAT case, you have to use the chain rule. Which means that Emily's answer has no point at all, because all it's saying is what I'm saying above.
And for that question, the answer which was found by finding that f(r,\theta) = r^2 is in fact correct : if you apply the chain rule and calculate df/dx * dx/dr + df/dy * dy/dr, you'll find exactly that it equals 2r, which is what one should get anyway.
The total derivative can be used for calculating the partial derivatives of the function , if the function can't be expressed in terms of those hidden parameters alone. That doesn't mean what the OP did there was wrong : in fact, what they did was a cent percent correct. The only thing was that it didn't use the total derivative : because it didn't need to.
Once again, to repeat ; the OP in that question finds that f(r,\theta) = r^2 by simplification, so gets the answer 2r. What Emily does is different : instead of simplifying the function f, she treats it as a composition of two functions. One function is g(r,\theta) = (r cos theta,r \sin theta) and the other function is h(x,y) = x^2+y^2. Now, f(r,\theta) equals (h(g(r ,\theta))) and therefore one can use the chain rule. There's no difference in the approaches, but one can't talk about...
... df/dr when f(x,y) = x^2+y^2 and x,y are functions of r,theta, because f isn't explicitly a function of r (and a few other things).
With that, I'm done sharing my thoughts. I still think that the gradient and the total derivative are the same thing, just that one is a column vector while the other is just a vector.
@Bhavay We can have follow up doubts, but it's a tricky subject so it's possible that we still remain confused!
We are required to find $u_{2}$ and $u_{3}$ of the sequence $u_{n}$ whose z-transform is the function F(z):
$$F(z)= \dfrac{2z^2+5z+4}{(z-1) ^{4}}$$
We are required to find
Approach: I managed to split it into partial fractions:
$$\dfrac{2}{(z-1)^2} + \dfrac{9}{(z-1)^3} + \dfrac{11}{(z-1)^4}$$
and...
