qid
stringlengths 1
7
| Q
stringlengths 87
7.22k
| dup_qid
stringlengths 1
7
| Q_dup
stringlengths 97
10.5k
|
|---|---|---|---|
1950989
|
How can I show that [imath]\mathcal P(\mathbb N)[/imath] is uncountable.
I saw somewhere the the cantor set is in bijection with [imath]\mathcal P(\mathbb N)[/imath], so this mean that [imath]\mathcal P(\mathbb N)[/imath] is uncountable. But how can I show that [imath]\mathcal P(\mathbb N)[/imath] is uncountable ?
|
941055
|
Proof of the uncountability of reals using the diagonal argument—problem?
Consider a common proof of the uncountability of [imath](0,1][/imath], as presented here for example: We assume that the reals can be arranged in a sequence [imath]x_k[/imath], represent every number in [imath]x_k[/imath] by its nonterminating decimal expansion, use the diagonal argument to create a sequence of decimal digits that differs from the nonterminating decimal expansion of each number in [imath]x_k[/imath] and then claim that this sequence of digits so generated must be the decimal expansion of some number not in [imath]x_k[/imath]. My problem is with this last step. Could it not be the case that the new sequence of digits is just the terminating decimal expansion of some number already in [imath]x_k[/imath]? Would it not be more correct to use the sequence [imath]x_k[/imath] to generate a sequence of all (both terminating and non-terminating) decimal expansion of all numbers in [imath][0,1)[/imath] and then apply the diagonal argument?
|
1951021
|
Why in calculus the angles are measured in radians?
Why is the formula [imath]\lim\limits_{h\rightarrow 0}\frac{\sin h}{h}=1[/imath] not valid when [imath]h[/imath] is measured in degrees?
|
1339540
|
Why does the derivative of sine only work for radians?
I'm still struggling to understand why the derivative of sine only works for radians. I had always thought that radians and degrees were both arbitrary units of measurement, and just now I'm discovering that I've been wrong all along! I'm guessing that when you differentiate sine, the step that only works for radians is when you replace [imath]\sin(dx)[/imath] with just [imath]dx[/imath], because as [imath]dx[/imath] approaches [imath]0[/imath] then [imath]\sin(dx)[/imath] equals [imath]dx[/imath] because [imath]\sin(\theta)[/imath] equals [imath]\theta[/imath]. But isn't the same true for degrees? As [imath]dx[/imath] approaches [imath]\theta[/imath] degrees then [imath]\sin(dx \,\text{degrees})[/imath] still approaches [imath]0[/imath]. But I've come to the understanding that [imath]\sin(dx \,\text{degrees})[/imath] approaches [imath]0[/imath] almost [imath]60[/imath] times slower, so if [imath]\sin(dx \,\text{radians})[/imath] can be replaced with [imath]dx[/imath] then [imath]\sin(dx \,\text{degrees})[/imath] would have to be replaced with [imath](\pi/180)[/imath] times [imath]dx[/imath] degrees. But the question remains of why it works perfectly for radians. How do we know that we can replace [imath]\sin(dx)[/imath] with just [imath]dx[/imath] without any kind of conversion applied like we need for degrees? It's not good enough to just say that we can see that [imath]\sin(dx)[/imath] approaches [imath]dx[/imath] as [imath]dx[/imath] gets very small. Mathematically we can see that [imath]\sin(.00001)[/imath] is pretty darn close to [imath]0.00001[/imath] when we're using radians. But let's say we had a unit of measurement "sixths" where there are [imath]6[/imath] of them in a full circle, pretty close to radians. It would also look like [imath]\sin(dx \,\text{sixths})[/imath] approaches [imath]dx[/imath] when it gets very small, but we know we'd have to replace [imath]\sin(dx \,\text{sixths})[/imath] with [imath](\pi/3) \,dx[/imath] sixths when differentiating. So how do we know that radians work out so magically, and why do they? I've read the answers to this question and followed the links, and no, they don't answer my question.
|
1949125
|
2 conjectured recursion limits for [imath]e[/imath] and [imath]\pi[/imath].
Consider the following recursions [imath] x_{n+2} = x_{n+1} + \frac{x_n}{n} [/imath] [imath]y_{n+2} = \frac{ y_{n+1}}{n} + y_n [/imath] I have been toying around with different starting values ( complex Numbers ) , divergeance etc. But was not able to conclude much. However I noticed when [imath] x_1 = 0 [/imath] [imath]y_1 = 0 [/imath] [imath] x_2 = 1 [/imath] [imath] y_2 = 1 [/imath] We get the following limit recursions [imath] \lim_{n \to \infty} \frac{n}{x_n} = e [/imath] [imath] \lim_{n \to \infty} \frac{2 n}{y_n ^2} = \pi [/imath] How to prove these ?? And how about the divergeance / convergeance for other complex initial values ? Edit : a partial answer occurs here Mirror algorithm for computing [imath]\pi[/imath] and [imath]e[/imath] - does it hint on some connection between them? http://www.pi314.net/eng/miroir.php But the issue of other starting values is not resolved yet. ( so this is not a complete duplicate ) For the first recursion we have an answer ( see below ) but at the time of posting , the second has no answer with respect to variable initial conditions yet.
|
1664933
|
Mirror algorithm for computing [imath]\pi[/imath] and [imath]e[/imath] - does it hint on some connection between them?
Benoit Cloitre offered two 'mirror sequences', which allow to compute [imath]\pi[/imath] and [imath]e[/imath] in similar ways: [imath]u_{n+2}=u_{n+1}+\frac{u_n}{n}[/imath] [imath]v_{n+2}=\frac{v_{n+1}}{n}+v_{n}[/imath] [imath]u_1=v_1=0[/imath] [imath]u_2=v_2=1[/imath] [imath]\lim_{n \to \infty} \frac{n}{u_n}=e[/imath] [imath]\lim_{n \to \infty} \frac{2n}{v_n^2}=\pi[/imath] The formulation and the proof can be seen here. What do you think - is it just a coincidence or is there some deeper meaning in this mirror algorithm about the connection of two constants? By @EricStucky in the comment, the better question: Is there any connection between [imath]e[/imath] and [imath]π[/imath] which is essentially different than Euler's formula? Of course, I expect an answer related to my own question about this 'mirror sequence' If, on the other hand, someone shows a clear relation between this sequence and Euler's formula, that's fine too
|
1949989
|
Can we use propositional logic to prove set identities?
For instance, to prove that [imath]A \cup (B \cap C)=(A \cup B)\cap(A\cup C) [/imath], can we use the definition of union and intersection, namely [imath]A \cup B=\{x\mid x \in A \lor x \in B\}[/imath] and [imath]A \cap B=\{x\mid x \in A \land x \in B\}[/imath], and derive the set or propositions [imath]A(x)=x \text{ is in }A , B(x)=x \text{ is in } B, C(x)=x \text{ is in } C[/imath]. Then, using logic, would it be sufficient to prove the following? [imath]A(X) \lor (B(x) \land C(x)) \equiv (A(x) \lor B(x)) \land (A(x) \lor C(x)) [/imath] Thanks!
|
1098542
|
How can I prove this relation between the elementary set theory and the elementary logic?
If you need to prove an equality like [imath]A\Delta B=(B\setminus C)\cup[C\cap (B\Delta A)][/imath] we can first prove [imath]p\underline{\lor} q\Longleftrightarrow (q\land\overline{r})\lor(q\underline{\lor}p)[/imath] (with a truth table for example) and then use it whith [imath]p=x\in A[/imath], [imath]q=x \in B[/imath] and [imath]r=x \in C[/imath]. So, if we now that [imath]p\underline{\lor} q\Longleftrightarrow (q\land\overline{r})\lor(q\underline{\lor}p)[/imath] then we now that [imath]A\Delta B=(B\setminus C)\cup[C\cap (B\Delta A)][/imath]. But my question is. is true in general (with a general set equation) the reverse implication? that is [imath]A\Delta B=(B\setminus C)\cup[C\cap (B\Delta A)][/imath] implies [imath]p\underline{\lor} q\Longleftrightarrow (q\land\overline{r})\lor(q\underline{\lor}p)[/imath].
|
1951446
|
Given two open sets, show their sum is open
Show that if either [imath]A[/imath] and [imath]B[/imath] is open, then [imath]A + B[/imath] is open. Attempt: Suppose [imath]A[/imath] is open, then for every [imath]a \in A, [/imath] there exists [imath]\epsilon > 0[/imath] such [imath]B_{\epsilon}(a) \subset A[/imath]. Similarly suppose [imath]B[/imath] is open ,then for every [imath]b \in B, [/imath] there exists [imath]\epsilon > 0[/imath] such [imath]B_{\epsilon}(b) \subset B[/imath]. So [imath]A + B[/imath] is open since for every [imath]a + b \in A + B[/imath] there is [imath]\epsilon > 0[/imath] such that [imath]B_{\epsilon}(a+ b) \subset A + B[/imath]. Can someone please verify this is correct? Any better approach or feedback would really help.
|
1624033
|
Sum of two open sets is open?
how to prove that if [imath]A[/imath] and [imath]B[/imath] are open in [imath](\mathbb{R},|.|)[/imath] then [imath]A+B[/imath] is open ? Where [imath]A+B=\{a+b\mid a\in A,b\in B\}[/imath] Thank you
|
1950842
|
How many solutions does the equation [imath]x+y+z=17[/imath] have where [imath]x,y,z[/imath] are non negative integers?
I'm not able to understand the solution to this problem. It says selecting [imath]17[/imath] items from a set of [imath](x,y,z)[/imath] Final answer: [imath](17+3-1)C17[/imath]
|
1463137
|
How many solutions does the equation [imath]x_1 + x_2 + x_3 = 14[/imath] have, where [imath]x_1[/imath] , [imath]x_2[/imath] , [imath]x_3[/imath] are non-negative integers.
I actually got this part and I got [imath]16[/imath] choose [imath]2[/imath], which would be [imath]120[/imath]. The part I didn't get which wouldn't fit into the title was in how many of these solutions is [imath]x_1 \geq 1[/imath], [imath]x_2 \geq 2[/imath],and [imath]x_3 \geq 3[/imath]? I'm not very good at combinatorics so I don't really know. I think the way to approach it would be saying how many solutions where [imath]x < 1 +[/imath] solutions where [imath]x < 2 +[/imath] solutions where [imath]x < 3[/imath] and subtract that from [imath]120[/imath], but I don't know how to do that.
|
1951536
|
show that a function is decreasing
let [imath]f(x)=xa^{\frac{1}{x}}-x[/imath] for positive [imath]x,a\in \mathbb{R}[/imath] Show that [imath]f(x)[/imath] is decreasing on [imath](0, \infty)[/imath]. I am trying to show that [imath]f'(x)= a^{\frac{1}{x}}(1-\frac{ln(a)}{x})-1[/imath] is always negative. To do that, I am thinking about showing that finding a max/min for [imath]f'(x)[/imath] by solving for [imath]f''(x) = 0[/imath] then prove that what we found was indeed a maximum by calculating [imath]f'''(x)[/imath] but then things get ugly. So I'm wondering if there is a easier way.
|
1853575
|
[imath]x(a^{1/x}-1)[/imath] is decreasing
Prove that [imath]f(x)=x(a^{1/x}-1)[/imath] is decreasing on the positive [imath]x[/imath] axis for [imath]a\geq 0[/imath]. My Try: I wanted to prove the first derivative is negative. [imath]\displaystyle f'(x)=-\frac{1}{x}a^{1/x}\ln a+a^{1/x}-1[/imath]. But it was very difficult to show this is negative. Any suggestion please.
|
1817833
|
Are there general methods that can be applied when using the Borel-Cantelli Lemma, to get a statement about a sequence of random variables?
I hope the title in itself is clear, if not allow me to give an example. In Class my Professor did the following: Given a sequence [imath](X_n)_{n \in \mathbb{N}}[/imath] of non-negative i.i.d. RV [imath]X_n \sim X[/imath] with [imath]E(X)< \infty[/imath] then [imath]\frac{X_n}{n} \to 0 [/imath] almost surely To prove this, it is easy to show that [imath]\sum P (X_n/n > \epsilon) < \infty[/imath] and thus conclude by Borel-Cantelli Lemma that [imath]P ( \limsup \lbrace X_n/n > \epsilon \rbrace ) =0 \iff P ( \lim \inf \lbrace X_n /n \leq \epsilon \rbrace )=1 [/imath] But one indeed looks for the answer that [imath]\exists \Lambda \subset \Omega[/imath] with [imath]P( \Lambda)=1[/imath] and for all realizations [imath]\omega \in \Lambda[/imath] we have that [imath]X_n(\omega)/n \to 0[/imath]. In the question Is it correct to say that ([imath]\color{red}{(} \limsup |W_k|/k\color{red}{)} \le 1) \supseteq \limsup \color{red}{(}|W_k|/k \le 1\color{red}{)}[/imath]? Daniel Fischer gives an answer on how to get an inclusion of the form [imath]\liminf \lbrace ... \rbrace \subset \lbrace \limsup ... \rbrace[/imath]. I do believe that he makes use of the (complementary) statement which says that for [imath]\epsilon >0[/imath] [imath] \lbrace \lim \sup | X_n - X| > \epsilon \rbrace \subset \lim \sup \lbrace | X_n -X | \geq \epsilon \rbrace \subset \lbrace \lim \sup |X_n - X | \geq \epsilon \rbrace \tag{*} [/imath] which gives [imath] \lbrace \lim \sup |X_n -X | < \epsilon \rbrace \subset \lim \inf \lbrace | X_n - X | < \epsilon \rbrace \subset \lbrace \lim \sup |X_n -X | \leq \epsilon \rbrace [/imath] My question(s): 1) Are there more chains of inclusions as in (*) that I should know in order to give statements about the [imath]\limsup[/imath] or [imath]\liminf[/imath] of sequence of Random Variables rather than the [imath]\limsup, \liminf[/imath] of the sequence of Events? 2) In the above chain of inclusions, does the [imath]\epsilon >0[/imath] actually depend on [imath]k[/imath]? The Definition gives [imath]\limsup \lbrace X_n/n > \epsilon \rbrace = \bigcap_{n=0}^\infty \bigcup_{k=n} A_k \text{ where } A_k:= \lbrace X_k/k > \epsilon \rbrace [/imath] so intuitively I would say that [imath]\epsilon=\epsilon(k)[/imath] varies as [imath]k[/imath] increases. 3) Are there any general techniques you'd recommend to make a statement about the limit of a sequence of random variables using Borel-Cantelli Lemmas?
|
1818345
|
Revision of Borel-Cantelli, [imath](X_n)_{n \geq 1}[/imath] sqn of [imath]\geq 0[/imath] identical RVs with [imath]E(X) < \infty[/imath] then [imath]X_n/n \to 0[/imath] a.s., are my arguments correct?
I care to understand the concept behind the Borel-Cantelli Lemma better, hence I would appreciate it if someone could take the time to check if my arguments below are clear and rigorous. Statement: Let [imath](X_n)_{n \geq 1}[/imath] be a sequence of non-negative random variables that are identically distributed [imath]X_n \sim X \geq 0 [/imath] with [imath]E(X)< \infty[/imath], show that [imath] X_n/n \to 0[/imath] almost surely Proof: Since [imath]X_n \sim X[/imath] for some non negative random variables we can easily compute that for [imath]\epsilon >0[/imath] we have [imath] \infty>E(X/\epsilon) = \int_0^\infty P(X/ \epsilon > t) dt = \sum_{n=1}^\infty \int_{n-1}^n P(X/ \epsilon > t ) dt \geq \sum_{n=1}^{\infty} P(X/ \epsilon >n) [/imath] Thanks to the Borel-Cantelli Lemma we obtain that [imath] P ( \limsup_{n \to \infty} \lbrace X_n/n > \epsilon \rbrace ) = 0 \iff P ( \liminf_{n \to \infty} \lbrace X_n/n \leq \epsilon \rbrace) = 1 [/imath] Question: Does the [imath]\epsilon[/imath] in the above depend on [imath]k[/imath]? I think it does and I would argue as follows [imath]P( \liminf_{n \to \infty} \lbrace X_n/n \leq \epsilon \rbrace )= P \left( \bigcup_{n=0}^\infty \bigcap_{k=n} \lbrace X_k/k \leq \epsilon (k) \rbrace \right) [/imath] Now using the chain of inclusions that state that we have for [imath]\epsilon >0[/imath] [imath] \lbrace \limsup |X_n-X| < \epsilon \rbrace \subset \liminf \lbrace |X_n-X| < \epsilon \rbrace \subset \lbrace \limsup |X_n-X| \leq \epsilon \rbrace [/imath] We conclude that [imath] P ( \lbrace \limsup_{n \to \infty } X_n/ n \leq \epsilon(k) \rbrace)=1 [/imath] Where I choose the same [imath]k \in \mathbb{N}[/imath] as above, and then choose [imath]k^*[/imath] large enough such that [imath]\epsilon(k) \leq \frac{1}{k^*}[/imath], define the set [imath] \Lambda_{k^*}:= \lbrace \limsup_{n \to \infty} X_n/n \leq 1/k^* \rbrace \implies P(\Lambda_{k^*})=1[/imath] and also we obtain that [imath]\Lambda:=\bigcap_{j \geq k^*} \Lambda_{k^*}[/imath] is still an almost sure event, i.e. [imath]P(\Lambda)=1[/imath] but for all realizations [imath]\omega \in \Lambda[/imath] we have [imath]\limsup_{n \to \infty} X_n( \omega)/n \leq 0 \implies X_n( \omega)/n \to 0[/imath] for all [imath]\omega \in \Lambda[/imath] that is[imath]X_n/n \to 0[/imath] almost surely [imath]\Box[/imath] Are my arguments given above correct and more importantly complete? Are there steps where you'd use different arguments?
|
1264077
|
Irreducible is prime in the ring [imath]\mathbb{Z} + X\, \mathbb{Q}[X][/imath]
If a ring is a UFD, its irreducible elements are exactly its prime elements. Show that the reverse is not true. Give a nontrivial counterexample. Hint: Consider the ring [imath]\mathbb{Z} + X\, \mathbb{Q}[X][/imath]. Note: in the following solution "denominator" of a rational number always means the denominator of its fraction simplified to lowest terms. Let [imath]R:= \mathbb{Z} + X\, \mathbb{Q}[X][/imath], whose only units are [imath]R^\times = \{+1, -1\}[/imath]. We show that if [imath]a[/imath] is reducible in [imath]\mathbb{Q}[X][/imath], which means that [imath]a = q r[/imath] with [imath]\deg(q), \deg(r) \ge 1[/imath], then it is reducible in [imath]R[/imath]. \begin{equation*} a(X) = \Bigl(\sum_{i=1}^{m} q_i X^i\Bigr) \cdot \Bigl(\sum_{j=1}^{n} r_j X^j\Bigr) \quad \text{with [imath]m, n \ge 1[/imath]}\, . \end{equation*} Since [imath]q_0 r_0 = a_0 \in \mathbb{Z}[/imath], there is an [imath]c \in\mathbb{Q}[/imath] with [imath]c q_0, \frac{r_0}{c} \in \mathbb{Z}[/imath]. To be more precise [imath]c = \frac{\text{Denominator of [/imath]q_0[imath]}}{\text{Denominator of [/imath]r_0[imath]}}[/imath]. Then \begin{equation*} a(X) = \Bigl(\sum_{i=1}^{m} c q_i X^i\Bigr) \cdot \Bigl(\sum_{j=1}^{n} \frac{r_j}{c} X^j\Bigr) \, , \end{equation*} so [imath]q, r \in R[/imath] and not units, so [imath]a[/imath] is reducible in [imath]R[/imath]. Let [imath]a, b, p \in R[/imath] with [imath]p[/imath] irreducible and [imath]p \mid a b[/imath]. Now we look at the problem in [imath]\mathbb{Q}[X][/imath]. As shown, [imath]p[/imath] cannot be reducible [imath]\mathbb{Q}[X][/imath], so it must be a unit or irreducible. We first treat the case that it is irreducible in [imath]\mathbb{Q}[X][/imath]. Since we are now in an UFD if [imath]p \mid a b[/imath] it follows that [imath]p \mid a[/imath] or [imath]p \mid b[/imath]. Assume without loss of generality that [imath]p \mid a[/imath]. So there is an [imath]u\in\mathbb{Q}[X][/imath] with [imath]p u =a[/imath]. It follows that [imath]p_0 \cdot u_0 = a_0 \in \mathbb{Z}[/imath]. If [imath]u_0 \not\in \mathbb{Z}[/imath], there is [imath]c \in \mathbb{Z}[/imath] (the denominator of [imath]u_0[/imath]) for which [imath]\frac{p_0}{c}\in\mathbb{Z}[/imath] and so [imath]\frac{p}{c}\in R[/imath], which contradicts the assumption that [imath]p[/imath] is irreducible in [imath]R[/imath]. But if [imath]u_0 \in \mathbb{Z}[/imath], also [imath]u\in R[/imath] and we have in [imath]R[/imath] (!) that [imath]p[/imath] divides [imath]a[/imath]. Now we treat the case that [imath]p[/imath] is a unit in [imath]\mathbb{Q}[X][/imath]. Since [imath](\mathbb{Q}[X])^\times = \mathbb{Q}\setminus\{0\}[/imath] and [imath]R \cap \mathbb{Q} = \mathbb{Z}[/imath], it follows that [imath]p\in\mathbb{Z}[/imath] and since [imath]p[/imath] is irreducible, it has to be a prime number. If [imath]p\mid ab[/imath], it follows that [imath]p_0 \mid a_0 b_0[/imath] (again with [imath]a_0, b_0\in \mathbb{Z}[/imath]), so [imath]p_0\mid a_0[/imath] or [imath]p_0\mid b_0[/imath] and therefore [imath]p \mid a[/imath] or [imath]p \mid b[/imath]. We have proven that all irreducible elements of [imath]R[/imath] are prime. But [imath]R[/imath] is not an UFD, for example [imath]\frac{X^2}{6} = \frac{X}{2} \cdot \frac{X}{3} = \frac{X}{6}\cdot X \, .[/imath] Can somebody have a look if this solution is ok? I think the solution seems far too complicated. Either I've missed something obvious or there is a theorem which would make it much easier. PS: What would be a trivial counterexample?
|
1263741
|
Let [imath]R = \mathbb{Z} + x\mathbb{Q}[x][/imath]. Find all the irreducibles in [imath]R[/imath].
Let [imath]R = \mathbb{Z} + x\mathbb{Q}[x] \subset \mathbb{Q}[x][/imath]. Find the irreducibles of [imath]R[/imath]. Show that the irreducible elements in [imath]R[/imath] are [imath]\pm p[/imath] for prime integers [imath]p[/imath] and the irreducible polynomials [imath]p(x) \in \mathbb{Q}[x][/imath] whose constant coefficient is [imath]\pm 1[/imath]. Prove these irreducibles are prime in [imath]R[/imath]. The first part is pretty clear. ([imath]\pm p[/imath] for prime integers [imath]p[/imath]), if [imath]\alpha \in R[/imath] s.t. [imath]\deg(\alpha)=0[/imath] then [imath]\alpha[/imath] is a constant polynomial thus in [imath]\mathbb{Z}[/imath]. If [imath]\alpha[/imath] is a prime then it is irreducible. Having trouble showing that for [imath]p(x) \in R[/imath] s.t. [imath]\deg(p(x)) > 0[/imath] with constant term [imath]\pm 1[/imath] are irreducible.
|
1952552
|
Integral of [imath]\int \frac{\sqrt{9-x^2}}{x}[/imath]
I don't understand how the integral [imath]\int \frac{\sqrt{9-x^2}}{x}=\sqrt{9-x^2}-3 \ln(\sqrt{9-x^2}+3)+3 \log(x)+c[/imath] I keep getting [imath]-3/x +c[/imath] as the answer.
|
867085
|
Integration practice of [imath]\int \frac{\sqrt{25-y^2}}{y}dy[/imath]
I need to solve [imath]\int \frac{\sqrt{25-y^2}}{y}dy[/imath]. I originally thought IBP, but that led to a very large and confusing algebra problem. Then I started to look at the [imath]\sqrt{25-y^2}[/imath] and started to consider doing a [imath]u[/imath]-substituion, and letting [imath]u=\sin\theta[/imath] and [imath]du=\cos\theta d\theta[/imath]. Once I started playing a bit, the intuition was lost. Any and all help is appreciated.
|
1952834
|
How to analyze the following non-linear ODE asymptotically,
[imath]y\frac{\text{d}^2y}{\text{d}x^2}-2\left(\frac{\text{d}y}{\text{d}x}\right)^2+xy^3\frac{\text{d}y}{\text{d}x}=0.[/imath] I want to analyze the above non-linear ODE asymptotically. Do you have an idea?
|
1952785
|
How to get solutions in elementary functions to the following non-linear ODE
[imath]y\frac{\text{d}^2y}{\text{d}x^2}-2\left(\frac{\text{d}y}{\text{d}x}\right)^2+xy^3\frac{\text{d}y}{\text{d}x}=0[/imath]. I want to get solution in elementary functions in order to do asymptotic analysis. I tried the problem with Mathematica. It did not give any solutions.
|
1748106
|
Find the latus rectum of the Parabola
Let [imath]y=3x-8[/imath] be the equation of tangent at the point [imath](7,13)[/imath] lying on a parabola, whose focus is at [imath](-1,-1)[/imath]. Evaluate the length of the latus rectum of the parabola. I got this question in my weekly test. I tried to assume the general equation of the parabola and solve the system of equations to calculate the coefficients with the help of these given conditions. But this way it becomes very lengthy and tedious. Can anyone provide an elegant solution? Thanks.
|
2206630
|
Length of latus rectum of parabola given the equation of the tangent to the parabola, the point of tangency and focus
Problem Statement:- Let [imath]y=3x-8[/imath] be the equation of tangent at the point [imath](7,13)[/imath] lying on the parabola whose focus is at [imath](-1,-1)[/imath]. If the latus rectum of parabola is [imath]k[/imath] then find the value of [imath]\lfloor{k^2}\rfloor[/imath]. I could not think of much on this, but I did the find that the image of focus of the parabola w.r.t any tangent lies on the directrix by messing around in geogebra. Some hints to this problem would be great.
|
1952702
|
Proving the upper- and lowerbound of a probability measure set-intersection, given certain conditions
So I've been going in circles over the last few hours of this seemingly simple problem: "Let [imath](X,F,\mu)[/imath] be a probability space. Let [imath]A,B,C[/imath] be sets contained in [imath]F[/imath], with [imath]\mu(A \cap B) \ge \frac{1}{2}[/imath], [imath]\mu(A \cap C) \ge \frac{1}{2}[/imath], [imath]\mu(B \cap C) \ge \frac{1}{2}[/imath]. Show that [imath]\frac{1}{4} \le \mu(A \cap B \cap C) \le \min\{\mu(A \cap B), \mu(A \cap C), \mu(B \cap C)\}[/imath]". Honestly, I'm plain simply stuck. I've been trying to see if I could get something useful out of De Morgan's identities as well as the distributive laws, but I seem to either be going in circles or end up with something not very useful due to unions I can not identify or the lack of information about [imath]A[/imath], [imath]B[/imath] and [imath]C[/imath]. Same thing with the strongly additive property of a measure, which is only stated for the intersection of two sets in my book. I feel like this should be simple, but I'm staring myself blind at this problem. Can anyone offer some guidance?
|
1952113
|
Show [imath]\frac{1}{4}\leq \mu(A\cap B \cap C)[/imath]
Let [imath]\{X,\mathcal{A},\mu\}[/imath] be a probability space. Let [imath]A[/imath], [imath]B[/imath], [imath]C[/imath] be sets in [imath]\mathcal{A}[/imath], which satisfy [imath]\mu (A\cap B)\geq \frac{1}{2}, \quad \mu (A\cap C)\geq \frac{1}{2}, \quad \mu (B\cap C)\geq \frac{1}{2}.[/imath] Show that [imath]\frac{1}{4}\leq \mu(A\cap B \cap C)[/imath]. I've managed to show in a previous assignment that [imath]\mu(A\cap B \cap C)\leq \min\{\mu (A\cap B), \mu (A\cap C), \mu (B\cap C)\},[/imath] but now I'm stuck with the above inequality. I've tried to draw it, but still no luck. Any hints?
|
1952774
|
It is possible to apply a derivative to both sides of a given equation and maintain the equivalence of both sides?
I have an equation with this shape, where [imath]k,t_1,r \in \Bbb N[/imath]: [imath]2^{(x+1)^2}=k+t_1(x^2+r)[/imath] And I noticed that I can find [imath]t_1[/imath] in terms of [imath]x[/imath] as follows: [imath] \frac{{\rm d} }{{\rm d}x} 2^{(x+1)^2} = \frac{{\rm d} }{{\rm d}x} \left( k+t_1(x^2+r) \right)[/imath] And then I would continue as follows: [imath] 2(x+1)\cdot2^{(x+1)^2}\cdot\ln(2) = t_12x[/imath] [imath] (x+1)\cdot2^{(x+1)^2}\cdot\ln(2) = t_1x[/imath] So finally: [imath] t_1=\frac{(x+1)\cdot2^{(x+1)^2}\cdot\ln(2)}{x}[/imath] Is it possible to use a derivative in such case? I think that if both functions [imath]2^{(x+1)^2}[/imath] and [imath]t_1(x^2+r)[/imath] are equivalent and the derivative exists it might be possible, but I am not very sure about the validity of that step. Thank you!
|
1310176
|
Can we differentiate equations without changing the solutions?
Suppose we have [imath]x^2+4x+3=0[/imath] We differentiate both sides to get [imath]2x+4=0[/imath]. So [imath]x=-2[/imath] But [imath](-2)^2+4(-2)+3=0[/imath] [imath]4-8+3=-1[/imath] is not equal to [imath]0[/imath]. Which tells that we cannot differentiate equalities. But in many proofs, like in integral calculus, differentiating both sides occurs. So, where is the fallacy?
|
393686
|
Prove that for all positive integer n, the inequality [imath]2n\choose n[/imath] [imath]<4^n[/imath] holds
How do I prove that for all positive integer n, the inequality [imath]2n\choose n[/imath][imath]<4^n[/imath] holds? Thank you!
|
931306
|
Inequality [imath]\binom{2n}{n}\leq 4^n[/imath]
I would like to prove the following inequality, for [imath]n=0,1,2,...[/imath], [imath] \binom{2n}{n}\leq 4^n.[/imath] I already proved it by induction, and I'm looking for another proof.
|
1953773
|
Prove that [imath]a:\mathbb{Z}_+\to \mathbb{R}[/imath], with [imath]a_n = \frac{1}{n}[/imath] is a Cauchy sequence.
I am not allow to use the fact if a sequence is convergent then its Cauchy which is the only way I can solve this problem.
|
1373814
|
Proving [imath]\left\{ \frac{1}{n} \right\}[/imath] is Cauchy
I am trying to understand better Cauchy sequence in the book that I am reading he is trying to provide counter argument that in general metric spaces we can find sequence that are Cauchy but aren't convergent for example [imath]\frac{1}{n}[/imath] is Cauchy but doesn't converge in the space T = (0,1]. I am trying to prove that the sequence is Cauchy to begin with if we consider [imath]n \geq m > 0[/imath] Consider [imath]|a_n - a_m| = \frac{m - n}{mn} < \frac{m}{mn} = 1/n < \epsilon[/imath]. So if we choose N = [imath]\frac{1}{\epsilon}[/imath] that should work since for [imath]n, m > N = \frac{1}{\epsilon}[/imath] we have the following: [imath]|a_n - a_m| = \frac{m - n}{mn} < \frac{m}{mn} = \frac{1}{n} < \epsilon[/imath]. what do you guys think any the proof is good or is there anything to make it better ? It is always good to learn from the extra bit of details as it makes my brain understand the subject more !
|
1953757
|
Square root of complex number
How do I find the square root of complex number [imath]7-(6\sqrt2)i[/imath]? I hope there's someone who can show me the method. Thanks in advance.
|
313111
|
Square root of a complex number
I was trying to find the square root of a complex number [imath]w=u+iv[/imath], I assumed [imath]z^2=(x+iy)^2=w[/imath] Now solving this equation I got two values of [imath]x[/imath] and two values of [imath]y[/imath] in terms of [imath]u[/imath] and [imath]v[/imath]. So, I get a possible of 4 solutions. But fundamental theorem of algebra says it has exactly two roots. What are these two roots? My work [imath]x=\pm \sqrt {\sqrt {u^2+v^2}+u}[/imath] and [imath]y=\pm \sqrt {\sqrt {u^2+v^2}-u}[/imath]
|
1954488
|
Find a function [imath]f(x)[/imath] which is defined at every real number but is continuous at [imath]0[/imath] and is not continuous at every other number
Find a function [imath]f(x)[/imath] which is defined at every real number but is continuous at [imath]0[/imath] and is not continuous at every other number. This was a bonus question for our Calculus 1 homework on limits a few months ago and no one could figure it out.
|
215126
|
Give an example of a function [imath]f: \mathbb{R} \to \mathbb{R}[/imath] which is continuous only at [imath]0[/imath].
I do not know an example. Will ask question if in doubt of the proofs provided thank you!!
|
63756
|
Tail sum for expectation
In Pitman's Probability, the tail sum formula for expectation is introduced for a nonnegative (0,1,...) discrete random variable [imath]X[/imath]: [imath]E(X) = \sum_{i=0}^\infty P(X > i).[/imath] I wonder if there is a similar formula for nonnegative continuous random variable [imath]X[/imath]: [imath]E(X) = \int_0^\infty P(X > x) dx?[/imath] If no, are there some conditions for it to hold? And how can it be proved? Here is my thought: If the cdf [imath]F[/imath] of [imath]X[/imath] is bijective, then [imath]X=F^{-1}(U)[/imath] for some random variable [imath]U[/imath] uniformly distributed over [imath][0,1)[/imath]. So [imath]E(X) = \int_0^1 F^{-1}(u) du.[/imath] To prove the tail sum formula, it suffices to prove [imath]\int_0^1 F^{-1}(u) du = \int_0^\infty P(X > x) dx.[/imath] But I am stuck here. What's more, is the condition that the cdf [imath]F[/imath] of [imath]X[/imath] is bijective really necessary for tail sum formula to hold? Can tail sum formula be generalized to a random variable that is not necessarily nonnegative? Thanks!
|
2307746
|
a puzzle on probability theory.
Let [imath]X[/imath] be a random variable.I want to prove: [imath]E[x]=\int_{-\infty}^{\infty}{P[X > x]dx}[/imath]...(*) Here is my proof.let [imath]f[/imath] be the PDF(probability density function) of [imath]X[/imath].[imath]\int_{-\infty}^{\infty}{P[X > x]dx}=\int_{-\infty}^{\infty}{(1-\int_{-\infty}^{x}{f(t)dt)}dx}[/imath].For the right hand size,we integral by part. [imath]\int_{-\infty}^{\infty}{(1-\int_{-\infty}^{x}{f(t)dt)}dx}=(x-x\int_{-\infty}^{x}{f(t)dt})\mid_{-\infty}^{\infty}+\int_{-\infty}^{\infty}xf(x)dx[/imath].Since the first item [imath](x-x\int_{-\infty}^{x}{f(t)dt})\mid_{-\infty}^{\infty}[/imath] equals zero,we have proved [imath]E[x]=\int_{-\infty}^{\infty}{P[X > x]dx}[/imath]. However I find my proof is [imath]\textbf{WRONG}[/imath] because PDF [imath]f\quad\textbf{May Not be continous}[/imath].How can I handle this case?Is (*) always right,or exists some counterexample?Well I am not good at probability theory,I search for help.
|
1949791
|
Show that with probability one the events cease to occur ultimately under certain conditions
Let [imath]A_n[/imath] be a sequence of independent events with [imath]lim_{n \to \infty}P(A_n)=0.[/imath] Suppose that [imath]\sum_{n=1}^{\infty}P(A_n^c \cap A_{n+1}) < \infty.[/imath] Will the events [imath]A_n[/imath] cease to occur eventually with probability one? I think that the question is asking to show that [imath]P(A_n\ i.o.)=0[/imath] And I tried to prove that [imath]\sum_{n=1}^{\infty}P(A_n)<\infty[/imath] So that I can use the Borel-Cantelli Lemma. But I did not get nowhere. I am kind of confused about all these operations includes limits, sets, and interplays between them, as well as the [imath]limsup[/imath] and [imath]liminf[/imath] stuff. Hope someone could help me.
|
42037
|
A variation of Borel Cantelli Lemma
If [imath]P(A_n) \rightarrow 0[/imath] and [imath]\sum_{n=1}^{\infty}{P(A_n^c\cap A_{n+1}})<\infty[/imath] then [imath]P(A_n \text{ i.o.})=0[/imath]. How to prove this? Thanks.
|
1954836
|
Limit of an infinite product
I want to evaluate this limit : [imath]\lim_{n\to \infty}\frac {\left((2n+1)(2n+2).....(2n+n)\right)^{1/n}}n[/imath] What I did was that I bought the [imath]\frac 1n[/imath] inside and divided throughout by [imath]n^n[/imath] which left me with the answer as [imath]2[/imath]. But the correct answer is [imath]\frac {27}{4e}[/imath]. How? PS : forgive me I dont know MathJax or LaTex so :P
|
475786
|
How to compute [imath]\lim_{n\rightarrow\infty}\frac1n\left\{(2n+1)(2n+2)\cdots(2n+n)\right\}^{1/n}[/imath]
If [imath]\displaystyle f(n)=\frac1n\Big\{(2n+1)(2n+2)\cdots(2n+n)\Big\}^{1/n}[/imath], then [imath]\lim\limits_{n\to\infty}f(n)[/imath] equals: [imath] \begin{array}{} (\mathrm{A})\ \frac4e\qquad&(\mathrm{B})\ \frac{27}{4e}\qquad&(\mathrm{C})\ \frac{27e}{4}\qquad&(\mathrm{D})\ 4e \end{array} [/imath] I couldn't get the right way to start off with this problem. But, as the options include the constant [imath]e[/imath] I think I will have to work out with logarithms. So, this is what I did. [imath] \lim_{n \to \infty} f(n)\\ =\mathrm{exp}\left(\ln\left(\lim_{n \to \infty} f(n)\right)\right)\\ =\mathrm{exp}\left(\lim_{n \to \infty}\left[\ln\left(f(b)\right)\right]\right)\\ =\mathrm{exp}\left(\lim_{n \to \infty}\left[\frac{1}{n^2}\left(\ln(2n+1)+\ln(2n+2)+\ldots+\ln(2n+n)\right)\right]\right) [/imath] I'm not able to proceed further. In case my method is correct please give me hints on proceeding further and in case it is wrong give me the same on another method.
|
1950999
|
Can we drop the requirement of a smooth manifold to be a topological manifold from the definition?
I've seen a definition of a manifold as a set [imath]M[/imath] with an atlas, that is, a collection of pairs [imath](U_\alpha,\phi_\alpha)[/imath] where all [imath]U_\alpha[/imath] cover [imath]M[/imath] and the coordinate maps [imath]\phi_\alpha[/imath] are compatible. The only notion of topology in this definition is that [imath]\phi_\alpha(U_\alpha)[/imath] is an open set in [imath]\mathbb R^n[/imath] (I guess using the standard topology). In this definition we could identify [imath]U_\alpha[/imath] as an open set, so this defines a topology in [imath]M[/imath]. But most definitions require [imath]M[/imath] to have a topology previously defined so [imath]U_\alpha[/imath] are open sets from the beggining. How are these two definitions related? It seems that the second definition allows many possible manifolds if we change the topology. Maybe the former is a particular case of the latter?
|
1945789
|
topological structure on smooth manifolds
In John Lee's Introduction to Smooth Manifolds, a smooth manifold is defined as a topological manifold with a smooth structure. In do Carmo's Riemannian Geometry, a differentiable (smooth) manifold is defined by giving the smooth structure on merely a set [imath]M[/imath] and the author makes a remark that such smooth structure induces a natural topology on [imath]M[/imath]. Here is my question: In Lee's definition, what is the relation between the topological structure and the smooth structure? Must the topology of the manifold (in Lee's definition) induced by the smooth structure in the way that do Carmo mentions? The following are the definition and the remark by do Carmo mentioned above.
|
1954977
|
Can't solve integral by partial fractions
Seems this should work via partial fractions; however, doesn't solve out for me. I recoginize that [imath]e^{2x}-1[/imath] is a difference of squares. Still doesn't decompose for me with partial fractions. numerator is [imath]e^x[/imath] demominotor is [imath](e^{2x}-1)(e^x+1)[/imath] [imath]\int \frac{e^x}{(e^2x-1) (e^x+1)}dx[/imath]
|
1954969
|
Integrate by Partial Fractions or Substitution
Evaluate [imath]\int\frac{e^xdx}{(e^{2x}-1)(e^x+1)}[/imath] I tried to decompose the integrand into A and B partial fractions. No success.
|
1955076
|
Help in solving exponential equation
Solve the following equation: [imath]\frac{8^x + 27^x}{12^x + 18^x} = \frac{7}{6}[/imath] All I managed to do is rewrite the given equation in a simpler form: [imath]\frac{4^x}{6^x + 9^x} + \frac{9^x}{6^x + 4^x} = \frac{7}{6}[/imath] I don't know what should be done next. Thank you in advance!
|
384090
|
Find all real numbers [imath]x[/imath] for which [imath]\frac{8^x+27^x}{12^x+18^x}=\frac76[/imath]
Find all real numbers [imath]x[/imath] for which [imath]\frac{8^x+27^x}{12^x+18^x}=\frac76[/imath] I have tried to fiddle with it as follows: [imath]2^{3x} \cdot 6 +3^{3x} \cdot 6=12^x \cdot 7+18^x \cdot 7[/imath] [imath] 3 \cdot 2^{3x+1}+ 2 \cdot 3^{3x+1}=7 \cdot 6^x(2^x+3^x)[/imath] Dividing both sides by [imath]6[/imath] gives us [imath]2^{3x}+3^{3x}=7 \cdot 6^{x-1}(2^x+3^x)[/imath] Is this helpful? If so, how should I proceed form here? If not any hints would be greatly appreciated.
|
577968
|
Given [imath]abc(a+b+c)=3[/imath] prove [imath](a+b)(b+c)(c+a)>8[/imath]
This is an exceptionally strange question. I have tried with all the methods I know. There is an equality which comes pretty close [imath](a+b)(b+c)(c+a)>8abc[/imath]. Please help me!!
|
1401650
|
Prove that [imath](a+b)(b+c)(c+a) \ge8[/imath]
Given that [imath]a,b,c \in \mathbb{R}^{+}[/imath] and [imath]abc(a+b+c)=3[/imath], Prove that [imath](a+b)(b+c)(c+a)\ge8[/imath]. My attempt: By AM-GM inequality, we have [imath]\frac{a+b}{2}\ge\sqrt {ab} \tag{1}[/imath] and similarly [imath]\frac{b+c}{2}\ge\sqrt {bc} \tag{2},[/imath] [imath]\frac{c+d}{2}\ge\sqrt {cd} \tag{3}.[/imath] Multiplying [imath](1)[/imath], [imath](2)[/imath] and [imath](3)[/imath] together we reach [imath](a+b)(b+c)(c+a) \ge8abc.[/imath] Now, I need to show that [imath]abc = 1[/imath]. Again by AM-GM inequality we have [imath]\frac{a+b+c}{3}\ge\sqrt[3]{abc} \implies \frac{\frac{3}{abc}}{3}\ge\sqrt[3]{abc} \implies abc\le1[/imath] Now if we show somehow that [imath]abc\ge1[/imath], we are done and that's where I am stuck. Can someone please explain how to show that? Other solutions to the above question are also welcomed.
|
1955241
|
Difference between [imath]\mathbb{Z}/p^e\mathbb{Z}[/imath] and the finite field [imath]\mathbb{F}_{p^e}[/imath]?
Q: Let [imath]p[/imath] be a prime and let [imath]e \ge 2[/imath]. The quotient ring [imath]\mathbb{Z}/p^e\mathbb{Z}[/imath] and the finite field [imath]\mathbb{F}_{p^e}[/imath] are both rings and have the same number of elements. Describe some ways in which they are intrinsically different. My textbook confirms my understanding that [imath]\mathbb{Z}/p\mathbb{Z}[/imath] and [imath]\mathbb{F}_{p}[/imath]? are the same thing. With [imath]\mathbb{Z}/p^e\mathbb{Z}[/imath], there will be elements that don't have multiplicative inverses, so it is not a field. What is [imath]\mathbb{F}_{p^e}[/imath] if it's not the same as [imath]\mathbb{Z}/p^e\mathbb{Z}[/imath]?
|
2369
|
Construction of [imath]p^n[/imath] field
I heard that one can construct field based on [imath]p^n[/imath] elements where [imath]p[/imath] is prime. I tried with [imath]p = n = 2[/imath]. It seemed easy as there are 2 groups which [imath]\{0,1,2,3\}[/imath] can form and 1 group formed by [imath]\{1,2,3\}[/imath]. However each time I get into contradiction. Is there anything I missed or it is not possible to form [imath]p^n[/imath] element group?
|
879114
|
If G is a group and N is normal in G with index d, then [imath]x^d \in N[/imath]
I want to show the statement in the title. If [imath]G[/imath] is a group and [imath]N[/imath] is normal in [imath]G[/imath] with [imath][G:N]=d[/imath], then [imath]x^d \in N[/imath] for all [imath]x \in G[/imath]. I want to consider the image [imath]xN[/imath] of [imath]x[/imath] in [imath]G/N[/imath]. [imath]G/N[/imath] has order [imath]d[/imath]. Could I now obtain that [imath](xN)^d=e[/imath]? Why?
|
510023
|
If the index [imath]n[/imath] of a normal subgroup [imath]K[/imath] is finite, then [imath]g^n\in K[/imath] for each [imath]g[/imath] in the group.
Let [imath]K \unlhd G[/imath] be a normal subgroup of some group [imath]G[/imath] and let [imath]|G/K|=n<\infty[/imath]. I want to show that [imath]g^n\in K[/imath] for all [imath]g\in G[/imath]. Let [imath]g\in G[/imath], if [imath]g\in K[/imath], then [imath]g^n\in K[/imath] and we are done. If [imath]g^n\notin K[/imath] then consider the set of left cosets [imath] C=\{K, gK,g^2K,...,g^{n-1}K\} [/imath] I want to show that these cosets are all disjoint and hence [imath]C=K[/imath], then I want to show that [imath]g^nK=K[/imath], so [imath]g^n\in K[/imath]. Suppose [imath] g^lK=g^mK [/imath] for some [imath]m,l<n[/imath], then [imath]g^{m-l}\in K[/imath]. I am not sure how to proceed from there.
|
1954966
|
Consider the matrix of the transformation
I need help with solving this hw problem for my matrix algebra class. Consider the transformation [imath]T: \Bbb R^3 \to \Bbb R^2[/imath] defined by: [imath] T(x) = T \pmatrix{x_1\\x_2\\ x_3} = (2x_1 + x_3) \pmatrix{1\\2} + (x_2 - 3x_3)\pmatrix{-1\\1} [/imath] 1a) determine the matrix of the above transformation 1b) determine the reduced row echelon form of the matrix found in 1a 1c) based on your answer to part 1b, is the transformation T onto? 1d) based on your answer to part 1b, is the transformation T one-to-one? 1e) based on your answer to part 1b, determine the set of vectors x in R^3 for which T(x) = 0. Write your answer in parametric vector form Would sincerely appreciate any help.
|
313798
|
Find the standard matrix for a linear transformation
If T: [imath]\Bbb R[/imath]3→ [imath]\Bbb R[/imath]3 is a linear transformation such that: [imath] T \Bigg (\begin{bmatrix}-2 \\ 3 \\ -4 \\ \end{bmatrix} \Bigg) = \begin{bmatrix} 5\\ 3 \\ 14 \\ \end{bmatrix}[/imath] [imath]T \Bigg (\begin{bmatrix} 3 \\ -2 \\ 3 \\ \end{bmatrix} \Bigg) = \begin{bmatrix}-4 \\ 6 \\ -14 \\ \end{bmatrix}[/imath] [imath] T\Bigg (\begin{bmatrix}-4 \\ -5 \\ 5 \\ \end{bmatrix} \Bigg) = \begin{bmatrix} -6\\ -40 \\ -2 \\ \end{bmatrix} [/imath] Then the standard matrix for T is... I'm not exactly sure how to approach this problem. Could anyone explain how to solve this problem?
|
1955037
|
Cycle decomposition for elements of order 2 in [imath]S_4[/imath]
I'm asked to write out the cycle decomposition of each element of order 2 in [imath]S_4[/imath]. However, from my understanding, a cycle decomposition is the product of disjoint cycles. But the elements in [imath]S_4[/imath] of order 2 are already products of disjoint cycles. For example, How do I simplify (1 2) or (1 2)(3 4)? Isn't the only way to write (1 2) as a product (1 2)=(1 2)(1 2)? Which is not disjoint and hence not a cycle decomposition? And isn't (1 2)(3 4) already a cycle decomposition?
|
311680
|
Finding the number of elements of order two in the symmetric group [imath]S_4[/imath]
Find the number of elements of order two in the symmetric group [imath]S_4[/imath] of all permutations of the four symbols {[imath]1,2,3,4[/imath]}. the order two elements are two cycles.number of [imath]2[/imath] cycles are [imath]6[/imath].but the given answer is [imath]9[/imath].where I am wrong?can anybody help me. more generally is there any formula for the problem "Find the number of elements of order [imath]r[/imath] in the symmetric group [imath]S_n[/imath]"
|
1953808
|
Question on big O notation for two small parameters of the same order
Suppose I have two small parameters [imath]\epsilon[/imath] and [imath]\delta[/imath] in some perturbation problem: [imath]0< \epsilon <<1[/imath] and [imath]0<\delta<<1[/imath], with [imath]\epsilon[/imath] and [imath]\delta[/imath] the same order of magnitude, [imath]\epsilon = O(\delta)[/imath]. Is it correct to state [imath]\frac{\epsilon}{\delta} = c[/imath] Where [imath]c[/imath] is some constant, with [imath]c = O(1)[/imath]?
|
1952708
|
Simple question on perturbation theory for a function with two small parameters
Suppose I have an analytic function [imath]f(x;\epsilon, \delta)[/imath] that depends on two small parameters that are of the same order, [imath]0 < \epsilon << 1[/imath] and [imath]0< \delta <<1[/imath] . For example in the context of odes, [imath]\dot{x} = f(x; \epsilon, \delta)[/imath], and so expanding [imath]f[/imath] in Taylor series, we have [imath]f(x;\epsilon, \delta) = f(x;0,0) + \epsilon \frac{\partial f}{\partial \epsilon}(x;0,0) + \delta \frac{\partial f}{\partial \delta}(x;0,0) + O(\epsilon \delta)[/imath] In the consequent computations with [imath]f[/imath], and any other related functions that arise that also depend on [imath]\epsilon[/imath] and [imath]\delta[/imath], is it justifiable to denote [imath]\frac{\epsilon}{\delta} = c[/imath] for some constant [imath]c[/imath] that is [imath]O(1)[/imath]? Or do I need to impose some more restrictions on [imath]\epsilon[/imath], [imath]\delta[/imath]?
|
1955637
|
Show that it is not a Banach Space
The vector space of all finite sequences [imath]x = (x_n)[/imath] (i.e. [imath]x_n = 0[/imath] for all but finitely many indices [imath]n \epsilon N[/imath]) is a normed space with respect to [imath]||x||_{\infty} := sup_{n \epsilon N} |x_n|[/imath]. I haven't really understood Banach spaces well, So I'm really stuck here!
|
478483
|
Completeness of sequence of reals with finitely many nonzero terms
Let [imath]S[/imath] be the set of all sequences [imath]x=\{x_n\}[/imath] of real numbers such that only a finite number of the [imath]x_n[/imath] are nonzero. Define [imath]d(x,y)=\max|x_n-y_n|[/imath]. Is the space complete? Completeness means that any Cauchy sequence converges to a point in the space. Suppose we have a sequence [imath]\{\textbf{a}_i\}_{i=0}^\infty[/imath], where [imath]\textbf{a}_i=(a_{i1},a_{i2},\ldots)[/imath]. Since [imath]\{\textbf{a}_i\}[/imath] is Cauchy, for any fixed [imath]k[/imath], the real sequence [imath]a_{1k},a_{2k},\ldots[/imath] is also Cauchy, hence converges to a real number [imath]b_i[/imath]. I'm not sure, however, whether the sequence [imath]\{\textbf{a}_i\}[/imath] converges to the sequence [imath]\textbf{b}=(b_1,b_2,\ldots)[/imath], or whether the sequence [imath]\textbf{b}[/imath] belongs to [imath]S[/imath].
|
1955330
|
Is a set of continuous and bounded functions defined on [0,1] compact? Is it bounded?
Suppose that [imath]S=\{f\in C[0,1]|\|f\|<=M\}[/imath] with [imath]\|f\|=\max_{x\in[0,1]}|f(x)|[/imath]. Is set [imath]S[/imath] a compact set? Is set [imath]S[/imath] a bounded set?
|
292350
|
Unit ball in [imath]C[0,1][/imath] not sequentially compact
This question is taken from Saxe K -Beginning Functional Analysis. Show that the closed unit ball in [imath]C[0,1][/imath] is not compact by proving that it is not sequentially compact. (It's assumed that we are using the uniform norm). I've been working on this for ages but I could not come up with any sequence [imath]\{f_n\}[/imath] in the unit ball such that there exists [imath]N\in \mathbb{N}[/imath] such that for all [imath]m,n\geq N[/imath] we have that [imath]d(f_n,f_m)>c[/imath]. Should be a nice example of this, please help me!
|
1955292
|
Strictly increasing function and its derivative
We have "[imath]f'(x)>0[/imath] in interval [imath]I\implies f(x)[/imath] strictly increasing in interval [imath]I[/imath]". However, the converse is not true. A frequently cited example is the function [imath]f(x)=x^3[/imath], which is strictly increasing but [imath]f'(0)=0[/imath]. Here comes my question: What is the necessary and sufficient condition of [imath]f'(x)[/imath] to make [imath]f(x)[/imath] strictly increasing? My guess is: [imath]f'(x)>0[/imath] almost everywhere. This is clearly a sufficient condition, as we can use Lebesgue integration and "drop the word 'almost'". But is this a necessary condition? Is there a weaker condition? A full solution or a little hint will both be appreciated. Thank you in advance.
|
1845927
|
Monotone functions and non-vanishing of derivative
The following result is well known: If [imath]f[/imath] is continuous on [imath][a, b][/imath], differentiable on [imath](a, b)[/imath] and [imath]f'[/imath] is non-zero on [imath](a, b)[/imath] then [imath]f[/imath] is strictly monotone on [imath][a, b][/imath]. However if the derivative vanishes at a finite number of points in [imath](a, b)[/imath] and apart from these derivative maintains a constant sign in [imath](a, b)[/imath] then also the function is strictly monotone on [imath][a, b][/imath] (just split the interval into finite number of intervals using these points where derivative vanishes and [imath]f[/imath] is strictly monotone in same direction in each of these intervals). Let's suppose now that [imath]f[/imath] is strictly monotone and continuous in [imath][a, b][/imath] and differentiable in [imath](a, b)[/imath]. What can we say about set of points [imath]A = \{x \mid x \in (a, b), f'(x) = 0\}[/imath] Can it be infinite? Can it be uncountable? How large the set [imath]A[/imath] can be?
|
1146891
|
If the sequence satisfies the property lim[imath]_{n\to \infty}(a_n-a_{n-2})=0[/imath], prove that lim[imath]_{n\to \infty}\frac{a_n-a_{n-1}}{n}=0[/imath].
If the sequence satisfies the property lim[imath]_{n\to \infty}(a_n-a_{n-2})=0[/imath], prove that lim[imath]_{n\to \infty}\frac{a_n-a_{n-1}}{n}=0[/imath]. I am experiencing difficulty showing this rather obvious result. By definition, we know that for any [imath]\epsilon \gt 0[/imath], there is some [imath]N[/imath] such that if [imath]n \gt N[/imath] then [imath]|a_n-a_{n-2}|\lt \epsilon[/imath] My guess is that I need to write [imath]|a_n-a_{n-1}|[/imath] as sums of [imath]|a_n-a_{n-2}|[/imath] from [imath]n[/imath] to [imath]N[/imath]. However, I am stuck with forming an inequality here. I would appreciate any solutions or suggestions.
|
1956109
|
With a sequence [imath]a_n[/imath] that satisfies [imath]\lim\limits_{n\to\infty}(a_{n} - a_{n-2})=0[/imath], prove or disprove this statement
I am stuck finding a way to prove/disprove this limit of a sequence. Assuming that [imath]\lim\limits_{n\to\infty}(a_{n} - a_{n-2})=0[/imath] for the sequence [imath]a_n[/imath], prove or disprove that [imath]\lim\limits_{n\to\infty}\frac{a_{n} - a_{n-1}}{n}=0.[/imath] An explanation on how to work this through would be very appreciated.
|
1956240
|
a submodule of a noetherian module is also noetherian?
I was studing some algebra by my own and I was wondering if this propiety is true: Let [imath]M[/imath] be a module over a ring [imath]A[/imath] and [imath]S\subseteq M[/imath] a submodule, if [imath]M[/imath] is Noetherian, then [imath]S[/imath] is Noetherian. I think I could use this prop: Let [imath]0\rightarrow M^\prime \rightarrow M \rightarrow M^{\prime\prime}\rightarrow 0[/imath] be an exact sequence of [imath]A[/imath]-modules. Then: [imath]M[/imath] is Noetherian if and only if, [imath]M^\prime[/imath] and [imath]M^{\prime\prime}[/imath] are Noetherian. and clearly I have [imath]S\rightarrow M\rightarrow 0[/imath] if I consider inclusion. But I can't find the other side homomorphism. Thanks for your help.
|
1061400
|
Exact sequence and Noetherian modules
Let [imath]R[/imath] be a ring, [imath]X,Y,Z[/imath] and [imath]T[/imath] four [imath]R-[/imath] modules such that there exists a short exact sequence [imath]0 \rightarrow X \xrightarrow{f_1} Y \xrightarrow{f_2} Z \xrightarrow{f_3} T \rightarrow 0 [/imath] Prove the following: (i) If [imath]X,Y[/imath] and [imath]T[/imath] are Noetherian modules, then [imath]Z[/imath] is Noetherian. (ii) If [imath]X,Z[/imath] and [imath]T[/imath] are Noetherian modules, then [imath]Y[/imath] is Noetherian. I couldn't prove any of them, I assume both parts have similar proofs so I'll write what I've tried to do in (i) and I would appreciate some help with one of the two so I can complete the rest by myself. In (i), there are alternative ways to show that [imath]Z[/imath] is Noetherian. I chose to show that every ascending chain is stationary. So let [imath](Z_k)_{k \in \mathbb N}[/imath] be an ascending chain in [imath]Z[/imath]. Then [imath](f_3(Z_k))_{k \in \mathbb N}[/imath] is an ascending chain in T. The same for [imath]({f_2}^{-1}(Z_k))_{k \in \mathbb N}[/imath] in Y. By hypothesis both of these chains are stationary. Let [imath]N \in \mathbb N[/imath] be such that [imath]f_3(Z_n)=f_3(Z_N)[/imath] for all [imath]n \geq N[/imath] and [imath]{f_2}^{-1}(Z_n)={f_2}^{-1}(Z_N)[/imath] for all [imath]n \geq N[/imath]. Take [imath]x \in Z_n[/imath] for [imath]n \geq N[/imath], I would like to show [imath]x \in Z_N[/imath]. We have [imath]f(x) \in f_3(Z_n)=f_3(Z_N)[/imath], so there is [imath]y \in Z_N[/imath] such that [imath]f_3(y)=f_3(x)[/imath]. But then [imath]f_3(y-x)=0[/imath], so [imath]x-y \in Ker(f_3)=Im(f_2)[/imath]. Here I got stuck, there is [imath]w \in Y[/imath] such that [imath]f_2(w)=x-y[/imath], but in which submodule of [imath]Y[/imath] is [imath]w[/imath]? And I don't see where to use that [imath]X[/imath] is also Noetherian. Any help would be greatly appreciated.
|
1956956
|
Complex Numbers, Understanding Square Roots
Since [imath]i^2 = -1[/imath], then doesn't [imath]i = \pm \sqrt[2]{i}[/imath]? How does [imath]i[/imath] only equal the plus part?
|
1955626
|
De Moivre Complex Roots: Definition of [imath]i[/imath] or [imath]\sqrt {-1}[/imath]
Does [imath]\sqrt{-1}=\pm\sqrt{-1}[/imath] since it has 2 solutions for roots? Note the link says there are two complex square roots for -1 which are [imath]i[/imath] and [imath]-i[/imath] https://en.m.wikipedia.org/wiki/Imaginary_unit.
|
1957031
|
Proof of odd number of divisors of a perfect square
How do we prove that a perfect square has an odd number of divisors? Eg 1024 is a perfect square and it has [imath]11[/imath] divisors comprising of [imath]2^0[/imath],[imath]2^1[/imath],[imath]2^2[/imath],...[imath]2^10[/imath]
|
349717
|
[imath]m[/imath] is a perfect square iff [imath]m[/imath] has an odd number of divisors?
Is this proof that [imath]m[/imath] is a perfect square iff [imath]m[/imath] has an odd number of divisors correct? [imath]\Rightarrow)[/imath] If [imath]m[/imath] is a perfect square there is an [imath]x[/imath] such that [imath]x = m/x[/imath]. The rest of the divisors come in pairs [imath]d, m/d[/imath], so [imath]m[/imath] has an odd number of divisors. [imath]\Leftarrow)[/imath] Again, divisors come in pairs [imath]m, m/d[/imath] If there are an odd number of divisors, then necessarily [imath]m/d = d[/imath] for some divisor [imath]d[/imath], so [imath]m[/imath] is a square.
|
1957023
|
For which rings can elements be written as a finite product of irreducible elements?
For which rings [imath]A[/imath] can elements be written as a finite product of irreducible elements? I feel like this is true for almost all rings I have ever encountered, but when do I know that I can write any nonzero nonunit element of [imath]A[/imath] as a finite product of irreducible elements? (I am not assuming the product is unique here.) In particular, if [imath]A[/imath] is an affine [imath]k[/imath]-algebra can I always do this? ([imath]k[/imath] algebraically closed field) Thanks!
|
714719
|
In a noetherian integral domain every non invertible element is a product of irreducible elements
I want to prove that in a noetherian ring [imath]R[/imath] which is also an integral domain, every non invertible element can be expressed as product of irreducible elements. I really do not know where to start. Can someone give me hint how to prove this?
|
99333
|
[imath]x[/imath] conjugate to [imath]y[/imath] in a group [imath]G[/imath] is an equivalence relation on [imath]G[/imath]
This is a question from the free Harvard online abstract algebra lectures. I'm posting my solutions here to get some feedback on them. For a fuller explanation, see this post. This problem is from assignment 5. a) Prove that the relation [imath]x[/imath] conjugate to [imath]y[/imath] in a group [imath]G[/imath] is an equivalence relation on [imath]G[/imath]. b) Describe the elements [imath]a[/imath] whose conjugacy class (= equivalence class) consists of the element [imath]a[/imath] alone. a) Let [imath]G[/imath] be a group and [imath]R[/imath] be a relation on [imath]G[/imath] defined by [imath]a\sim b[/imath] if [imath]a[/imath] is conjugate to [imath]b[/imath]. Then [imath]a\sim b[/imath] if there is a [imath]g\in G[/imath] such that [imath]a = gbg^{-1}[/imath]. Let [imath]a[/imath] be an element of [imath]G[/imath]. Then [imath]a=eae^{-1}[/imath]. So [imath]a\sim a[/imath] and [imath]R[/imath] is reflexive. Let [imath]a[/imath] and [imath]b[/imath] be elements of [imath]G[/imath] such that [imath]a\sim b[/imath]. Then there is a [imath]g\in G[/imath] such that [imath]a=gbg^{-1}[/imath]. Then [imath]b=g^{-1}ag[/imath]. Since [imath]g^{-1}\in G[/imath], [imath]b\sim a[/imath]. Hence, [imath]R[/imath] is symmetric. Let [imath]a,b[/imath], and [imath]c[/imath] be elements of [imath]G[/imath] such that [imath]a\sim b[/imath] and [imath]b\sim c[/imath]. Then there are elements [imath]g,g^\prime\in G[/imath] such that [imath]a=gbg^{-1}[/imath] and [imath]b=g^\prime cg^{\prime -1}[/imath]. Then [imath]a=g(g^\prime c g^{\prime -1})g^{-1}=(gg^\prime)c(g^{\prime -1}g^{-1})=(gg^\prime)c(gg^\prime)^{-1}[/imath]. Since [imath]gg^\prime\in G[/imath], [imath]a\sim c[/imath]. Hence [imath]R[/imath] is transitive. Therefore [imath]R[/imath] is an equivalence relation on [imath]G[/imath]. b) Let [imath]S[/imath] be the set of elements of [imath]G[/imath] such that, for [imath]s\in S[/imath], [imath]s=gsg^{-1}[/imath] for any [imath]g\in G[/imath]. Then [imath]sg=gs[/imath]. So [imath]S[/imath] is the set of elements that commute with every element of [imath]G[/imath]. In other words, [imath]S[/imath] is the center of [imath]G[/imath]. Again, I welcome any critique of my reasoning and/or my style as well as alternative solutions to the problem. Thanks.
|
658220
|
For a group [imath]G[/imath], show the relation [imath]x\sim y[/imath] defined by [imath]\exists a(y=axa^{-1})[/imath] is an equivalence relation on [imath]G[/imath].
Let G be a group. For [imath]x,y\in G[/imath], define [imath]x\sim y[/imath] if there exists some element [imath]a\in G[/imath] such that [imath]y=axa^{-1}[/imath]. Show that ~ defines an equivalence relation on [imath]G[/imath].
|
1956241
|
Show that the minimal polynomial [imath]f(x) [/imath] has [imath]\deg f(x)=5[/imath]
Let [imath]V,W[/imath] be finite dimensional vector spaces over [imath]\Bbb R[/imath] and let [imath]T_1:V\to V[/imath] and [imath]T_2:W\to W[/imath] be linear transformations whose minimal polynomial are given by [imath]f_1(x)=x^3+x^2+x+1[/imath] and [imath]f_2(x)=x^4-x^2-2[/imath]. Let [imath]T:V \oplus W \to V\oplus W [/imath] be the linear transformation defined by [imath]T(v,w)=(T_1(v),T_2(w))[/imath] for [imath](v,w)\in V\oplus W[/imath] and let [imath]f(x)[/imath] be the minimal polynomial of [imath]T[/imath]. Then Show that [imath]\text{nullity} (T)=0[/imath] and [imath]\deg f(x)=5[/imath] My try: Let [imath](x,y)\in \ker (V\oplus W)\implies T(v,w)=0\implies T_1(v)=0,T_2(w)=0\implies 0[/imath] is an eigen value of both [imath]T_1,T_2[/imath] which is false because minimal polynomials of both [imath]T_1,T_2[/imath] don't have [imath]0[/imath] as their root. Hence [imath]\text{nullity} (T)=0[/imath]. But I am unable to prove that [imath]\deg f(x)=5[/imath]. Please help me out here.
|
990562
|
Minimal polynomial of the operator [imath]T:V\oplus W\to V\oplus W[/imath]
Let [imath]V[/imath] and [imath]W[/imath] be two finite dimensional vector spaces over [imath]\Bbb R[/imath] and let [imath]T_{1}:V\to V[/imath] and [imath]T_{2}:W\to W[/imath] be two linear transformations whose minimal polynomials are given by [imath]f_{1}(x)=x^{3}+x^{2}+x+1[/imath] and [imath]f_{2}(x)=x^{4}-x^{2}-2[/imath]. Let, [imath]T:V\oplus W\to V\oplus W[/imath] be the linear transformation defined by [imath]T(v,w)=(T_{1}(v),T_{2}(w))[/imath] for [imath](v,w)\in V\oplus W[/imath] and let [imath]f(x)[/imath] be the minimal polynomial of [imath]T[/imath]. Then which of the followings are correct? (1) [imath]\deg(f(x))=7[/imath] (2) [imath]\deg(f(x))=5[/imath] (3) [imath]\mathrm{nullity}(T)=1[/imath] (4) [imath]\mathrm{nullity}(T)=0[/imath] .
|
1957799
|
If every cyclic subgroup of a group G be normal in G, prove that every subgroup of G is normal in G.
If every cyclic subgroup of a group G be normal in G, prove that every subgroup of G is normal in G. Attepmt Let G be a group. Let H be a Normal subgroup of G. Let [imath]K=\langle a \rangle[/imath] be a cyclic subgroup of G generated by [imath]a\in G[/imath]. We shall have to show that H is normal in G. That is to prove that for [imath]h\in H[/imath], [imath]ghg^{-1}\in G[/imath] for all [imath]g\in G[/imath]. If the steps I have written, please help me to solve the problem.
|
106072
|
If [imath]H[/imath] is a cyclic subgroup of [imath]G[/imath] and [imath]H[/imath] is normal in [imath]G[/imath], then every subgoup of [imath]H[/imath] is normal in [imath]G[/imath].
Exercise 11, page 45 from Hungerford's book Algebra. If [imath]H[/imath] is a cyclic subgroup of [imath]G[/imath] and [imath]H[/imath] is normal in [imath]G[/imath], then every subgroup of [imath]H[/imath] is normal in [imath]G[/imath]. I am trying to show that [imath]a^{-1}Ka\subset K[/imath], but I got stuck. What I am supposed to do now? Thanks for your kindly help.
|
1958066
|
Eigenvalues of [imath]A[/imath] if [imath]A^2=1[/imath]
Let [imath]A:V\to V[/imath] be an endomorphism of a vector space [imath]V[/imath] such that [imath]A^2=id[/imath]. Then the only possible eigenvalues of [imath]A[/imath] are [imath]\pm 1[/imath] and of course [imath]1[/imath] is an eigenvalue of [imath]A[/imath]. Are those in fact eigenvalues of [imath]A[/imath]?
|
496338
|
Diagonalizability of an involution over a Field
Suppose [imath]A[/imath] is a matrix over a field [imath]F[/imath] of characteristic [imath]\neq 2[/imath], with [imath]A\cdot A=I[/imath]. Show that [imath]A[/imath] is similar to a diagonal matrix whose entries are either [imath]1[/imath] or [imath]-1[/imath].
|
1958166
|
Show that the given linear transformation is invertble
Let [imath]T_1,T_2[/imath] be two linear transformations from [imath]\Bbb R^n[/imath] to [imath]\Bbb R^n[/imath].Let [imath]\{x_1,x_2\ldots x_n\}[/imath] be a basis of [imath]\Bbb R^n[/imath] .Suppose that [imath]T_1x_i\neq 0\forall i=1,2,.....n[/imath] and that [imath]x_i\perp \ker T_2\forall i=1,2,.....n[/imath]. Which of the following are true? [imath]T_1[/imath] is invertible [imath]T_2[/imath] is invertible [imath]T_2,T_1[/imath] is invertible Neither [imath]T_1[/imath] nor [imath]T_2[/imath] is invertible For [imath]T_1[/imath]: Define [imath]T_1:\Bbb R^2\to \Bbb R^2[/imath] by [imath]T_1(x_1)=x_1;T_1(x_2)=-x_1\implies T_1(x_1+x_2)=0\implies T _1[/imath] is not one-one and hence not invertible. For [imath]T_2[/imath]: Let [imath]x\in \ker T_2\implies Tx=0[/imath] where [imath]x=\sum_{i=1}^n c_ix_i[/imath] Now [imath]\langle x,x_i\rangle =0\forall i\implies c_i=0\forall i\implies x=0\implies \ker T_2=\{0\}[/imath] **But I am unable to show that [imath]T_2[/imath] is surjective. Let [imath]y\in \Bbb R^n\implies y=\sum c_ix_i[/imath].But I don't know what steps to follow next. Please give some hints.
|
1725702
|
How to Show that Linear transformation is invertible?
Let [imath]T_1,T_2[/imath] be two linear transformations from [imath]\mathbb{R}^n[/imath] to [imath]\mathbb{R}^m[/imath]. Let [imath]\{x_1,\cdots , x_n\}[/imath] be a basis of [imath]\mathbb{R}^n[/imath]. Suppose that [imath]T_1x_i \neq 0[/imath] for every [imath]i=1,2,\cdots, n[/imath] and that [imath]x_i \bot \ker T_2[/imath] for every [imath]i=1,2,\cdots, n[/imath]. Which of the following is/are necessarily true? [imath]T_1[/imath] is invertible. [imath]T_2[/imath] is invertible. both [imath]T_1,T_2[/imath] are invertible. neither [imath]T_1[/imath] nor [imath]T_2[/imath] is invertible. My Approch: Given dimension of [imath]\mathbb{R}=n[/imath] and [imath]x_i\perp Ker(T_2)[/imath] So, [imath]\text{Ker}T_2=0[/imath] Row space and null space are perpendicular to each other. From Rank Nullity Theorem: let T : V → W be a linear map. Then the rank of T is the dimension of the image of T and the nullity of T is the dimension of the kernel of T, so we have [imath]\text{dim}(\text{Im}(T))+ \text{dim}(\text{Ker}(T))=\text{dim}(V)[/imath] Applying above result we get, [imath]\text{dim}(\text{Im}(T))+ 0=n[/imath] Therefore, [imath]T_2[/imath] is one-one. I could not solve this problem completely. what to do next?
|
1930450
|
Both sides differentiation expression with integral
I have a problem with derivation of some function. This is related to computing the Poincare Map of Logistic Population model with Periodic harvesting. This function is: [imath]\phi(t,x_0) = x_0 + \int_{0}^{t} f(s,\phi(s,x_0)) ds \ (1)[/imath] where [imath]\phi(t,x_0)[/imath] is a function: [imath]\mathbb{R} \times \mathbb{R} \rightarrow \mathbb{R}[/imath]. I'm trying to differentiate equation of this function both sides with respect to [imath]x_{0}[/imath]. In book from this equation comes from the result of this differentiation is: [imath]\frac{\partial \phi}{\partial x_0} (t,x_0) = 1 + \int_0^{t} \frac{\partial f}{\partial x_0} (s, \phi(s,x_0)) \cdot \frac{\partial \phi}{\partial x_0} (s, x_0)ds[/imath]. I can't understand expression under the integral. In this book it is explained that it is clear from chain rule. When I'm trying to apply the chain rule I have expression under integral such as: [imath]\frac{\partial }{\partial x_0} f(s, \phi(s,x_0)) = \frac{\partial }{\partial \phi(s,x_0) } f(s, \phi(s,x_0)) \cdot \frac{\partial }{\partial x_0} \phi(s,x_0) [/imath] Could someone explain to me why in my expression first I have derivative respect by [imath]\partial \phi(s,x_0)[/imath] and then by [imath]\partial x_0?[/imath] In solution there are in both expressions [imath]\partial x_0[/imath] Why? Maybe important is here additional information that differentiable function [imath](1)[/imath] is solution of differential equation: [imath]\frac{\partial \phi}{\partial t} (t,x_0) = f(t,\phi,x_0))[/imath], where [imath]\phi(0,x_0)=x(0).[/imath] I will be grateful for your help Best regards
|
1929212
|
Both sides differentiation expression with integral
I have a problem with derivation of some function. This is related to computing the Poincare Map of Logistic Population model with Periodic harvesting. This function is: [imath]\phi(t,x_0) = x_0 + \int_{0}^{t} f(s,\phi(s,x_0)) ds [/imath], where [imath]\phi(t,x_0)[/imath] is a function: [imath]\mathbb{R} \times \mathbb{R} \rightarrow \mathbb{R}[/imath]. I'm trying to differentiate equation of this function both sides with respect to [imath]x_{0}[/imath]. In book from this equation comes from the result of this differentiation is: [imath]\frac{\partial \phi}{\partial x_0} (t,x_0) = 1 + \int_0^{t} \frac{\partial f}{\partial x_0} (s, \phi(s,x_0)) \cdot \frac{\partial \phi}{\partial x_0} (s, x_0)ds[/imath]. I can't understand expression under the integral. In this book it is explained that it is clear from chain rule.Could someone explain to me this transformation in easy way? I will be grateful for your help Best regards
|
1957653
|
How to rigorously prove that 1/t does not have a Laplace Transformation
I know that [imath]f(t)=\frac{1}{t}[/imath] does not have a Laplace transformation but I don't know the right(formal) way to prove it. Here is my way: Using the fact: [imath]L[\frac{f(t)}{t}]=\int_{s}^{\infty}F(t)dt[/imath], I have: [imath]L[\frac{1}{t}]=\int_{s}^{\infty}\frac{1}{s}dt = \frac{1}{s}|^{\infty}_{s}[/imath] this is divergent and therefore the Laplace transformation of [imath]f(t)=\frac{1}{t}[/imath] does not exist. However, I think my way is not rigorous. I hope I can receive a explicit proof of it (Maybe by definition to integrate). Thank you so much! Edit: Actually, I do not know how to actually prove that [imath]\int_{0}^{\infty}e^{-st}\frac{1}{t}dt[/imath] diverges. Please help me with this!
|
147126
|
Laplace transform of [imath]1/t[/imath]
Does the laplace transform of [imath]1/t[/imath] exist? If yes, how do we calculate it? Putting it in [imath]\int_0^\infty (e^{-st}/t) dt[/imath] won't solve. Is there any other way? If not, why? Thanks!
|
1958383
|
Verification of proof that two sequences converge to the same value
Consider the sequences [imath]\{x_n\}_{n=1}^\infty[/imath] and [imath]\{y_n\}_{n=1}^\infty[/imath] satisfying the following properties: [imath]0<y_1<x_1[/imath] [imath]x_{n+1}=(x_n+y_n)/2[/imath] for all [imath]n\ge2[/imath] [imath]y_{n+1}=2x_ny_n/(x_n+y_n)[/imath] for all [imath]n\geq 2[/imath] I would like to prove that [imath]\lim_{n\to\infty}x_n=\lim_{n\to\infty}y_n=\sqrt{x_1y_1}[/imath]. Here's my proof. Is it correct? [imath]x_2[/imath] is the arithmetic mean of [imath]x_1[/imath] and [imath]y_1[/imath], while [imath]y_2[/imath] is their harmonic mean. Since the former mean is always larger, [imath]x_1>x_2>y_2>y_1>0[/imath]. By extension: [imath]x_1>x_2>x_3>\dots x_n>x_{n+1}>y_{n+1}>y_n>\dots y_3>y_2>y_1>0[/imath] Thus [imath]\{x_n\}[/imath] and [imath]\{y_n\}[/imath] are monotone and bounded; by the monotone convergence theorem they converge to a finite limit. Let [imath]\lim_{n\to\infty}x_n=x[/imath] and [imath]\lim_{n\to\infty}y_n=y[/imath]. Then [imath]x=(x+y)/2[/imath] as [imath]n[/imath] goes to infinity, so [imath]x=y=L[/imath]. Moreover, [imath]x_1y_1=x_2y_2=x_3y_3=\dots=x_ny_n=x_{n+1}y_{n+1}=L^2[/imath], so [imath]L=\sqrt{x_1y_1}[/imath]. [imath]\blacksquare[/imath]
|
1192349
|
Proof the the Arithmetic-Harmonic Mean is expressible as the Geometric Mean
We define the Arithmetic-Harmonic mean of [imath]a,b \in \mathbb{R_+}[/imath] such that \begin{gather*} a_{n+1} = \frac{1}{2}(a_n + b_n) \\ b_{n+1} = \frac{2a_{n}b_{n}}{a_{n} + b_{n}} \end{gather*} Let us also assume that [imath]a \neq b[/imath]. I am trying to prove that the limit of these sequences exist, they are equal to each other, and as [imath]n \rightarrow \infty[/imath], the limit is equal to [imath]\sqrt{ab}[/imath], the Geometric Mean of a,b. This is stated by mathworld here: http://mathworld.wolfram.com/Arithmetic-HarmonicMean.html Since I have proven that AM [imath]\geq[/imath] HM, with equality only holding at a = b, we can see that [imath]a_1 \geq a_{2} \geq ... \geq a_n[/imath]. In the same vein we have [imath]b_1 \leq b_{2} \leq ... \leq b_n[/imath]. How can I rigourously show that these are monotonic sequences bounded above and below resepctively and therefore converge to each other? Secondly, can anyone give me a hint as to how I can show that this sequence converges to the Geometric Mean?
|
1958151
|
Proof explanation on a group of order [imath]595[/imath] having a normal Sylow [imath]17[/imath]-subgroup.
Prove that a group of order 595 has a normal Sylow 17-subgroup. The proof is as follows: By Sylow, [imath]n_{17} = 1[/imath] or [imath]35[/imath]. Assume [imath]n_{17} = 35[/imath]. Then the union of the Sylow [imath]17[/imath]-subgroups has [imath]561[/imath] elements. By Sylow, [imath]n_5 = 1[/imath]. Thus, we may form a cyclic subgroup of order [imath]85[/imath] (from a previous theorem) But then there are [imath]64[/imath] elements of order [imath]85[/imath]. This gives too many elements. My question is: Where does this [imath]64[/imath] come from?
|
602310
|
A group of order [imath]595[/imath] has a normal Sylow 17-subgroup.
Proof Verification: A group of order [imath]595[/imath] has a normal Sylow 17-subgroup. [imath]|G|=595=5.7.17[/imath] The divisors of [imath]595[/imath] are [imath]1,5,7,17,35,85,119,595[/imath]. [imath]17|n_{17}-1\implies n_{17}=1,35\\7|n_7-1\implies n_7=1,85,119\\5|n_5-1\implies n_5=1[/imath] If possible let [imath]n_{17}=35.[/imath] Then there is at least [imath]35.17-34=561[/imath] element of order [imath]17.[/imath] Then [imath]n_7=1[/imath] for otherwise the number of elements of order [imath]17,7[/imath] will exceed [imath]595![/imath] Let [imath]H_5,H_7[/imath] be the normal Sylow 5 and Sylow [imath]7[/imath] subgroups of [imath]G.[/imath] Then [imath]H_5H_7\le G[/imath] and since [imath]H_5\cap H_7=(e),G=H_5\times H_7\simeq H_5\oplus H_7.[/imath] Thus none of the [imath]35[/imath] elements in [imath]H_5H_7[/imath] is of order [imath]17.[/imath] Hence [imath]G[/imath] has at least [imath]561+35=596[/imath] elements! Please tell me whether the proof is right?
|
1959207
|
Elementary Number Theory GCD
If [imath](a,b)[/imath], [imath](a,y)[/imath] and [imath](b,x)[/imath] = [imath]1[/imath], show that [imath](ab, ax + by)[/imath] = [imath]1[/imath] I tried to do this by showing that [imath](a,by)[/imath]=[imath]1[/imath] and [imath](b,ax)[/imath]=[imath]1[/imath]. But then I got completely stuck. I tried to use Bezout's theorem, but ended up with nothing.
|
1954663
|
If [imath]\gcd(a,b)=1[/imath], [imath]\gcd(a,y)=1[/imath] and [imath]\gcd(b,x)=1[/imath] then prove that [imath]ax+by[/imath] is prime to [imath]ab[/imath]
If [imath]\gcd(a,b)=1[/imath], [imath]\gcd(a,y)=1[/imath] and [imath]\gcd(b,x)=1[/imath] then prove that [imath]ax+by[/imath] is prime to [imath]ab[/imath]. I tried assuming Diophantine Equations for all the relations and representing everything in terms of [imath]a[/imath], but that didn't lead me anywhere. There must be an intuitive and elegant approach right?
|
1959333
|
Existential proof with number theory
If [imath]a[/imath] is not a multiple of a prime [imath]p[/imath], then prove that there is an integer [imath]b[/imath] such that [imath]p^b-1[/imath] is a multiple of [imath]a[/imath].
|
507657
|
Proving if [imath]a[/imath] is not a multiple of prime [imath]p[/imath], then an integer [imath]b[/imath] exists such that [imath]p^b - 1[/imath] is a multiple of [imath]a[/imath]
I tried many approaches but none of them really worked I treated [imath]p^b-1[/imath] as a Geometric progression but it didn't work and that is about as far as I have been able to go I have no clue how to move forward
|
1958770
|
How to show that [imath]\displaystyle{\cot z-\frac{1}{z}}[/imath] is bounded on the given circle
[imath]f(z)=\cot z - \frac{1}{z}=\frac{z \cos z - \sin z}{z ~ \sin z}[/imath] having poles at [imath]z=n \pi[/imath] with residue [imath]1[/imath] at each pole. Also [imath]f(z)[/imath] has a removable singularity at [imath]z=0.[/imath] All these are asked to calculate and I've done that. If [imath]C_N[/imath] is the circle enclosing all the poles having the center at the origin and radius [imath]R_N,[/imath] how can I show that [imath]f[/imath] is bounded on [imath]C_N[/imath] for a suitable [imath]R_N[/imath] ? Any help is much appreciated. Edit: Note that
|
362044
|
Is [imath]\cot(z)-\frac1z[/imath] bounded on a given circle?
I am getting stuck with the problem given from the book Schaum's outlines p.240 Q.110 Given [imath]f(z)=\cot(z)-\frac1z[/imath] where [imath]z[/imath] lies on a circle of radius [imath]R=(N+\frac12)\pi[/imath] and centered at the origin. Prove that [imath]|f(z)|\le M[/imath] where [imath]M[/imath] is independent of [imath]N[/imath].
|
1959497
|
Equation [imath](p-1)! + 1 = p^n[/imath] for some [imath]p > 5[/imath] and [imath]n \in \mathbb{N}[/imath]
Wilson's Theorem states that a natural number [imath]p > 1[/imath] is a prime if and only if [imath](p-1)![/imath] is congruent to [imath]-1[/imath] modulo [imath]p[/imath]. A natural question to ask is: For which [imath]p[/imath] is [imath](p-1)![/imath] congruent to [imath]-1[/imath] modulo [imath]p^2[/imath]? About these numbers, called Wilson primes, not much is known: as far as I know, the only known Wilson primes are [imath]5[/imath], [imath]13[/imath] and [imath]563[/imath] and it is conjectured that there are infinitely many Wilson primes. Here, I would like to focus on the following more special question. Are there some prime [imath]p>5[/imath] and some integer [imath]n > 1[/imath] such that [imath](p-1)! + 1 = p^n[/imath]? Remarks: By Wilson's Theorem, it is necessary that [imath]p[/imath] is a prime. For [imath]p = 5[/imath], the assertion is satisfied with [imath]n = 2[/imath]. For [imath]n = 1[/imath], one can easily solve the equation and see that [imath]p=2[/imath] and [imath]p=3[/imath] are the only possible numbers in this case. What I've tried: First of all, I suspect that the equation does not have a solution in [imath]p > 5[/imath], [imath]n > 1[/imath], but I will suppose that the equation holds to get some necessary conditions on [imath]p[/imath] and [imath]n[/imath], and maybe a contradiction in the end. The equation in the question is equivalent to [imath](p-2)! = 1+...+p^{n-1}[/imath]. As the left hand side is divisible by [imath]5[/imath], the right hand side is also. By the same token, the left hand side is even, so the right hand side is also even. Whence, as [imath]p[/imath] is odd, [imath]n[/imath] has to be even. The statement does not really depend on [imath]p[/imath] being a prime. By Wilson's Theorem one just knows that [imath]p[/imath] being a prime is necessary, but maybe one does not really need it and can do some kind of induction on [imath]p[/imath] to see that the equation cannot be solved. However, I was not able to do this. As [imath]p^p \geq p! \geq (p-1)! + 1 = p^n[/imath], we see that [imath]p \geq n[/imath]. To show that the equation does not have any solution, it suffices to show that for a fixed [imath]p[/imath] and an arbitrary positive [imath]n > 1[/imath], either (i) [imath](p-1)! + 1 < p^n[/imath], or (ii) [imath](p-1)! + 1 > p^n[/imath]. I suspect that (i) holds if [imath]n > \frac{p}2[/imath] and that (ii) holds if [imath]n \leq \frac{p-1}2[/imath]. If this is true, it would suffice to prove the two statements (i') [imath](p-1)! + 1 < p^{(p+1)/2}[/imath], and (ii') [imath](p-1)! + 1 > p^{(p-1)/2}[/imath] instead. Look at it analytically: The equation becomes [imath]\Gamma(x) + 1 = x^n[/imath]. The plots suggest that the equation does not have any solution for [imath]x > 5[/imath] as the left hand side increases much faster than the right hand side. One could try to do some kind of approach with derivatives, but the Gamma-Function is the thing that disturbs me here. I would be thankful for any approach regarding this problem, thank you!
|
1882567
|
integers [imath]n[/imath] such that [imath](n-1)!+1[/imath] is a power of [imath]n[/imath]
Please, find all positive integers n for which (n-1)! +1 is a power of n.
|
1959991
|
How to compute the limit of the following product?
How to compute [imath]\lim_{n \to \infty}\frac{1}{n}\frac{2}{n}...\frac{n}{n} [/imath] ? Thanks,
|
1959929
|
show that [imath]\limsup(\frac{n!}{n^n})=0[/imath]
Let [imath]a_n=\frac{n!}{n^n}[/imath] be a sequence. Prove that [imath]\limsup(a_n)=0[/imath]. I know that the series [imath]\sum^{\infty}_{n=1} \frac{n!}{n^n}[/imath] converges, so the nth term must converge to 0. But I cannot use the convergence of series to prove this. Basically what I have is: The sequence [imath]a_n[/imath] is decrescent and bounded by 0 [imath](a_n>0 \; \forall n)[/imath], therefore is convergent and it's limit is the [imath]\limsup(a_n)[/imath]. So I must show that [imath]\lim\frac{n!}{n^n}=0[/imath] Let [imath]\epsilon>0[/imath], prove that exists [imath]n_0 \in \mathbb{N}[/imath] such that, if [imath]n>n_0[/imath] [imath]\left|\frac{n!}{n^n}-0\right|<\epsilon \Rightarrow -\epsilon<\frac{n!}{n^n}<\epsilon[/imath] I know that [imath]\frac{n!}{n^n}>=-\epsilon[/imath] because [imath]\frac{n!}{n^n}>0[/imath]. How can I prove that [imath]\frac{n!}{n^n}<\epsilon[/imath]?
|
1959713
|
Divisibility Proof [imath]8\mid (x^2 - y^2)[/imath] for [imath]x[/imath] and [imath]y[/imath] odd
[imath]x,y \in\Bbb Z[/imath]. Prove that if [imath]x[/imath] and [imath]y[/imath] are both odd, then [imath]8\mid (x^2 - y^2)[/imath]. My Proof Starts: Assume [imath]x[/imath] and [imath]y[/imath] are both odd. So, [imath]x = 2k + 1[/imath] and [imath]y = 2l +1[/imath] for some integers [imath]k[/imath] and [imath]l[/imath]. Thus, \begin{align} x^2 - y^2 &= (2k + 1)^2 - (2l + 1)^2 \\ &= 4k^2 + 4k + 1 - (4l^2 + 4l + 1) \\ &= 4k^2 + 4k - 4l^2 - 4l \end{align} My two concerns: 1) Is this correct so far? 2) How would I deal with the “[imath]8\;\mid[/imath]” part?
|
397830
|
Show that [imath]8 \mid (a^2-b^2)[/imath] for [imath]a[/imath] and [imath]b[/imath] both odd
If [imath]a,b \in \mathbb{Z}[/imath] and odd, show [imath]8 \mid (a^2-b^2)[/imath]. Let [imath]a=2k+1[/imath] and [imath]b=2j+1[/imath]. I tried to get [imath]8\mid (a^2-b^2)[/imath] in to some equivalent form involving congruences and I started with [imath]a^2\equiv b^2 \mod{8} \Rightarrow 4k^2+4k \equiv 4j^2+4j \mod{8}[/imath] [imath]\Rightarrow k^2+k-j^2-j=2m[/imath] for some [imath]m \in \mathbb{Z}[/imath] but I am not sure this is heading anywhere that I can tell. Second attempt: Use Euler's Theorem and as [imath]\gcd(a,8)=\gcd(b,8)=1[/imath] and [imath]\phi(8)=4[/imath], [imath]a^4 \equiv b^4 \equiv 1 \mod 8[/imath] so [imath]a^4-b^4\equiv 0 \mod{8}[/imath]. I haven't gotten too much further are there any hints?
|
984632
|
Algebraic Proofs in Combinatorics [imath]\binom{n + m}{2} = nm + \binom{n}{2} + \binom{m}{2}[/imath]
Prove the following identity using an Algebraic Proof. [imath]\binom{n + m}{2} = nm + \binom{n}{2} + \binom{m}{2}[/imath] I have no idea where to begin on this problem or let alone finish it.
|
711311
|
How to algebraically prove [imath]\binom{n+m}{2} = nm + \binom{n}{2} + \binom{m}{2}[/imath]?
Need help trying to prove this problem algebraically. [imath]\binom{n+m}{2} = nm + \binom{n}{2} + \binom{m}{2}[/imath] The farthest I've got is simplifying the RHS to [imath]nm + \frac{n(n-1)}{2!} + \frac{m(m-1)}{2!}[/imath] but not sure what to do after that.
|
1960294
|
Every subgroup [imath]H[/imath] of [imath]G[/imath] with [imath]H\ne G[/imath] is cyclic. Is [imath]G[/imath] cyclic?
Let [imath](G,*)[/imath] be a group. Suppose that every subgroup [imath]H[/imath] of [imath]G[/imath] with [imath]H\ne G[/imath] is cyclic. Is [imath]G[/imath] cyclic? Justify your answer. What i tried I mentioned that [imath]G[/imath] is not necessarily cyclic. Using Lagrange theorem every subgroup of [imath]G[/imath] has an order which is the divisor of the order of [imath]G[/imath]. Say if the order of [imath]G[/imath] is [imath]6[/imath] then every subgroup must have an order of [imath]1[/imath], [imath]2[/imath], or [imath]3[/imath] for this case. And since all the subgroup are cyclic, then this implies that [imath]G[/imath] must have elements of order of [imath]1[/imath] , [imath]2[/imath] and [imath]3[/imath] but [imath]G[/imath] does not necessarily need to have an element of order [imath]6[/imath] which is the necessary condition for [imath]G[/imath] to be cyclic, since for [imath]G[/imath] to be cyclic the order of an element in [imath]G[/imath] need to be equal to the order of the group and this provide a counterexample. However i still could not think of a specific counterexample with this conditions, Could anyone help me with this. Thanks
|
1444723
|
Prove or Disprove: If every nontrivial subgroup of a group [imath]G[/imath] is cyclic, then [imath]G[/imath] is cyclic.
Prove or Disprove: If every nontrivial subgroup of a group [imath]G[/imath] is cyclic, then [imath]G[/imath] is cyclic. This is a question from page [imath]181[/imath], Chapter [imath]3.4[/imath] Elements of Modern Algebra ([imath]8^{th}[/imath] Edition) by Linda Gilbert. My intuition tells me that the given statement is false, however, I cannot seem to find a rigorous argument (or counterexample) to disprove the statement. Can anyone please point me in the right direction? I know that if a group [imath]G[/imath] is cyclic, then every subgroup is cyclic. This is not an if and only if statement - hence my intuition telling me the above mentioned statement is false.
|
1931798
|
Do limits evaluated at infinity exist?
Here is some limit: [imath]\lim_{x \to b} f(x)[/imath] We know that for a limit to exist, we must have [imath]\lim_{x \to b+} f(x) = \lim_{x \to b-} f(x)[/imath] So I am confused because, when [imath]b=+\infty[/imath] we can only evaluate this limit from the left side and not the right side. We can't approach infinity from a higher infinity. Does this mean that limits evaluated at infinity don't exist and therefore none of the limit laws like addition apply to limits evaluated at infinity? EDIT: So does that mean I can use the limit laws such as addition, composition, etc on limits evaluated at infinity as long as the limits tend to a finite value? I.e. [imath]\lim_{x \to \infty} [f(x) + g(x)] = \lim_{x \to \infty} f(x) + \lim_{x \to \infty} g(x)[/imath] as long as both of the separate limits are some finite value? And so on, for multiplication, composition, etc?
|
2412623
|
Doubt about the existence of a limit at infinity
From what I know, limits only exist if both side limits exist and are equal: [imath]{\lim_{x\to a}f(x) = L}[/imath] [imath]if[/imath] [imath]{\lim_{x\to a^+}}f(x) = {\lim_{x\to a^-}f(x) = L}[/imath] But can this be applied to limits at infinity? In that case: [imath]{\lim_{x\to \infty}f(x) = L}[/imath] [imath]if[/imath] [imath]{\lim_{x\to +\infty} f(x)} = {\lim_{x\to -\infty}f(x) = L}[/imath] Is this correct or [imath]{\lim_{x\to \infty}f(x)}[/imath] should be taken as [imath]{\lim_{x\to +\infty}} f(x)[/imath] ?
|
1960414
|
Proof about divisibility and greatest common divisor that doesn't use Bezout's formula
Suppose the [imath]\gcd(a,b) = 1[/imath] and [imath]c[/imath] divides [imath]a + b[/imath]. Prove that [imath]\gcd (a,c) = 1 = \gcd (b,c)[/imath]. This is relatively easy using Bezout's formula. See here. But is there a way to do it without Bezout?
|
513965
|
Suppose the gcd (a,b) = 1 and c divides a + b. Prove that gcd (a,c) = 1 = gcd (b,c)
I am lost. So far... If [imath]\gcd (a,b) = 1[/imath], by Bezout's Formula [imath]ax + by = 1[/imath] If [imath]c|(a+b)[/imath], then [imath]cf = a+b[/imath] Then, [imath]a (x-y) + cfy = 1[/imath] [imath]b(yx) + cfx = 1[/imath] Am I on the right track? Any suggestions?
|
1960072
|
Uniqueness of QR decomposition
Let [imath]M[/imath] be a square [imath]N\times N[/imath] matrix with linearly independent columns. I was wondering why the [imath]QR[/imath] decomposition is unique in this case and how to show it? The linked question below does not explain why [imath]Q[/imath] is also unique and I didn't quite understand its explanation of why [imath]R[/imath] is unique also.
|
635016
|
Is the matrix [imath]R[/imath] in the [imath]QR[/imath] decomposition unique?
I'd like to know for positive diagonal elements why is [imath]R[/imath] in [imath]QR[/imath] decomposition unique. My guess is it must have something to do with linearly independence of the column of [imath]R[/imath], but then I can think of a property that lead me to uniqueness of [imath]R[/imath].
|
1956291
|
Evaluation of [imath]\lim_{n\rightarrow \infty}\binom{2n}{n}^{\frac{1}{n}}[/imath]
Evaluation of [imath]\lim_{n\rightarrow \infty}\binom{2n}{n}^{\frac{1}{n}}[/imath] without using Limit as a sum and stirling Approximation. [imath]\bf{My\; Try:}[/imath] Using [imath]\binom{2n}{n} = \sum^{n}_{r=0}\binom{n}{r}^2[/imath] Using [imath]\bf{Cauchy\; Schwarz}[/imath] Inequality [imath]\left[\sum^{n}_{r=0}\binom{n}{r}^2\right]\cdot \left[\sum^{n}_{r=0}1\right]\geq \left(\sum^{n}_{r=0}\binom{n}{r}\right)^2 = 2^{2n} = 4^n[/imath] So [imath]\frac{4^n}{n+1}<\sum^{n}_{r=0}\binom{n}{r}^2=\binom{2n}{n}[/imath] But i did not understand how can i calculate upper bound such that i can apply the Squeeze theorem.
|
2524655
|
prove that [imath]\lim_{n\to\infty} \sqrt[n]{\binom{2n}{n}}=4[/imath]
prove that [imath]\lim_{n\to\infty} \sqrt[n]{\binom{2n}{n}}=4[/imath] I proved this using induction and the squeeze theorem I didn't found any one that solve it this way so i wanted to share my solution.
|
1960854
|
How prove this [imath]\sum_{1\le k\le n-1,\gcd{(k,n)}=1}k=\frac{1}{2}n\varphi{(n)}[/imath]
Let postive integer [imath]n[/imath] is not a power of a prime. Prove that [imath]\sum_{1\le k\le n-1,\gcd{(k,n)}=1}k=\dfrac{1}{2}n\varphi{(n)}\tag{1}[/imath] where [imath]\varphi{(n)}[/imath] is the Euler totient function I kown [imath]\sum_{1\le k\le n-1,\gcd{(k,n)}=1}=\varphi{(n)}[/imath]But I can't prove question [imath](1)[/imath]
|
456273
|
sum of all positive integers less than [imath]n[/imath] and relatively prime to [imath]n[/imath]
could any one tell me how to find the sum of all positive integers less than [imath]n[/imath] and relatively prime to [imath]n[/imath]? [imath]n>1[/imath] For [imath]n=7[/imath], I have [imath]\phi(n)=6[/imath] and the sum is [imath]{6(6+1)\over 2}={n\over 2}.\phi(n)[/imath], is that true for any [imath]n[/imath]?
|
1961278
|
Prove the derivative of [imath]f(x)=\left( \left| x \right| \right) ^{3}[/imath] exist.
i got a problem with this exercise. I need to prove the derivative of this function: [imath]f(x)=\left( \left| x \right| \right) ^{3}[/imath] exist. Well, i make via definition:[imath]\lim _{h\rightarrow 0}{\frac { \left( \left| x+h \right| \right) ^{3 }- \left( \left| x \right| \right) ^{3}}{h}}[/imath] I know this: [imath]|x|^3=|x|^2|x|=x^2|x|[/imath] Now, [imath]\lim _{h\rightarrow 0}{\frac { \left( \left| x+h \right| \right) ^{3 }- \left( \left| x \right| \right) ^{3}}{h}}[/imath] = [imath]1/2\,{\frac { \left| x \right| x \left( 5\,\overline{x}+x \right) }{ \overline{x}}}[/imath] But i think my answer is too bad, please help ):
|
1948692
|
Can [imath]f(x)=|x|^3[/imath] be differentiated
Can [imath]f(x)=|x|^3[/imath] be differentiated? If yes, how? If no, why? What I've tried: lim as [imath]h[/imath] tends to [imath]0[/imath] of [imath]\frac{f(x+h)-f(x)}{h} = \lim_{h \to 0} (3(x+h)^2|x=h|-3x^2|x|)/h[/imath] However, I have no idea how to proceed from here. Please help :( Thanks!
|
1960862
|
prove that [imath]\frac{a^3}{bc}+\frac{b^3}{ca}+\frac{c^3}{ab}\geq a+b+c[/imath]
If [imath]a,b,c>0[/imath], Then prove that [imath]\frac{a^3}{bc}+\frac{b^3}{ca}+\frac{c^3}{ab}\geq a+b+c[/imath] [imath]\bf{My\; Try::}[/imath] Using Cauchy- Schwarz Inequality [imath]\frac{a^4}{abc}+\frac{b^4}{abc}+\frac{c^4}{abc}\geq \frac{a^2+b^2+c^2}{3abc}[/imath] Now How can i solve after that , Help required, Thanks
|
679544
|
Proof of one inequality [imath]a+b+c\leq\frac{a^3}{bc}+\frac{b^3}{ca}+\frac{c^3}{ab}[/imath]
How to prove this for positive real numbers? [imath]a+b+c\leq\frac{a^3}{bc}+\frac{b^3}{ca}+\frac{c^3}{ab}[/imath] I tried AM-GM, CS inequality but all failed.
|
1960956
|
Inequality in normed space with normalization
Is the following true for [imath]\|x\|,\|y\|\geq 1[/imath] and [imath]x,y\in X[/imath], where [imath]X[/imath] is a normed space [imath]\left\| \frac{x}{\|x\|} + \frac{y}{\|y\|} \right\| \leq \| x+y\|.[/imath]
|
1960799
|
[imath]\left|\frac{x}{|x|}-\frac{y}{|y|}\right|\leq |x-y|[/imath], for [imath]|x|, |y|\geq 1[/imath]?
Let [imath](V,|\cdot|)[/imath] be a normed space. Let [imath]x, y\in V.[/imath] If [imath]|x|,|y|\geq 1[/imath], then does it follow that [imath]\left|\frac{x}{|x|}-\frac{y}{|y|}\right|\leq |x-y|\;?[/imath] Actually I got a hint in form of a picture, but I failed to use it.
|
1960701
|
How many non negative integer solutions are there for the equation [imath]x_1 + x_2 + x_3 + x_4 + x_5 + x_6 = 36[/imath] with restrictions?
So I had a few enquiries with this question: Consider the non negative integer solutions for the equation [imath]x_1 + x_2 + x_3 + x_4 + x_5 = 36[/imath] a) How many distinct solutions are there? b) How many distinct solutions are there if [imath]x_1 \geq 12[/imath]? c) How many distinct solutions are there if [imath]x_1 < 18[/imath]? d) How many distinct solutions are there if [imath]x_1 < 18[/imath] and [imath]x_2 < 6[/imath]? I was wondering if distinct is the same as unique? as in there cannot be solutions that have the same numbers but rearranged? Also, how would one approach this question. Is there a principle that would be best applied to this question?
|
1960684
|
Finding distinct solutions for the equation
Consider the non negative integer solutions for the equation [imath] x_1 + x_2 + x_3 + x_4 + x_5 = 36. [/imath] (a) How many distinct solutions are there? (b) How many distinct solutions are there if [imath]x_1 \ge 12[/imath] ? (c) How many distinct solutions are there if [imath]x_1 < 18[/imath] ? (d) How many distinct solutions are there if [imath]x_1 < 18[/imath] and [imath]x_2 < 6[/imath] ? My working: a) [imath]\frac{5!}{1!} = 150 [/imath] Not sure how to approach this question.
|
1493253
|
Prove using induction that from a set of [imath]n+1[/imath] numbers from [imath]\{1,2,..2n\}[/imath], at least one number will evenly divide another.
Given a set of [imath]n+1[/imath] numbers out of the first [imath]2n[/imath] natural numbers, [imath]\{1,2,\ldots,2n\}[/imath], prove that there are two numbers in the set, one of which divides the other. I can't tell if I'm reducing the problem in the right way. Would it be something like: Induction Hypothesis: A set of [imath]n+1[/imath] numbers from the first [imath]2n[/imath] natural numbers contains two numbers, one of which evenly divides the other. Base Case: [imath]n=1[/imath], [imath]\{1,2\}[/imath], 2 is divisible by 1. Consider a set of [imath]n+2[/imath] numbers, [imath]A[/imath], from the first [imath]2n+2[/imath] natural numbers. There are three cases: Each number in [imath]A[/imath] is less than or equal to [imath]2n[/imath]. The proof follows from the induction hypothesis. One number in [imath]A[/imath] is greater than [imath]2n[/imath]. The remaining set of [imath]n+1[/imath] numbers must each be less than or equal to [imath]2n[/imath]. The proof follows from the induction hypothesis. [imath]A[/imath] contains both [imath]2n+1[/imath] and [imath]2n+2[/imath]. I'm not sure how to solve this case. I'm working through Udi Manber's Introduction To Algorithms: A Creative Approach for personal development, so in keeping with the spirit of the book, I'm trying to use induction. Without induction, the question is identical to Prove two numbers of a set will evenly divide the other
|
2077929
|
Induction to prove that a set of [imath]n+1[/imath] integers between [imath]1[/imath] and [imath]2n[/imath] has at least one number which divides another number in the set
Question states: "Show that if [imath]n+1[/imath] integers between [imath]1[/imath] and [imath]2n[/imath] (inclusive) are chosen, the set of chosen integers will contain at least one number which divides another member of the same set." I have found a non-inductive proof online, which uses pigeonhole principle and expressing each number of the chosen set as [imath]2^{aj}[/imath] where [imath]j[/imath] is an odd number from [imath]\{1,3,5,...,2n-1\}[/imath]. (so there are [imath]n[/imath] boxes of [imath]j[/imath]) So pretty much divide a number repeatedly by 2 until that number becomes odd. For example, we can express [imath]56[/imath] as [imath]2^3 \times 7[/imath]. we have [imath]n+1[/imath] numbers which can fit into [imath]n[/imath] boxes, and by the Pigeon Hole Principle we conclude that there are two numbers in a given [imath]j[/imath] boxes; one of them divides the other. However, my book says that I can prove the question by using induction. Obviously, the base case holds. Here is my progress so far: Assume true for an integer [imath]n[/imath]. We need to prove that an arbitrary set of [imath]n+2[/imath] integers from [imath]\{1,...,2n,2n+1,2n+2\}[/imath] have a number which divides another number from the chosen set. If either one or none of the two new numbers ([imath]2n+1[/imath] and [imath]2n+2[/imath]) is chosen, we are done, since we must pick at least [imath]n+1[/imath] elements from the rest of the set [imath]\{1,...,2n\}[/imath] . A problem arises if BOTH of the two new numbers are chosen. Then we are required to prove that an arbitrary set of [imath]n[/imath] numbers from the set [imath]\{1,...,2n\}[/imath] contains two numbers, one of which divide another, or that the same arbitrary set of numbers must contain a number that divides either [imath]2n+1[/imath] or [imath]2n+2[/imath]. If the arbitrary set of [imath]n[/imath] numbers contain either [imath]1[/imath], [imath]2[/imath], or [imath]n+1[/imath], then we are done. So we must prove that an arbitrary set of [imath]n[/imath] numbers from the set [imath]\{3,4,...,n-1,n,n+2,...,2n\}[/imath] contains two numbers, one of which divide another, or that the same arbitrary set of numbers must contain a number that divides either [imath]2n+1[/imath] or [imath]2n+2[/imath]. How do I show this?
|
1961926
|
Given a short exact sequence, is the following true?
Say [imath]0 \to M_1 \to M_2 \to M_3 \to 0[/imath] is a short exact sequence and each [imath]M_i[/imath] is an [imath]R[/imath]-module. If [imath]M_2[/imath] is finitely generated, then so also is [imath]M_1[/imath]?
|
485000
|
Is the submodule of a finitely generated free module finitely generated?
[imath]R[/imath] is a commutative ring with unit and [imath]M[/imath] is a finitely generated free [imath]R[/imath]-module. Is the submodule of [imath]M[/imath] also finitely generated?
|
1961951
|
How to prove the sum of triangular numbers in the form n(n+1)(n+2)/6?
Show that the sum of the first [imath]n[/imath] triangular numbers is [imath]n[/imath]([imath]n + 1[/imath])([imath]n + 2[/imath])/6
|
376284
|
Sum of the first [imath]n[/imath] triangular numbers - induction
Question: Prove by mathematical induction that [imath](1)+(1+2)+(1+2+3)+\cdots+(1+2+3+\cdots+n)=\frac{1}{6}n(n+1)(n+2)[/imath] is true for all positive integers n. Attempt: I did the the induction steps and I got up to here: [imath]RTP:\frac{1}{6}n(n+1)(n+2)+(1+2+3+\cdots+n+(n+1))=\frac{1}{6}(n+1)(n+2)(n+3)[/imath] Where do I go from here? Thank you very much.
|
1961760
|
Normal Subgroups, quotient groups
I am working on a problem in my Algebra homework and I am having some problems. The question is: if [imath]H_1[/imath] and [imath]H_2[/imath] are normal subgroups of [imath]G_1[/imath] and [imath]G_2[/imath] respectively and we know that [imath]G_1[/imath] is isomorphic to [imath]G_2[/imath] and [imath]G_1/H_1[/imath] is isomorphic to [imath]G_2/H_2[/imath], can we conclude from this that [imath]H_1[/imath] is isomorphic to [imath]H_2[/imath]? I need to prove that is true or give a counterexample if it is false. My first idea was to use the idea behind the proof of the 1st isomorphism theorem. For this I have considered the inclusion homomorphisms [imath]i_1: H_1 \to G_1[/imath] and [imath]i_2:H_2 \to G_2[/imath], which are injective homomorphisms. These homomorphisms admit left inverses. Then I considered the surjective homomorphisms [imath]f_1: G_1 \to G_1/H_1[/imath] and [imath]f_2: G_2 \to G_2/H_2[/imath] which admit right inverses. And the last the isomorphism [imath]\varphi:G_1/H_1 \to G_2/H_2[/imath] that exists by assumption. Then, we can construct the homomorphism [imath]\psi[/imath] from [imath]H_1[/imath] to [imath]H2[/imath] as [imath]\psi = i_2^{-1}f_2\varphi f_1^{-1}i_1.[/imath] But, I dont know if I am tired enough to confuse these things but I think that my beautiful homomorphism takes everybody in [imath]H_1[/imath] to the identity in [imath]H_2[/imath], 'cause if [imath]h \in H_1[/imath], then [imath]i_1(h)=h \in G_1[/imath], then, [imath]f_1^{-1}(h)=hH_1=H_1[/imath], then [imath]\varphi(H_1)=H_2[/imath] and [imath]f_2(H_2)=1[/imath] and [imath]i_2^{-1}(1)=1.[/imath] I don't know if this reasoning can take me somewhere. Can anybody give me a hint to solve this problem? Any hint, any reference, any idea is gonna be welcome for me. Thank you so much, everyone.
|
1584568
|
Isomorphic quotient groups [imath]\frac{G}{H} \cong \frac{G}{K}[/imath] imply [imath]H \cong K[/imath]?
I know that given a group [imath]G[/imath] and two normal subgroups [imath]H,K \subset G[/imath] then it is not true that: "if [imath]H \cong K[/imath] then [imath] \frac{G}{H} \cong \frac{G}{K} [/imath] (the counterexample is quite easy with products of cyclic groups) " My question is: Is the converse true? i.e. Given that [imath]\frac{G}{H} \cong \frac{G}{K}[/imath] then [imath]H \cong K[/imath] ? I feel that the answer is no, but I can't think of an example.
|
1962541
|
How to prove [imath]\mu(X \cup Y) \geq \mu(X) +\mu(Y) [/imath] for this measure.
Let [imath]\mu[/imath] be the Lebesgue outer measure I want to prove [imath]\mu(X \cup Y) = \mu(X) +\mu(Y) [/imath] so far I have used properties of this measure to show [imath]\mu(X \cup Y) \leq \mu(X) +\mu(Y) [/imath] so what remains is to show [imath]\mu(X \cup Y) \geq \mu(X) +\mu(Y) [/imath] however I am unsure how to proceed. All I have that I have yet to use is that [imath]X,Y[/imath] are subsets of real numbers and that they are disjoint and non empty. Any help??
|
1962374
|
Is finite additivity a property of the Lebesegue outer measure?
In the Lebesgue outer measure we have the following properties: the length of the empty set is [imath]0[/imath], monotonicity and countable subadditivity. My question is, is it possible to use the above properties to show finite additivity? I have no idea where to begin (or even if it's true) so any hints are definitely appreciated.
|
1963303
|
Combinatorial Proof of [imath]\sum_{k=1}^{n-1} k\cdot k!= n!-1 [/imath]
For [imath]n ≥ 2[/imath], [imath]\sum_{k=1}^{n-1} k \cdot k!= n!-1 [/imath] On the left-hand side, we could be choosing ordered subteams from [imath]n-1[/imath] people (let's say for some reason one of these people cannot be qualified). the range of k would be the size of subteams and [imath]k![/imath] would be ordering them. [imath]k[/imath] could also be [imath]{k \choose 1}[/imath], which means we could be choosing a team leader for each ordered subteam. On the right-hand side, we order all [imath]n[/imath] people and then take out one of these cases but what I can't figure out is how to equate these two sides. Could you give any hints that could help me progress here? Note: I apologize that this was a duplicate. Thanks for everyone who helped!
|
1198735
|
Proving [imath]\sum_{k=1}^n k k!=(n+1)!-1[/imath]
Prove: [imath]\displaystyle\sum_{k=1}^n k k!=(n+1)!-1[/imath] (preferably combinatorially) It's pretty easy to think of a story for the RHS: arrange [imath]n+1[/imath] people in a row and remove the the option of everyone arranged to height from shortest to highest, but it doesn't hold up for the LHS. Alternatively, trying to visualize the LHS, I noticed that it's like a right angle tetrahedra: 1 2!+2! 3!+3!+3! ... But it doesn't help to see a connection to the RHS. Note: no integrals or gamma function nor use of other identities without proving them nor generating functions.
|
1961938
|
When stating a theorem in textbook, use the word "For all" or "Let"?
(Some report that my question is similar to another post. However, that post is talking about writing the "proof", rather than "stating" the theorem. "Proving" a theorem is NOT of the same structure and situation as "stating" a theorem. So this question is not duplicated to the other! Do not let it to be closed! And by the way, I'm also the OP of that question...) In writing a textbook, when we need to state a theorem that is a universal quantification, we can use the word "for all ..."(or equivalently "for every", "for any", "for arbitrary", "for each") or "let ...", Which of these ways is more ideal? Why? Although I think writing as "for all" is the more natural way to reflect the logical structure, that is a universal quantifier [imath]\forall[/imath], the popular style I have seen tends to use "let". Any theoretical aspect or experience is welcome. Example set 1. For all natural numbers [imath]n[/imath], if [imath]n[/imath] is even, then [imath]n[/imath] squared is even. Let [imath]n[/imath] be a natural number. If [imath]n[/imath] is even, then [imath]n[/imath] squared is even. Example set 2. Let [imath]A,B[/imath] be two sets. If for all [imath]x\in A[/imath], [imath]x\in B[/imath], then we say [imath]A[/imath] is a subset of [imath]B[/imath]. For all pairs [imath]A,B[/imath] of sets, if for all [imath]x\in A[/imath], [imath]x\in B[/imath], then we say [imath]A[/imath] is a subset of [imath]B[/imath]. Example set 3. Let [imath]Y[/imath] be a subspace of [imath]X[/imath]. Then [imath]Y[/imath] is compact if and only if every covering of [imath]Y[/imath] by sets open in [imath]X[/imath] contains a finite subcollection covering [imath]Y[/imath]. (Munkres Topology Lemma 26.1) For all subspaces [imath]Y[/imath] of [imath]X[/imath], [imath]Y[/imath] is compact if and only if every covering of [imath]Y[/imath] by sets open in [imath]X[/imath] contains a finite subcollection covering [imath]Y[/imath]. Example set 4. For every [imath]f:X\to Y[/imath] being a bijective continuous function, if [imath]X[/imath] is compact and [imath]Y[/imath] is Hausdorff, then [imath]f[/imath] is a homeomorphism. (adapted by me, maybe ill-grammared?) For every bijective continuous function [imath]f:X\to Y[/imath], if [imath]X[/imath] is compact and [imath]Y[/imath] is Hausdorff, then [imath]f[/imath] is a homeomorphism. (adapted by me.) Let [imath]f:X\to Y[/imath] be a bijective continuous function. If [imath]X[/imath] is compact and [imath]Y[/imath] is Hausdorff, then [imath]f[/imath] is a homeomorphism. (Munkres Topology Theorem 26.6) New added example set 5 (I skipped the quantification on [imath]E,f:E\to\mathbb{R},L,c[/imath], just focus on the key part here.) If "[imath]\forall\varepsilon>0,\exists\delta>0,\forall x\in E,0<|x-c|<\delta\rightarrow |f(x)-L|<\varepsilon[/imath]", then we say [imath]f(x)[/imath] converges to [imath]L[/imath] when [imath]x[/imath] approaches [imath]c[/imath]. If, for all [imath]\varepsilon>0[/imath], there exists [imath]\delta>0[/imath] such that for all [imath]x\in E[/imath], if [imath]0<|x-c|<\delta[/imath] then [imath]|f(x)-L|<\varepsilon[/imath]", then we say [imath]f(x)[/imath] converges to [imath]L[/imath] when [imath]x[/imath] approaches [imath]c[/imath]. (Using "for all") If let [imath]\varepsilon>0[/imath], there is [imath]\delta>0[/imath], such that let [imath]x\in E[/imath], if [imath]0<|x-c|<\delta[/imath] then [imath]|f(x)-L|<\varepsilon[/imath]", then we call [imath]f(x)[/imath] converges to [imath]L[/imath] when [imath]x[/imath] approaches [imath]c[/imath]. (Using "let". I think this type is not natural. But I can't tell why.)
|
1956909
|
While proving a universal quantification, we begin by "Let ... in ..." or "Arbitrary choose ..."?
While proving a universal quantification, say [imath]\forall a\in D,~P(x)\rightarrow Q(x)[/imath]. I see (at least) two type of people, one begin the proof by (1) Way 1 Let [imath]a[/imath] in [imath]D[/imath], (or let [imath]a[/imath] be any element in [imath]D[/imath]) [imath]\cdots\cdots[/imath] (some arguments) [imath]\therefore P(a)\rightarrow Q(a)[/imath] Since we made no special assumptions on [imath]a[/imath], any [imath]a[/imath] in [imath]D[/imath] also deduce [imath]P(a)\rightarrow Q(a)[/imath]. Thus we have shown that [imath]\forall a\in D,~P(x)\rightarrow Q(x)[/imath]. (This is exactly what the U.G. rule says.) one by: (2) Way 2 Arbitrary choose [imath]a[/imath] in [imath]D[/imath], [imath]\cdots\cdots[/imath] (some arguments) [imath]\therefore P(a)\rightarrow Q(a)[/imath] Since what we choose is arbitrary, so every element in [imath]D[/imath] also holds true. (some blur here.. Even I guess whether it is "exactly" a usage of U.G. rule) I'm not ask for a elementary problem that how to prove a universal quantification. Instead, I'm curious which formulation of such a proof is more proper, logical and formal logical structure. I think way 2 is somewhat not clear since yes, we had indeed arbitrary chosen a element [imath]a[/imath] in [imath]D[/imath], but someone may intensionally argue that "hey, it is because you're lucky that you randomly chose the right [imath]a[/imath] at the very begin that makes the predicate true in the set! If you want to show that this is true for any element in the set, you should not have use the word choose in the proof's begin; or at least, you should say at end that since our intermediate steps do not use the special assumption of [imath]a[/imath], so it suits for all elements. But in this way, this becomes slightly complex than way 1, why not just use way 1?" The reason that way II is valid is somewhat of a meta-level, not indicate by the language itself.
|
1963930
|
Prove [imath][0, 1][/imath] and [imath](0, 1][/imath] are equinumerous by use of bijection
What my thought was that the bijection was a piecewise function: [imath]f(x) = 1/n+1[/imath] if [imath]x = 1/n[/imath] for some [imath]n \in \Bbb N[/imath] and [imath]x=1/n[/imath] if [imath]x \neq1/n[/imath] for some [imath]n \in \Bbb N[/imath]. However, the textbook says this is incorrect. I don't actually see why. Was it a matter of brackets or the open variable [imath](0,1][/imath], which is what I tend to think.
|
1964290
|
Prove [imath][0, 1][/imath] and [imath](0, 1][/imath] are equinumerous
I am having a bit of trobule here. THis post was marked duplicate HOWEVER the open and closed segments were different. My thought is this: What my thought was that the bijection was a piecewise function: [imath]f(x) = 1/(n+1)[/imath] if [imath]x = 1/n[/imath] for some [imath]n \in \Bbb N[/imath] and [imath]x=1/n[/imath] if [imath]x \neq1/n[/imath] for some [imath]n \in \Bbb N[/imath]. However, the textbook says this is incorrect. I don't actually see why. Was it a matter of brackets or the open variable [imath](0,1][/imath], which is what I tend to think. How could I amend these two functions to get the proper answer?
|
1964107
|
Prove by induction [imath]\sum_{i=1}^{n} \frac{1}{\sqrt{i}}[/imath] [imath]< 2\sqrt{n}[/imath] for [imath]n \geq 1[/imath]
So far I have this. I have a feeling I'm getting off track with the last two steps. We want to prove [imath]\sum_{i=1}^{n} \frac{1}{\sqrt{i}}[/imath] [imath]< 2\sqrt{n}[/imath] for [imath]n \geq 1[/imath] Base Case Prove P(1): [imath]\sum_{i=1}^{1} \frac{1}{\sqrt{1}}[/imath] [imath]< 2\sqrt{1}[/imath]. We get [imath]1 < 2[/imath]. Induction Hypothesis [imath]\sum_{i=1}^{n} \frac{1}{\sqrt{n}}[/imath] [imath]< 2\sqrt{n}[/imath] is true for [imath]n \geq 1[/imath] Induction Step Prove P(k + 1): [imath]\sum_{i=1}^{k + 1} \frac{1}{\sqrt{k + 1}} < 2\sqrt{k + 1} [/imath] LHS = [imath]\sum_{i=1}^{k + 1} \frac{1}{\sqrt{k + 1}}[/imath] = [imath]\sum_{i=1}^{k} \frac{1}{\sqrt{k}}[/imath] - 1 + [imath]\frac{1}{\sqrt{k + 1}}[/imath] + [imath]\frac{1}{\sqrt{k + 2}}[/imath] [imath]< 2\sqrt{k}[/imath] - 1 + [imath]\frac{1}{\sqrt{k + 1}}[/imath] + [imath]\frac{1}{\sqrt{k + 2}}[/imath]
|
1963565
|
Proving [imath]\sum_{k=1}^n\frac1{\sqrt k}<2\sqrt n[/imath] by induction
I am just starting out learning mathematical induction and I got this homework question to prove with induction but I am not managing. [imath]\sum_{k=1}^n\frac1{\sqrt k}<2\sqrt n[/imath] Please help!
|
1963960
|
Does irreducibility in [imath]\mathbb Q[X][/imath] always imply the irreducibility in [imath]\mathbb Z[X][/imath]?
From Wikipedia page: Eisenstein Criterion: I thought the irreducibility in [imath]\mathbb Q[X][/imath] automatically implies the irreducibility in [imath]\mathbb Z[X][/imath], because [imath]\mathbb Z\subset \mathbb Q[/imath]. Could someone give a counterexample of a polynomial [imath]f\in \mathbb Z[X][/imath] which is irreducible over [imath]\mathbb Q[X][/imath] and reducible over [imath]\mathbb Z[X][/imath]?
|
1581835
|
Irreducible polynomials over [imath]\mathbb Q[/imath] and [imath]\mathbb Z [/imath]
When I read "Contemporary Abstract Algebra" by Joseph gallian, under the topic irreducible polynomials, his first example is the polynomial [imath]2x^2+4=0[/imath] is reducible over [imath]\mathbb Z[/imath] but irreducible over [imath] \mathbb Q[/imath]. I dont know how is this possible? Since it is of degree 2, we can see the roots of the polynomial, where it lies? if it lies in [imath]\mathbb Z[/imath] then it would lie in [imath] \mathbb Q[/imath], then how can it be reducible over [imath]\mathbb Z[/imath] when the roots are complex numbers? pls explain
|
854180
|
Determinant of a non-square matrix
I wrote an answer to this question based on determinants, but subsequently deleted it because the OP is interested in non-square matrices, which effectively blocks the use of determinants and thereby undermined the entire answer. However, it can be salvaged if there exists a function [imath]\det[/imath] defined on all real-valued matrices (not just the square ones) having the following properties. [imath]\det[/imath] is real-valued [imath]\det[/imath] has its usual value for square matrices [imath]\det(AB)[/imath] always equals [imath]\det(A)\det(B)[/imath] whenever the product [imath]AB[/imath] is defined. [imath]\det(A) \neq 0[/imath] iff [imath]\det(A^\top) \neq 0[/imath] Does such a function exist?
|
2689010
|
How to find the determinant of two non-square matrices?
I have to find the determinant of a non-square matrix. But because we can have the determinant of square matrices only. Then, how can I transform a [imath]2 \times 3 [/imath] matrix to be a square one? EDIT: I have two matrices, the first one of size [imath]2 \times 3 [/imath] and the second one of size [imath]3 \times 2 [/imath], I want to find the determinant of their product without finding their product.
|
1964039
|
Why do positive definite matrices have to be symmetric?
Definitions of positive definiteness usually look like this: A symmetric matrix [imath]M[/imath] is positive definite if [imath]x^T M x > 0[/imath] for all vectors [imath]x \neq 0[/imath]. Why must [imath]M[/imath] be symmetric? The definition seems to make sense for general square matrices.
|
1954167
|
Do positive semidefinite matrices have to be symmetric?
Do positive semidefinite matrices have to be symmetric? Can you have a non-symmetric matrix that is positive definite? I can't seem to figure out why you wouldn't be able to have such a matrix, but all my notes specify positive definite matrices as "symmetric [imath]n \times n[/imath] matrices." Can anyone help me with an example of a non-symmetric positive definite matrix, or some insight into a proof for why it would need to be symmetric should that be the case? Thanks!
|
1964803
|
Prove that the matrix multiplication of a set of linearly independent vectors produces a set of linearly independent vectors
If B is a mxn matrix and V the solution space of the homogeneous linear system Bx=0, and there exists a set of linearly independent vectors [imath]u_1[/imath], [imath]u_2[/imath],...,[imath]u_k[/imath] in [imath]R^n[/imath], Prove that if V intersects span{[imath]u_1[/imath], [imath]u_2[/imath], ..., [imath]u_k[/imath]} = {0}, then [imath]Bu_1[/imath], [imath]Bu_2[/imath],..., [imath]Bu_k[/imath] are linearly independent. And if V intersects span{[imath]u_1[/imath], [imath]u_2[/imath], ..., [imath]u_k[/imath]} is not equals to {0}, are [imath]Bu_1[/imath], [imath]Bu_2[/imath], ..., [imath]Bu_k[/imath] linearly independent?
|
1959114
|
Linear independence of images by [imath]A[/imath] of vectors whose span trivially intersects [imath]\ker(A)[/imath]
I know that any subset of V with more than k vectors is always linearly dependent, but how do I include matrix A into the vectors given to prove the statement? Also, for part (b), does the statement mean that Ax=0 only has the trivial solution since V intersects span{u1, u2, ..., uk}={0}? Any help would be appreciated! Thank you!
|
1964730
|
Prove that for any two integers [imath]a[/imath] and [imath]b[/imath] [imath]ab(a^{60}-b^{60}) [/imath] is divisible by [imath]56786730[/imath]
Prove that for any two integers [imath]a[/imath] and [imath]b[/imath], [imath]ab(a^{60}-b^{60}) [/imath] is divisible by [imath]56786730[/imath] [imath]56786730=2\cdot3\cdot5\cdot7\cdot11\cdot13\cdot31\cdot61[/imath] It is easy to prove that [imath]2,31,61|\ ab(a^{60}-b^{60}) [/imath] using Fermat's little theorem for 61 and 31 and taking cases for 2. But I can't figure out a way to solve this for all the other factors. Can anybody help me solve this. Except this, I would also appreciate if someone could share another method to solve this question which doesn't involve modular congruence as I haven't really studied that yet.
|
331538
|
[imath]\forall m,n \in \Bbb N[/imath] : [imath]\ 56786730\mid mn(m^{60}-n^{60})[/imath]
how to prove : [imath]\forall m,n \in \Bbb N[/imath] : [imath]56786730\mid mn(m^{60}-n^{60})[/imath] my effort: [imath]56786730=2.3.5.7.11.13.1841[/imath] -Is [imath]1841[/imath] prime? we must be prove: [imath]2|m n(m^{60}-n^{60})[/imath],...,[imath]13|mn(m^{60}-n^{60})[/imath] but how? by using Fermat theorem we have if [imath]\gcd(m,i)=1 , \gcd(n,i)=1 , (n)m^{i-1} \equiv 1 \pmod i, i=2,3,5,7,11,13[/imath]so [imath]2|mn(m^{60}-n^{60})[/imath],...,[imath]13|mn(m^{60}-n^{60})[/imath] because [imath]i|0, i=2,3,5,7,11,13[/imath] and [imath]1,2,4,6,10,12 [/imath] divide [imath]60[/imath] for other value of [imath]m,n[/imath] ? and factorization of $1841?
|
1965051
|
A Systematic Way to find Counter Examples? via Linear Algebra
So I had a test today, and one of the questions was: "Give an example of a set, S contained in [imath]R^3[/imath] that is closed under scalar multiplication but isn't closed under addition" And I was just wondering -- is there a systematic way to go about this rather than search for examples? Perhaps some sort of algebraic way to create such a set that satisfies these conditions without any further thought?
|
1607647
|
example of a nonempty subset is closed under scalar multiplication but not a subspace
Could anyone provide an example of a nonempty subset [imath]U[/imath] of [imath]R^2[/imath] such that [imath]U[/imath] is closed under scalar multiplication, but [imath]U[/imath] is not a subspace of [imath]R^2[/imath]
|
1966252
|
[imath]3X^3+4Y^3+5Z^3[/imath] has no rational solutions?
How to show [imath]3X^3+4Y^3+5Z^3[/imath] has no rational solutions ? I managed to bring it down to the elliptic curve [imath]Y^2=X^3-2^43^360^2[/imath] but now I'm stuck to show group of rational points of this curve is trivial. Please help anyone. Or could any one suggest me some different proof without using elliptic curves ?
|
55119
|
On the equation [imath]3x^3 + 4y^3 + 5z^3 = 0[/imath]
How can I show that the equation [imath]3x^3 + 4y^3 + 5z^3 = 0[/imath] has nonzero solutions modulo every integer but not in [imath]\mathbb{Z}[/imath]?
|
1966537
|
Hausdorff dimension of a Cantor set.
I have a rather basic question but I couldn't find a straight answer anywhere. Consider [imath]C \subset [0,1][/imath] a Cantor set, that is a closed, nowhere dense and perfect subset of [imath][0,1][/imath]. Is the Hausdorff dimension of [imath]C[/imath] necessarily positive? And even more basic: is it possible to find [imath]A[/imath] an uncountable subset of [imath][0,1][/imath] whose Hausdorff dimension is [imath]0[/imath]? Obviously a negative answer to the former question would provide such an example. I'd be very satisfied with dry answers with references if that's all you can provide :)
|
73547
|
Uncountable sets of Hausdorff dimension zero
Let [imath]A \subset \mathbb{R}[/imath] be a countable set. It is easy to see that [imath]A[/imath] has Hausdorff dimension [imath]\dim_H(A) = 0[/imath]. Do there exist uncountable sets [imath]A \subset \mathbb{R}[/imath] with [imath]\dim_H(A) = 0[/imath]?
|
1966641
|
Find the limit [imath]\lim_{n\to\infty} \bigl(1+\frac{1}{n}\bigr)^{n^2}\cdot\frac{1}{e^n}[/imath]
Find [imath]\displaystyle\lim_{n\to\infty} \Bigl(1+\dfrac{1}{n}\Bigr)^{n^2}\cdot\dfrac{1}{e^n}[/imath] We have: [imath] \lim_{n\to\infty} \Bigl(1+\dfrac{1}{n}\Bigr)^{n^2}\cdot\dfrac{1}{e^n}= \lim_{n\to\infty}\dfrac{e^{n^2\ln\bigl(1+\frac{1}{n}\bigr)}}{e^n}= \lim_{n\to\infty} e^{n^2\ln\bigl(1+\frac{1}{n}\bigr)-n} [/imath] Then I have no idea. Can anyone help me please?
|
360510
|
What is [imath] \lim_{n\to\infty}\frac{1}{e^n}\Bigl(1+\frac1n\Bigr)^{n^2}[/imath]?
How to solve the following limit question? [imath]\lim_{n\to\infty}\frac{1}{e^n}\Bigl(1+\frac1n\Bigr)^{n^2}[/imath] Thanks a lot.
|
1966964
|
Why the discriminant of a system of linear equation=0 implies it has infinity many solution?
Given a system of linear equations: [imath](S):\begin{cases} a_1x+b_1y+c_1z=d_1\\ a_2x+b_2y+c_2z=d_2\\ a_3x+b_3y+c_3z=d_3 \end{cases}[/imath] And we use matrix to represent: [imath]\left( \begin{array}{ccc|c} a_1 & b_1 & c_1 & x\\ a_2 & b_2 & c_2 & y\\ a_3 & b_3 & c_3 & z\\ \end{array}\right)= \begin{pmatrix} d_1\\ d_2\\ d_3\\ \end{pmatrix} [/imath]Let [imath]M=\begin{pmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2\\ a_3 & b_3 & c_3 \\ \end{pmatrix}[/imath] [imath]\begin{pmatrix} x\\ y\\ z\\ \end{pmatrix}=M^{-1}\begin{pmatrix} d_1\\ d_2\\ d_3\\ \end{pmatrix}[/imath] For (S) to have a unique solution, [imath]det M\ne0[/imath], and no solution for [imath]det M=0[/imath] But I am wondering why [imath]det M=0[/imath] also implies it has infinity many solution.
|
640554
|
Set of Linear equation has no solution or unique solution or infinite solution?
For the system [imath] \left\{ \begin{array}{rcrcrcr} x &+ &3y &- &z &= &-4 \\ 4x &- &y &+ &2z &= &3 \\ 2x &- &y &- &3z &= &1 \end{array} \right. [/imath] what is the condition to determine if there is no solution or unique solution or infinite solution? Thank you!
|
1966697
|
summation of Newton symbol product
Anyone has idea how to simplify this summation: [imath]\sum\limits_{k=0}^{N}\binom{M-1+k}{k}\binom{M-1+N-k}{N-k}[/imath]
|
890770
|
Sum of products of binomial coefficients: [imath] \sum_{l=0}^{n-j} \binom{M-1+l}{l} \binom{n-M-l}{n-j-l} = \binom{n}{j} [/imath]
In a proof I've come across the following identity: [imath] \sum_{l=0}^{n-j} \binom{M-1+l}{l} \binom{n-M-l}{n-j-l} = \binom{n}{j} [/imath] I see that it's right, when plugging in numbers, but I don't see the algebraic or combinatorial proof behind it. Can anyone help me with that?
|
1967070
|
The remainder when this determinant is divided by 5
The question is find the remainder when [imath]\begin{vmatrix} { 2014 }^{ 2014 } & { 2015 }^{ 2015 } & { 2016 }^{ 2016 } \\ { 2017 }^{ 2017 } & { 2018 }^{ 2018 } & { 2019 }^{ 2019 } \\ { 2020 }^{ 2020 } & { 2021 }^{ 2021 } & { 2022 }^{ 2022 } \end{vmatrix}[/imath] is divided by 5. While googling, I found this answer Is there a quick way to find the remainder when this determinant is divided by [imath]5[/imath]? but unfortunately I don't have the skill level to understand that answer. It says I'll have to use Fermat's little theorem(https://en.wikipedia.org/wiki/Fermat%27s_little_theorem) but I can't figure out how to apply it in this question.
|
1931365
|
Is there a quick way to find the remainder when this determinant is divided by [imath]5[/imath]?
Find the remainder when the determinant [imath]\begin{vmatrix} { 2014 }^{ 2014 } & { 2015 }^{ 2015 } & { 2016 }^{ 2016 } \\ { 2017 }^{ 2017 } & { 2018 }^{ 2018 } & { 2019 }^{ 2019 } \\ { 2020 }^{ 2020 } & { 2021 }^{ 2021 } & { 2022 }^{ 2022 } \end{vmatrix}[/imath] is divided by [imath]5[/imath]. I'm aware that this problem has a number-theoretic solution involving congruence relations. But considering that this was asked as a multiple-choice question in a test, what should be the best way to approach problems like this? The options were [imath](a)\quad1\quad (b)\quad2\quad (c)\quad3\quad (d)\quad 4[/imath]
|
1967323
|
Why [imath] e^{-2\pi} \neq 1? [/imath]
Consider the following: [imath] e^{2 \pi i} = 1 [/imath] [imath] \left(e^{2 \pi i}\right)^i = 1^i [/imath] [imath] e^{2 \pi i^2} = 1 [/imath] [imath] e^{-2 \pi} = 1 [/imath] but calculator tells me that: [imath]\quad e^{-2 \pi} = 0.0018674427317...[/imath] Can anyone explain what is going on?
|
463898
|
Why is [imath]\left(e^{2\pi i}\right)^i \neq e^{-2 \pi}[/imath]?
Here's my (obviously flawed) proof that [imath]1=e^{-2 \pi}[/imath]: [imath] 1^i=1\\ e^{2 \pi i} = 1\\ \left(e^{2\pi i}\right)^i = 1^i\\ e^{-2 \pi} = 1 [/imath] What's the issue? I understand that exponentiation is not injective (and thus [imath]-1 \neq 1[/imath] even though [imath](-1)^2 = 1^2[/imath]), but I don't think that's an issue here: I'm only raising things to the power of [imath]i[/imath], which I don't think is multi-valued.
|
1822810
|
Geometric interpretation of the dual cone of [imath]l^1[/imath] is [imath]l^\infty[/imath]?
I just noticed somewhere in Convex Optimization that the dual cone of [imath]l^1[/imath] is [imath]l^\infty[/imath]! (A diamond in [imath]\mathbb{R}^2[/imath] for [imath]l^1[/imath] is a square in [imath]\mathbb{R}^2[/imath] for [imath]l^\infty[/imath].) In fact I cannot imagine that. Can you please explain it geometrically by the definition of the dual cone? [Ref. Convex Optimization book, Stephen Boyd] [imath]K = \{(x,t): \Vert x\Vert_1 \le t\} \Rightarrow K^* = \{(x,t): \Vert x\Vert_\infty \le t\}[/imath] Definition: [imath]K[/imath] is a cone, then the dual cone is : [imath]K^* = \{y: x^T y \geq 0 \ \text{for all} \ x \in K\}[/imath] I would be glad if you have any comment about that. For simplicity you can discuss about that in [imath]\mathbb{R}^2[/imath].
|
1110093
|
Dual cone of [imath]K = \{(x,t) \mid \| \boldsymbol{x} \|_1 \le t \}[/imath]
This slide shows the dual cone of [imath]K = \{(x,t) \mid \| \boldsymbol{x} \|_1 \le t \}[/imath] is [imath]K^{*} = \{(x,t) \mid \| \boldsymbol{x} \|_{\infty} \le t\}[/imath]. Is it right? How is it proved?
|
1967754
|
How is: [imath](n+1)!(n+1)+(n+1) = (n+2)![/imath]?
i want to show that [imath](n+1)!(n+1)+(n+1)! = (n+2)![/imath] i put it into wolfram and it showed this to be true, but im not sure why. I need to show this as a part of a induction proof.
|
1966279
|
Factorials, Simplify the addition and multiplication of factorials.
I have an equation containing factorials here [imath](k+1)!+(k+1)(k+1)![/imath] yet I am having a hard time understanding how to simplify it using algebra. A simple search on wolfram gets me a reduced form of [imath](k + 2)! [/imath] This would be a great refresher to such problems, sadly I don't know the elementary operations to reduce it.
|
1966886
|
How do I see that [imath]a^2 \equiv 1 \mod{( p \cdot q)}[/imath] has four incongruent solutions?
How do I see that [imath]a^2 \equiv 1 \mod{( p \cdot q)}[/imath] has four congruent solutions? Studying cryptography, I've seen that given [imath]p[/imath] and [imath]q[/imath] are primes, the equation above should have 4 incongruent solutions. I see immediately, that [imath]\pm 1[/imath] are solutions, but I don't know how to prove existence of other two. Also, are there circumstances for which [imath]a^2 \equiv x \mod{( p \cdot q)}[/imath] has four incongruent solutions, e.g. does it always hold for [imath]\gcd({x, p\cdot q}) = 1[/imath]?
|
1139488
|
Prove that if the equation [imath]x^{2} \equiv a\pmod{pq}[/imath] has any solutions, then it has four solutions.
Suppose [imath]n = pq[/imath] with [imath]p[/imath] and [imath]q[/imath] distinct odd primes. Suppose that [imath]\gcd(a,pq)=1[/imath]. Prove that if the equation [imath]x^{2} \equiv a\pmod n[/imath] has any solutions, then it has four solutions. Proof: Suppose [imath]n = pq[/imath] with [imath]p[/imath] and [imath]q[/imath] distinct odd primes and that gcd([imath]a,pq[/imath]) = 1. Let us have the equation [imath]x^{2} \equiv a\pmod n[/imath]. Then, [imath]x^{2} \equiv a[/imath] (mod [imath]pq[/imath]), [imath]p \not = q[/imath]. By definition, we can separate the equation into two equations such that [imath]y^{2} \equiv a \equiv b[/imath] (mod [imath]p[/imath]) and [imath]z^{2} \equiv a \equiv c[/imath] (mod [imath]q[/imath]). Let [imath]g_{p}[/imath] be a primitive root modulo [imath]p[/imath] and [imath]g_{q}[/imath] be a primitive root modulo [imath]q[/imath]. Then, [imath]b[/imath] is equal to some power of [imath]g_{p}[/imath] and [imath]c[/imath] is equal to some power of [imath]g_{q}[/imath]. With the fact that [imath]b[/imath] has a square root modulo [imath]p[/imath] (i.e. [imath]r^{2} \equiv b[/imath](mod [imath]\ p)[/imath]) and [imath]c[/imath] has a square root modulo [imath]q[/imath] (i.e. [imath]t^{2} \equiv c[/imath](mod [imath]\ q)[/imath]), there is an even power of [imath]g_{p}[/imath] and of [imath]g_{q}[/imath] such that [imath]b = g_{p}^{2k_{1}}[/imath](mod [imath]\ p)[/imath] and [imath]c = g_{q}^{2k_{2}}[/imath](mod [imath]\ q)[/imath] for some [imath]k_{1}, k_{2} \in Z[/imath]. By computing, we have the following: [imath]r^{2} \equiv b[/imath](mod [imath]\ p)[/imath] [imath]\equiv b^{(p + 1) / 2}([/imath]mod [imath]\ p)[/imath] [imath]\equiv (g_{p}^{2k_{1}})^{(p + 1) / 2}([/imath]mod [imath]\ p)[/imath] [imath]\equiv (g_{p}^{p + 1})^{k_{1}}([/imath]mod [imath]\ p)[/imath] [imath]\equiv g_{p}^{2k_{1} + (p - 1)k_{1}}([/imath]mod [imath]\ p)[/imath] [imath]\equiv b \cdot g_{p}^{(p - 1)k_{1}}([/imath]mod [imath]\ p)[/imath] [imath]\equiv b[/imath] and [imath]t^{2} \equiv c[/imath](mod [imath]\ q)[/imath] [imath]\equiv c^{(q + 1) / 2}([/imath]mod [imath]\ q)[/imath] [imath]\equiv (g_{q}^{2k_{2}})^{(q + 1) / 2}([/imath]mod [imath]\ q)[/imath] [imath]\equiv (g_{q}^{q + 1})^{k_{2}}([/imath]mod [imath]\ q)[/imath] [imath]\equiv g_{q}^{2k_{2} + (q - 1)k_{2}}([/imath]mod [imath]\ q)[/imath] [imath]\equiv c \cdot g_{q}^{(q - 1)k_{2}}([/imath]mod [imath]\ q)[/imath] [imath]\equiv c[/imath] Hence, [imath]r[/imath] is a square root of a modulo [imath]p[/imath] and [imath]t[/imath] is a square root of a modulo [imath]q[/imath], which means there are two solutions for each [imath]p[/imath] and [imath]q[/imath]. Since [imath]p \not = q \in \mathbb{Z}_{n}[/imath], we have isomorphism such that [imath]\mathbb{Z}_{n} \simeq \mathbb{Z}_{p} \cdot \mathbb{Z}_{q}[/imath]. Therefore, if the equation [imath]x^{2} \equiv a[/imath] (mod [imath]n[/imath]) has any solutions, it must have four solutions. [imath]\blacklozenge[/imath] What do you think about the proof I wrote?
|
1966971
|
Surjective function from [imath]\mathbb{N}[/imath] to [imath]\mathbb{Q}[/imath]
I want to find some simple surjective function from [imath]\mathbb{N}[/imath] to [imath]\mathbb{Q}[/imath]. Since [imath]\mathbb{Q}[/imath] is countable, it should be possible. Can someone find such function?
|
1690403
|
Find a surjective function [imath]f:\mathbb{N}\to \mathbb{Q}[/imath]
I'm trying to find a surjective function [imath]f:\mathbb{N}\to \mathbb{Q}[/imath]; I know that at least one such function must exist since [imath]\mathbb{Q}[/imath] is countable, but I haven't been able to find one. Can someone show me one such function? Best regards, lorenzo
|
1958470
|
L'Hopital rule variation
If [imath] \lim_{x \rightarrow a} f(x)= \infty\quad \lim_{x \rightarrow a} g(x)=\infty[/imath] and [imath] \lim_{x \rightarrow a} \frac{f'(x)}{g'(x)}=L [/imath] then [imath] \lim_{x \rightarrow a} \frac{f(x)}{g(x)}=L [/imath] Is this correct? Any response would be appreciated.
|
147356
|
Proof of the L'Hôpital Rule for [imath]\frac{\infty}{\infty}[/imath]
I ask for the proof of the L'Hôpital rule for the indeterminate form [imath]\frac{\infty}{\infty}[/imath] utilizing the rule for the form [imath]\frac{0}{0}[/imath]. Theorem: Let [imath]f,g:(a,b)\to \mathbb{R}[/imath] be two differentiable functions such as that: [imath]\forall x\in(a,b)\ \ g(x)\neq 0\text{ and }g^{\prime}(x)\neq 0[/imath] and [imath]\lim_{x\to a^+}f(x)=\lim_{x\to a^+}g(x)=+\infty[/imath] If the limit [imath]\lim_{x\to a^+}\frac{f^{\prime}(x)}{g^{\prime}(x)}[/imath] exists and is finite, then [imath]\lim_{x\to a^+}\frac{f(x)}{g(x)}=\lim_{x\to a^+}\frac{f^{\prime}(x)}{g^{\prime}(x)}[/imath] My attempt: Since [imath]\lim_{x\to a^+}f(x)=+\infty[/imath], [imath]\exists \delta>0:a<x<a+\delta<b\Rightarrow f(x)>0\Rightarrow f(x)\neq 0[/imath] Let [imath]F,G:(a,a+\delta)\to \mathbb{R}[/imath], [imath]F(x)=\frac{1}{f(x)}[/imath], [imath]G(x)=\frac{1}{g(x)}[/imath]. Then by the hypothesis [imath]\lim_{x\to a^+}F(x)=\lim_{x\to a^+}G(x)=0[/imath], [imath]\forall x\in(a,b)\ \ G(x)\neq 0\text{ and }G^{\prime}(x)=-\frac{1}{g^2(x)}g^{\prime}(x)\neq 0[/imath] The question is, does the limit [imath]\lim_{x\to a^+}\frac{F^{\prime}(x)}{G^{\prime}(x)}=\lim_{x\to a^+}\frac{-\frac{1}{f^2(x)}f^{\prime}(x)}{-\frac{1}{g^2(x)}g^{\prime}(X)}=\lim_{x\to a^+}\frac{g^2(x)f^{\prime}(x)}{f^2(x)g^{\prime}(x)}[/imath] exist? The limit [imath]\lim_{x\to a^+}\frac{f^{\prime}(x)}{g^{\prime}(x)}[/imath] exists by the hypothesis but we don't know if the limit [imath]\displaystyle\lim_{x\to a^+}\frac{g^2(x)}{f^2(x)}[/imath] exists to deduce that the limit [imath]\lim_{x\to a^+}\frac{F^{\prime}(x)}{G^{\prime}(x)}[/imath] exists to use the L'Hôpital Rule for the form [imath]\frac{0}{0}[/imath]. EDIT: After discussing it with other users in the site, we came to the conclusion that this proof is only partial and can't logically be continued to yield the Theorem. As a result, the rule for the [imath]\frac{0}{0}[/imath] form can't be used to proove the rule for the [imath]\frac{\infty}{\infty}[/imath] form. Mr. Tavares and myself have already given two different proofs (with the pretty much the same main idea) of the Theorem in question using Cauchy's Mean Value Theorem. You can read them below. You can also read the proof Rudin gives for a stronger version of the Theorem (that does not suppose that [imath]\lim_{x\to a^+}f(x)=+\infty[/imath]) in his book Principle of Meathematical Analysis. If you have any objections in either proofs please let me know. Thank you.
|
1968585
|
Finding A Bijection Between Two Finite Intervals
I've received a problem in my course that has confused me a bit and I'm not sure what my instructor means or what I am allowed to assume. The problem reads like this: "*Find a bijection between the following sets: any two finite intervals of the type [imath][a,b][/imath] and [imath][\alpha,\beta][/imath] with [imath]a<b[/imath] and [imath]\alpha<\beta[/imath]". If I understand correctly, there can't be a bijection between two finite sets unless they have the same amount of elements (or the same cardinality, but we haven't defined that yet) and if I were to assume this it would be trivially simple because then I could just say that there are [imath]n[/imath] elements in both and map [imath]a[/imath] to [imath]\alpha[/imath] and [imath]b[/imath] to [imath]\beta[/imath] and so forth and I'd be done. This seems almost too simple and I feel as though I've missed something that would make the exercise more difficult. Have I solved it or is there more to it? How do I see it and how do I go about creating the bijection?
|
1534311
|
Prove two sets have same Cardinality by writing down bijection
a. Prove that the interval [imath]A = [1,3][/imath] has the same cardinality as [imath]B = [1,5][/imath] by writing down a bijection from [imath]A \to B[/imath]. Don't prove it is a bijection. b. Consider the following infinite set: [imath]A = {1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4},..., \frac{1}{n}}[/imath], Prove the set [imath]A[/imath] has the same cardinality as the integers by writing down a . bijection from [imath]A[/imath] onto [imath]Z[/imath]. I don't know how to find a function that is a bijection from one set to another. Can anyone help by explaining the thought process behind it, I've been having trouble with these types of problems? Thanks.
|
1968727
|
Integrating [imath]sin^n(x)[/imath]
I tried different methods of integration, but I can not solve this problem. [imath]\int \sin^n (x) dx[/imath]
|
112687
|
Integrating [imath]\int \sin^n{x} \ dx[/imath]
I am working on trying to solve this problem: Prove: [imath]\int \sin^n{x} \ dx = -\frac{1}{n} \cos{x} \cdot \sin^{n - 1}{x} + \frac{n - 1}{n} \int \sin^{n - 2}{x} \ dx[/imath] Here are the steps that I follow in the example that I am reading: [imath]u = \sin^{n - 1}{x}[/imath] [imath]du = (n - 1) \cdot \sin^{n - 2}{x} \cdot \cos{x} \ dx[/imath] [imath]v = -\cos{x}[/imath] [imath]dv = \sin{x} \ dx[/imath] [imath]\int \sin^n{x} \ dx = \sin^{n - 1}{x} \cdot \sin{x} \ dx[/imath] [imath]\int \sin^n{x} \ dx = \underbrace{\sin^{n - 1}{x}}_{u} \cdot \underbrace{-\cos{x}}_{v} - \int \underbrace{-\cos{x}}_{v} \cdot \underbrace{(n - 1) \cdot \sin^{n - 2}{x} \cdot \cos{x} \ dx}_{du}[/imath] [imath]\int \sin^n{x} \ dx = -\cos{x} \cdot \sin^{n - 1}{x} + (n - 1)\int \sin^{n - 2}{x} \cdot \cos^{2}{x} \ dx[/imath] [imath]\int \sin^n{x} \ dx = -\cos{x} \cdot \sin^{n - 1}{x} + (n - 1)\int \sin^{n - 2}{x} \cdot \left(1 - \sin^{2}{x}\right) \ dx[/imath] Here is where I get lost. How did we go from [imath]\int \sin^{n - 2}{x} \cdot \left(1 - \sin^{2}{x}\right) \ dx[/imath] to [imath]\int \sin^{n - 2}{x} \ dx - (n - 1) \int \sin^{n}{x} \ dx[/imath]? Even more specifically, where did [imath]\sin^{n}{x}[/imath] come from? [imath]\int \sin^n{x} \ dx = -\cos{x} \cdot \sin^{n - 1}{x} + (n - 1)\int \sin^{n - 2}{x} \ dx - (n - 1) \int \sin^{n}{x} \ dx[/imath] I get this part. [imath]n\int \sin^n{x} \ dx = -\cos{x} \cdot \sin^{n - 1}{x} + (n - 1)\int \sin^{n - 2}{x} \ dx[/imath] [imath]\int \sin^n{x} \ dx = -\frac{1}{n} \cos{x} \cdot x \ \sin^{n - 1}{x} + \frac{n - 1}{n} \int \sin^{n - 2}{x} \ dx[/imath] Could someone please explain what I am missing? Thank you for your time.
|
1968935
|
Find all functions [imath]f:\mathbb R \to \mathbb R[/imath] that satisfy [imath]f(x) + 3 f\left( \frac {x-1}{x} \right) = 7x[/imath] for all nonzero [imath]x[/imath].
Find all functions [imath]f:\mathbb R \to \mathbb R[/imath] that satisfy [imath] f(x) + 3 f\left( \frac {x-1}{x} \right) = 7x [/imath] for all nonzero [imath]x[/imath]. I don't know what's going on today, my brain is not moving! I'm stuck on this problem. Solutions are greatly appreciated.
|
1968871
|
Real domain and range function to find all functions with nonzero x.
Find all functions [imath]f:\mathbb R \to \mathbb R[/imath] that satisfy [imath] f(x) + 3 f\left( \frac {x-1}{x} \right) = 7x [/imath] for all nonzero [imath]x[/imath]. My thoughts so far are to plug in [imath](x - 1)/x[/imath] for [imath]x,[/imath] but it always comes out so complicated, so I don't really know where to go from there.
|
1968173
|
the exponent of the highest power of p dividing n!
The formula for the exponent of the highest power of prime [imath]p[/imath] dividing [imath]n![/imath] is [imath]\sum \frac{n}{p^k}[/imath], but the question is [imath]n=1000![/imath] (really, it has the factorial) and [imath]p=5[/imath]. When I use Wolfram Alpha , I panicked because the number has [imath]2,567[/imath] decimal digits. I think if I write this number I'd need paper all the way to the Amazon. Perhaps I misunderstand the formula?
|
1965074
|
Power of 2 & 5 in product of consecutive numbers
Is there any way to calculate the powers of 2 and 5 in a product of consecutive [imath]n[/imath] numbers, if given [imath]l[/imath] and [imath]r[/imath]. i.e Suppose [imath]x = l \times (l+1) \times (l+2) \times ...... \times (r-1) \times r[/imath], Then if [imath]x = 2^{p_1}.3^{p_2}.5^{p_3}.7^{p_4}....[/imath], So I am interested a direct way of calculating [imath]p_1[/imath] and [imath]p_3[/imath] with just [imath]l[/imath] and [imath]r[/imath] I am just curious about it. I just want [imath]2[/imath] & [imath]5[/imath], but if there is a general solution for any prime number, It would be great if you mention that.
|
1967121
|
Can anyone help me with this do Carmo question?
(Geodesic frame). Let [imath]M[/imath] be a Riemannian manifold of dimension [imath]n[/imath] and let [imath]p \in M[/imath]. Show that there exists a neighborhood [imath]U \subset M[/imath] of [imath]p[/imath] and [imath]n[/imath] vector fields [imath]E_1,...,E_n \in X(U)[/imath], orthonormal at each point of [imath]U[/imath], such that, at [imath]p[/imath], [imath]\nabla_{E_i}E_j(p) = 0[/imath]. Such a family [imath]E_i, i = 1,..., n,[/imath] of vector fields is called a (local) geodesic frame at p. Question: I don't know how I can to prove the step: [imath]\nabla_{E_i}E_j(p)=0[/imath]. I am using [imath]\exp(te_i)[/imath] as a geodesic for this.
|
796608
|
If [imath]\Gamma^k_{ij}(p)=0[/imath], then [imath]\nabla_{E_i}E_j (p)=0?[/imath]
I'm having the same problem as it was questioned here. I can't get throught the step where I need to show that [imath]\nabla_{E_i}E_j (p)=0[/imath]. It only leads to [imath] \nabla_{E_i}E_j(p)=\sum_{lk}^n a_{il}(0)\dfrac{\partial b_jk}{\partial x_l}(0)\partial x_k(p) [/imath] which I can't show that equals zero. On the linked question, there is also the indication that [imath]\nabla_{E_i}E_j (E_k)=\Gamma^k_{ij}[/imath]. This could lead that [imath]\nabla_{E_i}E_j (E_k)(p)=\Gamma^k_{ij}(p)=0[/imath], but [imath]\nabla_{E_i}E_j (E_k)(p)[/imath] is not equal to [imath]\nabla_{E_i}E_j (p)[/imath] which is the one I need. Can someone help me?
|
1968596
|
Natural Deduction stuck
[imath]\vdash (( p → q) → p) → p[/imath] how would I go about solving this with no premises. to get the implication i know i need [imath](p → q) → p[/imath] as my first assumption but im stuck to where to go from there
|
1545970
|
How can I solve this logic question using propositional logic (Natural deduction)?
[imath]\big((P\rightarrow Q)\rightarrow P\big) \rightarrow P[/imath] I need to solve this using simple natural deduction rules these can be hypothesis, [imath]\rightarrow[/imath] intro, [imath]\rightarrow[/imath] elim, conj and disjunct intro and elim, and negation intro (Reductio ad absurdum) and negation elimination (Double)
|
1969412
|
Show that [imath](\sqrt{2}+1)^n = \sqrt{m}+\sqrt{m-1}[/imath] for some positive integer [imath]m[/imath]
Let [imath]n[/imath] be a positive integer. Show that [imath](\sqrt{2}+1)^n = \sqrt{m}+\sqrt{m-1}[/imath] for some positive integer [imath]m[/imath]. I was wondering what would be the easiest way of proving this. Expanding [imath](\sqrt{2}+1)^n[/imath] we get [imath](\sqrt{2}+1)^n = \binom{n}{0}(\sqrt{2})^n+\binom{n}{1}(\sqrt{2})^{n-1}+\cdots+\binom{n}{n} = a+b\sqrt{2}[/imath] and so either [imath]\sqrt{m} = a[/imath] and [imath]\sqrt{m-1} = b\sqrt{2}[/imath] or [imath]\sqrt{m} = b\sqrt{2}[/imath] and [imath]\sqrt{m-1} = a[/imath]. How do we continue from here?
|
1869954
|
Expansion of [imath](1+\sqrt{2})^n[/imath]
I was asked to show that [imath]\forall n\in \mathbb N[/imath] there exist a [imath]p\in \mathbb N^\ast[/imath] such that [imath](1+\sqrt{2})^n = \sqrt{p} + \sqrt{p-1}[/imath] I used induction but it wasn't fruitful, so I tried to use the binomial expansion of [imath](1+\sqrt{2})^n[/imath] but it seems I lack some insight to go further. Any hint is welcomed.
|
1969123
|
Show the cirucumcircles of all triangles [imath]PAB[/imath] have the same radius
Given [imath]2[/imath] circles [imath]W_1[/imath] and [imath]W_2[/imath] which intersect at points [imath]X[/imath],[imath]Y[/imath]. Let [imath]P[/imath] be an arbitrary point on [imath]W_1[/imath]. Suppose the lines [imath]PX[/imath],[imath]PY[/imath] meet [imath]W_2[/imath] again at points [imath]A[/imath],[imath]B[/imath] respectively. Prove that the circumcircles of all triangles [imath]PAB[/imath] have the same radius. MY TRY:- I tried to use coordinate geometry, but it didn't work. Please help me.
|
1961875
|
Prove that the circumradius of [imath]\triangle PAB[/imath] is constant
Given two circle [imath]\omega_1[/imath] and [imath]\omega_2[/imath] which intersect at points [imath]X[/imath] and [imath]Y[/imath]. Let [imath]P[/imath] be an arbitrary point on [imath]\omega_1[/imath]. Suppose that the lines [imath]PX[/imath] and [imath]PY[/imath] meet [imath]\omega_2[/imath] again at [imath]A[/imath] and [imath]B[/imath] respectively. Prove that the circumcircles of all triangles [imath]PAB[/imath] have equal radius. What I tried is not worth adding here.
|
1969720
|
Proving that a sequence converges if [imath]\limsup=\liminf[/imath]
Okay so here I've tried proving that : Let [imath]x_n[/imath] be a sequence, if [imath]\limsup{x_n}=\liminf{x_n}[/imath] then [imath]x_n[/imath] converges. Proof. By definition, [imath]\limsup{x_n}[/imath] and [imath]\liminf{x_n}[/imath] are limits of subsequences that are formed from [imath]x_n[/imath]. Let these subsequences be [imath]a_n[/imath] and [imath]b_n[/imath] such that: [imath]\lim{a_n}=\limsup{x_n}=a \\ \lim{b_n}=\liminf{x_n}=b[/imath] Okay let's analyse this : There exists and [imath]ɛ_1[/imath] such that [imath]|a_n-a|<ɛ_1[/imath] for [imath]n>N^{'}[/imath] There exists and [imath]ɛ_2[/imath] such that [imath]|b_n-b|<ɛ_2[/imath] for [imath]n>N{''}[/imath] Let [imath]N=max(N{'},N{''})[/imath] so [imath]|a_n-a|+|b_n-b|<ɛ_1+ɛ_2[/imath] for [imath]n>N[/imath] By triangle inequality :: [imath]|a_n+b_n-(a+b)|<ɛ_1+ɛ_2[/imath] So we have found that when we add this subsequences together, they converge to [imath]a+b[/imath], what if [imath]a=b[/imath] Then : [imath]|a_n+b_n-(2a)|<ɛ_1+ɛ_2[/imath] So both [imath]a_n[/imath] and [imath]b_n[/imath] have infinite traces around the point [imath]a[/imath] Since their trace is a subset of the actual sequence [imath]x_n[/imath] we can considered that they are approaching the open ball [imath]B(a,ɛ)[/imath]. when they approach to the ball, for [imath]n>N[/imath] sequences equal out [imath]a_n=b_n[/imath] and we can also say that [imath]a_n=b_n=x_n[/imath] for [imath]n>N[/imath] So that when [imath]\liminf=\limsup[/imath] the sequence [imath]x_n[/imath] converges. What do you think about my proof? IS it correct? What else should I add, what should I adopt? I'm new to analysis so help me out please. Thanks.
|
122755
|
Sequence converges iff [imath]\limsup = \liminf[/imath]
I want to prove that a sequence of real numbers [imath]\{s_n\}[/imath] converges to [imath]s[/imath] if and only if [imath]\limsup_{n \to \infty} s_n = \liminf_{n \to \infty} s_n = s[/imath]. Here are my definitions: For any sequence of real numbers [imath]\{s_n\}[/imath], let [imath]E[/imath] be the set of all subsequential limits of [imath]\{s_n\}[/imath], including possibly [imath]+\infty[/imath] and/or [imath]-\infty[/imath] if any subsequence of [imath]\{s_n\}[/imath] diverges to infinity. Then [imath]\limsup_{n \to \infty} s_n = \sup E[/imath], and [imath]\liminf_{n \to \infty} s_n = \inf E[/imath]. I know the theorem that a sequence converges to a point if and only if every one of its subsequences converge to that same point, so one direction of this proof is easy: If [imath]\{s_n\}[/imath] converges to some point [imath]s \in \mathbb{R}[/imath], then every subsequence of [imath]\{s_n\}[/imath] converges to [imath]s[/imath]. So the set [imath]E[/imath] of every subsequential limit of [imath]\{s_n\}[/imath] consists of the single point [imath]s[/imath], so [imath]\limsup_{n \to \infty} s_n = \sup \{s\} = s = \inf \{s\} = \liminf_{n \to \infty} s_n[/imath] But the other direction seems more tricky... If [imath]\limsup_{n \to \infty} s_n = \liminf_{n \to \infty} s_n = s[/imath], then every convergent subsequence converges to the same point [imath]s[/imath]. Also, there can be no subsequences which diverge to infinity (otherwise [imath]\limsup_{n \to \infty} s_n[/imath] would be [imath]+\infty[/imath], or [imath]\liminf_{n \to \infty} s_n[/imath] would be [imath]-\infty[/imath]). But can't there be subsequences which diverge otherwise? And wouldn't that throw off the convergence of [imath]\{s_n\}[/imath]? EDIT: I'd also be willing to accept a solution which makes use of the "Pinching Theorem" (if [imath]a_n \leq s_n \leq b_n[/imath] for every [imath]n \in \mathbb{N}[/imath], and if [imath]a_n \to s[/imath] and [imath]b_n \to s[/imath], then [imath]s_n \to s[/imath]).
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.