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• Today, 16:29
All good. (Nod) It may suffice to simply state that an $m\times m$ matrix of rank $m$ is invertible. It's a property of matrices: If A is a...
5 replies | 38 view(s)
• Today, 15:53
How about multiplying by $A^{-1}$? $A$ is invertible isn't it? (Thinking)
5 replies | 38 view(s)
• 5 replies | 52 view(s)
• Today, 13:51
Hey evinda!! I don't think we can draw that conclusion. Consider $A=\begin{pmatrix}1&0\\1&0\end{pmatrix}$, what is $A(A-I)$? (Worried)
5 replies | 38 view(s)
• Today, 13:45
Then we have: don't we? (Thinking) It follows that the limit of $f''(x)$ is also $0$, doesn't it? Then there is no need to start with the...
23 replies | 576 view(s)
• Today, 13:06
It is correct and you are correct too. Note that $\frac 1{1-x}=1+x+x^2+x^3+\ldots$. Written with $y$ we have: $(1-y)^{-1}=1+y+y^2+y^3+\ldots$ ...
9 replies | 306 view(s)
• Today, 03:50
That works. (Nod) Note that $f(x)=\frac{1}{3}(x-2)^2e^{x/3}$ is hard to match with a line, since it has an exponential function in it. However,...
5 replies | 52 view(s)
• Yesterday, 17:20
Hey mathmari!! We could, but perhaps the intersection points are "nice" points. What are the potentially nice points? (Wondering)
5 replies | 52 view(s)
• Yesterday, 11:06
The problems with MHF were always about personnel and authority structure. If the leaders of MHF had been willing to appoint multiple admins, like we...
6 replies | 230 view(s)
• Yesterday, 09:27
Yep. (Nod)
4 replies | 76 view(s)
• April 7th, 2020, 17:57
When I throw it at Octave online, I get: So it seems it is none of the above, but the answer $3$ is close. For reference, I have re-encoded...
1 replies | 76 view(s)
• April 6th, 2020, 11:34
Hey mathmari!! This is an example where Newton-Raphson can have problems if we are not careful. If we pick a starting value that is too far from...
4 replies | 76 view(s)
• April 5th, 2020, 10:37
I substituted $n=\frac 1{x^2}$, which also means that $x^2=\frac 1 n$. So we get $\Big(1 + \frac 32 x^2\Big)^{1/x^2} = \Big(1 + \frac 32 \cdot \frac... 9 replies | 306 view(s) • April 5th, 2020, 09:14 It's like this: $$\Big(1-\frac 12 x^2 + \ldots\Big)\Big(1+\frac 12 (2x)^2 - \ldots\Big)\\ =1\cdot 1 -\frac 12 x^2 \cdot 1+1\cdot \frac 12 (2x)^2 -... 9 replies | 306 view(s) • April 5th, 2020, 07:04 The series expansion of \cos x = 1 - \frac 12x^2 + \ldots. And the expansion of \frac 1{1-x} = 1+x+x^2+\ldots So:$$\frac{\cos(x)}{\cos(2x)}... 9 replies | 306 view(s) • April 3rd, 2020, 22:31 I am reading Aisling McCluskey and Brian McMaster: Undergraduate Topology, Oxford University Press, 2014... ... and am currently focused on Chapter... 1 replies | 252 view(s) • April 3rd, 2020, 15:47 I have obtained access to the full solutions manual, after contacting Wiley about it. I will not type up the solution in full, but simply note a few... 1 replies | 308 view(s) • April 2nd, 2020, 15:47 In the mid-1990's, an electrical engineer/computer scientist by the name of Judea Pearl started to change the world by greatly improving our... 0 replies | 268 view(s) • April 2nd, 2020, 12:25 Klaas van Aarsen replied to a thread f in Calculus It appears the answer is: dkm Acronym for "don't kill me". Often used when somebody says something that somebody else finds really funny,... 5 replies | 364 view(s) • April 1st, 2020, 15:04$\newcommand{\doop}{\operatorname{do}}$Problem: (This is from Study question 4.3.1 from Causal Inference in Statistics: A Primer, by Pearl,... 1 replies | 308 view(s) • March 30th, 2020, 05:31 Let's take a look at a couple of examples. If$f(x)=x$, then$f'(x)=1$and$f''(x)=0$. So$\lim\limits_{x\to +\infty}f'(x)\ne 0$, isn't it?... 23 replies | 576 view(s) • March 30th, 2020, 04:26 Yep. (Nod) And no, nothing more specific. 23 replies | 576 view(s) • March 30th, 2020, 03:32 I believe so yes. Consider$f(x)=\ell$. It satisfies all conditions, doesn't it? (Wondering) And it is monotone instead of strictly monotone. To... 23 replies | 576 view(s) • March 29th, 2020, 14:38 Indeed. (Thinking) 23 replies | 576 view(s) • March 29th, 2020, 14:26 Yep. (Nod) 23 replies | 576 view(s) • March 29th, 2020, 13:57 Then the inequality also holds yes. What if fill in, say,$f'(y)=-1\$ in the inequality? Would it satisfy it? (Wondering)
23 replies | 576 view(s)
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