AsafHaas
• Member for 4 years, 10 months
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• Israel

By taylor's theorem, it holds that $\cos(x) = 1 + R_1(x)$. In the Lagrange's form of the remainder, there exists $c$ such that $R_1(x) = \frac {(-\cos(c))} {2!} \cdot x^2$. $|R_1(x)| = |\frac {(-\cos(... View answer Accepted answer 4 votes $$\log_a(x) = \frac {\log_b(x)} {\log_b(a)}$$ for any$b$. therefore:$\frac 1 2 \log_{\sqrt 2}(x-2) = \frac {log_2(x - 2)} {2 \log_2(\sqrt 2)} = \frac {log_2(x - 2)} {2 \times \frac 1 2} = \log_2(x-...

You can prove that $f$ is not bounded, since then it will follow that $f$ is increasing and unbounded and therefore diverge to $\infty$ as $x \to \infty$. There is a sneek peek at the proof's outline ...

Recall that $\sin(x) = \cos(\frac \pi 2 - x)$. Therefore your equation is: $$\cos(x) = \cos(\frac \pi 2 - x)$$ And from here we get: $x = \frac \pi 2 - x + 2\pi k \Rightarrow 2x = \frac \pi 2 + 2 \pi ... View answer 3 votes Notice that both the numerator and the denominator tends to 0 as$x \to 0$. You can use L'Hôpital's rule, and then the numerator's derivative is... Good luck! :) View answer Accepted answer 2 votes You can use the Chinese Remainder Theorem. Since$3$and$4$are coprime, the set of equations: $$x = 1 \mod 3 \\ x = 2 \mod 4$$ has a solution which is unique modulo$3\cdot4 = 12$By observing we ... View answer 2 votes Notice that the cardinality of polynomials of degree$0$(Only free coefficients) is$|\mathbb{R}| = \mathfrak{c}$(We just map such polynomials to their free coefficients). The cardinality of ... View answer Accepted answer 2 votes If I understood your question correctly, you want the Riemann sum of$f(t)$for the partition$P = \{0, 1/100, 2/100, ..., 99/100, 1\}$This is, by definition: $$\sum_{i=1}^{100} f(c_i) \frac 1 {100}... View answer Accepted answer 2 votes The transformation is a linear transformation. In order for a transformation T to be a linear transformation, it has to implement 2 conditions:$$(1) \hspace{0.2cm} T(v + w) = T(v) + T(w) \\ (2) \... View answer Accepted answer 2 votes The Lauren remainder in Lagrange form is $$R_n(x) = \frac {f^{(n+1)}(c)} {(n+1)!}x^{n+1}$$ for some point$c$. So seeking for an$n$for which$|R_n(x)| \lt 0.1$means:$|\frac {{cos}^{(n+1)}(c)} {(...

Notice that $\cos(n\pi) = (-1)^n$ and therefore $\cos^2(n\pi) = 1$, and try using the Limit Comparison Test.

An easy conclusion form Lagrange's theorem is that if $|G| = n$ and $a \in G$ then $a^n = e$. Proof: Since $<a> \le G$, By Lagrange's theorem $|<a>| = o(a) \mid |G| = n$. Therefore $n = \... View answer 1 votes Well as you said you can take $$A = \begin{bmatrix} 1 & 0 \\ 1 & 0 \\ 0 & 1 \end{bmatrix}$$ and then$Row(a) = span\{(1,0), (0, 1)\}$is the entire space, ... View answer Accepted answer 1 votes No, it does not suffice to say that$e^x$yields only positive numbers, or that$\mathcal{B}$yields only elements in$U$. You need to show that every element in$U$is a linear combination of the ... View answer Accepted answer 1 votes This is saying in English: "if there exist an$i$in the set$[m]$such that$x_i = x$, then$hS(x) = y_i$. Otherwise,$hS(x) = 0$." View answer 1 votes For the derivation, notice that there are 4 possible states of$a$and$b$: $$a = 0, b = 0 \\ a=0, b=1 \\ a=1,b=0 \\ a=1, b=1$$ Now, the expression$(a*b)$is true for the fourth state, the ... View answer 1 votes The equivalence class of a$x \in A$($A$is a set) is defined as: $$[x] = \{a \in A | aRx \}$$ Meaning it is the set of all items in$A$related to$x$. Your question is to find all of those ... View answer 1 votes No. Take $$f(x) = \begin{cases} 1 & x \in \mathbb{Q} \\ 0 & \text{otherwise} \end{cases}$$ And$C =1$. Now your$g(x)$is exactly$f(x)$which is the Dirichlet function, and is not ... View answer 1 votes So the characteristic polynomial is: $$p(\lambda) = \begin{vmatrix} -1-\lambda & 1 & 1 \\ 1 & -1-\lambda & 1 \\ 1 & 1 & -1-\lambda \\ \end{... View answer 1 votes First, recall this facts about coordinate vectors:$$[v_1 + v_2]_B = [v_1]_B + [v_2]_B[\alpha v]_B = \alpha[v]_Bv = 0 \iff [v]_B = 0$$Now we can prove your claim. Let \alpha_1, \alpha_2, .... View answer 1 votes You can use the root test for sequences: Take the sequence in absolute value: The limit of the sequence with an n'th root is$$L = \lim_{n \to \infty}\sqrt[n] {|a_n|} =\lim_{n \to \infty} |a| (1+\... View answer 1 votes First I must say that$z = x^2 + y^2$is a paraboloid and not a cone, just for being exact. Now for the problem: You need to calculate the volume between the paraboloid and the sphere in the domain ... View answer Accepted answer 1 votes As hinted in the comments, consider the matrix$A^*A$. Let us calculate its trace:$tr(A^*A) = \sum_{i=1}^n \sum_{j=1}^n ((A)_{i,j}(A^*)_{j,i}) = \sum_{i=1}^n \sum_{j=1}^n a_{i,j} * \overline {a_{i,j}}...

It seems that the integral does not converge on $[1, \infty]$ for example, therefore not on $[0, \infty]$: $\lim_{x \to \infty} \frac {\frac {\log(x)\arctan(x)} {\sqrt x}} {\frac 1 {\sqrt x}} = \lim_{... View answer 0 votes It is equivalent. If$\limsup_{n \to \infty} a_n = \liminf_{n \to \infty} a_n = L \in \mathbb{R}$, then the set of partial limits will only contain$L$(If it contains anything else: say bigger than$...

Consider the isomorphism $\varphi: \{0, 2, 4, 6, 8\} \to \mathbb{Z}_5$ given by $0 \to 0, 2 \to 2, 4\to4,6\to1,8\to3$.

Eventually I figured out a proof without using the Fundamental Theorem of Arithmetic: Denote $\gcd(m, \alpha ) = u$. $u|m \Rightarrow \gcd(u, n)=1$. $u|\alpha \Rightarrow \alpha = \beta u$. Now ...

Inflection points are defined as passing from convexity to concavity, which is equivalent as passing from an increasing $f^\prime$ to a decreasing $f^\prime$ for differentiable functions, its extremas....
You can show again the contrapositive: assume $\exists \epsilon_0 > 0: B(x, \epsilon_0) \cap M = \emptyset$. can you proceed showing that $x \notin \overline{M}$? (Hint: you need to show that $\... View answer Accepted answer 0 votes I afraid that your answer for (a) is incorrect... Check for example that the word$w = 111$is accepted by your automata, but$n_1(w) = 0 \pmod{3}$. Label the states in your answer to (a) as:$\$q_0: ...