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Articles by André Nicolas
TFEB triggers a matrix degradation and invasion program in triple-negative breast cancer cells upon mTORC1 repression
Keywords extracellular matrix MT1-MMP invadopodia mTORC1 TFEB triple-negative breast cancer tumor invasion patient-derived xenograft mTOR inhibitors Introduction The phosphatidylinositol 3-kinase (PI3K)/AKT/mammalian target of rapamycin (mTOR) signaling pathway is often hyperactivated in breast cancers (BCs), including in triple-negative BCs (TNBCs), providing therapeutical options for this poor prognosis subtype, which is immune to endocrine therapy.1,2 Under nutrient-replete conditions,...
Ortner, S. (1979). ¿Es la mujer con respecto al hombre lo que la naturaleza con respecto a la cultura? En Harris, O., & K. Young, Antropología y feminismo (109-131). Anagrama
Ortner, S. (1979). ¿Es la mujer con respecto al hombre lo que la naturaleza con respecto a la cultura? En Harris, O., & K. Young, Antropología y feminismo (109-131). Anagrama Munive Rodríguez, Andrea Irene [1] ; Azurín Icaza, André Nicolás [1] [1] Pontificia Universidad Católica del Perú Localización: Anthropía, ISSN-e 2076-2704, ISSN 2076-0574, Nº. 19, 2022, págs. 223-231 Idioma: español Enlaces Texto completo
La experiencia peatonal en espacios públicos: el caso de residentes de distritos periféricos de Lima Norte y El Callao
AUGÉ, M., 2000 – Los “no lugares”. Espacios del anonimato: una antropología de la sobremodernidad, 125 pp.; Barcelona: Gedisa Editorial. CRESSWELL, T., 2010 – Towards a politics of mobility. Environment and Planning. Society and Space, 28: 17-31. DAMMERT, L., 2002 – La inseguridad urbana en Argentina. Diagnóstico y perspectivas. In: Seguridad ciudadana, ¿espejismo o realidad? (F. Carrión, ed.): 283-315; Quito: Facultad Latinoamericana de Ciencias Sociales (FLACSO). DEGREGORI, C., BLONDET, C.
How can this expression be simplified?
Another proof. By Laplace expansion (along the last row): $$ -\sum_{cyc}a^2(b-c) = \det\begin{pmatrix} 1 & 1 & 1 \\ a & b & c \\ a^2 & b^2 & c^2 \end{pmatrix},\tag{1}$$ But the RHS is a Vandermonde matrix, whose determinant is well-known.
How can this equation be simplified?
Another proof. By Laplace expansion (along the last row): $$ -\sum_{cyc}a^2(b-c) = \det\begin{pmatrix} 1 & 1 & 1 \\ a & b & c \\ a^2 & b^2 & c^2 \end{pmatrix},\tag{1}$$ But the RHS is a Vandermonde matrix, whose determinant is well-known.
Opposite of Fermat's Last Theorem?
Clearly we must have $x, y<z$. Suppose $x\not=y$; WLOG, suppose $x<y$. Then this can be rewritten as $$a^x(1+a^{y-x})=a^xa^{z-x},$$ with $z-x, y-x>0$. This gives $$1+ a^{y-x}=a^{z-x};$$ but the right hand side is divisible by $a$, while the left hand side is not. So we must have $x=y$. But then this yields $$2a^x=a^z,$$ which in turn yields $$2=a^{z-x}$$ with $z-x>0$. So there are no nontrivial solutions.
Where did I go wrong in my evaluation of the integral of cosine squared?
As noted above, your answer is actually correct, you just have a duplication that cancels. Alternatively, $\int (1-\sin^2x)\,dx = \int\cos^2 x\,dx$ So: $$\int(\cos^2 x + 1 -\sin^2 x)\,dx = 2\int\cos^2x\,dx.$$ Now, $\cos^2 x -\sin^2 x = \cos 2x$. So the left hand side is: $$\int(1+\cos(2x))\,dx = x + \frac{1}{2}\sin(2x) + C$$ Dividing by $2$ and substituting for $\sin(2x)=2\sin x\cos x$, we get: $$\frac{x+\sin(x)\cos(x)}{2}+C'$$
Proving that all integers are even or odd
If you really want to avoid using general facts about division with remainder, you can use mathematical induction on $n$ to prove it for nonnegative integers: Base case: $0$ is even because $0=2\cdot 0$. Induction step: Assume that $n$ is odd or even; then we must prove that $n+1$ is also either odd or even. First subcase: $n$ is even, so $n=2k$ for some $k$. Then $n+1=2k+1$ and so it is by definition odd. Second subcase: $n$ is odd, so $n=2k+1$ for some $k$.
Is it true that if two open sets in a topological space interset each other and share the same boundary, they are the same?
As pointed by Josh Keneda, $A=\{(x,y):x^2+y^2<1\}$ and $B=\mathbb{R}^2\setminus\partial A$ give a counter-example. You may have a more convoluted conter-example by considering the basins of attraction, in $\mathbb{C}$, for the Newton's map applied to $p(x)=x^3-1$.
Halting problem is solvable
"Not solvable" here means there is no algorithm that will determine, given a Turing machine $M$ and an input $i$, whether $M$ will halt given input $i$. This has no direct connection with axioms, or with incompleteness. As to axiomatization, in a suitable formal system, we could in principle add axioms that say $M$ halts on input $i$ when it does, and does not when it does not. Indeed it would be enough to deal with the non-halting cases. But this set of axioms could not be recursive.
Calculus: Finding Arc Length--Squaring the Derivative Where did the -1/2 come from?
Recall the identity $$(a-b)^2=a^2-2ab+b^2.$$ Take $a=x^4$ and $b=\frac{1}{4x^4}$. We have $2ab=\frac{1}{2}$. Remark: The fun part is that when you add $1$ (arclength formula), the $-\frac{1}{2}$ turns into $\frac{1}{2}$, and our new expression is $(a+b)^2$. So now we can take the square root, getting $a+b$, that is, $x^4+\frac{1}{4x^4}$. This is simple to integrate. The numbers in this arclength problem were carefully chosen to make the final integration easy.
Math Competition Problem- Geometry/Calculus
So I tried the good old Calculus 1 approach and turned this into an optimization problem. The equations got REALLY hairy, but it was okay since this was the graphing calculator section of the exam. I called the longer part of the horizontal diagonal $x_2$, the shorter part of the horizontal diagonal $x_1$, and half of the vertical diagonal $v$.
Prime factorization number theory
The numbers $\frac{n}{d_1}$, $\frac{n}{d_2}$, and so on range (backwards) over all the divisors of $n$. So multiplying by $n$ we find that $\sigma(n)=2n$, where $\sigma$ is the sum of divisors function. So $n$ is perfect and has $6$ divisors. Recall that the even perfect numbers have the shape $2^{p-1}(2^p-1)$ where $2^p-1$ is a Mersenne prime. The only way we get $6$ divisors is if $p=3$, that is, $n=28$. But maybe $n$ could be an odd perfect number.
Possible alternative way of expressing continuity of a function?
The epsilon and the delta are understood to be numbers. For the epsilon-delta definition to be rigorous, there must be a meaning of the differences f(x)-f(a) and x-a as numbers. For example if there is a metric on the underlying spaces. In more general cases the definition is: The inverse image of an open set is open, Since open sets are necessary, the spaces must be topological. Otherwise there is no notion of continuity.
Find the roots of the summed polynomial
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Count the number of integer solutions of a linear equation
Assuming that they are nonnegative integers: The number of solutions to $x_1 + x_2 + x_3 = k$ is $\dbinom{k+2}{2}$ by the Stars and Bars approach. Therefore, we can add up the number of ways for every combination of $k$ that works - if the first three add to $k$, the second three add to $24-k$. In addition, we let $k$ go from $13$ to $24$, so that the inequality holds.
Finding the roots of a different Quadratic equation from the roots of a Given Quadratic equation
Let $\alpha = \dfrac{-b+\sqrt{b^2-4ac}}{2a}$ and $\beta = \dfrac{-b-\sqrt{b^2-4ac}}{2a}$. Note that $\alpha + \beta = -\dfrac{b}{a}$ and $\alpha\beta = \dfrac{c}{a}$. We want to calculate: $$\dfrac{2c-b\pm \sqrt{(b-2c)^2 - 4(a-b+c)c}}{2(a-b+c)} $$ In terms of $\alpha$ and $\beta$.
Fermat's little theorem
In problem (a), use Fermat's little theorem, which says (or a rather, a very slightly different version says) that for any prime number $p$, and any integer $n$ that's not divisible by $p$, we have $$n^{p-1}\equiv 1\bmod p$$ In particular, use $n=2$ and $p=17$. Keep in mind that $2017=(126\times 16)+1$. In problem (b), note that $30\equiv 61\equiv -1\bmod 31$ (you don't even have to use Fermat's little theorem here).
Finding a bijective function between an open disk and the open square
One domain is $\{\rho_1(x,y)<1\}$, the other is $\{\rho_2(x,y) <1\}$, where $\rho_1(x,y) =\sqrt{x^2+y^2}$, and $\rho_2(x,y) = \max(|x|,|y|)$. Find a map $F\colon (x,y) \mapsto (x',y')$ so that $\rho_1(x,y) = \rho_2(x',y')$. It should be linear on each line through the origin.
Two different trigonometric identities giving two different solutions
Using two different sum-difference trigonometric identities gives two different results in a task where the choice of identity seemed unimportant. The task goes as following: Given $cos 2x =-\frac {63}{65} $ , $cosy=\frac {7} {\sqrt{130}}$, under the condition $0<x,y<\frac {\pi}{2}$, calculate $x+y$. The first couple of steps are the same: finding $sin$ and $cos$ values for both $x$ and $y$.
Ring of real-valued convergent sequences
Recall that a ring element $r$ is idempotent if $r^2=r$. The ring $R$ of convergent sequences has only countably many idempotents. For a sequence $(x_n)$ is idempotent precisely if any $x_i$ is $0$ or $1$. If the idempotent sequence $(x_n)$ converges, all but finitely many of the $x_i$ are $0$, or all but finitely many are $1$. The ring $S$ of all sequences has uncountably many idempotents. Thus $R$ and $S$ cannot be isomorphic.
How many 12-bit strings with more 1’s than 0’s?
I am not sure how to solve this, but this is my best guess. Since we want to know how many 12 bit strings have more 1's than 0's, start with 5 since it is one less than 12/2=6. Then we proceed with: $${12 \choose 5} + {12 \choose 4} + {12 \choose 3} + {12 \choose 2}+ {12 \choose 1}+ {12 \choose 0}=1586$$ Is my reasoning correct? I am not sure if we should use the choose function here, but I believe we do.
Compare two powers of numbers without common divisor
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A supposed to be easy calculus problem
Find the values of $m$ if the line $y=mx+2$ is a tangent to the curve $x^2-2y^2=1$. First we differentiate $x^2-2y^2=1$ with respect to $y$ to get the gradient. We get $y^2=\frac{1}{2}x^2-\frac{1}{2}\implies y=\pm\sqrt{\frac{1}{2}x^2-\frac{1}{2}}$.
How many binary strings of length 2n + 1 have more 1's than 0's? Use bijection to prove.
NOTE: This is a homework question, so I only ask for hints and suggestions to nudge me in the correct direction. Let $n \in \mathbb{N}$. How many binary strings of length 2n + 1 have more 1's than 0's? Use bijection to prove. We need a set of known size, B, along with a bijective function $f: A \rightarrow B$, where A is the set with binary strings of length $2n + 1$ and where each element in A has more 1's than 0's.
A coin is flipped ten times. What is the probability that the first three are heads if an equal number of heads and tails are flipped?
I understand the question but I am not sure how to solve it. For example, if we flip HHHTTTTT then the next three must be heads because of the question. This however seems counterintuitive. I believe that there are $2^{10}$ possible strings, but I am unsure of how to count all possible strings that begin with HHH. It is: When given the counts of heads and tails resulting from the flips, what is the probability that the order of the results has heads in the first three places?
Solving a cubic equation involving trigonometric functions
Hint: We want $$(\sin x-\cos x)(\sin^2 x+\sin x\cos x+\cos^2 x).$$ You should be able to find $\sin x\cos x$ by squaring $\sin x-\cos x$, which you know. Remarks: $1$. We used the identity $a^3-b^3=(a-b)(a^2+ab+b^2)$. $2$. I have not checked your calculations. However, from expressions for $\sin x$ and $\cos x$, you can compute $\sin^3 x-\cos^3 x$ by cubing. More painful, but doable.
Pseudo-Cauchy sequence
I have never seen this terminology before, so I will provide the given definition. A Pseudo-Cauchy sequence is : A sequence $(a_n)$ if for any $\epsilon > 0$ there exists $N \in \mathbb{N}$ such that $|a_{n+1} - a_n | \leq \epsilon \space \forall \space n \geq N$ So then my question is that is a pseudo-cauchy sequence always converging?
Show two numbers are relatively prime
The following is not the most natural way to prove the coprimality, but we will find integers $x$ and $y$ such that $(2n+1)x+(4n^2+1)y=1$. Note that $x_0=-(2n-1)$, $y_0=1$ almost works, except that $(2n+1)x_0+(4n^2+1)y_0=2$. There is an easy fix. For $$(2n+1)(x_0+4n^2+1)+(4n^2+1)(y_0-(2n+1))=2.$$ Let $x=\frac{x_0+4n^2+1}{2}=2n^2-n+1$ and $y=\frac{1+(-(2n+1))}{2}=-n$.
Overlapping intervals of a Poisson arrival process
Calls arrives according to a Poisson arrival process with rate lambda = 15. Find E(N(2,4]N(3,5]) My thoughts: E(N(2,4]) = E(N(3,5]) = lambda * t = 15 * 2 = 30 However, I cannot figure out the next step, since (2,4] and (3,5] are NOT non-overlapping, so they are NOT independent and I cannot get the resulting expectation by simply multiplying them. I'm really stuck in this. Can someone please explain this problem? Many thanks.
Number of sequences formed of $k$ pairwise disjoint subsets of a set of $n$ elements is $(k+1)^n$.
For each $a \in \{1,2,\ldots,n\}$, there are $k+1$ possibilities: $a$ can be element of $S_1$, of $S_2$, ..., of $S_n$ or of none of the $S_i$. It is not possible that $a$ is an element of more than one of the $S_i$, since that would contradict the condition $S_i \cap S_j$ for $i \neq j$. Multiplying these $k+1$ possibilities for all $a \in \{1,2,\ldots,n\}$ yields the answer of $(k+1)^n$ possibilities.
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