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11

First, note that: $$\int_0^kf(x)\,dx=\int_0^kf(k-x)\,dx$$ Therefore, we now have that: $$\begin{equation} \begin{split} \int_0^\frac{\pi}{2}\sin^{n}x-\cos^{n}x\,dx & =\int_0^\frac{\pi}{2}\sin^{n}\left(\dfrac{\pi}{2}-x\right)-\cos^{n}\left(\dfrac{\pi}{2}-x\right)\,dx \\ & =\int_0^\frac{\pi}{2}\cos^{n}x-\sin^{n}x\,dx \\ & ...


10

(assuming $p>2$) odd squares are all $1$ mod($4$), your expression is $3$ mod($4$).


6

If $2\vert x^2-1$, then $x$ must be odd, so write $x=2y+1, y\in\mathbb{Z}$. Then $x^2-1=4y^2+4y=8\frac{y(y+1)}{2}$. Note that at least one of $y,y+1$ is even, so the fraction on the right is always an integer. Thus we have that $8\vert x^2-1$.


4

Much easier: as $p|rs, p\text{ prime}\implies p|r\text{ or }p|s$: $$p|a^n\text{ and }p|b^k\implies p|a\text{ and }p|b.$$


4

You want to prove the intersection of the closed balls of radius $3$ around $(1,1,1)$ and $(-1,-1-1)$ has infinite cardininality. It suffices to prove the intersection of the open balls of radius $3$ around $(1,1,1)$ and $(-1,-1-1)$ is non-empty (because non-empty open sets have infinite cardinality). To see this notice $(0,0,0)$ is in this intersection.


4

Great questions! Here are my answers: You're right that we would end up with an infinite list -- a countably infinite list! But this doesn't have to equal the set $S$. For example, consider the infinite set $\Bbb Z$. Now, let $s_{1} = 2$, $s_{2} = 4$, $s_{3} = 6$, $\dots$. Then clearly $\{s_{n}\}$ is infinite, but does it equal our original set $\Bbb ...


4

For all $x\in A$ there is $y\in B$ such that $f(x)=y$; thus $f^{-1}(y)=x$. Therefore $f^{-1}$ is surjective.


4

Suppose $2p+1=k^2$. Clearly $p=2$ fails so we may assume $p$ is odd. We may rewrite it as $2p=(k-1)(k+1)$. Now since the LHS is even, so must the RHS. Note now that $k+1$ and $k-1$ must always have the same parity. Thus $k$ must be odd, so $k+1$ and $k-1$ are even. But this means that $(k+1)(k-1)$ must be divisible by $4$, and the LHS is clearly not ...


4

Hint: if $a=\cos\alpha$, $\sqrt{1-a^2}=\sin\alpha$, $b=\cos\beta$ and $\sqrt{1-b^2}=\sin\beta$, then you need to show that $$ \lvert\cos\alpha\sin \beta+\cos \beta\sin\alpha-\sqrt3 \sin\alpha\sin\beta+\sqrt3 \cos\alpha\cos\beta| =\lvert\sin(\alpha+\beta)-\sqrt3\cos(\alpha+\beta)|\le2, $$ that is, $$ ...


4

Just for kicks, here is a completely different way to prove that $f(x)=x^2$ is continuous. The identity function is continuous (easy to see, think for example inverse images of open sets). Baby Rudin 4.4 teaches us that if $\lim_{x\to p} f(x)=A$ and $\lim_{x\to p} g(x)=B$, then $\lim_{x\to p} f(x)g(x)=AB$. From this we can infer that if $f$ is continuous ...


4

For the first one, you can proceed directly: Notice that if the expression is equal to some rational $r$, then $$x + b = r(x + a) \implies x(1 - r) = a - b$$ Now the right side is rational, but the left side is irrational unless..... For the second, I'd suggest proceeding similarly. Write $$x^2 + x + \sqrt 2 = r(y^2 + y + \sqrt 2)$$ and rearrange to ...


4

In general, we cannot expect that $$\int_{\Omega} |f_n-f| \cdot |f_n| \, d\mu \to 0;$$ roughly, the reason is that $L^1$-convergence does not imply $L^2$-convergence. (Counter)Example Consider $((0,1),\mathcal{B}((0,1)))$ endowed with the Lebesgue measure restricted to $(0,1)$ and $$f_n(x) := n^{3/4} 1_{[0,1/n]}(x).$$ Then $$\int |f_n(x)| \, dx = ...


4

Actually, you need both to hold; in classical logic, $P \vee Q \to R$ is equivalent to $(P \to R) \wedge (Q \to R)$. The intuition is as follows: assuming $P \vee Q$, you know at least one of $P$ or $Q$ is true, but you don't know which. Thus, to deduce $R$ from the assumption $P \vee Q$, you need to be able to deduce $R$ from whichever one is true—since ...


4

Assume $x > 1$. Let $a > 0$. Then \begin{align*} g(x + a) - g(x) & = (x + a)^2 - 2(x + a) + 1 - (x^2 - 2x + 1)\\ & = x^2 + 2ax + a^2 - 2x - 2a + 1 - x^2 + 2x - 1\\ & = 2ax + a^2 - 2a\\ & = 2a(x - 1) + a^2 \end{align*} Since $x > 1$ and $a > 0$, $2a(x - 1) > 0$ and $a^2 > 0$. ...


3

Compute the derivative of the function: $g'(x)=2x-2.$ We can see that $\forall x>1$, $g'(x)>0$ by computing the second derivative $g''(x)=2$. This implies that $g$ has rate of change greater than $0$ for all $x$ greater than $1$ because $g'(1)=0$. So $x>1\implies g'(x)>0$, that is, this function is strictly increasing. It is also monotone ...


3

Yes. "$\iff$" is used to denote logical equivalence, or necessary and sufficient conditions. It is read "if and only if," of which "iff" is a common abbreviation. $p \iff q$ is true if $p$ is true whenever $q$ is true and $q$ is true whenever $p$ is. Assume $\sqrt{5}$ is a rational number. By the definition of rational number, $\sqrt{5} = \frac{p}{q}$, ...


3

You're wrong about it being a "big no no". You know that the equation $ax^2+bx+c=0$ holds for every value of $x$. If it holds for every value of $x$ then in particular it must hold for $x=1$, $x=2$, and $x=1204$. This yields three linear equations whose solution is $a=b=c=0$. If the hypothesis holds, then the coefficients don't have a choice but to be $0$ ...


3

The number of paths of length $2$ from $u$ to $v$ is equal to the sum over $w\in V$ of all the ways to get from $u$ to $w$ multiplied by all the ways to get from $w$ to $v$, in one step. So it is the sum over $w\in G$ of $A_{u,w}A_{w,v}$. This is exactly $A^2_{uv}$.


2

To answer your question without resorting to any explanation beyond what you have provided: No. This part is seriously off "Assume 2|x−1 and 4|x+1" without any explanation. When you say "assume this", "assume that", you have to cover all options. 2 is prime so it must divide either (x+1) or (x-1), that essential part is missing. Now, if (x+1) divides 2, ...


2

Yeah, that's fine. Remember that you're allowed to use words in proofs; feel free to use as many as necessary in order to be clear. Maybe the following would be a good example: Consider the binomial expansion of $(\sqrt{n} + 1)^8$, where the first term is $n^4$. So take $(n^4 + \ldots + 1)$ and (...your next steps here)


2

If you were to apply the result you quote directly, what you'd actually obtain is that if $2 \mid x^2-1$ then $8 \mid (x^2-1)^2$. Specifically, you're taking $a=2$, $c=4$ and $b=d=x^2-1$ in the result you mention. However, it is true that if $2 \mid x^2-1$ then $8 \mid x^2-1$; it's just true for different reasons. Specifically, if $2 \mid x^2-1$ then $x-1$ ...


2

Lemma: Lagrange identity Let $x=(x_1,...,x_n),y=(y_1,...,y_n)\in\mathbb{R}^n$. Then $\|x\|^2 \|y\|^2-\langle x,y\rangle^2=\sum_{i<j}(x_iy_j-x_jy_i)^2$. Before prove this formula, note that it means $$\sum_{i=1}^nx_i^2\sum_{j=1}^ny_i^2=\sum_{i<j}(x_iy_j-x_jy_i)^2+\left(\sum_{i=1}^nx_iy_i\right)^2.$$ Thus, if we take $y_i=1$, then ...


2

Consider the following cases: $n\equiv0\pmod3 \implies n^2+n\equiv0^2+0\equiv0\pmod3$ $n\equiv1\pmod3 \implies n^2+n\equiv1^2+1\equiv2\pmod3$ $n\equiv2\pmod3 \implies n^2+n\equiv2^2+2\equiv0\pmod3$


2

It is not clear what tools are allowed, so I will guess. Let $b=1+t$ where $t$ is positive. Then any positive integer $n$, by the Bernoulli Inequality, we have $(1+t)^n \ge 1+nt\gt nt$. It follows that $$b^{-n}=\frac{1}{(1+t)^n}\lt \frac{1}{nt}.$$ For suitable $n$, we have $\frac{1}{nt}\lt c$.


2

The intuition is that a point $p$ and a vector $v$ define a hyperplane in $\mathbb R^d$ through $p$ perpendicular to $v$. The points $q$ such that the inner product of the vector $q - p$ with $v$ is positive, $\langle v, q-p \rangle > 0$, are all on one side of this hyperplane, the points $q$ such that $\langle v, q-p \rangle < 0$ are all on the ...


2

Hint: If $M_C$ is the midpoint of $AB$, then $A M_CS$ and $M_CBS$ have equal bases and share the same height. Can you proceed for the first part? For the second part, a similar approach, except consider the triangles $CM_CB$ and $CM_CA$. Are these areas equal? Express each of the areas in terms of $X, Y$ and $Z$. Can you finish up for the second part?


2

Suppose $A$ and $B$ are any sets and $A \subseteq B.$ We must show that $ x\in A\cup B \implies x\in B$. Let $x \in A\cup B$. We must show that $x\in B$. By definition of union, $x \in A$ or $x\in B$. In case $x\in A$ then since $A\subseteq B$ $x\in B$. In case x ∈ B, then clearly x ∈ B. So in either case, $x\in B$ as was to be show.


2

Hint: For the second portion, try a proof by contradiction: For non-zero divisors $a, b \in R$, suppose $(ab)c=0$ for nonzero $c$, then by associativity $a(bc)=0$...


2

How is $\Pr(A_3) = \Pr(A_1 ∩ A_2 ∩ A_3)$? If GS 1, 2, 3, and 4 are in different groups, then GS 1, 2, and 3 are in different groups; and GS 1 and 2 are in different groups. Therefore, if $A_3$, then $A_2$ and $A_1$. In set-theoretic terms, $A_3 \subseteq A_2 \subseteq A_1$. The intersection of a set with any one of its subsets is the subset. ...


2

Well, it depends on the book. The statement of this theorem seems like it's assuming some other types of analysis/ general topology (esp. since many of these theorems are generalized in topology.) However, that's the trouble with specific math courses, they tend to actually assume some mathematical maturity from other courses you have studied. So, check out ...



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