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https://cartesianproduct.wordpress.com/2012/07/15/more-on-error-correction/ | # More on error correction
Considering the chances of a decoding error (ie having more errors than our error correction code can handle)…
$P(no errors) = (1 -p)^n$ where p is the probability of a bit flip and n the length of the code.
So in our case that gives $P(no errors) = (0.9)^4 = 0.6561$
But we can also work out the possibility of k bit flips, using the binomial distribution:
$P(k_{errors}) = _nC_k p^k (1-p)^{n-k}$
So what are the prospects of a decoding error. This is the lower bound (only the lower bound because – as the table in the previous post showed – some errors might be detected and some not for a given Hamming distance):
$P(total errors)$ = $\sum_{k = d}^n$ $_nC_kp^k(1-p)^{n-k}$
For us $d=4, n=4$, so therefore as $0! = 1$, the lower bound in our case is $0.1^4 (0.9)^{4 - 4} = 0.0001$ which isn’t bad even for such a noisy channel.
But what is the guaranteed success rate?
Here we are looking at:
$\sum_{k=0}^{\lfloor\frac{d - 1}{2}\rfloor}$ $_nC_k p^k (1 - p)^{n - k}$
(Recalling $d \geq 2v +1$ for v bits of error correction)
In our case this gives:
$_4C_0 p^0 (1 -p)^4 + _4C_1 p^1 (1 -p)^3$ $= 0.6561 + 0.2916 = 0.9477$
This shows the power of error correction – even though there is a 10% chance of an individual bit flipping, we can actually keep the error down to just over 5%. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 14, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8726085424423218, "perplexity": 421.0703857078309}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-50/segments/1480698541529.0/warc/CC-MAIN-20161202170901-00253-ip-10-31-129-80.ec2.internal.warc.gz"} |
https://physics.stackexchange.com/questions/69828/equivalence-theorem-of-the-s-matrix?noredirect=1&lq=1 | # Equivalence Theorem of the S-Matrix
as far as I know the equivalence theorem states, that the S-matrix is invariant under reparametrization of the field, so to say if I have an action $S(\phi)$ the canonical change of variable $\phi \to \phi+F(\phi)$ leaves the S-matrix invariant.
In Itzykson's book there is now an exercise in which you have to show, that the generating functional $$Z^\prime(j)=\int \mathcal{D}[\phi] \exp{\{iS(\phi)+i\int d^4x\hspace{0.2cm} j(x)(\phi +F(\phi)\}}$$ gives the same S-matrix as the ordinary generating functional with only $\phi$ coupled to the current due to the vanishing contact terms. He then writes, that this proves the equivalence theorem, which I do not fully understand.
Suppose I take this canonical change of variable, then I get a new action $S^\prime(\phi)=S(\phi+F(\phi))$ and a generating functional $$Z(j)=\int \mathcal{D}[\phi] \exp\{iS(\phi+F(\phi))+i\int d^4x\hspace{0.2cm} j(x)\phi\}$$ If I now "substitute" $\phi+F(\phi)=\chi$ I get $$Z(j)=\int \mathcal{D}[\chi] \det\left(\frac{\partial \phi}{\partial \chi}\right) \exp\{iS(\chi)+i\int d^4x\hspace{0.2cm} j(x)\phi(\chi)\}$$ with $\phi(\chi)=\chi + G(\chi)$ the inverse of $\chi(\phi)$. Therefore comparing $Z(j)$ and $Z^\prime(j)$ I get an extra jacobian determinant.
Where is my fallacy, or why should the determinant be 1?
• To clarify, the function $F(\phi)(x)$ is a function only of $\phi(x)$, or is it allowed to depend on local derivatives, or is it a general smooth functional? Jul 2 '13 at 17:54
• Echoing @BebopButUnsteady's comment, what is your definition of a "canonical change of variables" in a Lagrangian theory? Jul 2 '13 at 18:06
• I mean a change of variables, which is invertible and has therefore a nonvanishing jacobi determinant. It should further be of the kind $x \to x + F(x)$ so to say a point transformation. The function $F(\phi)$ should only depend on $\phi(x)$ here. Jul 2 '13 at 18:18
• There is a discussion in Zee (page 68, Appendix 2 : Field Redefinition) Jul 3 '13 at 8:49
• @gaugi: I agree with you that this at least somewhat more subtle than the texts imply. I will try to write an (incomplete) answer summarizing what I've figured out. Jul 3 '13 at 14:55
First, the equivalence theorem refers to S-matrix elements rather than off-shell n-point functions, or their generator $Z[j]$, which are generally different. What you have to study is the LSZ formula that gives the relation between S-matrix elements and expectation values of time-ordered product of fields (off-shell n-point functions, what one gets after taking derivatives of $Z[j]$ and setting $j=0$). You will see that even thought these time-ordered products are different, the S-matrix elements are equal just because the residues of these products in the relevant poles are "equal" (they are strictly equal if the matrix elements of the fields between vacuum and one-particle states ( $\langle p|\phi|0\rangle$) are equal, if they are not equal, but both of them are different from zero, one can trivially adapt the LSZ formula to give the same results).
Second, the generating functional
$$Z[j]=\int \mathcal{D}[\phi] \exp{\{iS(\phi)+i\int d^4x\hspace{0.2cm} j(x)\phi(x) \}}$$
is not valid for all actions functionals $S$. I will illustrate this with a quantum-mechanical example—the generalization to quantum field theory is trivial. The key point is to notice that the "fundamental" path integral is the phase-space or Hamiltonian path integral, that is, the path integral before integrating out momenta.
Suppose an action $S[q]=\int L (q, \dot q) \, dt=\int {\dot q^2\over 2}-V(q)\, dt$, then the generating of n-point functions is:
$$Z[j]\sim\int \mathcal{D}[q] \exp{\{iS(q)+i\int dt\hspace{0.2cm} j(t)q(t) \}}$$
The Hamiltonian that is connected with the action above is $H(p,q)={p^2\over 2}+V(q)$ and the phase-space path integral is: $$Z[j]\sim \int \mathcal{D}[q]\mathcal{D}[ p] \exp{\{i\int p\dot q - H(p,q)\;dt+i\int dt\hspace{0.2cm} j(t)q(t) \}}$$ Now, if one performs a change of coordinates $q=x+G(x)$ in the Lagrangian: $$\tilde L(x,\dot x)=L(x+G(x), \dot x(1+G(x)))={1\over 2}\dot x^2 (1+G'(x))^2-V(x+G(x))$$ the Hamiltonian is: $$\tilde H={\tilde p^2\over 2(1+G'(x))}+V\left( x+G(x)\right)$$ where the momentum is $\tilde p={d\tilde L\over d\dot x }=\dot x \; (1+G'(x))^2$. A change of coordinates implies a change in the canonical momentum and the Hamiltonian. And now the phase-space path integral is: $$W[j]\sim \int \mathcal{D}[x]\mathcal{D}[\tilde p] \exp{\{i\int \tilde p\dot x - \tilde H(\tilde p,x)\;dt+i\int dt\hspace{0.2cm} j(t)x(t) \}}\,,$$ as you were probably expecting. However, when one integrates the momentum, one obtains the Langrangian version of the path integral: $$W[j]\sim\int \mathcal{D}[x]\;(1+G'(x)) \exp{\{iS[x+G(x)]+i\int dt\hspace{0.2cm} j(t)x(t) \}}$$ where $(1+G'(x))$ is just $\det {dq\over dx}$. Thus, your second equation is wrong (if one assumes that the starting kinetic term is the standard one) since the previous determinant is missing. This determinant cancels the determinant in your last equation. Nonetheless, $Z[j]\neq W[j]$, since changing the integration variable in the first equation of this answer $$Z[j]\sim\int \mathcal{D}[x]\;(1+G'(x)) \exp{\{iS[x+G(x)]+i\int dt\hspace{0.2cm} j(t)(x(t)+G(x)) \}}$$ which does not agree with $W[j]$ due to the term $j(t)(x(t)+G(x))$. So that, both generating functional of n-point functions are different (but the difference is not the Jacobian), although they give the same S-matrix elements as I wrote in the first paragraph.
Edit: I will clarify the questions in the comments
Let $I=S(\phi)$ be the action functional in Lagrangian form and let's assume that the Lagrangian generating functional is given by $$Z[j]=\int \mathcal{D}[\phi] \exp{\{iS(\phi)+i\int d^4x\hspace{0.2cm} j\phi \}}$$
Obviously, we may change the integration variable $\phi$ without changing the integral. So that, if $\phi\equiv \chi + G(\chi)$, one obtains:
$$Z[j]=\int \mathcal{D}[\chi]\,\det(1+G'(\chi)) \exp{\{iS(\chi +G(\chi))+i\int d^4x\hspace{0.2cm} j(\chi + G(\chi))\}}$$
If we want to use this generating functional in terms of the field variable $\chi$, the determinant is crucial. If we had started with the action $S'(\chi)=S(\chi +G(\chi))=I$ — without knowing the existence of the field variable $\phi$ —, we would had derived the following Lagrangian version of the generating functional: $$Z'[j]=\int \mathcal{D}[\chi]\,\det(1+G'(\chi)) \exp{\{iS'(\chi )+i\int d^4x\hspace{0.2cm}j \chi\}}$$ Note that $Z'[j]\neq Z[j]$ (but $Z[j=0]=Z'[j=0]$) and therefore the off-shell n-point functions are different. If we want to see if these generating functional give rise the same S-matrix elements, we can, as always, perform a change of integration variable without changing the functional integral. Let's make the inverse change, that is, $\chi\equiv\phi+F(\phi)$: $$Z'[j]=\int \mathcal{D}[\phi]\, \det(1+F'(\phi)) \det(1+G'(\chi)) \exp{\{iS'(\phi+F(\phi) )+i\int d^4x\hspace{0.2cm} j(\phi + F(\phi))\}}=\int \mathcal{D}[\phi]\, \exp{\{iS(\phi)+i\int d^4x\hspace{0.2cm} j(\phi + F(\phi))\}}$$
So that, one has to introduce the n-point functions connected with $Z[j]$ and $Z'[j]$ in the LSZ formula and analyze if they give rise to same S-matrix elements, even though they are different n-point functions.
(Related question: Scalar Field Redefinition and Scattering Amplitude)
• My second formula was from the viewpoint, that I have an action $S(\phi)$ for which I do not know, that it is a somehow transformed other action, e.g. for the free case. Then I naively write my generating functional as my second formula, as I only know the action $S(\phi)$. I have transformed nothing there. Why is it wrong then? If I now transform this action I wrote to the free case I get the third formula which coincides with the transformation law you gave, but not with the formula from Itzykson due to the missing determinant. Jul 4 '13 at 12:49
• By the way, I do understand that terms like $j(t)G(x)$ do not contribute in the S-Matrix due to LSZ, as they are contact terms and that I must not compare Green´s functions. Jul 4 '13 at 12:51
• @drake: I believe we are on the same page. I was essentially trying to explain the second paragraph of your answer, which claims that if we were handed the non linear $S$ we would know to right down the det in the measure. I understand this as being the fact that a Lagrangian does not unambiguously define correlators, because contact terms are singular. Only certain prescriptions for these terms will lead to a coherent theory. The det is a manifestation of the fact that our usual prescriptions are not coherent for this lagrangian. Jul 5 '13 at 4:24
• @gaugi: I think the point is that writing simply $W[J] = \int\exp(i\int\mathcal{L} +J\chi)$ gives pathological results when there are derivatives in the interaction. You need to start from the Hamiltonian prescription. So if someone hands you a Lagrangian $\mathcal{L}$ you should get the Hamiltonian, write down the path integral and then integrate out the momenta, which will get you the determinant. Jul 5 '13 at 13:54
• @BebopButUnsteady Thank you! If the Hamiltonian density is $T_{ij}(q)p_ip_j+W_i(q)p_i+V(q)$, then the integral over momenta gives $(\det (T(q)))^{-1/2}$. $T_{ij}$ is often (but not always) a constant and thus it does not have any implication. Jul 5 '13 at 21:41
I) Ref. 1 never mentions explicitly by name the following two ingredients in its proof:
1. The pivotal role of the Lehmann-Symanzik-Zimmermann (LSZ) reduction formula $$\left[ \prod_{i=1}^n \int \! d^4 x_i e^{ip_i\cdot x_i} \right] \left[ \prod_{j=1}^m \int \! d^4 y_j e^{-ik_j\cdot y_i} \right] \langle \Omega | T\left\{ \phi(x_1)\ldots \phi(x_n)\phi(y_1)\ldots \phi(y_m )\right\}|\Omega \rangle$$ $$~\sim~\left[ \prod_{i=1}^n \frac{i\langle \Omega |\phi(0)|\vec{\bf p}_i\rangle }{p_i^2-m^2+i\epsilon}\right] \left[ \prod_{j=1}^m \frac{i\langle \vec{\bf k}_j |\phi(0)|\Omega\rangle }{k_j^2-m^2+i\epsilon}\right] \langle \vec{\bf p}_1 \ldots \vec{\bf p}_n|S|\vec{\bf k}_1 \ldots \vec{\bf k}_m\rangle$$ $$\tag{A} +\text{non-singular terms}$$ $$\text{for each} \quad p_i^0~\to~ E_{\vec{\bf p}_i} , \quad k_j^0~\to~ E_{\vec{\bf k}_j}, \quad i~\in~\{1, \ldots,n\}, \quad j~\in~\{1, \ldots,m\}.$$ [Here we have for simplicity assumed that spacetime is $\mathbb{R}^4$; that interactions take place in a compact spacetime region; that asymptotic states are well-defined; that there is just a single type of scalar bosonic field $\phi$ with physical mass $m$.]
2. That eqs. (9-102), (9-103), (9-104a) and (9-104b) on p.447 are just various versions of the Schwinger-Dyson (SD) equations. [The SD equations can be proved either via integration by part, or equivalently, via an infinitesimal changes in integration variables, in the path integral. The latter method is used in Ref. 1.]
On the middle of p. 447, Ref. 1 refers to a field redefinition $\varphi\to \chi$ as canonical if
[...] the relation $\varphi\to \chi$ may be inverted (as a formal power series).
This is certainly not standard terminology. Also it is a somewhat pointless definition, since any reader would have implicitly assumed without being told that field redefinitions are invertible. Note in particular, that Ref. 1 does not imply a Hamiltonian formulation with the word canonical.
II) The Equivalence Theorem states that the $S$-matrix $\langle \vec{\bf p}_1 \ldots \vec{\bf p}_n|S|\vec{\bf k}_1 \ldots \vec{\bf k}_m\rangle$ [calculated via the LSZ reduction formula (A)] is invariant under local field redefinitions/reparametrizations.
In this answer, we will mainly be interested in displaying the main mechanism behind the Equivalence Theorem at the level of correlation functions (as opposed to carefully tracing the steps of Ref. 1 at the level of the partition function).
In the LSZ formula (A), let us consider an infinitesimal, local field redefinition
$$\tag{B} \phi~ \longrightarrow ~\phi^{\prime}~=~ \phi +\delta \phi$$
without explicit space-time dependences; i.e., the transformation
$$\tag{C} \delta \phi(x)~=~ f\left(\phi(x), \partial\phi(x), \ldots, \partial^N\phi(x)\right)$$
at the spacetime point $x$ depends on the fields (and their spacetime derivatives to a finite order $N$), all evaluated at the same spacetime point $x$. [If $N=0$, the transformation (C) is called ultra-local.]
One may now argue that near the single particle poles, this will only lead to a multiplicative rescaling on both sides of the LSZ formula (A) with the same multiplicative constant, i.e., the $S$-matrix is invariant. This multiplicative rescaling is known as wave function renormalization or as field-strength renormalization in Ref. 3.
III) Finally, let us mention that Vilkovisky devised an approach, where $1$-particle-irreducible (1PI) correlation functions are invariant off-shell under field reparametrizations, cf. Ref. 4.
References:
1. C. Itzykson and J-B. Zuber, QFT, (1985) Section 9.2, p. 447-448.
2. A. Zee, QFT in a Nutshell, 2nd ed. (2010), Chapter 1, Appendix B, p. 68-69. (Hat tip: Trimok.)
3. M.E. Peskin and D.V Schroeder, (1995) An Introduction to QFT, Section 7.2.
4. G.A. Vilkovisky, The Unique Effective Action in QFT, Nucl. Phys. B234 (1984) 125.
• This is an old answer but I had a couple of questions: 1. Aren't field redefinitions involving derivatives non-invertible? 2. Is the assumption that field redefinition be local necessary? Jul 19 '17 at 2:25
• @Qmechanic How to proof your statement "One may now argue that near the single particle poles, this will only lead to a multiplicative rescaling on both sides of the LSZ formula (A) with the same multiplicative constant" ? Oct 10 at 9:10 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 1, "x-ck12": 0, "texerror": 0, "math_score": 0.9534968137741089, "perplexity": 366.9290937236878}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323587719.64/warc/CC-MAIN-20211025154225-20211025184225-00460.warc.gz"} |
https://www.physicsforums.com/threads/hohmann-transfer.266141/ | Hohmann transfer
1. Oct 22, 2008
garyman
A Hohmann transfer orbit is used to send a spacecraft to neptune. However the positions of the other planets were not taken into consideration. The craft approaches Jupiter at an angle of 75 degrees to Jupiter's orbit. Calculate (a) the orbital velocity of Jupiter, (b) the probe's velocity upon reaching Jupiter.
Not really sure how to start this question. Does anyone else have any ideas?
2. Oct 25, 2008
alphysicist
Hi garyman,
I would say the starting point is to calculate the parameters of the elliptical Hohmann orbit using the beginning and ending orbital radii, and using the equations for the velocity impulses needed. However, it's not clear to me what quantities are to be considered as "given". What quantities are you allowed to look up?
Have something to add?
Similar Discussions: Hohmann transfer | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.93502277135849, "perplexity": 680.9390141939799}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-50/segments/1480698541910.39/warc/CC-MAIN-20161202170901-00248-ip-10-31-129-80.ec2.internal.warc.gz"} |
http://whitecraneeducation.com/classrooms/classroom.php?id=7&cid=78&tab=4 | x2
Definitions and Simplifying
( )
Subtracting Rational Expressions
Fundamentally, the process for subtracting two rational expressions follows the same steps as the process for adding only with subtraction in step four. We've found that students often have challenges with the subtraction process so we've included it as a separate section to give us a chance to look at it in more detail.
The process for adding two rational expressions is almost identical to the process for adding two fractions.
1. Reduce all of the fractions. You don't have to do this but it'll make your life easier later on.
2. Find the least common denominator.
3. Make each rational expression into an equivalent expression with the least common denominator.
4. Make a new fraction by subtracting the numerators and keeping the least common denominator.
5. Reduce the fraction from step four.
Example 1
Simplify $$\frac{x - 1}{x - 3}-\frac{x}{x + 4}$$
Both of those fractions are already reduced so we can go right to finding the least common denominator. In this case, that's going to be $(x - 3)(x + 4)$. To get each fraction with the same denominator we have to multiply the numerator and denominator of the first expression by $x + 4$ and the second one by $x - 3$.
$$\frac{(x + 4)(x - 1)}{(x + 4)(x - 3)}-\frac{x(x - 3)}{(x + 4)(x - 3)}$$ $$\frac{x^2 + 3x - 4}{(x + 4)(x - 3)}-\frac{x^2 - 3x}{(x + 4)(x - 3)}$$ $$\frac{x^2 + 3x - 4 - (x^2 - 3x)}{(x + 4)(x - 3)}$$ $$\frac{x^2 + 3x - 4 - x^2 + 3x}{(x + 4)(x - 3)}$$ $$\frac{6x - 4}{(x + 4)(x - 3)}$$
If you aren't clear on how I got the first numerator on the second line, take a look at our page on subtracting polynomials.
The numerator factors to 2(3x - 2). Since there's neither a 2 nor a $3x - 2$ in the denominator, there's nothing we can cancel here. This makes our final answer:
$$\frac{6x - 4}{x^2 + 4x - 12}$$
Example 2
Simplify $$\frac{x^2 + x - 2}{2x^2 + 5x + 2}-\frac{x^2 - 5x - 6}{2x^2 - 7x - 4}$$
First we need to factor all of the polynomials in both expressions.
$$\frac{(x + 2)(x - 1)}{(x + 2)(2x + 1)}-\frac{(x - 6)(x + 1)}{(2x + 1)(x - 4)}$$
Since the numerator and denominator of the first expression both have a $x + 2$ we can reduce the expression by canceling them both.
$$\frac{x - 1}{2x + 1}-\frac{(x - 6)(x + 1)}{(2x + 1)(x - 4)}$$
Now, looking at the factors in the denominator, the least common denominator is $(2x + 1)(x - 4)$. The second expression already has that as its denominator so all we need to do is multiply the numerator and denominator of the first expression by $x - 4$.
$$\frac{(x - 4)(x - 1)}{(x - 4)(2x + 1)}-\frac{(x - 6)(x + 1)}{(2x + 1)(x - 4)}$$ $$\frac{x^2 - 5x + 4}{(x - 4)(2x + 1)}-\frac{x^2 - 5x - 6}{(2x + 1)(x - 4)}$$ $$\frac{x^2 - 5x + 4 - (x^2 - 5x - 6)}{(x - 4)(2x + 1)}$$ $$\frac{x^2 - 5x + 4 - x^2 + 5x + 6}{(x - 4)(2x + 1)}$$ $$\frac{10}{(x - 4)(2x + 1)}$$
The numerator can't be factored any further and there's nothing that goes into 10 evenly in the denominator so that expression is our final answer. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9635194540023804, "perplexity": 267.3182241120769}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-51/segments/1544376824338.6/warc/CC-MAIN-20181213010653-20181213032153-00619.warc.gz"} |
https://en.wikibooks.org/wiki/HSC_Extension_1_and_2_Mathematics/2-Unit/Preliminary/Trigonometric_ratios | # HSC Extension 1 and 2 Mathematics/2-Unit/Preliminary/Trigonometric ratios
Trigonometric ratios deals with sin, cos and tan, which are the three main trigonometric functions. Here we define them in two equivalent ways, exploring regions where they are positive and negative, and various identities (things which are always true) about them.
## Definitions
Here we introduce two definitions of the trigonometric ratios. We present the right-angle triangle definition first, because it is conceptually easier to understand, and is more useful in the geometrical and physical applications. However, whereas the first definition is only applicable for angles between 0° and 90°, the second definition is more general, being valid for all angles, including those greater than 360° and those less than 0°.
### Right-angled triangle
Diagram of a right-angled triangle, with adjacent, opposite, and hypotenuse labeled.
Consider a right-angled triangle like the one shown here. We choose one of corners (not the right-angle) and name the angle there ${\displaystyle \theta }$. Then, we label the sides, according to whether they are opposite (it doesn't touch the angle), the hypotenuse or the other adjacent side. We then define sin, cos and tan to be functions of ${\displaystyle \theta }$ (pronounced and written 'theta') such that
{\displaystyle {\begin{aligned}\sin(\theta )&={\frac {\mbox{Opposite}}{\mbox{Hypotenuse}}}\\\cos(\theta )&={\frac {\mbox{Adjacent}}{\mbox{Hypotenuse}}}\\\tan(\theta )&={\frac {\sin(\theta )}{\cos(\theta )}}={\frac {\mbox{Opposite}}{\mbox{Adjacent}}}\end{aligned}}}
Note that these are functions of ${\displaystyle \theta }$, not just a constant multiplied by ${\displaystyle \theta }$. Also, note that these are used so commonly that we normally omit the parentheses:
${\displaystyle \sin(\theta )=\sin \theta \,}$
and similarly for cos and tan.
#### Limitations of this definition
Since this is a right-angled triangle, and the angle sum of a triangle is 180°, ${\displaystyle \theta }$ may only range from 0° to 90°. To define sin, cos and tan for other ranges, we look to a better definition, as below.
### Unit circle
${\displaystyle x^{2}+y^{2}=1\;}$
The unit circle, radius: 1, center: (0, 0):
The unit circle is a very good way for defining the trigonometric functions. If you make an angle t with the x-axis and the radius, the sine value of that angle is the y-value of the intercept between the radius and the circle, and the cosine value is the x-value of the intercept between the radius of the circle. So for any angle t, the point on the graph where the radius meets the circle has the coordinates (cost, sint)
This is because the radius can form a right-angled triangle with the x-axis with one corner on the origin, the other corner on a point on the graph either above or below the x-axis and right or left of the y-axis, and the right-angled corner somewhere on the x-axis below or above the other corner.
## Trigonometric ratios of: – θ, 90° – θ, 180° ± θ, 360° ± θ.
The relation sin2θ + cos2θ = 1, and those derived from it, should be known, as well as ratios of – θ, 90° – θ, 180° + θ, 360° + θ in terms of the ratios of q. Once familiarity with the trigonometric ratios of angles of any magnitude is attained, some practice in solving simple equations, of the type likely to occur in later applications, should be discussed.
## The exact ratios.
Ratios for 0°, 30°, 45°, 60°, 90° should be known as exact values. The exercises given on this section of work should emphasize the use of the exact ratios.
## Bearings and angles of elevation.
The compass bearing, measured clockwise from the North and given in standard three-figure notation (e.g. 023°) should be treated, as well as common descriptions such as ‘due East’, ‘South–West’, etc. Angles of elevation and depression should both be defined, and their use illustrated.
## Sine and cosine rules for a triangle. Area of a triangle, given two sides and the included angle.
The formulae
${\displaystyle {\frac {a}{\sin A}}={\frac {b}{\sin B}}={\frac {c}{\sin C}}}$
${\displaystyle a^{2}=b^{2}+c^{2}-2bc\cos A\;}$
should be proved for any triangle. The expression for the area, 1/2bc sinA, should also be proved.
In applications of these formulae, systematic ‘solution of triangles’ is not required. (This is the type of exercise where the sizes of (say) two sides and one angle of a triangle are given and the sizes of all other sides and angles must be found). The applications should be a means of fixing the results in the pupil’s mind, and should be restricted to simple twodimensional problems requiring only the above formulae. Attention must be given to interpreting calculator output where obtuse angles are required. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 10, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9466018676757812, "perplexity": 689.9622500703814}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-44/segments/1476988721387.11/warc/CC-MAIN-20161020183841-00242-ip-10-171-6-4.ec2.internal.warc.gz"} |
https://basic2tech.com/physics-and-measurement/ | PHYSICS AND MEASUREMENT
The word Physics originates from the Greek word
Physis, which means nature. Physics in raw terms
is the study of everything around us. Physics is
one of the oldest subjects (unknowing) invented by
humanity. Possibly the oldest discipline in Physics
could be astronomy.
The goals of Physics or Physicist is to express
everyday happenings in a concise mathematical
formula. These formulas are then used by other
Physicist and engineers to predict results of their
experiments. For example Isaac Newton (1642 –
1727) found the laws behind the motion of bodies,
we now use these laws to design rockets that
travel to moon and other planets.
Another major thing that Physicists do is to revise
the laws from time to time depending on
experimental results. Isaac Newton found laws of
motion in the 17th century, these laws worked at
normal speeds, but when a object’s speed is
comparable to that of speed of light, these laws
fails. Albert Einstein (1879 – 1955), put forward the
theory of relativity which gives the same result of
Newton’s laws of motion at slow speeds and far
accurate results to speeds that go up to the speed
of light.
Definition: “MEASUREMENT”is the determination of
the size or magnitude of something “Or” The
comparison of unknown quantity with some
standard quantity of the same rates is known as
measurement
Measurement
Measurement is integral part of Physics like any
other scientific subject. Measurement is a integral
part of human race, without it there will be no
trade, no statistics. You can see the philosophy of
measurement in little kids who don’t even know
what math is. Kids try to compare their height,
size of candy, size of dolls and amount of toys
they have. All these happen even before they know
math. Math is built into our brains even before we
start to learn it.Math provides a great way to study
about anything, that’s why we see computers
involved in almost anything because they are good
at math.
Scale
Scales are used to measure. One would know a
simple ruler or tape could be used to measure
small distances, your height and possibly much
more in Physics we do have certain scales for
certain quantities which we would see very shortly.
Length, Mass and Time
The current system of units has three standard
units: The meter, kilogram, and second. These
three units form the mks-system or the metric
system .
A meter is a unit of length, currently defined as the
distance light travels within 1/299782458th of a
second.
A kilogram is a unit of mass. While it was
previously defined as a specific volume of water
(e.g. 1 Liter or a 10cm^3 cube), it’s current
definition is based on a prototype platinum-iridium
cylinder.
A second is a unit of time. Originally defined as
the amount of time the earth needs to make
1/86400 of a rotation, it is now defined as
9192631770 oscillations of a Cesium-133 atom.
Dimensional and Unit Analysis
Dimensional analysis to determine if an equation is
dimensionally correct. When you are presented
with an equation, dimensional analysis is
performed by stripping the numerical components
and leaving only the unit types (such as Length,
Mass, or Time). It may also be used to determine
the type of unit used for an unknown variable. For
example, the force of gravity may appear as the
following:
It gets converted to the following:
and as such, the unit of force involves multiplying
length and mass, and dividing by the square of the
time.
Unit analysis is similar to dimensional analysis,
except that it uses units instead of the basic
dimensions. The same principle applies; the
numbers are removed, and the units are verified to
be equal on both sides of the equation.
Density Formula
The formula for density is Density Formula
d = density m = mass v = volume
Density
Density is the amount of mass per volume. The
quantity of mass per unit volume of a
substance.The density, or more precisely, the
volumetric mass density, of a substance is its
mass per unit volume. The symbol most often
used for density is ρ (the lower case Greek letter
rho). Mathematically, density is defined as mass
divided by volume:[1]
\rho = \frac{m}{V},
where ρ is the density, m is the mass, and V is the
volume. In some cases (for instance, in the United
States oil and gas industry), density is loosely
defined as its weight per unit volume,[2] although
this is scientifically inaccurate – this quantity is
more specifically called specific weight.
For a pure substance the density has the same
numerical value as its mass concentration.
Different materials usually have different densities,
and density may be relevant to buoyancy, purity
and packaging. Osmium and iridium are the
densest known elements at standard conditions for
temperature and pressure but certain chemical
compounds may be denser.
To simplify comparisons of density across different
systems of units, it is sometimes replaced by the
dimensionless quantity “relative density” or
“specific gravity”, i.e. the ratio of the density of the
material to that of a standard material, usually
water. Thus a relative density less than one means
that the substance floats in water.
The density of a material varies with temperature
and pressure. This variation is typically small for
solids and liquids but much greater for gases.
Increasing the pressure on an object decreases the
volume of the object and thus increases its
density. Increasing the temperature of a substance
(with a few exceptions) decreases its density by
increasing its volume. In most materials, heating
the bottom of a fluid results in convection of the
heat from the bottom to the top, due to the
decrease in the density of the heated fluid. This
causes it to rise relative to more dense unheated
material.
The reciprocal of the density of a substance is
occasionally called its specific volume, a term
sometimes used in thermodynamics. Density is an
intensive property in that increasing the amount of
a substance does not increase its density; rather it
increases its mass.
Conversion of Units
How many kilometers are in 20 miles? To find out,
you will have to convert the miles into kilometers.
A conversion factor is a ratio between two
compatible units.
You may also see conversion factors between
weight (e.g. pounds) and mass (e.g. kilograms).
These factors rely on equivalence (e.g. 1 kilogram
is “close enough” to 2.2 pounds) based on
external factors. While that cannot apply in all
situations, these factors may be used in some
limited scopes.
Estimates and Order-of-Magnitude calculation
The order of magnitude gives the approximate idea
of the powers of 10 .Any number in the form a*10b
[ here a multiplied by 10.. And 10raised to the
power b]if a >or = (10)^0.5 the a become 1 and b
is not changed but when a>(10)^0.5 then a is
taken as 10 so power of b increases by 1.
Significant Figures
A significant figure is a digit within a number that
is expected to be accurate. In contrast, a doubtful
figure is a digit that might not be correct.
Significant figures are relevant in measured
numbers, general estimates or rounded numbers.
As a general rule, any non-zero digit shown is a
significant figure. Zeros that appear after the
decimal point and are at the end of the number are
also significant. Zeros at the end of the number but
before the decimal point are not included as
significant figures (although exceptions may
occur.)
In general, an operation performed on two
numbers will result in a new number. This new
number should have the same number of
significant digits as the least accurate number. If
an exact number is used, it should have the same
number of digits as the estimated number. If both
numbers are exact, the new number should be
calculated fully (within reason).
When doing calculations, you should only keep at
most 1 doubtful digit; while it is acceptable to
keep them when using a handheld calculator or
reflect the correct number of significant digits.
Other units
The current metric system also includes the
following units:
An ampere (A) is a measure for electric current.
A kelvin (K) is a measure for temperature.
A mole (mol) is the amount of substance (based
on number of atoms rather than mass.)
A candela (cd) is a measure for luminous
intensity.
The Lumen (lm) is a measure unit for total
amount light visible for the human eye emitted
by a source.
The lux (lx) is a measure unit for luminous flux
per unit area. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8800451755523682, "perplexity": 3318.702714018063}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764499801.40/warc/CC-MAIN-20230130034805-20230130064805-00465.warc.gz"} |
http://blog.computationalcomplexity.org/2005/07/majority-is-stablest.html | ## Wednesday, July 27, 2005
### Majority is Stablest
Consider the following two voting schemes to elect a single candidate.
1. Majority Vote.
2. A Majority of Majorities (think an electoral college system with states of equal size).
Which of these voting systems are more stable, i.e., less likely to be affected by flipping a small number of votes?
In an upcoming FOCS paper, Elchanan Mossel, Ryan O'Donnell and Krzysztof Oleszkiewicz prove the "Majority is Stablest" conjecture that answers the above question and in fact shows that majority is the most stable function among balanced Boolean functions where each input has low influence. To understand this result we'll need to define the terms in the statement of the theorem.
• Balanced: A Boolean function is balanced if it has the same number of inputs mapping to zero as mapping to one.
• The influence of the ith variable is the expectation over a random input of the variance of setting the ith bit of the input randomly. The conjecture requires the influence of each variable to be bounded by a small constant.
• Stability: The noise stability of f is the expectation of f(x)f(y) where x and y are chosen independently.
The majority is stablest conjecture has applications for approximation via the unique games conjecture.
1. I believe stability is defined with a parameter \epsilon
as E[f(x)f(y)] where for each i,
with probability 1-\epsilon y_i=x_i and otherwise y_i is chosen uniformly.
2. Also, it is an asymptotic theorem--the influences have to go to zero for it to say something interesting.
3. I keep seeing results about one-time voting systems, including instant-runoff voting or approval voting, but how about results where parties get to vote in multiple elections? (For example, real runoff elections.)
4. http://www.econ.boun.edu.tr/papers/pdf/wp-98-01.pdf
A Degree and an Efficiency of Manipulation of Known Social Choice Rules (Fuad Aleskerov, Eldeniz Kurbanov)
This paper compares some 24 voting methods.
fyi...
5. This result is related to some work in the study of voting power in political science; see here for some discussion and here for a discussion of various voting models and their interpretation. A key issue turns out to be modeling the correlations of votes for people who are near each other. (As far as I know, the results in pure math and cs assume independent voters.) | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8751825094223022, "perplexity": 1148.6586922039144}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-10/segments/1393999677213/warc/CC-MAIN-20140305060757-00070-ip-10-183-142-35.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/the-bar-is-in-static-equilibrium-w-15n-na-11n.431786/ | # Homework Help: The bar is in static equilibrium W = 15N, Na = 11N
1. Sep 24, 2010
### indiangeek
1. The problem statement, all variables and given/known data
The bar is in static equilibrium W = 15N, Na = 11N and Nb which is perpendicular to the bar,Nb = 5.66N.What is the magnitude of tension forceT?
2. Relevant equations
3. The attempt at a solution
W - Na - Nb*sin(theta) = 0;
15 - 11 - 5.66*sin(theta) = 0
theta = 45 deg
Nb*cos(theta) = T
T = 5.66*cos(45)
T = 4 N
But the options given are a)17.79N b)25N c)12N d)21N e)0.9 m^2
solve this problem,,,,,,,,,,,,,
#### Attached Files:
• ###### untitled.bmp
File size:
162.8 KB
Views:
134
2. Sep 24, 2010
### PhanthomJay
Re: statics
3. Sep 25, 2010
### pongo38
Re: statics
This is not a well-defined problem because distances have not been given. You have used the law of equilibrium for equating forces in two directions, but you have not used the principle of sum of moments is zero. Wht happens if you try to check your answer wiith T=4? That you cannot do without distances (or at least the ratio of the distances on the rod)
4. Sep 25, 2010
### PhanthomJay
Re: statics
The bar is at a 45 degree angle. Although the distances are not given, W must be located a certain fraction of the bar length from one end, about half way up ,based on the values given.
5. Sep 27, 2010
### pongo38
Re: statics
I would encourage indiangeek to check statics problems using all the equations of equilibrium available. You might be able to determine the exact location of Nb (in relation to W and Na by using the principle of moments. Having done that, the alleged solution T=4 can be checked. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8098949193954468, "perplexity": 2172.427912573879}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-39/segments/1537267158958.72/warc/CC-MAIN-20180923020407-20180923040807-00328.warc.gz"} |
http://mathhelpforum.com/calculus/212338-equivalence-norms.html | ## Equivalence of Norms
I'm somewhat stuck with the simple proof of the following:
Let the (Lebesgue-)measure of some $\displaystyle \Omega$ be finite and $\displaystyle 1 \le p \le q \le \infty$.
Then for all $\displaystyle u \in L^q(\Omega)$ it is also true that $\displaystyle u \in L^p(\Omega)$, whereby
$\displaystyle ||u||_p \le \text{meas}(\Omega)^{\frac{1}{p}-\frac{1}{q}}||u||_q$;
for $\displaystyle q=\infty$ set $\displaystyle \frac{1}{q}:=0$.
Proof: If $\displaystyle q=\infty$, then $\displaystyle ||u||^p_p=\int_{\Omega}|u(x)|^p\,dx \le \text{meas}(\Omega) \sup_{\Omega \setminus N}|u|^p$, where N ist a large enough null set.
If $\displaystyle q<\infty$, then the Hölder inequality should help, but I'm somewhat confused by the suitable choice of exponents. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9782485961914062, "perplexity": 366.5674595256455}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-43/segments/1539583511761.78/warc/CC-MAIN-20181018084742-20181018110242-00299.warc.gz"} |
http://mathhelpforum.com/geometry/63767-width-height-solid-print.html | # Width and Height of Solid
• December 7th 2008, 10:09 AM
magentarita
Width and Height of Solid
The length of a rectangular solid is 7. The width of the solid is 2 more than the height. The volume of the solid is 105. Find the width and the height of the solid.
• December 7th 2008, 10:33 AM
Moo
Hello,
Quote:
Originally Posted by magentarita
The length of a rectangular solid is 7. The width of the solid is 2 more than the height. The volume of the solid is 105. Find the width and the height of the solid.
Let l, w, h be respectively the length, the width and the height of the solid.
We know that the volume of such a shape is defined as being : V=h*l*w
We know that w=2h ("the width is 2 more than the height)
So $105=7*2h*h=14h^2$
Hence $h^2=\frac{105}{14}=\frac{15}{2}$
this is the height. Multiply by 2 to get the width
• December 9th 2008, 09:51 PM
magentarita
ok....
Quote:
Originally Posted by Moo
Hello,
Let l, w, h be respectively the length, the width and the height of the solid.
We know that the volume of such a shape is defined as being : V=h*l*w
We know that w=2h ("the width is 2 more than the height)
So $105=7*2h*h=14h^2$
Hence $h^2=\frac{105}{14}=\frac{15}{2}$
this is the height. Multiply by 2 to get the width | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 4, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9512919783592224, "perplexity": 524.5824481673943}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2013-48/segments/1386164848402/warc/CC-MAIN-20131204134728-00075-ip-10-33-133-15.ec2.internal.warc.gz"} |
https://infoscience.epfl.ch/record/149757 | Infoscience
Journal article
# An experimental investigation of laminar-turbulent transition in complex fluids
An experimental study of laminar-turbulent transition in complex diluted fruit purees flows in circular ducts, is presented in this work. Data measured using a rectilinear pipe viscosimeter are analyzed to single out useful critical values of both the wall shear stress and the flow rate at which transition-to-turbulence occurs for a given dilution degree. A comparison between the Dodge-Metzner-Reed methods and the classical Mishra & Tripathi and Hanks correlations to estimate the critical generalized Reynolds number, is also discussed. The emerging discrepancies can reasonably be attributed to viscoelastic effects, which probably become important near the point where transition-to-turbulence occurs. This analysis therefore has important practical implications for the prevention of the breakdown of the structure due to turbulent mechanical stresses [J. Food Engng. 52 (2002) 397]. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9490177631378174, "perplexity": 2023.1027097034241}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-17/segments/1492917127681.84/warc/CC-MAIN-20170423031207-00404-ip-10-145-167-34.ec2.internal.warc.gz"} |
https://pure.uai.cl/en/publications/measurement-of-quarkonium-production-in-protonlead-and-protonprot | # Measurement of quarkonium production in proton–lead and proton–proton collisions at 5.02TeV with the ATLAS detector
ATLAS Collaboration
Research output: Contribution to journalArticlepeer-review
77 Scopus citations
## Abstract
The modification of the production of J/ ψ, ψ(2 S) , and Υ(nS) (n= 1 , 2 , 3) in p+Pb collisions with respect to their production in pp collisions has been studied. The p+Pb and pp datasets used in this paper correspond to integrated luminosities of 28nb-1 and 25pb-1 respectively, collected in 2013 and 2015 by the ATLAS detector at the LHC, both at a centre-of-mass energy per nucleon pair of 5.02 TeV. The quarkonium states are reconstructed in the dimuon decay channel. The yields of J/ ψ and ψ(2 S) are separated into prompt and non-prompt sources. The measured quarkonium differential cross sections are presented as a function of rapidity and transverse momentum, as is the nuclear modification factor, Rp Pb for J/ ψ and Υ(nS). No significant modification of the J/ ψ production is observed while Υ(nS) production is found to be suppressed at low transverse momentum in p+Pb collisions relative to pp collisions. The production of excited charmonium and bottomonium states is found to be suppressed relative to that of the ground states in central p+Pb collisions.
Original language English 171 European Physical Journal C 78 3 https://doi.org/10.1140/epjc/s10052-018-5624-4 Published - 1 Mar 2018
## Fingerprint
Dive into the research topics of 'Measurement of quarkonium production in proton–lead and proton–proton collisions at 5.02TeV with the ATLAS detector'. Together they form a unique fingerprint. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9980831146240234, "perplexity": 4616.956327183187}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446710771.39/warc/CC-MAIN-20221130192708-20221130222708-00317.warc.gz"} |
http://physicshelpforum.com/special-general-relativity/15119-acceleration-contradiction.html | Special and General Relativity Special and General Relativity Physics Help Forum
Feb 23rd 2019, 08:24 PM #1 Junior Member Join Date: Feb 2019 Posts: 7 Acceleration contradiction? Let there be three inertial frames, F0, F1 and F2. F1 is moving in the negative x direction relative to F0 with speed V = sqrt(3)/2 * c. F2 is moving in the positive x direction relative to F0 with speed V = sqrt(3)/2 * c. In frame F1 there are two BB pellets at rest spaced 10 meters apart as measured in F1. F0, per Einstein, measures the separation between the two BB pellets to be 5 meters. At time t0 in frame F0, frame F0 simultaneously starts both BB pellets accelerating in the positive x direction. The two BB pellets each accelerate in the identical pattern until they have zero velocity relative to frame F2 at which time F0 simultaneously stops the accelerations. An observer travels with one of the BB pellets accelerating identically as the adjacent BB pellet accelerates. That observer measures that the two BB pellets were initially 10 meters apart. When the BB pellets and the traveling observer have zero velocity with respect to frame F0, the traveling observer measures the BB pellets to be only 5 meters apart, just as every other observer in frame F0 does. When the BB pellets and traveling observer reach frame F2 and stop accelerating, the traveling observer once again measures the separation between the two BB pellets to be 10 meters. Since the acceleration is always in the positive x direction, and since the pattern is identical for the two BB pellets (we can say make the acceleration constant as measured by the traveling observer), why does the accelerating observer say that the identical acceleration in the positive direction sometime causes the two BB pellets to move toward each other (the first part of the journey) and sometimes causes the two BB pellets to move away from each other (the second part of the journey)? Thanks David Seppala Bastrop TX
Feb 25th 2019, 06:47 AM #2 Senior Member Join Date: Jun 2016 Location: England Posts: 845 You are hopping between reference frames in a carefree manner! One always has to be very careful when moving between reference frames that all relevant terms are considered. I haven't had time (yet) to trace your arguments slowly and carefully through each of your jumps, so I can't comment (yet) on the correctness of your conclusions. __________________ ~\o/~
Feb 26th 2019, 04:45 AM #3 Senior Member Join Date: Jun 2016 Location: England Posts: 845 It was a little confusing because you have a miss-typing in your post: The pellets start accelerating from Frame F1 (not F0 as your post seems to indicate). The Pellets are now in their own (accelerating) frame (FP). At all times the observer in Frame FP will observe the pellets to be 10m apart. But observers in other frames will observe the distance between the pellets to change as their velocities relative to FP change. __________________ ~\o/~
Feb 26th 2019, 05:16 AM #4 Junior Member Join Date: Feb 2019 Posts: 7 Woody wrote: "The Pellets are now in their own (accelerating) frame (FP). At all times the observer in Frame FP will observe the pellets to be 10m apart." No, your statement is incorrect. When the accelerating frame FP has zero velocity with respect to frame F0 they measure the same distance between the BB pellets as all other observers in frame F0 do. All F0 observers measure the distance between the BB pellets to be 5 meters from the start to the finish of the acceleration since frame F0 measures the initial distance between the BB pellets to be 5 meters (per Einstein), and the acceleration of each BB pellet started simultaneously from frame F0 point of view. David Seppala Bastrop TX
Feb 26th 2019, 06:04 AM #5
Senior Member
Join Date: Jun 2016
Location: England
Posts: 845
When the accelerating frame FP has zero velocity with respect to frame F0 they measure the same distance between the BB pellets as all other observers in frame F0 do.
Agreed.
All F0 observers measure the distance between the BB pellets to be 5 meters from the start to the finish of the acceleration since frame F0 measures the initial distance between the BB pellets to be 5 meters (per Einstein), and the acceleration of each BB pellet started simultaneously from frame F0 point of view.
Wrong.
The whole point of the reference frames is that all objects (and observers) in that frame are moving at the same speed.
Any observations between two (or more) frames (moving a different speeds)
will produce differences to the same observations taken within a single frame.
The pellets (FP) are initially in Frame F1 but are accelerating through Frame F0 (and eventually on to Frame F2).
When they are in Frame F1, observers in F1 will see them 10m apart,
observers in F0 will see them as being 5m apart.
When the pellets accelerate enough to reach Frame F0,
observers in F0 will see them as being 10m apart,
but the observers in F1 will now see them as being 5m apart!
The observers in FP will always see them as 10m apart.
__________________
~\o/~
Last edited by Woody; Feb 26th 2019 at 06:11 AM.
Feb 26th 2019, 07:23 AM #6 Junior Member Join Date: Feb 2019 Posts: 7 As stated in the original post, F0 measures the distance between the pellets to be 5 meters when they are in F1. F0 starts the acceleration of each pellet simultaneously, and each pellet accelerates in the identical pattern. Therefore, F0 always measures the distance between the two pellets to be 5 meters through out the entire journey. With identical accelerations and with each acceleration starting simultaneously as measured in frame F0, the distance between pellets as measured in F0 never changes. David Seppala Bastrop TX
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Similar Physics Forum Discussions Thread Thread Starter Forum Replies Last Post magnetismman Advanced Electricity and Magnetism 2 Apr 8th 2014 01:42 PM ginarific Periodic and Circular Motion 1 Jan 31st 2010 07:43 AM | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8392706513404846, "perplexity": 2157.3825336642494}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-22/segments/1558232257699.34/warc/CC-MAIN-20190524164533-20190524190533-00318.warc.gz"} |
http://mathhelpforum.com/calculus/166822-epsilon-proof-x-infinity-currently-defeating-me-print.html | # Epsilon-proof as x->-infinity currently defeating me
• Dec 23rd 2010, 04:31 PM
Grep
Epsilon-proof as x->-infinity currently defeating me
I think Epsilon-delta proofs and I should spend some quality time together.
The problem is I need to prove that:
$\lim_{x \to -\infty} a^x = 0$
So, I need to show that, for every $\epsilon > 0$, there is a $\delta$ such that $|f(x) - L| < \epsilon$ whenever $x < \delta$.
So, given $\epsilon > 0$, I need to find $\delta$ such that:
$|a^x - 0| < \epsilon$ whenever $x < \delta$
And here's where I am stuck. I think I'll have to take a log in there somewhere. Wish my textbook wasn't so scarce on examples (Stewart's Calculus 4th edition). I sort of get limits where x->c, but I'm totally lost on the ones with infinities, and the differences between handling + and - infinities. Can't find any examples online of ones as x-> -infinity, which surely might help. Nor are the two videos at Khan Academy useful in this case.
I bet you guys get a lot of these. Surprisingly confusing for what seems a relatively simple concept, at first. Help, hints, pointers, nudges and taunts all appreciated.
• Dec 23rd 2010, 04:40 PM
Ackbeet
If you think of $\delta$ as a large negative number, and getting more and more negative, you'll be on the right track; also, you have to assume that $a>0,$ right? Otherwise, you have complex numbers floating around.
How can you simplify $|a^{x}-0|<\epsilon?$
If you can find a $\delta=\delta(\epsilon)$ that works, you'll be done. How could you do that?
• Dec 23rd 2010, 05:18 PM
Grep
Quote:
Originally Posted by Ackbeet
If you think of $\delta$ as a large negative number, and getting more and more negative, you'll be on the right track; also, you have to assume that $a>0,$ right? Otherwise, you have complex numbers floating around.
How can you simplify $|a^{x}-0|<\epsilon?$
Well, first and obviously, $|a^{x} - 0| < \epsilon \Rightarrow |a^{x}| < \epsilon$.
Then, I can assume that $a^{x} > 0$ so I have just:
$a^{x} < \epsilon$ whenever $x < \epsilon$.
Quote:
Originally Posted by Ackbeet
If you can find a $\delta=\delta(\epsilon)$ that works, you'll be done. How could you do that?
We want:
$a^{x} < \epsilon$ whenever $x < \epsilon$
Or:
$log_{a}(a^{x}) = log_{a}(\epsilon) \Rightarrow x = log_{a}(\epsilon)$
So I should try $\delta = log_{a}(\epsilon)$.
Given $\epsilon > 0$, we choose $\delta = log_{a}(\epsilon)$. Let $x < \epsilon$. Then
$|a^{x} - 0| = a^{x} < a^{\delta} = a^{log_{a}(\epsilon)} = \epsilon$
Thus
$|a^{x} - 0| < \epsilon$
Is that right? Egads, I think I just did it. Hey, I think I smell burnt toast! (Wink)
• Dec 23rd 2010, 07:32 PM
chisigma
Quote:
Originally Posted by Grep
I think Epsilon-delta proofs and I should spend some quality time together.
The problem is I need to prove that:
$\lim_{x \to -\infty} a^x = 0$
...
... that's true if $a >1$... of course...
http://digilander.libero.it/luposaba...ato[1].jpg
Merry Christmas from Italy
$\chi$ $\sigma$
• Dec 23rd 2010, 07:45 PM
Grep
Quote:
Originally Posted by chisigma
... that's true if $a >1$... of course...
Good point, the problem description says "Let a > 1", which I should have stated. Apologies for leaving out an important part of the problem description. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 35, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.958623468875885, "perplexity": 975.0117672990782}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-40/segments/1474738660882.14/warc/CC-MAIN-20160924173740-00070-ip-10-143-35-109.ec2.internal.warc.gz"} |
https://feynmand.com/ml-vader/ | Want create site? Find Free WordPress Themes and plugins.
Let’s see how?
Let’s pick up from where we last left and continue our journey into the realms of Supervised Learning Algorithms – ‘The One Labels’. We will discuss our very first algorithm, called, Linear Regression.
Remember this:
[i]
Consider you are a Commander on the Death Star and you have just received Darth Vader’s command to present him a working system that allows him to choose the optimal battleship for defeating the Rebellion. Everyone else at work, you decide to take it upon yourself to develop this system.
So, let get going!
Machine Learning is all based on data. So, consider the following data set:
Size of Darth Vader’s Ship(sq. m.) Rounds of Lasers 2500 500 1695 256 2400 450 1500 240 3000 720 2650 520 . . . . . .
The above data can be plotted as follows:
The question: Given data like this, how can we learn to predict the number of laser rounds Darth Vader can fire, as a function of the size of a ship?
Before we tread any further, we need to establish some notations as the math is about to get messy. We’ll use x(i) to denote the “input” variables (Size of Vader’s ship in this example), also called input features, and y(i) to denote the “output” or target variable that we are trying to predict (laser rounds). A pair (x(i), y(i)) is called a training example, and the dataset that we’ll be using to learn —a list of m training examples {(x(i), y(i)); i = 1, . . . ,m}—is called a training set. Note that the superscript “(i)” in the notation is simply an index into the training set, and has nothing to do with exponentiation. We will also use X to denote the space of input values (ship size), and Y the space of output values (laser count). In this example, X = Y = R (Real Numbers).
A general overview of how a Supervised Learning Algorithm works can be drawn from this image:
[ii]
Here, training data is the data we have. New data refers to different sizes of ships that can be entered into the trained model and Prediction is the estimated Laser Rounds the ship can carry.
When the target variable(y) that we’re trying to predict is continuous, such as in our example, we call the learning problem a “regression problem”. When y can take on only a small number of discrete values (such as if, given the size of ship, we wanted to predict if it is a Jet or a Battle Cruiser), we call it a classification problem.
# How does it Work?
A regression algorithm fits data by drawing out features from available training example (in our case, the size of ship). It can have more features such as the Fuel Tank size, canon size and many more. It is up to the programmer/researcher/data scientist to choose the number of features.
To perform supervised learning, we must decide how we’re going to represent functions/hypotheses h in a computer. As an initial choice, let us say we decide to approximate y as a linear function of x (keep track of what x and y denote here):
$$h_\theta(x) = \theta_0+\theta_1x_1+\theta_2x_2$$ ..(1)
Here, the θi’s are the parameters (also called weights/features) parameterizing the space of linear functions mapping from X to Y. The hypothesis above takes three features into consideration. To simplify our notation, we also introduce the convention of letting x0 = 1 (this is the intercept term), so that the hypothesis can be written as:
$$h_\theta(x) = \sum_{i=0}^n \theta_ix_1 = \theta^Tx$$ ..(2)
Above, equation (1) is condensed into a shorter form. The right-hand part of eq.1 can easily be written as a summation of θ and x. Now you might wonder how the summation gets converted to matrices in equation 2. Here it is: if we put all the values of θ and x in two individual vectors (vectors are 1-dimensional matrices of order 1n or n1) and use transpose (T in eq.2 stands for transpose) on one of them, the product of the two will be equal to right-hand side of eq.1(If you have doubts, pick up a pencil and a paper and see for yourself).
Now, our task is to, given the training set, learn the parameters θ that contribute to the size of ship. On proper thought (go on; use those brains) this can be achieved if the hypothesis computed by out algorithm are close to y. So, we need to calculate
For each set of features, how close is the hypothesis to y. Hence, we define, another equation (don’t worry, it’s the last one for today), the Cost Function:
$$j(\theta)=\frac{1}{2}\sum_{i=0}^n(h_\theta(x^{(i)})-y^{(i)})$$ ..(3)
To those of you wondering what gibberish this is, this equation is called the Ordinary Least-Square cost function and to the brainiacs out there who are laying waste on their scalp over why not simply use the difference between h and y, the squared function is used so that the difference is always positive. And no, I didn’t just put it here to confuse you. It is all through a very natural and intuitive process that this equation comes into play (the grand design is at work). We can discuss more on this later.
Let’s move on! The boss fight where we get to know how this algorithm improves its outputs.
We need to find θ that minimizes the cost function (the lower the difference between h and y, the better the prediction). For this, we need to consider the Gradient Descent algorithm which is just a simple update for θ :
$$\theta_k=\theta_k-\alpha\frac{\partial}{\partial \theta}j(\theta)$$ ..(4)
This equation optimizes θ for all the values from (1, 2….k). k is the order of the feature vector θ and signifies learning rate (another one of those algorithm you don’t need to care about.) The algorithm is very natural in the sense that it repeatedly takes its steps in the direction of steepest descent, i.e., the direction in which θ the most unless the value for is so large that values overshoot and the function doesn’t find its minima or it is so small that it take very long to converge. You must avoid both of these conditions.
Upon solving the partial differentiation part in eq. 4, we get which when plugged into eq.4 gives up our final update rule:
Repeat until convergence: {
$$\theta_k=\theta_k+\alpha\sum_{i=0}^n(y^{(i)}-h_\theta(x^{(i)}))x_j^{(i)}$$
(for every j) }
The process is repeated n times for every j. For every value of j, the algorithm looks though all of the training data and then makes an update. This process is called the Batch Gradient Descent.
Another instance of this algorithm that works as follows:
for i=1 to n:
$$\theta_k=\theta_k+\alpha(y^{(i)}-h_\theta(x^{(i)}))x_j^{(i)}$$
(for every j)
It is called the Stochastic Gradient Descent. This process, too, works fine and if you have large data set to work on, this would be a better choice over Batch Gradient Descent due to its less convergence time.
After all this hard work, what is the outcome:
We find the values for θ0 = 90.30, θ1 = 0.1592, θ2 = −6.738 and the plot comes out as:
Finally, Darth Vader has a system that can tell him which ship size to choose to defeat Luke!
[iii]
Applications:
The applications of Curve-fitting are ubiquitous. From optimizing biological parameters to astronomy and from the stock market analysis to weather data analysis, Linear Regression is widely to estimate results.
I think this should satiate you for the day. Use the comments section for any queries. Until next time, this is Pratyush signing off.
Keep Hacking!
References:
[i] Image Source: http://starwars.wikia.com/wiki/Death_Star
[ii] Image Source:http://sebastianraschka.com/Articles/2014_intro_supervised_learning.html
[iii] Image Source: http://www.starwarswavelength.com/category/dark-side-thoughts/
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### Posted by Pratyush Kumar
I am a homo sapien currently residing on planet Earth. I strive to understand things clearly and help others do the same.
### One Comment
1. […] and calculated the gradients [remember how we calculated gradients for the Cost Function in the previous article] we now need to update the parameters in such a way that the model accurately maps all the […] | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8430355191230774, "perplexity": 842.6386848838557}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-51/segments/1512948581033.57/warc/CC-MAIN-20171216010725-20171216032725-00425.warc.gz"} |
http://math.stackexchange.com/questions/292376/verify-that-a-random-variable-is-a-stopping-time | # Verify that a random variable is a stopping time
Let $\lbrace X_{n}\rbrace$ be a stochastic process adapted to filtration $\lbrace \mathcal{F}_{n}\rbrace$. Let $B\subset \mathbb{R}$ be closed. Then $$\tau(\omega):=\mathtt{inf}\lbrace n\in\mathbb{N}:X_{n}(\omega)\in B\rbrace$$ is a stopping time.
My solution. Fix $k\in\mathbb{N}$. Then $$\lbrace\tau=k\rbrace=\lbrace\omega:X_{k}(\omega)\in B, X_{n}(\omega)\notin B,n<k\rbrace=\lbrace \omega:X_{k}(\omega)\in B\rbrace\cap\bigcap_{n=0}^{k}\lbrace\omega:X_{n}(\omega)\notin B\rbrace$$ Now, $X_{k}$ is $\mathcal{F}_{k}$-measurable, $B$ is a Borel set, so $X_{k}^{-1}(B)\in \mathcal{F}_{k}$. Since $B$ closed, the complement of $B$ is open, so also Borel, and for $n<k$ we have, by the same logic, $X_{n}^{-1}(B^{c})\in\mathcal{F}_{n}\subset \mathcal{F}_{k}$. The intersection of finitely many elements of $\mathcal{F}_{k}$ lies in $\mathcal{F}_{k}$, so for any natural $k$ it's true that $\lbrace\tau=k\rbrace\in\mathcal{F}_{k}$. Now, is this solution correct? Is it necessary to assume the closedeness of $B$? I have a feeling it's sufficient to assume that $B$ be Borel, isn't it?
-
Your solution is good, though your intersection should end at index $k-1$, not $k$. – Byron Schmuland Feb 1 '13 at 22:13
As ByronSchmuland already mentioned your intersection should end at index $k-1$, i.e.
$$\{\tau=k\} = \ldots = \{\omega; X_k(\omega) \in B\} \cap \bigcap_{n=0}^{k-1} \{\omega; X_n(\omega) \notin B\}$$
The closedness of $B$ is not necessary for the given proof. Since $B$ is a Borel set we also have that $B^c$ is a Borel set (it's a sigma-algebra!), hence $X_n^{-1}(B^c) \in \mathcal{F}_n$ by the $\mathcal{F}_n$-measurability of $X_n$.
If you consider a stochastic process $(X_t)_{t \geq 0}$, it wouldn't be that easy to prove that $\tau$ is a stopping time - in this case one often assumes that $B$ is open (resp. closed) and left- or right-continuity of the paths $t \mapsto X_t(\omega)$. But in this case it work's fine, because it's a process in discrete time. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9711276292800903, "perplexity": 121.9945153892792}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-23/segments/1405997892648.47/warc/CC-MAIN-20140722025812-00016-ip-10-33-131-23.ec2.internal.warc.gz"} |
https://figshare.shef.ac.uk/articles/dispersion_m_-_A_MatLab_script_for_phase_angle_and_amplitude_correction_of_pressure_bar_signals/3996876/1 | ## dispersion.m - A MatLab script for phase angle and amplitude correction of pressure bar signals
2016-10-12T12:04:38Z (GMT)
In processing the signals from split Hopkinson pressure bar (SHPB) experiments it is often assumed that longitudinal stress waves in the pressure bars propagate one-dimensionally at a common velocity c_0, and so measurements taken at the strain gauges are often simply translated to the end of the bar using a suitable time delay. In reality, stress waves propagate at a specific phase velocity, c_p, which is a function of frequency and the bar's diameter, one-dimensional wave speed and Poisson's ratio. Phase velocity decreases as the frequency of a wave increases, leading to dispersion of a signal as it propagates down the bar. Dispersion of the stress pulse is accompanied by a frequency-dependent variation in stress and strain across the bar cross-section, so that a signal recorded on the surface of the bar at some distance from the specimen will not accurately describe the stresses the specimen was subjected to, and hence cannot be used to accurately determine the specimen response.<p>This script uses an implementation of the dispersion-correction method described by Tyas and Pope (2005) to ensure that the inferred measurements of axial stress and strain accurately represent the specimen behaviour. In this method:</p><ol><li>The time-domain strain signal is converted into the frequency domain using the fast Fourier transform (FFT)</li><li>A correction is applied to the phase angle of each frequency component to account for the dispersion over the distance between the strain gauge and the bar end, using Bancroft's equation.</li><li>A correction is applied to the amplitude of each frequency component using the factors M_1 and M_2, which account for the variation of strain and Young's modulus across the bar cross section, respectively. These are derived from Davies' analysis of the radial effects in a cylindrical pressure bar.</li><li>The signal is transformed back into the time domain using the inverse FFT.</li></ol><p>Dispersion.m uses a pre-calculated, normalised look-up table of phase velocity, M1 and M2 to improve calculation time. A lookup table for a Poisson's ratio of 0.29 has been provided in the .zip file, and other tables can be constructed using the relationships defined in Tyas and Pope (2005).<br></p><p><br></p><p>Further information on the operation of the script is also available in Barr (2016), linked below.</p> | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8668830990791321, "perplexity": 809.4774457039198}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-09/segments/1550247484648.28/warc/CC-MAIN-20190218033722-20190218055722-00204.warc.gz"} |
https://documen.tv/question/in-a-sample-of-oygen-gas-at-room-temperature-the-average-kinetic-energy-of-all-the-balls-stays-c-24003142-16/ | ## In a sample of oxygen gas at room temperature, the average kinetic energy of all the balls stays constant. Which postulate of kinetic molecu
Question
In a sample of oxygen gas at room temperature, the average kinetic energy of all the balls stays constant. Which postulate of kinetic molecular theory best explains
how this is possible?
Attractive forces between gas particles are negligible because the particles of an ideal gas are moving so quickly
O Collisions between gas particles are elastic, there is no net gain or loss of kinetic energy
Gases consist of a large number of small particles, with a lot of space between the particles
O Gas particles are in constant, random motion, and higher kinetic energy means faster movement
in progress 0
2 months 2021-07-22T19:29:57+00:00 1 Answers 10 views 0 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8264042139053345, "perplexity": 537.7988691398342}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-39/segments/1631780057388.12/warc/CC-MAIN-20210922193630-20210922223630-00492.warc.gz"} |
https://proofwiki.org/wiki/Power_of_Identity_is_Identity | Power of Identity is Identity
Theorem
Let $\struct {M, \circ}$ be a monoid whose identity element is $e$.
Then:
$\forall n \in \Z: e^n = e$
Proof
Since $e$ is invertible, the power of $e$ is defined for all $n \in \Z$.
We prove the case $n \ge 0$ by induction.
Basis for the Induction
By definition of power of monoid element:
$e^0 = e$
so the theorem holds for $n = 0$.
This is our basis for the induction.
Induction Hypothesis
Our induction hypothesis is that the theorem is true for $n = k$:
$e^k = e$
Induction Step
In the induction step, we prove that the theorem is true for $n=k+1$.
We have:
$\displaystyle e^{k + 1}$ $=$ $\displaystyle e^k \circ e$ Definition of Power of Element of Monoid $\displaystyle$ $=$ $\displaystyle e^k$ Definition of Identity Element $\displaystyle$ $=$ $\displaystyle e$ Induction Hypothesis
Therefore, by Principle of Mathematical Induction:
$\forall n \in \Z_{\ge 0} : e^n = e$
$\Box$
Now we prove the case $n < 0$.
We have:
$\displaystyle e^n$ $=$ $\displaystyle \paren {e^{-n} }^{-1}$ Definition of Power of Element of Monoid $\displaystyle$ $=$ $\displaystyle e^{-1}$ since $-n > 0$ $\displaystyle$ $=$ $\displaystyle e$ Inverse of Identity Element is Itself
Thus:
$\forall n \in \Z : e^n = e$
$\blacksquare$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9985595345497131, "perplexity": 288.6233582382585}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-10/segments/1581875141749.3/warc/CC-MAIN-20200217055517-20200217085517-00198.warc.gz"} |
https://www-fourier.univ-grenoble-alpes.fr/?q=fr/content/generalization-kolyvagins-trace-relations | 100, rue des maths 38610 Gières / GPS : 45.193055, 5.772076 / Directeur : Louis Funar
# Generalization of Kolyvagin's trace relations.
Mercredi, 14 Décembre, 2005 - 15:00
Prénom de l'orateur :
Dmitry
Nom de l'orateur :
LOGACHEV
Résumé :
In 1989 -- 1991 Kolyvagin proved finiteness of Tate -- Shafarevich group of elliptic curves $E$ over $Bbb Q$ of analytic rank 0 and 1 which are quotients
of modular curves $X_0(N)$. There is a natural problem to extend this result to the case when an analog of $X_0(N)$ is any Shimura variety $X$, and an analog of $E$ is a quotient motive of some $H^i(X)$.
The purpose of the present series of lectures is to consider some particular steps of this general problem. No preliminary knowledge of the original Kolyvagin's proof is necessary, a survey of the proof will be given. The whole subject contains a lot of open problems, some cases are not treated completely yet. These reseach problems will be stated explicitly. Since the technique of solution of these problems in some cases is reduced to elementary matrix calculations, the subject can be
interesting for young researchers.
Trace relations describe the image of the action of a Hecke correspondence $T_p$ on a subvariety $V$ of a Shimura variety $X$. We find irreducible components of $T_p(V)$
and Galois action on them. Firstly, we give an answer in terms of reductive groups $G_X$, $G_V$ defining $X$, $V$ respectively. It is not clear beforehand thatthe original formulas are suitable for calculations.
We consider the case when $X$ is a Siegel variety of any genus $g$ and $V$ its subvariety parametrizing abelian $g$-folds with multiplication by an imaginary
quadratic field, i.e. we have $G_X=GSp_{2g}$ and $G_V=GU(r,s)$, $r+s=g$.
Further, we show how to apply the general formulas to this situation. The results have a simple geometric interpretation in terms of geometry of some Grassmann
variety over $Bbb F_p$. Practically, we define some partitions on these varieties.
ATTENTION: horaire exceptionnel.
Institution de l'orateur :
Universidad Simón Bolívar (Caracas, Venezuela
Thème de recherche :
Théorie des nombres
Salle :
04 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.865272581577301, "perplexity": 764.1229812227001}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296948620.60/warc/CC-MAIN-20230327092225-20230327122225-00394.warc.gz"} |
http://weblib.cern.ch/collection/ALICE%20Preprints?ln=sk&as=1 | # ALICE Preprints
Posledne pridané:
2022-11-24
04:02
Real and virtual direct photon measurements with ALICE / Scheid, Sebastian In this contribution the latest measurements of real and virtual photons in Pb-Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV from the ALICE Collaboration are presented. [...] arXiv:2211.11934. - Fulltext
2022-11-23
19:35
Multiplicity dependence of charged-particle production in pp, p-Pb, Xe-Xe and Pb-Pb collisions at the LHC / ALICE Collaboration, CERN Multiplicity ($N_{\mathrm {ch}}$) distributions and transverse momentum ($p_{\mathrm T}$) spectra of inclusive primary charged particles in the kinematic range of $|\eta| < 0.8$ and 0.15 GeV/$c$ $< p_{\mathrm T} < 10$ GeV/$c$ are reported for pp, p-Pb, Xe-Xe and Pb-Pb collisions at centre-of-mass energies per nucleon pair ranging from $\sqrt{s_\mathrm{NN}} = 2.76$ TeV up to 13 TeV. A sequential two-dimensional unfolding procedure is used to extract the correlation between the transverse momentum of primary charged particles and the charged-particle multiplicity of the corresponding collision. [...] CERN-EP-2022-266.- Geneva : CERN, 2022 Draft (restricted): PDF;
2022-11-23
04:30
Exploring jet interactions in the quark-gluon plasma using jet substructure measurements in Pb-Pb collisions with ALICE / Ehlers, Raymond Jets are generated in hard interactions in high-energy nuclear collisions. [...] arXiv:2211.11800. - 6 p. Fulltext
2022-11-23
04:27
Commissioning and first performances of the ALICE MID RPCs / Terlizzi, Livia ALICE (A Large Ion Collider Experiment) at the CERN Large Hadron Collider (LHC) is designed to study p-p and Pb-Pb collisions at ultra-relativistic energies. [...] arXiv:2211.11254. - Fulltext
2022-11-18
22:59
Enhanced deuteron coalescence probability in jets / ALICE Collaboration, CERN /ALICE The transverse-momentum ($p_{\rm T}$) spectra and coalescence parameters $B_2$ of (anti)deuterons are measured in pp collisions at $\sqrt{s} = 13$ TeV in and out of jets. In this measurement, the direction of the leading particle with the highest $p_{\rm T}$ in the event ($p_{\rm T}^{\rm{ lead}} > 5$ GeV/$c$) is used as an approximation for the jet axis [...] CERN-EP-2022-264; arXiv:2211.15204.- Geneva : CERN, 2022 - 11 p. Draft (restricted): PDF; Fulltext: PDF;
2022-11-18
19:07
Light (anti)nuclei production in Pb-Pb collisions at $\sqrt{s_{\mathrm{NN}}}=5.02$ TeV / ALICE Collaboration, CERN /ALICE The measurement of the production of deuterons, tritons and ${^{3}\mathrm{He}}$ and their antiparticles in Pb-Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV is presented in this article. The measurements are carried out at midrapidity ($|y| < 0.5$) as a function of collision centrality using the ALICE detector. [...] CERN-EP-2022-263; arXiv:2211.14015.- Geneva : CERN, 2022 - 16 p. Draft (restricted): PDF; Fulltext: PDF;
2022-11-18
14:29
Pseudorapidity densities of charged particles with transverse momentum thresholds in pp collisions at $\sqrt{s}=$ 5.02 and 13 TeV / ALICE Collaboration, CERN The pseudorapidity density of charged particles with minimum transverse momentum ($p_{\rm T}$) thresholds of 0.15, 0.5, 1, and 2 GeV$/c$ was measured in pp collisions at centre-of-mass energies of $\sqrt{s}=$ 5.02 and 13 TeV with the ALICE detector. The study is carried out for inelastic collisions with at least one primary charged particle having a pseudorapidity ($\eta$) within $\pm0.8$ and $p_{\rm T}$ larger than the corresponding threshold [...] CERN-EP-2022-262.- Geneva : CERN, 2022 Draft (restricted): PDF;
2022-11-18
13:54
First measurement of $\mathbf{\Lambda}_\mathbf{c}^\mathbf{+}$ production down to $p_\mathbf{T} = \mathbf{0}$ in pp and p-Pb collisions at $\sqrt{s_\mathbf{NN}}= \mathbf{5.02 TeV}$ / ALICE Collaboration, CERN The production of prompt ${\mathrm {\Lambda_{c}^{+}}}$ baryons has been measured at midrapidity in the transverse momentum interval $0 < p_{\rm T} < 1$ GeV/$c$ for the first time, in pp and p-Pb collisions at a centre-of-mass energy per nucleon-nucleon collision $\sqrt{s_\mathrm{NN}} = 5.02$ TeV. The measurement was performed in the decay channel $\mathrm{\rm \Lambda_{c}^{+} \to p K^{0}_{S}}$ by applying new decay reconstruction techniques using a Kalman-Filter vertexing algorithm and adopting a machine-learning approach for the candidate selection. [...] CERN-EP-2022-261; arXiv:2211.14032.- Geneva : CERN, 2022 - 21 p. Draft (restricted): PDF; Fulltext: PDF;
2022-11-17
04:09
Two-particle transverse momentum correlations in pp and p-Pb collisions at LHC energies / Collaboration, Alice Two-particle transverse momentum differential correlators, recently measured in Pb-Pb collisions at LHC energies, provide an additional tool to gain insights into particle production mechanisms and infer transport properties, such as the ratio of shear viscosity to entropy density, of the medium created in Pb-Pb collisions. [...] CERN-EP-2022-230 ; arXiv:2211.08979. - 21 p. Fulltext
2022-11-17
04:09
Production of ${\rm K}^{0}_{\rm{S}}$, $\Lambda$ ($\bar{\Lambda}$), $\Xi^{\pm}$ and $\Omega^{\pm}$ in jets and in the underlying event in pp and p$-$Pb collisions The production of strange hadrons (K$^{0}_{\rm S}$, $\Lambda$, $\Xi^{\pm}$, and $\Omega^{\pm}$), baryon-to-meson ratios ($\Lambda/{\rm K}^{0}_{\rm S}$, $\Xi/{\rm K}^{0}_{\rm S }$, and $\Omega/{\rm K}^{0}_{\rm S}$), and baryon-to-baryon ratios ($\Xi/\Lambda$, $\Omega/\Lambda$, and $\Omega/\Xi$) associated with jets and the underlying event were measured as a function of transverse momentum ($p_{\rm T}$) in pp collisions at $\sqrt{s} = 13$ TeV and p-Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV with the ALICE detector at the LHC. [...] CERN-EP-2022-218 ; arXiv:2211.08936. - 28 p. Fulltext | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9989325404167175, "perplexity": 4051.936223679923}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446710813.48/warc/CC-MAIN-20221201121601-20221201151601-00141.warc.gz"} |
https://www.enotes.com/homework-help/topic/calculus1 | # Calculus
• Calculus
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• Calculus
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• Calculus
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• Calculus
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• Calculus
You want the limit of y=(1-cos 2x)/x^2 while x approaches 0. y = (1-cos 2x)/x^2 => [1 - (1 - 2*(sin x)^2)]/x^2 => 2*(sin x)^2/x^2 => 2*(sin x / x)^2 lim x--> 0 (sin x / x) = 1 Using...
• Calculus
We need to determine the integral of (cos x)^7 * sin x. Int [(cos x)^7 * sin x dx] let cos x = u => - du = sin x dx => Int [ -u^7 du] => -u^8 / 8 + C substitute u = cos x => - (cos...
• Calculus
We have to determine the value of lim x--> 0[(cos x - cos 3x) / x*sin x If we substitute x = 0, we get (1- 1) / 0 = 0/0 As this is an indeterminate form we use l'Hopital's rule and replace the...
• Calculus
The extreme values of a function occur at the points where the derivative is equal to 0. f(x) = 2x^3 + 3x^2 - 12x + 5 => f'(x) = 6x^2 + 6x - 12 6x^2 + 6x - 12 = 0 => x^2 + x - 2 = 0 => x^2...
• Calculus
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• Calculus
First we need to determine the points of intersection between lnx and ln^2 x ==> ln x = ln^2 x ==> ln^2 x - ln x = 0 ==> lnx ( lnx -1) =0 ==> lnx = 0 ==> x = 1 ==> lnx-1 = 0...
• Calculus
We have to find the area enclosed between y=x^2 - 2x + 2 and y = -x^2 + 6. First lets find the points of intersection x^2 - 2x + 2 = -x^2 + 6 => 2x^2 - 2x - 4 = 0 => x^2 - x - 2 = 0 => x^2...
• Calculus
The area of the region bounded by the curve y = sqrt (x - 1), the y- axis and the lines y = 1 and y = 5 is the limited integral of the expression of x in terms of y, between y = 5 and y = 1. y =...
• Calculus
We have to verify that lim x-->0 [ ln(1+x)/x] = 1. substituting x = 0, we get the indeterminate form 0/0, therefore we can use the l'Hopital's rule and substitute the numerator and denominator...
• Calculus
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• Calculus
We have to find the limit of f(x)=(sin x-cos x)/cos 2x for x--> 45 degrees. We know that cos 2x = (cos x)^2 - (sin x )^2 lim x--> 0 [(sin x-cos x)/cos 2x] => lim x--> 0 [(sin x-cos...
• Calculus
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• Calculus
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• Calculus
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• Calculus
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• Calculus
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• Calculus
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• Calculus
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• Calculus
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• Calculus
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• Calculus
We have to find the value of lim x--> 0 [ ln(1+x)/(sinx+sin3x)] substituting x = 0, we get the indeterminate form 0/0. Therefore we can use l'Hopital's Rule and substitute the numerator and...
• Calculus
Hi, djshan, Sorry, but I'm not too sure what you want us to do here. Are we going to graph this? Find the intercepts? Find the zeros? Something else? I would assume we are graphing it. To...
• Calculus
We have to find Int [e^2x * cos 3x dx] Here the best way to solve would be to use integration by parts. Int [u dv] = u*v – Int [v du] take u = e^2x, du = 2*e^2x dx dv = cos 3x dx, v = (1/3)* sin...
• Calculus
We have to find the antiderivative of y = x / sqrt ( x^2 - 9) Let u = x^2 - 9 => du / dx = 2x => x dx = du/2 Int [ x / sqrt ( x^2 - 9) dx] => Int [(1/2)*(1/ sqrt u) du] => Int [ (1/2)*...
• Calculus
We have to find y' for y = (2-x)^(sqrt x) Use natural logariths for both the sides ln y = ln[ (2-x)^(sqrt x)] use the property ln a^x = a*ln x => ln y = (sqrt x)*ln ( 2 - x) Do implicit...
• Calculus
To find the slant asymptote of x^3 / (x + 2)^2 we have to divide x^3 by (x + 2)^2 (x^2 + 4x + 4) | x^3...........................................| x - 4 ...........................x^3 + 4x^2 + 4x...
• Calculus
We have to find the derivative of y = arc sin x/(1-x^2). We use the quotient rule here: y' = [(arc sin x)'*(1 - x^2) - ( arc sin x)*(1 - x^2)']/(1 - x^2)^2 => [sqrt(1-x^2)*(1 - x^2) + 2x*(arc...
• Calculus
We have to find the value of lim x--> 90[ (1- sin x)/(cos x)^2] substituting x = 90 degrees, we get the indeterminate form 0/0, so we can use l'Hopital's rule and substitute the numerator and...
• Calculus
We have to find the derivative of y = (10 + lg (x^10) + e^10x)^10. We use the chain rule to find the derivative of y. y' = 10 * (10 + lg (x^10) + e^10x)^9 * (10 / x + 10*e^10x) => 10 * (10 + lg...
• Calculus
We have to find the integral of f'(x)=11e^x/(11+e^x) f'(x)=11e^x/(11+e^x) let 11 + e^x = y e^x dx = dy Int [ 11e^x/(11+e^x) dx] => Int [ 11dy/y] => 11*ln |y| + C substitute y = 11 + e^x =>...
• Calculus
We have the function y = sin x + cos 3x. The derivative of sin x is cos x and the derivative of cos x is -sin x. Also, for a function of the form y= f(g(x)), the derivative of y or y' is given by...
• Calculus
We have to find the integral of 1/(16x^2+24x+9) 1/(16x^2+24x+9) => 1/(4x+3)^2 let 4x + 3 = u => du/dx = 4 => dx = du/4 Int[ 1/(16x^2+24x+9) dx] => Int [ 1/u^2 du/4] => (1/4) Int [...
• Calculus
We have to find the value of lim x--> pi[ sin 5x / sin x] We see that substituting x with pi gives us the form 0/0 which is indeterminate. We can use therefore use l'Hopital's rule and use the...
• Calculus
We have f(x) = (3x + 1)*e^-x We use the product rule to find f'(x) f'(x) = (3x + 1)'*e^-x + (3x + 1)*(e^-x)' => 3*e^-x - (3x +1)e^-x => 3*e^-x - f(x) f''(x) = -3e^-x - f'(x) => -3e^-x -...
• Calculus
We need to find f(x) given that f'(x) = sin 2x /((sin x)^2 - 4) let ((sin x)^2 - 4) = y dy = 2*sin x * cos x dx =>dy = sin 2x dx Int [ sin 2x /((sin x)^2 - 4) dx] => Int [ 1/y dy] => ln...
• Calculus
We have to find lim x--> -2 [(x^2-2x-8)/(x^3+8)] using l'Hopital's rule. First we find out if the l'Hopital's Rule can be used here. Substituting x = -2 we get the indeterminate form 0/0;...
• Calculus
We have the function: f(x)=12x^4+24x^2+56 f(x) = 12x^4 + 24x^2 + 56 f'(x) = 48x^3 + 48x If f'(x) = 0 => 48x^3 + 48x = 0 => x^3 + 3 = 0 => x( x^2 + 1) = 0 x1 = 0 x2 = -sqrt (-1) => x2 =...
• Calculus
The area bound by the curve y = cosx/(sin^2x-4), the x axis and the lines x = 0 and x = pi/2 is the integral of y between the limits x = 0 and x = pi/2. y = cos x/ ((sin x)^2 - 4) let sin x = y... | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.941501796245575, "perplexity": 1535.3600014991607}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-51/segments/1575540569332.20/warc/CC-MAIN-20191213230200-20191214014200-00173.warc.gz"} |
https://jayryablon.wordpress.com/2008/05/01/heisenberg-and-schwinger-continued-draft-paper/ | # Lab Notes for a Scientific Revolution (Physics)
## May 1, 2008
### Heisenberg Uncertainty and Schwinger Anomaly Continued: Draft Paper
I have been writing a paper to rigorously develop the hypothesis I presented last week, in a post linked at Heisenberg Uncertainty and Schwinger Anomaly: Two Sides of the Same Coin?. I believe there is enough developed now, and I think enough of the kinks are now out, so you all may take a sneak preview. Thus, I have linked my latest draft at:
Heisenberg Uncertainty and the Schwinger Anomaly
Setting aside the hypothesized connection between the magnetic anomaly and uncertainty, Sections 4 through 7, which have not been posted in any form previously, stand completely by themselves, irrespective of this hypothesis. These sections are strictly mathematical in nature, and they provide an exact measure for how the uncertainty associated with a wavefunction varies upwards from $\hbar/2$ as a function of the potential, and the parameters of the wavefunction itself. The wavefunction employed is completely general, and the uncertainty relation is driven by a potential $V$.
This is still under development, but this should give you a very good idea of where this is headed.
Of course, I welcome comment, as always.
Best regards,
Jay.
## 1 Comment »
1. Good post. I learn something new and challenging on blogs I stumbleupon everyday.
It will always be interesting to read through content from other authors
and practice a little something from their
websites.
A blog page regarding technological innovation: 5.1 surround system – Danuta,
Comment by Danuta — August 27, 2013 @ 8:29 am
Create a free website or blog at WordPress.com. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 2, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8420376181602478, "perplexity": 1758.9937314672825}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-18/segments/1555578584186.40/warc/CC-MAIN-20190423015050-20190423041050-00234.warc.gz"} |
https://socratic.org/questions/how-do-you-simplify-4-98-0 | Algebra
Topics
# How do you simplify 4.98^0?
Jan 20, 2017
${4.98}^{0} = \textcolor{g r e e n}{1}$
#### Explanation:
Any number (excluding perhaps zero) to the power of $0$ is equal to $1$
Consider the following sequence for any number $n$ which is not zero:
$\textcolor{w h i t e}{\text{XXX}} {n}^{4} = n \times n \times n \times n$
$\textcolor{w h i t e}{\text{XXX}} {n}^{3} = \frac{{n}^{4}}{n} = n \times n \times n$
$\textcolor{w h i t e}{\text{XXX}} {n}^{2} = \frac{{n}^{3}}{n} = n \times n$
$\textcolor{w h i t e}{\text{XXX}} {n}^{1} = \frac{{n}^{2}}{n} = n$
$\textcolor{w h i t e}{\text{XXX}} {n}^{0} = \frac{{n}^{1}}{n} = \frac{n}{n} = 1$
...and we could continue on:
$\textcolor{w h i t e}{\text{XXX}} {n}^{- 1} = \frac{{n}^{0}}{n} = \frac{1}{n}$
$\textcolor{w h i t e}{\text{XXX}} {n}^{- 2} = \frac{{n}^{- 1}}{n} = \frac{1}{{n}^{2}}$
$\textcolor{w h i t e}{\text{XXX}} {n}^{- 3} = \frac{{n}^{- 2}}{n} = \frac{1}{{n}^{3}}$
...and so on...
Jan 20, 2017
Another way of writing ${4.98}^{0}$ is : $\frac{4.98}{4.98} = 1$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 14, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9884803295135498, "perplexity": 1387.9882853489826}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-47/segments/1573496664752.70/warc/CC-MAIN-20191112051214-20191112075214-00479.warc.gz"} |
http://mathhelpforum.com/calculus/63223-show-volume-pyramid-height-h-whose-base-equilateral-triangle-o.html | # Math Help - Show that the volume of a pyramid of height h whose base is an equilateral triangle o
1. ## Show that the volume of a pyramid of height h whose base is an equilateral triangle o
Show that the volume of a pyramid of height h whose base is an equilateral triangle of side s is equal to
{(√3)(h)(s^2)}
(12)
2. Originally Posted by derrickd
Show that the volume of a pyramid of height h whose base is an equilateral triangle of side s is equal to
{(√3)(h)(s^2)}
(12)
For, equilateral triangle of base,
height of equilateral triangle $= \sqrt{s^2-\left(\frac{s}{2}\right)^2}=\frac{s\sqrt{3}}{2}$
Area of equilateral triangle (base) $A= \frac{s\times \frac{s\sqrt{3}}{2}}{2}=\frac{s^2\sqrt{3}}{4}$
Volume= $\frac{A\times h}{3}=\frac{\frac{s^2\sqrt{3}}{4}h}{3}=\frac{hs^2\ sqrt{3}}{12}$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 3, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8717763423919678, "perplexity": 233.78508443944799}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-52/segments/1418802769642.136/warc/CC-MAIN-20141217075249-00030-ip-10-231-17-201.ec2.internal.warc.gz"} |
https://math.stackexchange.com/questions/3393578/cubic-polynomial-smoothly-connecting-two-circles | # Cubic polynomial smoothly connecting two circles
Given two circles with radii $$R_L$$ and $$R_R$$ and centers at $$(-(R_L+a),\,0)$$ and $$(R_R+a,\,0)$$, respectively, find a cubic polynomial $$p(x)=b+cx^2+dx^3$$ that smoothly connects the two circles.
$$b$$ is a parameter so $$p(0)=b$$ and the linear term of the polynomial is omitted because we want $$\frac{\mathrm{d}p}{\mathrm{d}x}\Big|_{x=0}=0$$.
My attempt at a solution. Let $$\mathrm{C}_{L,R}$$ the equations for the upper half of the $$L,R$$ circles. I formulate two equations relating $$\mathrm{C}_{L,R}$$ and $$p$$, and two equations relating $$\frac{\mathrm{d}}{\mathrm{d}x}\mathrm{C}_{L,R}$$ and $$\frac{\mathrm{d}p}{\mathrm{d}x}$$. Let $$x_{L,R}$$ be the points where $$p(x)$$ and $$\mathrm{C}_{L,R}(x)$$ intersect, then:
$$\mathrm{C}_L(x_L)-p(x_L) = 0$$ $$\mathrm{C}_R(x_R)-p(x_R) = 0$$ $$\frac{\mathrm{d}}{\mathrm{d}x}\mathrm{C}_L(x_L) - \frac{\mathrm{d}p}{\mathrm{d}x}(x_L) = 0$$ $$\frac{\mathrm{d}}{\mathrm{d}x}\mathrm{C}_R(x_R) - \frac{\mathrm{d}p}{\mathrm{d}x}(x_R) = 0$$ so I have a system of four nonlinear equations with four unknowns $$(x_L,\,x_R,\,c,\,d)$$. I coded a simple Newton's method for the system and it works well for some combinations of parameters $$(a,\,b,\,R_L,\,R_R)$$ when the initial guess is close enough, especially when $$|R_L-R_R|$$ is not too large and I use a constant damping for the Newton iterations. I can find initial guesses that I think are good via a graphical interface I coded. However, as $$|R_L-R_R|$$ gets larger the solver fails spectacularly to converge even with very close initial guesses and very small damping. (I should add that I'm actually taking the square of the equations to avoid square roots of negative numbers during the Newton iterations).
My question is threefold:
a) what other method or modification can I use to make the solver more stable?
b) this problem seems to me like it should be solved somewhere, do you know a reference?
c) more generally, is there a reason this should fail as horribly as it does when $$|R_L-R_R|>>1$$?
• I would try parametrizing this with only two unknowns $x_{L,R}$, fitting a cubic polynomial $a_0+a_1x+a_2x^2+a_3x^3$ to the points and tangents on the circles, and solving for $a_0=b,a_1=0$ instead. – user856 Oct 14 '19 at 16:48
• Thank you. I'm not sure I follow. If I did that, wouldn't $a_2$ and $a_3$ be unknowns as well? If I had two unknowns and four equations I'm not even sure what I could do. Also, setting $a_0=b$ and $a_1=0$ is exactly what I'm doing. By fitting do you mean I should try some optimisation technique? – mvaldez Oct 14 '19 at 17:03
• Do you mean that parameter $b$ si fixed : all curves must pass through point $(0,b)$ ? – Jean Marie Oct 14 '19 at 17:05
• yes, the parameters $(a,\,b,\,R_L,\,R_R)$ are fixed so the curve I want to find must pass through $(0,\,b)$ and have vanishing derivative there – mvaldez Oct 14 '19 at 17:07
• I mean treat only the points $x_{L,R}$ as unknowns. Compute a cubic polynomial $a_0+a_1x+a_2x^2+a_3x^3$ passing through them, which can be done in closed form. Then you get $c=a_2,d=a_3$. But you may not have $a_0=b,a_1=0$, so you need to choose $x_{L,R}$ to satisfy them. Thus you have two nonlinear equations in two unknowns. – user856 Oct 14 '19 at 17:09
Here is a method giving all solutions in a deterministic way.
Take a look at the following figure :
Fig. 1 : The exhaustive set of 8 solution curves in the case $$a=1, \ b=1, \ R_L=3, \ R_R=1$$. Please note that some tangencies are internal to the circles.
How has this result been obtained ? By using a Computer Algebra System with a rather simple system of 4 polynomial equations in the 4 unknowns $$c,d,x_L,x_R$$ where (your notations) the two last parameters are the abscissas of the tangency points of the cubic curve with equation
$$y=f(x)=b+cx^2+dx^3$$
with the Left and Right circle resp.
Here they are:
$$\begin{cases}(x_L+R_L+a)^2+f(x_L)^2&=&R_L^2& \ \ (i)\\ (x_R-R_R-a)^2+f(x_R)^2&=&R_R^2& \ \ (ii)\\ \dfrac{f(x_L)}{x_L+R_L+a}&=&- \dfrac{1}{f'(x_L)}& \ \ (iii)\\ \dfrac{f(x_R)}{x_R-R_R-a}&=&- \dfrac{1}{f'(x_R)}& \ \ (iv)\\ \end{cases}\tag{1}$$
The first two equations express that $$M_L=(x_L,f(x_L))$$ and $$M_R=(x_R,f(x_R))$$ belong to their resp. circles.
Let us now explain the third equation. Let $$C_L(-(R_L+a),0)$$ denotes the center of the first circle. The slope of the tangent in $$M_L$$ is $$f'(x_L)$$. $$\vec{C_LM_L}=\binom{x_L+R_L+a}{f(x_L)-0}$$, being orthogonal to this tangent, has a slope $$-\dfrac{1}{f'(x_L)}$$ (see there). Similar reasoning for the fourth equation.
Remarks :
1) I haven't found the limitations you mention in the case of a large gap between $$R_L$$ and $$R_R$$. For example when $$R_L=100$$ and $$R_R=1$$, (with $$a=1$$ and $$b=0$$), one finds 14 solutions...
2) The third equation can be written : $$f(x_L)f'(x_L)+x_L+R_L+a=0$$ : this avoids possible division by zero. Same thing for the fourth equation.
Another case (Fig. 2), this time with $$R_L=R_R$$ displaying some spurious solutions under the form of ... second degree curves, i.e., parabolas (one can indeed have $$d=0$$) :
Fig. 2.
Still another case, with four solutions (one of them looks to be a parabola but isn't):
Fig. 3.
Here is the Matlab program that has given figure 1
(running time : around 30 seconds on my rather slow computer). Please note the "isreal" conditions (we want the different unknowns to be real) : in fact in the first case, there are $$40$$ solutions, $$32$$ of them being spurious, i.e., with complex coefficients...
Final remark : In fact, there exists a different way to solve system (1) by doing it in two steps ; first, by expressing the fact that equations (i) and (iii) have a common root $$x_L$$ giving (using a "resultant" ) a first (non-linear) constraint between $$c$$ and $$d$$. Doing the same for equations (ii) and (iv) for common root $$x_R$$, gives a second constraint. Then, in a second step, solve the system of 2 (non-linear...) equations in 2 unknowns $$c$$ and $$d$$.
syms xL xR c d;%symbolic variables declaration
a=1;b=1;m=max([RL,RR]);p=2*m+a;
axis([-p+3,p+3,-p-3,p+3]);axis equal;
f=@(x)(b+c*x*x+d*x*x*x);%cubic function
fp=@(x)(2*c*x+3*d*x*x);%its derivative
%The system of 4 equations in 4 unknowns ; sol. in [XL,XR,C,D]
[XL,XR,C,D]=solve(...
(xL+RL+a)^2+f(xL)^2==RL^2,
(xR-RR-a)^2+f(xR)^2==RR^2,...
xL+f(xL)*fp(xL)==-RL-a,...
xR+f(xR)*fp(xR)==RR+a,...
xL,xR,c,d);% the 4 variables
plot([-p,p],[0,0]);\$x axis
plot([0,0],[-p,p]);%y axis
UC=exp(i*(0:0.01:2*pi));%unit circle
plot(-(RL+a)+RL*UC,'k');%circle C_L
plot((RR+a)+RR*UC,'k');%circle C_R
x=-p:0.01:p;%range of values for x
for k=1:length(XL)
c=C(k);d=D(k);xL=XL(k);xR=XR(k);
if isreal(c) && isreal(d) && isreal(xL) && isreal(xR)
plot(x,b+c*x.*x+d*x.*x.*x,'color',rand(1,3));hold on;
end;
end;
• Thank you Jean Marie, your answer was very helpful. This new set of equations seem to be much more stable – mvaldez Oct 22 '19 at 20:45
• Thanks. I have added a small remark (remark 2). – Jean Marie Oct 23 '19 at 7:06
• See as well the final remark I just added. – Jean Marie Oct 23 '19 at 7:42 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 54, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8862836360931396, "perplexity": 241.5879798348246}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-04/segments/1610703517966.39/warc/CC-MAIN-20210119042046-20210119072046-00773.warc.gz"} |
http://mathoverflow.net/questions/134417/two-variable-polynomials-irreducible-as-formal-power-series | # Two-variable polynomials, irreducible as formal power series
Let $k$ be a field and $f\in k[a,b]$ an irreducible two-variable polynomial, $B := k[a,b]/(f)$ and $C$ the integral closure of $B$ in its fraction field.
I call $f$ good if it is irreducible in the ring $k[[a,b]]$ of formal power series; equivalently (Nagata, Local Rings, p. 122, Ex. 1), if $C$ has exactly one prime ideal lying above $\mathfrak m = (a,b)B$ (hence, $f$ being good just means that the curve $B$ is analytically irreducible at the origin).
I'm looking for examples of "good" polynomials $f$ such that the residue field $L$ of $C$ is a proper extension of $k$.
I can show (at least if $k$ is infinite) that $[L:k] \cdot r = \mu(f)$, where $\mu(f)$ is the degree of the lowest-degree summand of $f$, and $r$ is the ramification index of $\mathfrak m$ in $C$. Hence, it is clear that $\mu(f)$ must be large enough if one wants interesting examples.
The only "generic" class of examples I could come up with is the one where $f$ is homogeneous: Then $f$ being irreducible implies $f$ is good, and $[L:k] = \deg f$.
- | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9748081564903259, "perplexity": 57.355991418237934}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-10/segments/1394011098060/warc/CC-MAIN-20140305091818-00005-ip-10-183-142-35.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/riesz-lemma.114238/ | # Riesz Lemma
1. Mar 14, 2006
### Castilla
Good Morning.
I am reading the first pages of the "Lessons of Functional Analysis" of Riesz and Nagy, because I learned that this book was the main source for Apostol chapter about the Lebesgue Integral, and I am trying to find a proof of the Fundamental Theorem of Calculus for Lebesgue Integrals worked out within this approach (not Measure Theory).
One of the first theorems of Riesz-Nagy's book states: "Every monotonic function f(x) posseses a finite derivative at every point x with the possible exception of the points x of a set of measure zero".
To proof this T they use this lemma:
"Let g(x) be a continuous function defined in the closed interval (a, b), and let E be the set of points x interior to this interval and such that there exists and e lying to the right of x with g(e) > g(x). Then the set E is either empty or an open set, i.e., it decomposes into a finite number or a denumerable infinity of open and disjoint intervals (a_k, b_k) (the underline is mine), and g(a_k) < g(b_k) for all these intervals".
Well I suppose I can follow the lemma's proof on the book but in this
statement there is something I did not know. Why to be E an opet set implies that it decomposes into a finite number or a denumerable infinity of open and disjoint intervals ??? | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9425721764564514, "perplexity": 451.83535358711987}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-51/segments/1544376826800.31/warc/CC-MAIN-20181215061532-20181215083532-00030.warc.gz"} |
https://www.physicsforums.com/threads/simple-calculus-question.195299/ | # Homework Help: Simple calculus question
1. Nov 1, 2007
### Darkiekurdo
1. The problem statement, all variables and given/known data
A reservoir of square cross-section has sides sloping at an angle of 45 degrees with the vertical. The side of the bottom is p feet in length, and water flows in the reservoir at the rate of c cubic feet per minute. Find an expression for the rate at which the surface of the water is rising at the instant its depth is h feet. Calculate this rate when p = 17, h = 4 and c = 35.
2. Relevant equations
Not stated
3. The attempt at a solution
I don't know how to begin, I'm sorry.
2. Nov 1, 2007
### Dick
Start by trying to find an expression for the volume V as a function of h and p.
3. Nov 1, 2007
### Darkiekurdo
What shape is the reservoir?
4. Nov 1, 2007
### Dick
Did you read the problem? It's a truncated pyramid with square cross section.
5. Nov 1, 2007
### Darkiekurdo
Hmm. I didn't know that. How did you come up with that?
6. Nov 1, 2007
### Dick
I read the first sentence of the problem you posted.
7. Nov 1, 2007
### d_leet
Did you even read the problem? You asked the question and it seems like you haven't even read it. The first sentence describes the shape of the resevoir.
8. Nov 1, 2007
### Darkiekurdo
Of course I read it but I didn't know what that was. Sorry.
9. Nov 1, 2007
### Darkiekurdo
Anyway, the volume of a truncated pyramid is
$$V = \frac{H}{3}(A + a + \sqrt{Aa})$$
right?
Last edited: Nov 1, 2007
10. Nov 1, 2007
### Dick
Right, if the A and a are the areas of the top and bottom.
11. Nov 1, 2007
### Darkiekurdo
What should I do next?
12. Nov 1, 2007
### Dick
Apply that formula to your problem. If the reservoir is filled to height h, what are the areas of the top and bottom?
13. Nov 1, 2007
### Darkiekurdo
So the area of A = p2 and a = (p+2h)2.
14. Nov 1, 2007
### Dick
Now you are cooking. What's the formula then for the volume V in terms of p and h?
15. Nov 1, 2007
### Darkiekurdo
Then I'll just substitute those in the formula for the volume:
$$\frac{H}{3}[p^2 + p(p + 2h) + (p + 2h)^2]$$
16. Nov 1, 2007
### Dick
H=h, right? Now to make life easier later, I would expand that out. Now you have V as a function of p and h. What next? Please don't say "I don't know".
17. Nov 1, 2007
### Darkiekurdo
Yeah, I should have written it as 1/3h instead of H/3 actually.
Thus,
$$p^2h + 2ph^2 + \frac{4}{3}h^3$$
18. Nov 1, 2007
### Dick
Better and better. What next?
19. Nov 1, 2007
### Darkiekurdo
I'm stuck. :(
20. Nov 1, 2007
### Darkiekurdo
Maybe we should take time in consideration? | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8567028045654297, "perplexity": 1610.1784570237094}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-34/segments/1534221213691.59/warc/CC-MAIN-20180818154147-20180818174147-00410.warc.gz"} |
http://mathhelpforum.com/discrete-math/74615-solved-proof-equal-sum-subsets-pigeonhole-principle-print.html | # [SOLVED] Proof for equal sum subsets (pigeonhole principle)
• February 19th 2009, 10:11 PM
sritter27
[SOLVED] Proof for equal sum subsets (pigeonhole principle)
Let S be a set of ten distinct positive integers, all less than 107. Prove that S has two distinct subsets that both have the same sum.
So I've been racking my mind over this and I cannot figure out how to write a proper proof for it. I was thinking the pigeon hole principle might help out somehow, but I'm not entirely sure how. I'd be rather grateful to anyone who can help point me in the right direction.
• February 20th 2009, 02:28 AM
tah
Quote:
Originally Posted by sritter27
Let S be a set of ten distinct positive integers, all less than 107. Prove that S has two distinct subsets that both have the same sum.
So I've been racking my mind over this and I cannot figure out how to write a proper proof for it. I was thinking the pigeon hole principle might help out somehow, but I'm not entirely sure how. I'd be rather grateful to anyone who can help point me in the right direction.
Are you sure that it's true, does it work for any set S ? In fact if S could be partition into two sets of the same sum, then sum of S has to be even !! but take S = {1,2,3,4,5,6,7,8,9,10} and sum S is 55, so this S could not be partitioned in such a way ...
• February 20th 2009, 02:37 AM
Sum of subsets
Hello everyone
Quote:
Originally Posted by tah
Are you sure that it's true, does it work for any set S ? In fact if S could be partition into two sets of the same sum, then sum of S has to be even !! but take S = {1,2,3,4,5,6,7,8,9,10} and sum S is 55, so this S could not be partitioned in such a way ...
The question doesn't say it's a partition - merely that the subsets are distinct. I don't know that we can assume that this even means that they are disjoint, simply that they are not equal.
Having said that, I don't know how to tackle this question! But if I have any thoughts, I'll post them!
• February 20th 2009, 02:54 AM
tah
Quote:
Hello everyoneThe question doesn't say it's a partition - merely that the subsets are distinct. I don't know that we can assume that this even means that they are disjoint, simply that they are not equal.
Having said that, I don't know how to tackle this question! But if I have any thoughts, I'll post them!
Ahh right i see.
• February 20th 2009, 04:43 AM
Subsets
Hello sritter27
Quote:
Originally Posted by sritter27
Let S be a set of ten distinct positive integers, all less than 107. Prove that S has two distinct subsets that both have the same sum.
So I've been racking my mind over this and I cannot figure out how to write a proper proof for it. I was thinking the pigeon hole principle might help out somehow, but I'm not entirely sure how. I'd be rather grateful to anyone who can help point me in the right direction.
A set with 10 elements has $2^{10}$ = 1024 distinct subsets.
Maximum sum = 106 + 105 + ... + 97 = ...
Minimum sum = 1 + 2 + ... + 10 = ...
No of different sums = ...
Can you supply the rest?
PS. Sorry, that's not quite right. The minimum sum is zero - if the subset is the empty set.
• February 20th 2009, 09:22 AM
sritter27
Quote:
A set with 10 elements has $2^{10}$ = 1024 distinct subsets.
Maximum sum = 106 + 105 + ... + 97 = ...
Minimum sum = 1 + 2 + ... + 10 = ...
No of different sums = ...
Can you supply the rest?
PS. Sorry, that's not quite right. The minimum sum is zero - if the subset is the empty set.
Ah, I think it makes sense now.
So if there are 1024 possible subsets of S and the maximum sum of S is 1015 then there are fewer possible sums than possible subsets. So, by the pigeonhole principle there must be two subsets of S that have the same sum? If that is true then if two subsets have the same sum but are not distinct (non-disjoint), they can be made distinct by removing the common elements between them and thus they would still have the same sum.
Is this right?
• February 20th 2009, 10:15 AM
Sum of subsets
Hello sritter
Quote:
Originally Posted by sritter27
Ah, I think it makes sense now.
So if there are 1024 possible subsets of S and the maximum sum of S is 1015 then there are fewer possible sums than possible subsets. So, by the pigeonhole principle there must be two subsets of S that have the same sum? If that is true then if two subsets have the same sum but are not distinct (non-disjoint), they can be made distinct by removing the common elements between them and thus they would still have the same sum.
Is this right?
Yes, you can certainly make the subsets disjoint in this way without affecting the equality of their sums, although the question doesn't actually say 'disjoint', merely distinct (which usually, of course, means 'not equal'). | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 2, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.908565878868103, "perplexity": 260.0645906775184}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-48/segments/1448398447906.82/warc/CC-MAIN-20151124205407-00023-ip-10-71-132-137.ec2.internal.warc.gz"} |
https://intl.siyavula.com/read/maths/grade-12/differential-calculus/06-differential-calculus-07 | We think you are located in United States. Is this correct?
Applications of differential calculus
6.7 Applications of differential calculus (EMCHH)
Optimisation problems (EMCHJ)
We have seen that differential calculus can be used to determine the stationary points of functions, in order to sketch their graphs. Calculating stationary points also lends itself to the solving of problems that require some variable to be maximised or minimised. These are referred to as optimisation problems.
The fuel used by a car is defined by $$f\left(v\right)=\frac{3}{80}{v}^{2}-6v+245$$, where $$v$$ is the travelling speed in $$\text{km/h}$$.
What is the most economical speed of the car? In other words, determine the speed of the car which uses the least amount of fuel.
If we draw the graph of this function we find that the graph has a minimum. The speed at the minimum would then give the most economical speed.
We have seen that the coordinates of the turning point can be calculated by differentiating the function and finding the $$x$$-coordinate (speed in the case of the example) for which the derivative is $$\text{0}$$.
${f}'\left(v\right)=\frac{3}{40}v-6$
If we set $${f}'\left(v\right)=0$$ we can calculate the speed that corresponds to the turning point:
\begin{align*} {f}'\left(v\right) & = \frac{3}{40}v-6 \\ 0 & = \frac{3}{40}v-6 \\ v & = \frac{6\times 40}{3} \\ & = 80 \end{align*}
This means that the most economical speed is $$\text{80}\text{ km/h}$$.
Finding the optimum point:
Let $$f'(x) = 0$$ and solve for $$x$$ to find the optimum point.
To check whether the optimum point at $$x = a$$ is a local minimum or a local maximum, we find $$f''(x)$$:
• If $$f''(a) < 0$$, then the point is a local maximum.
• If $$f''(a) > 0$$, then the point is a local minimum.
Worked example 21: Optimisation problems
The sum of two positive numbers is $$\text{10}$$. One of the numbers is multiplied by the square of the other. If each number is greater than $$\text{0}$$, find the numbers that make this product a maximum.
Draw a graph to illustrate the answer.
Analyse the problem and formulate the equations that are required
Let the two numbers be $$a$$ and $$b$$ and the product be $$P$$.
\begin{align*} a+b &=10 \ldots \ldots (1) \\ P &= a \times b^{2} \ldots \ldots (2) \end{align*}
Make $$b$$ the subject of equation ($$\text{1}$$) and substitute into equation ($$\text{2}$$):
\begin{align*} P &= a\left(10-a\right)^{2} \\ &=a \left( 100 - 20a + a^{2} \right) \\ \therefore P(a) &= 100a - 20a^{2} + a^{3} \end{align*}
Differentiate with respect to $$a$$
${P}'\left(a\right)= 100 - 40a + 3a^{2}$
Determine the stationary points by letting $$P'(a) = 0$$
We find the value of $$a$$ which makes $$P$$ a maximum:
\begin{align*} {P}'\left(a\right) & = 3a^{2} -40a + 100 \\ 0 &= (3a - 10)(a - 10) \\ \therefore a = 10 & \text{ or } a = \frac{10}{3} \end{align*}
Substitute into the equation ($$\text{1}$$) to solve for $$b$$:
\begin{align*} \text{If } a = 10 : \enspace b & = 10 - 10 \\ & = 0 \enspace (\text{but } b > 0 ) \\ \therefore \text{no solution} \\ \\ \text{If } a = \frac{10}{3} : \enspace b & = 10 - \frac{10}{3} \\ & = \frac{20}{3} \end{align*}
Determine the second derivative $$P''(a)$$
We check that the point $$\left(\frac{10}{3};\frac{20}{3}\right)$$ is a local maximum by showing that $${P}''\left(\frac{10}{3}\right) < 0$$:
\begin{align*} {P}''\left(a\right) & = 6a -40 \\ \therefore {P}''\left(\frac{10}{3}\right) & = 6\left(\frac{10}{3}\right) -40 \\ & = 20 -40 \\ & = -20 \end{align*}
The product is maximised when the two numbers are $$\frac{10}{3}$$ and $$\frac{20}{3}$$.
Draw a graph
To draw a rough sketch of the graph we need to calculate where the graph intersects with the axes and the maximum and minimum function values of the turning points:
Intercepts:
\begin{align*} P(a) &= a^{3} -20a^{2} + 100a \\ &= a(a - 10)^{2} \\ & \\ \text{Let } P(a) = 0: \enspace & (0;0) \text{ and } (10;0) \end{align*}
Turning points:
\begin{align*} P'(a) & = 0 \\ \therefore a = \frac{10}{3} & \text{ or } a = 10 \end{align*}
Maximum and minimum function values:
\begin{align*} \text{Substitute } \left( \frac{10}{3}; \frac{20}{3} \right): \enspace P &= ab^{2} \\ &= \left( \frac{10}{3} \right) \left( \frac{20}{3} \right)^{2} \\ &= \frac{4000}{27} \\ & \approx 148 \qquad (\text{Maximum turning point})\\ & \\ \text{Substitute } (0;10): \enspace P &= ab^{2} \\ &= \left( 10 \right) \left( 0 \right)^{2} \\ &= 0 \qquad (\text{Minimum turning point}) \end{align*}
Note: the above diagram is not drawn to scale.
Worked example 22: Optimisation problems
Michael wants to start a vegetable garden, which he decides to fence off in the shape of a rectangle from the rest of the garden. Michael has only $$\text{160}\text{ m}$$ of fencing, so he decides to use a wall as one border of the vegetable garden. Calculate the width and length of the garden that corresponds to the largest possible area that Michael can fence off.
Examine the problem and formulate the equations that are required
The important pieces of information given are related to the area and modified perimeter of the garden. We know that the area of the garden is given by the formula:
$\text{Area }= w \times l$
The fencing is only required for $$\text{3}$$ sides and the three sides must add up to $$\text{160}\text{ m}$$.
$160 = w + l + l$
Rearrange the formula to make $$w$$ the subject of the formula:
$w=160 -2 l$
Substitute the expression for $$w$$ into the formula for the area of the garden. Notice that this formula now contains only one unknown variable.
\begin{align*} \text{Area } &= l(160 - 2l) \\ &= 160l - 2l^{2} \end{align*}
Differentiate with respect to $$l$$
We are interested in maximising the area of the garden, so we differentiate to get the following:
\begin{align*} \frac{dA}{dl} = A' &= 160 - 4l \end{align*}
Calculate the stationary point
To find the stationary point, we set $${A}'\left(l\right)=0$$ and solve for the value(s) of $$l$$ that maximises the area:
\begin{align*} {A}'\left(l\right) & = 160 - 4l \\ 0 &= 160 - 4l \\ 4l &= 160 \\ \therefore l &= 40 \end{align*}
Therefore, the length of the garden is $$\text{40}\text{ m}$$.
Substitute to solve for the width:
\begin{align*} w & = 160 -2l \\ & = 160 -2\left(40 \right) \\ & = 160 - 80 \\ & = 80 \end{align*}
Therefore, the width of the garden is $$\text{80}\text{ m}$$.
Determine the second derivative $${A}''\left(l\right)$$
We can check that this gives a maximum area by showing that $${A}''\left(l\right) < 0$$:
${A}''\left(l\right) = -4$
A width of $$\text{80}\text{ m}$$ and a length of $$\text{40}\text{ m}$$ will give the maximum area for the garden.
Important note:
The quantity that is to be minimised or maximised must be expressed in terms of only one variable. To find the optimised solution we need to determine the derivative and we only know how to differentiate with respect to one variable (more complex rules for differentiation are studied at university level).
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Solving optimisation problems
Exercise 6.11
The sum of two positive numbers is $$\text{20}$$. One of the numbers is multiplied by the square of the other. Find the numbers that make this product a maximum.
Let the first number be $$x$$ and the second number be $$y$$ and let the product be $$P$$. We get the following two equations:
\begin{align*} x + y & = 20 \\ xy^{2} & = P \end{align*}
Rearranging the first equation and substituting into the second gives:
\begin{align*} P & = (20 - x)^{2}x \\ & = 400x - 40x^{2} + x^{3} \end{align*}
Differentiating and setting to $$\text{0}$$ gives:
\begin{align*} P' & = 400 - 80x + 3x^{2} \\ 0 & = 3x^{2} - 80x + 400 \\ & = (3x-20)(x-20) \end{align*}
Therefore, $$x=20$$ or $$x=\frac{20}{3}$$.
If $$x=20$$ then $$y=0$$ and the product is a minimum, not a maximum.
Therefore, $$x=\frac{20}{3}$$ and $$y=20-\frac{20}{3} = \frac{40}{3}$$.
Therefore the two numbers are $$\frac{20}{3}$$ and $$\frac{40}{3}$$ (approximating to the nearest integer gives $$\text{7}$$ and $$\text{13}$$).
A wooden block is made as shown in the diagram. The ends are right-angled triangles having sides $$3x$$, $$4x$$ and $$5x$$. The length of the block is $$y$$. The total surface area of the block is $$\text{3 600}\text{ cm^{2}}$$.
Show that $$y= \frac{\text{300} - x^{2}}{x}$$.
We start by finding the surface area of the prism:
\begin{align*} \text{Surface area} & = 2 \left(\frac{1}{2}b \times h\right) + 3xy + 4xy +5xy \\ \text{3 600} & = (3x \times 4x) + 12xy \\ & = 12x^{2} + 12 xy \end{align*}
Solving for $$y$$ gives:
\begin{align*} 12 xy & = \text{3 600} - 12x^{2} \\ y & = \frac{\text{3 600} - 12x^{2}}{12 x}\\ y& = \frac{\text{300} - x^{2}}{x} \end{align*}
Find the value of $$x$$ for which the block will have a maximum volume.
(Volume = area of base $$\times$$ height)
Start by finding an expression for volume in terms of $$x$$:
\begin{align*} V&=\text{ area of triangle } \times y \\ V & = 6x^{2} \times \frac{\text{300} - x^{2}}{x} \\ & = 6x ( \text{300} - x^{2})\\ & = \text{1 800}x - 6x^{3} \end{align*}
Now take the derivative and set it equal to $$\text{0}$$:
\begin{align*} V' & = \text{1 800} - 18x^{2} \\ 0 & = \text{1 800} - 18x^{2} \\ 18x^{2} & = \text{1 800} \\ x^{2} & = 100\\ x & = \pm 10 \end{align*}
Since the length can only be positive, $$x=10$$
$$\therefore x = \text{10}\text{ cm}$$
Determine the shortest vertical distance between the curves of $$f$$ and $$g$$ if it is given that: \begin{align*} f(x)&= -x^{2}+2x+3 \\ \text{and } g(x)&= \frac{8}{x}, \quad x > 0 \end{align*}
\begin{align*} \text{Let the distance } P(x) &= g(x) - f(x)\\ &=\frac{8}{x} - (-x^{2}+2x+3) \\ &=\frac{8}{x} +x^{2} - 2x - 3 \end{align*}
To minimise the distance between the curves, let $$P'(x) = 0:$$
\begin{align*} P'(x) &= - \frac{8}{x^2 } + 2x - 2 \qquad (x \ne 0) \\ 0 &= - \frac{8}{x^2 } + 2x - 2 \\ \therefore 0 &= - 8 + 2x^3 - 2x \\ 0 &= 2x^3 - 2x - 8 \\ 0 &= x^3 - x - 4 \\ 0 &= (x - 2)(x^2 + x + 2) \\ \therefore x =2 & \text{ or } x = \frac{-1 \pm \sqrt{(1)^2 - 4(1)(2)}}{2(1)} \\ & \quad = \text{ no real solutions } \\ \therefore x & =2 \end{align*}
Therefore, the shortest distance:
\begin{align*} P(2) &=\frac{8}{(2)} +(2)^{2} - 2(2) - 3 \\ &= 4 + 4 - 4 -3 \\ &= \text{1}\text{ unit} \end{align*}
The diagram shows the plan for a verandah which is to be built on the corner of a cottage. A railing $$ABCDE$$ is to be constructed around the four edges of the verandah.
If $$AB=DE=x$$ and $$BC=CD=y$$, and the length of the railing must be $$\text{30}\text{ m}$$, find the values of $$x$$ and $$y$$ for which the verandah will have a maximum area.
We need to determine an expression for the area in terms of only one variable.
The perimeter is:
\begin{align*} P & = 2x + 2y \\ 30 & = 2x + 2y \\ 15 & = x + y \\ y&=15-x \end{align*}
The area is:
\begin{align*} A & = y^{2} - (y-x)^{2} \\ & = y^{2} - (y^{2} - 2xy + x^{2}) \\ & = y^{2} - y^{2} + 2xy - x^{2} \\ & = 2xy - x^{2} \end{align*}
We use the expression for perimeter to eliminate the $$y$$ variable so that we have an expression for area in terms of $$x$$ only:
\begin{align*} A(x) & = 2x(15-x) - x^{2} \\ & = 30x - 2x^{2} - x^{2} \\ & = 30x - 3x^{2} \end{align*}
To find the maximum, we need to take the derivative and set it equal to $$\text{0}$$:
\begin{align*} A'(x) & = 30 - 6x \\ 0 & = 30 - 6x \\ 6x & = 30 \\ x & = 5 \end{align*}
Therefore, $$x=\text{5}\text{ m}$$ and substituting this value back into the formula for perimeter gives $$y=\text{10}\text{ m}$$.
A rectangular juice container, made from cardboard, has a square base and holds $$\text{750}\text{ cm}^{3}$$ of juice. The container has a specially designed top that folds to close the container. The cardboard needed to fold the top of the container is twice the cardboard needed for the base, which only needs a single layer of cardboard.
If the length of the sides of the base is $$x$$ cm, show that the total area of the cardboard needed for one container is given by: $A (\text{in square centimetres}) = \frac{\text{3 000}}{x} + 3x^{2}$
\begin{align*} V & = x^2h \\ 750 & = x^2h \\ \therefore h & = \frac{750}{x^2}\\ A & = \text{ area of sides } + \text{ area of base } + \text{ area of top } \\ &= 4xh + x^2 + 2x^2 \\ &= 4xh + 3x^2 \\ \text{Substitute } h &= \frac{750}{x^2}: \\ A &= 4x\left( \frac{750}{x^2} \right) + 3x^2 \\ &= \frac{3000}{x}+ 3x^2 \end{align*}
Determine the dimensions of the container so that the area of the cardboard used is minimised.
\begin{align*} A(x) &= \frac{3000}{x}+ 3x^2 \\ A'(x) &= - \frac{3000}{x^2}+ 6x \\ \therefore 0 &= - \frac{3000}{x^2}+ 6x \\ 6x &= \frac{3000}{x^2} \\ x^3 &= 500 \\ \therefore x &= \sqrt[3]{500} \\ &\approx \text{7,9}\text{ cm} \\ \therefore h & = \frac{750}{(\text{7,9})^2}\\ &\approx \text{12,0}\text{ cm} \end{align*}
Rates of change (EMCHK)
It is very useful to determine how fast (the rate at which) things are changing. Mathematically we can represent change in different ways. For example we can use algebraic formulae or graphs.
Graphs give a visual representation of the rate at which the function values change as the independent (input) variable changes. This rate of change is described by the gradient of the graph and can therefore be determined by calculating the derivative.
We have learnt how to determine the average gradient of a curve and how to determine the gradient of a curve at a given point. These concepts are also referred to as the average rate of change and the instantaneous rate of change.
$\text{Average rate of change } = \frac{f\left(x+h\right)-f\left(x\right)}{(x + h) - x}$ $\text{Instantaneous rate of change } = \lim_{h\to 0}\frac{f\left(x+h\right)-f\left(x\right)}{h}$
When we mention rate of change, the instantaneous rate of change (the derivative) is implied. When average rate of change is required, it will be specifically referred to as average rate of change.
Velocity is one of the most common forms of rate of change:
\begin{align*} \text{Average velocity } &= \text{Average rate of change } \\ & \\ \text{Instantaneous velocity } &= \text{Instantaneous rate of change } \\ &= \text{Derivative} \end{align*}
Velocity refers to the change in distance ($$s$$) for a corresponding change in time ($$t$$).
$v\left(t\right)=\frac{ds}{dt}={s}'\left(t\right)$
Acceleration is the change in velocity for a corresponding change in time. Therefore, acceleration is the derivative of velocity
$a\left(t\right)={v}'\left(t\right)$
This implies that acceleration is the second derivative of the distance.
$a\left(t\right)={s}''\left(t\right)$
Worked example 23: Rate of change
The height (in metres) of a golf ball $$t$$ seconds after it has been hit into the air, is given by $$H\left(t\right)=20t-5{t}^{2}$$. Determine the following:
1. The average vertical velocity of the ball during the first two seconds.
2. The vertical velocity of the ball after $$\text{1,5}$$ $$\text{s}$$.
3. The time at which the vertical velocity is zero.
4. The vertical velocity with which the ball hits the ground.
5. The acceleration of the ball.
Determine the average vertical velocity during the first two seconds
\begin{align*} {v}_{\text{ave}} & = \frac{H\left(2\right)-H\left(0\right)}{2-0} \\ & = \frac{\left[20\left(2\right)-5{\left(2\right)}^{2}\right]-\left[20\left(0\right)-5{\left(0\right)}^{2}\right]}{2} \\ & = \frac{40-20}{2} \\ & = \text{10}\text{ m.s$^{-1}$}\end{align*}
Calculate the instantaneous vertical velocity
\begin{align*} v\left(t\right) & = H'(t) \\ &= \frac{dH}{dt} \\ & = 20-10t \end{align*}
Velocity after $$\text{1,5}$$ $$\text{s}$$:
\begin{align*} v\left(\text{1,5}\right) & = 20-10\left(\text{1,5}\right) \\ & = \text{5}\text{ m.s$^{-1}$} \end{align*}
Determine the time at which the vertical velocity is zero
\begin{align*} v\left(t\right) & = 0 \\ 20-10t & = 0 \\ 10t & = 20 \\ t & = 2 \end{align*}
Therefore, the velocity is zero after $$\text{2}\text{ s}$$
Find the vertical velocity with which the ball hits the ground
The ball hits the ground when $$H\left(t\right)=0$$
\begin{align*} 20t-5{t}^{2} & = 0 \\ 5t\left(4-t\right) & = 0 \\ t=0& \text{ or } t=4 \end{align*}
The ball hits the ground after $$\text{4}$$ $$\text{s}$$. The velocity after $$\text{4}$$ $$\text{s}$$ will be:
\begin{align*} v\left(4\right) & = {H}'\left(4\right) \\ & = 20-10\left(4\right) \\ & = -\text{20}\text{ m.s$^{-1}$} \end{align*}
The ball hits the ground at a speed of $$\text{20}\text{ m.s^{-1}}$$. Notice that the sign of the velocity is negative which means that the ball is moving downward (a positive velocity is used for upwards motion).
Acceleration
\begin{align*} a=v'(t)&=H''(t) \\ &=-10 \end{align*}\begin{align*} \therefore a & = -\text{10}\text{ m.s$^{-2}$} \end{align*}
Just because gravity is constant does not mean we should necessarily think of acceleration as a constant. We should still consider it a function.
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Rates of change
Exercise 6.12
A pump is connected to a water reservoir. The volume of the water is controlled by the pump and is given by the formula: \begin{align*} V(d)&=64+44d-3d^{2} \\ \text{where } V&= \text{ volume in kilolitres}\\ d&= \text{ days} \end{align*}
Determine the rate of change of the volume of the reservoir with respect to time after $$\text{8}$$ days.
\begin{align*} \text{Rate of change }&= V'(d) \\ V'(d)&= 44 -6d \\ \text{After 8 days, rate of change will be:}\\ V'(8)&=44-6(8)\\ &= -\text{4}\text{ kℓ per day} \end{align*}
Is the volume of the water increasing or decreasing at the end of $$\text{8}$$ days. Explain your answer.
The rate of change is negative, so the function is decreasing.
After how many days will the reservoir be empty?
\begin{align*} \text{Reservoir empty: } V(d)&=0 \\ \therefore 64 + 44d -3d^{2}&=0 \\ (16-d)(4+3d)&=0\\ \therefore d = 16 \text{ or } & d = -\frac{4}{3} \\ \therefore \text{ It will be empty after } \text{16}\text{ days} \end{align*}
When will the amount of water be at a maximum?
Maximum at turning point.
\begin{align*} \text{Turning point where } V'(d)&= 0 \\ \therefore 44 - 6d&=0\\ d &=\frac{44}{6} \\ &= 7\frac{1}{3} \text{ days} \end{align*}
Calculate the maximum volume.
Maximum at turning point.
\begin{align*} \text{Maximum volume } &= V\left(\frac{22}{3}\right) \\ V\left(\frac{22}{3}\right) &= 64 + 44\left(\frac{22}{3}\right) - 3\left(\frac{22}{3}\right)^{2} \\ &= \text{225,3}\text{ kℓ} \end{align*}
Draw a graph of $$V(d)$$.
A soccer ball is kicked vertically into the air and its motion is represented by the equation: \begin{align*} D(t)&=1 + 18t -3t^{2} \\ \text{where } D &= \text{distance above the ground (in metres)} \\ t&= \text{ time elapsed (in seconds)} \end{align*}
Determine the initial height of the ball at the moment it is being kicked.
\begin{align*} D(t)&=1 + 18t - 3t^{2} \\ D(0)&=1 + 18(0) - 3(0)^{2} \\ &= 1 \text{ metre} \end{align*}
Find the initial velocity of the ball.
\begin{align*} \text{Velocity } = D'(t) &= 18 - 6t \\ \text{Initial velocity } &= D'(0) \\ D'(0) =18 - 6(0) &=\text{18}\text{ m.s$^{-1}$} \end{align*}
Determine the velocity of the ball after $$\text{1,5}$$ $$\text{s}$$.
\begin{align*} \text{Velocity after } \text{1,5}\text{ s}&=D'(\text{1,5}) \\ D'(\text{1,5})&=18-6(\text{1,5})^{2} \\ &=18-9 \\ &=\text{9}\text{ m.s$^{-1}$} \end{align*}
Calculate the maximum height of the ball.
Maximum height is at the turning point.
\begin{align*} \text{Turning point: } D'(t)&=0 \\ \therefore 18 - 6t&=0\\ 6t&=18 \\ t&=\text{3}\text{ s} \\ \text{Maximum height }&=D(3) \\ &=1+(18)(3)-(3)(3)^{2} \\ &=\text{28}\text{ metres} \end{align*}
Determine the acceleration of the ball after $$\text{1}$$ second and explain the meaning of the answer.
\begin{align*} \text{Acceleration }&= D''(t) \\ D''(t)&= -\text{6}\text{ m.s$^{-2}$} \end{align*}
Interpretation: the velocity is decreasing by $$\text{6}$$ metres per second per second.
Calculate the average velocity of the ball during the third second.
Average velocity during third second:
\begin{align*} &= \frac{D(3)-D(2)}{3-2} \\ &= \frac{1+18(3)-3(3)^{2}-\left[1+18(2)-3(2)^{2}\right]}{1} \\ &= \text{3}\text{ m.s$^{-1}$} \end{align*}
Determine the velocity of the ball after $$\text{3}$$ seconds and interpret the answer.
\begin{align*} \text{Instantaneous velocity}&= D'(3) \\ &= 18-6(3) \\ &= \text{0}\text{ m.s$^{-1}$} \end{align*}
Interpretation: this is the stationary point, where the derivative is zero. The ball has stopped going up and is about to begin its descent.
How long will it take for the ball to hit the ground?
\begin{align*} \text{Hits ground: } D(t)&=0 \\ -3t^{2}+18t+1&=0\\ t&=\frac{-18 \pm\sqrt{(18^{2}-4(1)(-3)}}{2(-3)} \\ t&=\frac{-18\pm\sqrt{336}}{-6} \\ \therefore t&=-\text{0,05} \text{ or } t=\text{6,05} \end{align*}
The ball hits the ground at $$\text{6,05}$$ $$\text{s}$$ (time cannot be negative).
Determine the velocity of the ball when it hits the ground.
\begin{align*} \text{Velocity after } \text{6,05}\text{ s}&= D'(\text{6,05}) \\ D'(\text{6,05})=18-5(\text{6,05})&= -\text{18,3}\text{ m.s$^{-1}$} \end{align*}
If the displacement $$s$$ (in metres) of a particle at time $$t$$ (in seconds) is governed by the equation $$s=\frac{1}{2}{t}^{3}-2t$$, find its acceleration after $$\text{2}$$ seconds.
We know that velocity is the rate of change of displacement. This means that $$\frac{dS}{dt} = v$$: \begin{align*} s &=\frac{1}{2}t^{3} - 2t \\ v &=\frac{3}{2}t^{2} - 2 \end{align*}
We also know that acceleration is the rate of change of velocity. This means that $$\frac{dv}{dt} = a$$: \begin{align*} v &=\frac{3}{2}t^{2} - 2 \\ a &= 3t \end{align*}
Substituting $$t=2$$ gives $$a=\text{6}\text{ m.s^{-2}}$$.
During an experiment the temperature $$T$$ (in degrees Celsius) varies with time $$t$$ (in hours) according to the formula: $$T\left(t\right)=30+4t-\frac{1}{2}{t}^{2}, \enspace t \in \left[1;10\right]$$.
Determine an expression for the rate of change of temperature with time.
We find the rate of change of temperature with time by differentiating: \begin{align*} T(t) &=30+4t-\frac{1}{2}t^{2} \\ T'(t) &= 4 - t \end{align*}
During which time interval was the temperature dropping?
We set the derivative equal to $$\text{0}$$: \begin{align*} 0 &= 4 - t \\ t &= 4 \end{align*} We look at the coefficient of the $$t^{2}$$ term to decide whether this is a minimum or maximum point. The coefficient is negative and therefore the function must have a maximum value. The interval in which the temperature is increasing is $$[1;4)$$. The interval in which the temperature is dropping is $$(4;10]$$. We can check this by drawing the graph or by substituting in the values for $$t$$ into the original equation. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 1.0000098943710327, "perplexity": 677.2172264593129}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-25/segments/1623487608702.10/warc/CC-MAIN-20210613100830-20210613130830-00328.warc.gz"} |
https://www.itdroplets.com/tag/2012/ | # Content Location shows the package as distributed but it’s not
I had this issues for a while and I ended up creating a ticket with Microsoft as I couldn’t get any answer to my issue. I’ve also posted this on technet, but nobody dared to answer 🙂
The reason why I wanted to open a ticket with Microsoft is that I have had a few issues since the migration to 2012 R2 and to be honest this kind of error it’s ok once, but what if it happens again? I could have just removed the package and started from scratch, that would have fixed it but this time I’ve been stubborn and I’ve learnt something new. Also, the package was 4GB, multiply that for all the Distribution Points. I don’t like to waste time if I don’t have to.
Here’s the post: SCCM Console is lying to me! Content shows as distributed but it isn’t.
### The issue
In my case, this happened with one of the Microsoft Office Deployment Packages I created.
When I clicked on the Deployment Package, the Status at the bottom of the console would tell me that the content was distributed (correctly) to 6 Distribution Points but when I actually went in the Deployment Package’s properties I could see like 10 Distribution Points in there. So basically Content Location shows the package as distributed but it’s not.
If I tried to Remove/Validate/Redistribute the content, nothing would happen. I couldn’t even treat this as a new distribution as the phantom Distribution Points would not appear as it happens when the content is already distributed.
I was able to distribute this content to other DPs. Updating the content did not help either.
I’ve also tried to either distribute (Start-CMContentDistribution) and remove (Remove-CMContentDistribution) the package from these DPs via PowerShell but also here no luck. When I tried to distribute it, it told me that the package was already distributed, when I removed it, nothing happened as when I did it through the console.
I then checked one of the Distribution Points through Content Library Explorer and I could see that the content was grayed out and the package ID had a * right after the ID. Also here I could not validate nor re-distribute the content.
### The cause
No idea! Microsoft went through the logs and found nothing. The engineer told me that he’s never seen such an issue before.
### The solution
Well, as there was no root cause, the only solution that Microsoft suggested was to manually delete the entries of these Distribution Points, associated to the Package ID, from the SQL server. (more…)
# Task sequence has failed with the error code 0x800700A1
I see lots of administrators struggling with this error especially when they replace an HDD on a machine and then try to re-image it. Task sequence has failed with the error code 0x800700A1 is an error that also come up when you’re imagining a brand new Virtual Machine.
What does the error mean? 0x800700A1 means BAD_PATHNAME. The reason why you get this error, most of the times, is due the fact that the task sequence isn’t able to find the path because the drive is RAW.
In order to fix this you will have to initialize and format that Hard Drive. If you are in the SCCM’s WinPE environment, simply press F8 to get to the command prompt and type the following:
(more…) | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8207522034645081, "perplexity": 1179.3010925110539}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570987822098.86/warc/CC-MAIN-20191022132135-20191022155635-00328.warc.gz"} |
https://math.stackexchange.com/questions/3462958/moore-penrose-pseudoinverse-of-a-matrix-product | # Moore-Penrose Pseudoinverse of a matrix product
$$X_{n \times p}$$ is a real, thin ($$n>p$$) rectangular matrix of rank $$p$$, so $$X^T X$$ is full rank. The Moore-Penrose pseudoinverse of $$X$$ is given by $$X^+=(X^TX)^{-1}X^T$$.
Let's now define $$W=XA$$
($$X$$ is linearly transformed by $$A_{p \times k}$$ to produce a $$W_{n \times k}$$). $$W$$ is of rank $$k$$ (so $$W^TW$$ is full rank).
I need to compute the Moore-Penrose Pseudoinverse of $$W$$ for many different $$A$$ matrices efficiently. Is there an identity or a decomposition that allows expressing $$W^+$$ as some function of $$X^+$$ or $$(X^TX)^{-1}$$?
due to distributive order reversal rule for transposing multiple and associativity of multiplication , $$W^+=(A^T (X^TX) A)^{-1} A^T X^T$$ not too much opportunity for pre-calculation it seems | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 19, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9808038473129272, "perplexity": 262.56170212359683}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320304572.73/warc/CC-MAIN-20220124155118-20220124185118-00159.warc.gz"} |
https://www.nag.com/numeric/py/nagdoc_latest/naginterfaces.library.ieee.create_infinity.html | # naginterfaces.library.ieee.create_infinity¶
naginterfaces.library.ieee.create_infinity(isign)[source]
create_infinity creates a signed infinite value.
For full information please refer to the NAG Library document for x07ba
https://www.nag.com/numeric/nl/nagdoc_27.1/flhtml/x07/x07baf.html
Parameters
isignint
Determines the sign of the infinity to be created.
If is greater than or equal to , a positive infinity is returned, otherwise a negative infinity is returned.
Returns
xfloat
The required infinite value.
Notes
create_infinity sets to be positive or negative infinity.
References
IEEE, 2008, Standard for Floating-Point Arithmetic (IEEE Standard 754-2008), IEEE, New York. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9118744134902954, "perplexity": 2738.012607640656}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-31/segments/1627046154163.9/warc/CC-MAIN-20210801061513-20210801091513-00427.warc.gz"} |
https://www.physicsforums.com/threads/crystal-lattice-and-unit-cell.110932/ | # Homework Help: Crystal lattice and unit cell
1. Feb 16, 2006
### asdf1
what's the difference between those two?
2. Feb 16, 2006
### assyrian_77
A lattice is a (idealized) structure of crystals where each lattice point represent an identical group of atoms. A group is sometimes called a basis. One can form basis vectors with which all lattice point positions can be expressed. The unit cell (I guess you mean the primitive cell) is the volume defined by the set of basis vectors. This is the smallest volume; there are no cells with smaller volumes.
Hopefully this was of help. If not, don't hesitate to ask.
3. Feb 16, 2006
### Gokul43201
Staff Emeritus
What does your textbook or class notes say ?
4. Feb 17, 2006
### asdf1
my textbook says:
unit cell: crystal structure that's the smallest group of atoms possessing
the symmetry of the crystal
lattice: orderly array of points in space
so does that mean that the unit cell is part of the lattice?
5. Feb 17, 2006
### Gokul43201
Staff Emeritus
Yes, it is the smallest "representative" part.
6. Feb 17, 2006
### asdf1
ok, thanks!!! | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.806404173374176, "perplexity": 2811.2401846716302}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-30/segments/1531676590329.25/warc/CC-MAIN-20180718193656-20180718213656-00400.warc.gz"} |
http://mathhelpforum.com/trigonometry/128324-worked-solution-provided-but-why-z-360-deg-rejected.html | # Math Help - Worked solution provided, but why is z = 360 deg rejected?
1. ## Worked solution provided, but why is z = 360 deg rejected?
Can any kind soul help me in the below?
Find z between 0 and 360 degrees which satisfy the equation.
1. sec^2 z = 4 sec z - 3
sec^2 z = 4 sec z - 3
sec^2 z - 4 sec z + 3 = 0
(sec z - 3)( sec z - 1) = 0
sec z = 3 or sec z = 1
cos z = 1/3 or cos z = 1
z = 70.5 deg, z = 0, 360 deg(rejected)
cos is positive => so in 1st or 4th quadrant
z = 70.5, 360-70.5
= 70.5, 289.5
My question: why is z = 360 deg rejected?
why is z = 0 deg not rejected?
2. Originally Posted by ppppp77
My question: why is z = 360 deg rejected?
why is z = 0 deg not rejected?
It is not clear why in this case, was your domain $[0,360)$ or $[0,360]$ ?
3. Originally Posted by ppppp77
Can any kind soul help me in the below?
Find z between 0 and 360 degrees which satisfy the equation.
1. sec^2 z = 4 sec z - 3
sec^2 z = 4 sec z - 3
sec^2 z - 4 sec z + 3 = 0
(sec z - 3)( sec z - 1) = 0
sec z = 3 or sec z = 1
cos z = 1/3 or cos z = 1
z = 70.5 deg, z = 0, 360 deg(rejected)
cos is positive => so in 1st or 4th quadrant
z = 70.5, 360-70.5
= 70.5, 289.5
My question: why is z = 360 deg rejected?
why is z = 0 deg not rejected?
oh my, pickslides, you are really awesome!! | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 2, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8651766180992126, "perplexity": 2381.8130912473725}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-41/segments/1410657130272.73/warc/CC-MAIN-20140914011210-00188-ip-10-196-40-205.us-west-1.compute.internal.warc.gz"} |
http://mathhelpforum.com/calculus/60110-another-fundamental-theorem-calculus-problem.html | # Thread: Another Fundamental Theorem of Calculus problem
1. ## Another Fundamental Theorem of Calculus problem
Does anyone know how to start or do this problem. I don't have an example in my book of it.
$\int_{-8}^ {6} f(x)= 2+e^x$
2. Originally Posted by McDiesel
Does anyone know how to start or do this problem. I don't have an example in my book of it.
Originally Posted by McDiesel
$\int_{-8}^ {6} f(x)= 2+e^x$
That is a nonsense problem. That is, it show the confusion of whom so ever wrote it
$\int_{-8}^ {6} f(x)$ is a definite integral, a number!
Whereas $2+e^x$ is a function of x.
Are you sure that you have given all the information?
3. I'm sorry,
Use a definite integral to find the area between $f(x)= 2+e^x$ and the x-axis over the integral [-8,6]
4. Originally Posted by McDiesel
I'm sorry,
Use a definite integral to find the area between $f(x)= 2+e^x$ and the x-axis over the integral [-8,6]
This is very simple since $\forall{x}\in\mathbb{R}~2+e^x>0$
So $\text{Area}=\int_{-8}^{6}2+e^xdx$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 8, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9948027729988098, "perplexity": 421.04259561905}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-09/segments/1487501170651.78/warc/CC-MAIN-20170219104610-00391-ip-10-171-10-108.ec2.internal.warc.gz"} |
http://elsenaju.eu/Determinant/Vector-Calculation.htm | # Vector Calculus
## History of vector calculation
The vector calculus goes back to H. G. Grassmann and parallel to Hamilton. Grassmann published in 1844 the "Lineale Ausdehnungslehre". As a precursor Descartes and Möbius apply. The Irish mathematician William Rowan Hamilton (1805 - 1865) developed the theory of quaternions, which are considered as precursors of the vectors. The term scalar goes back to Hamilton. 1888 the Italian mathematician Giuseppe Peano (1858-1932) developed an axiomatic definition of a vector space.
## Scalar
Quantities that can be represented by a real number. Examples for scalar values are temperature, mass, ...
## Vector
Vectors, in addition to its value require the specification of a direction. In physics, e.g. Velocity, field strength, ... In general, a vector is not limited to three dimensions. In general it is an n-tuple of real numbers that is often listed as a column vector.
$\stackrel{\to }{v}=\left(\begin{array}{c}{\mathrm{v1}}_{}\\ {v}_{2}\\ ⋮\\ {v}_{n}\end{array}\right)$
## Linear Dependence
Two vectors are called linearly dependent or collinear if one can be converted by multiplication by a scalar in the other. The cross product of collinear vectors disappears.
## Orthogonality
Two vectors are called orthogonal if the dot product of two vectors vanishes. Geometrically, the vectors are perpendicular to each other then that is the angle enclosed by the vectors is 90°.
## Unit vector
Vectors of length 1 are called unit vectors. Each vector can be converted by normalizing into the unit vector by the vector is divided by its length.
## Calculation rules for vectors
### Multiplication of a vector with a scalar
The multiplication of a vector by a scalar positive λ only changes the length of the vector and not direction. In the multiplication with a scalar, the negative direction of the vector changes in the opposite direction.
${\lambda }\cdot \stackrel{\to }{v}=\left(\begin{array}{c}{\lambda }\cdot {v}_{1}\\ {\lambda }\cdot {v}_{2}\\ ⋮\\ {\lambda }\cdot {v}_{n}\end{array}\right)$
For the multiplication of a vector by a scalar λ the distributive law holds.
${\lambda }\cdot \left(\stackrel{\to }{v}+\stackrel{\to }{w}\right)={\lambda }\cdot \stackrel{\to }{v}+{\lambda }\cdot \stackrel{\to }{w}$
The addition of vectors is done in Cartesian coordinates componentwise. The vector addition is commutative and associative.
$\stackrel{\to }{v}+\stackrel{\to }{w}=\left(\begin{array}{c}{v}_{1}\\ {v}_{2}\\ ⋮\\ {v}_{n}\end{array}\right)+\left(\begin{array}{c}{w}_{1}\\ {w}_{2}\\ ⋮\\ {w}_{n}\end{array}\right)=\left(\begin{array}{c}{v}_{1}+{w}_{1}\\ {v}_{2}+{w}_{2}\\ ⋮\\ {v}_{n}+{w}_{n}\end{array}\right)$
Geometrically, the resulting vector can be constructed by one of the vectors is parallel shifted to the other end point of the vector. The connection from the start point of the first vector to the end point of the second vector is the resultant of vector of vector addition.Graphical vector addition
### Subtraction of vectors
Subtraction of vectors is carried out as by the addition of the components are subtracted.
$\stackrel{\to }{v}-\stackrel{\to }{w}=\left(\begin{array}{c}{v}_{1}\\ {v}_{2}\\ ⋮\\ {v}_{n}\end{array}\right)-\left(\begin{array}{c}{w}_{1}\\ {w}_{2}\\ ⋮\\ {w}_{n}\end{array}\right)=\left(\begin{array}{c}{v}_{1}-{w}_{1}\\ {v}_{2}-{w}_{2}\\ ⋮\\ {v}_{n}-{w}_{n}\end{array}\right)$
Geometrically, the construction similar to the addition of only the vector will be mirrored to the negative sign in the direction.Graphical vector subtraction
### Dot Product (inner Product) of vectors
The dot product is defined as the product of the components and the sum of these products. The scalar product is not of the order-dependent (commutative). The scalar product name comes from the fact that the result is a scalar and not a vector.
$\stackrel{\to }{v}\cdot \stackrel{\to }{w}=\left(\begin{array}{c}{v}_{1}\\ {v}_{2}\\ ⋮\\ {v}_{n}\end{array}\right)\cdot \left(\begin{array}{c}{w}_{1}\\ {w}_{2}\\ ⋮\\ {w}_{n}\end{array}\right)$ $={v}_{1}\cdot {w}_{1}+{v}_{2}\cdot {w}_{2}+\dots +{v}_{n}\cdot {w}_{n}$$=\sum _{i=1}^{n}\left({v}_{i}\cdot {w}_{i}\right)$
The scalar product can be interpreted as the product of the geometric projection of a vector in the direction of the other vector. Physically it means that the product is formed only with the Kompnente of the vector that is effective in the direction of another vector.
The scalar product can be expressed geometrically. Here, φ the included angle of the vectors.Graphical inner product
$\stackrel{\to }{v}\cdot \stackrel{\to }{w}=|\stackrel{\to }{v}||\stackrel{\to }{w}|\mathrm{cos}\phi$
For the scalar product of the distributive law apply
$\stackrel{\to }{u}\cdot \left(\stackrel{\to }{v}+\stackrel{\to }{w}\right)=\stackrel{\to }{u}\cdot \stackrel{\to }{v}+\stackrel{\to }{u}\cdot \stackrel{\to }{w}$
and the commutative law
$\stackrel{\to }{v}\cdot \stackrel{\to }{w}=\stackrel{\to }{w}\cdot \stackrel{\to }{v}$
Dot Product calculator for vectors with 3 components.
$\stackrel{\to }{v}\cdot \stackrel{\to }{w}=\left(\begin{array}{c}{v}_{1}\\ {v}_{2}\\ {v}_{3}\end{array}\right)\cdot \left(\begin{array}{c}{w}_{1}\\ {w}_{2}\\ {w}_{3}\end{array}\right)$
Input fields for the vector elements.
v1= w1=
v2= w2=
v3= w3=
### Length of a vector
The magnitude of a vector geometrically corresponds to the length of the vector.
$\left|\stackrel{\to }{v}\right|=$ $|$ $\left(\begin{array}{c}{v}_{1}\\ {v}_{2}\\ ⋮\\ {v}_{n}\end{array}\right)$ $|$$=\sqrt{{v}_{1}^{2}+{v}_{2}^{2}+\dots +{v}_{n}^{2}}$$=\sqrt{\sum _{i=1}^{n}{v}_{i}^{2}}$
Calculator for the length of a vector with 3 components.
$|$ $\stackrel{\to }{v}$ $|$ $=$ $|$ $\left(\begin{array}{c}{v}_{1}\\ {v}_{2}\\ {v}_{3}\end{array}\right)$ $|$
Input fields for the vector elements.
v1=
v2=
v3=
### Cross Product (outer product) of vectors
The vector product is defined in three-dimensional Euclidean vector space as follows.
$\stackrel{\to }{v}\cdot \stackrel{\to }{w}=\left(\begin{array}{c}{v}_{1}\\ {v}_{2}\\ {v}_{3}\end{array}\right)⨯\left(\begin{array}{c}{w}_{1}\\ {w}_{2}\\ {w}_{3}\end{array}\right)$$=\left(\begin{array}{c}{v}_{2}{w}_{3}-{v}_{3}{w}_{2}\\ {v}_{1}{w}_{3}-{v}_{3}{w}_{1}\\ {v}_{1}{w}_{2}-{v}_{2}{w}_{1}\end{array}\right)$
The vector product provides as a result a vector which is perpendicular to the plane spanned by the two vectors and the length of which corresponds to the surface area of the clamped parallelogram.
The cross product can be expressed geometrically. Here, φ the included angle of the vectors and n is the vector perpendicular to the surface.
$v→⨯w→= |v→||w→|sinφn→$
The vector product is anti-commutative
$v→⨯w→ = -w→⨯v→$
Cross Product calculator for vectors with 3 components.
$\stackrel{\to }{v}\cdot \stackrel{\to }{w}=\left(\begin{array}{c}{v}_{1}\\ {v}_{2}\\ {v}_{3}\end{array}\right)⨯\left(\begin{array}{c}{w}_{1}\\ {w}_{2}\\ {w}_{3}\end{array}\right)$
Input fields for the vector elements.
v1= w1=
v2= w2=
v3= w3=
### Multiplication of a vector with a matrix
The product of a matrix by a vector is a linear mapping. Explains the multiplication when the number of columns of the matrix is equal to the number of elements of the vector. The result is a vector whose number of components equal to the number of rows of the matrix. That the example a matrix with two rows a vector always maps to a vector with two components.Graphical Matrix-Vector produkt
$A\cdot \stackrel{\to }{v}=\left(\begin{array}{cccc}{a}_{11}& {a}_{12}& \dots & {a}_{1m}\\ {a}_{21}& {a}_{22}& \dots & {a}_{2m}\\ & ⋮\\ {a}_{n1}& {a}_{n2}& \dots & {a}_{nm}\end{array}\right)\cdot \left(\begin{array}{c}{v}_{1}\\ {v}_{2}\\ ⋮\\ {v}_{m}\end{array}\right)=$
$\left(\begin{array}{c}{a}_{11}{v}_{1}+{a}_{12}{v}_{2}+\dots +{a}_{1m}{v}_{m}\\ {a}_{21}{v}_{1}+{a}_{22}{v}_{2}+\dots +{a}_{2m}{v}_{m}\\ ⋮\\ {a}_{n1}{v}_{1}+{a}_{n2}{v}_{2}+\dots +{a}_{nm}{v}_{m}\end{array}\right)$
## Multiple products
Multiple products of vectors are not associative in general. That the general rule is that the order in which the products are carried out is relevant.
In general is:
$\stackrel{\to }{u}\cdot \left(\stackrel{\to }{v}\cdot \stackrel{\to }{w}\right)\ne \left(\stackrel{\to }{u}\cdot \stackrel{\to }{v}\right)\cdot \stackrel{\to }{w}$
and
$\stackrel{\to }{u}⨯\left(\stackrel{\to }{v}⨯\stackrel{\to }{w}\right)\ne \left(\stackrel{\to }{u}⨯\stackrel{\to }{v}\right)⨯\stackrel{\to }{w}$
For the double vector product the development applies:
$\stackrel{\to }{u}⨯\left(\stackrel{\to }{v}⨯\stackrel{\to }{w}\right)=\stackrel{\to }{v}\cdot \left(\stackrel{\to }{u}\cdot \stackrel{\to }{w}\right)-\stackrel{\to }{w}\cdot \left(\stackrel{\to }{u}\cdot \stackrel{\to }{v}\right)$
The scalar triple product $\left(\stackrel{\to }{u}⨯\stackrel{\to }{v}\right)\cdot \stackrel{\to }{w}$ is equal to the volume of the plane defined by the three vectors parallelepiped. The scalar triple product is positive if the three vectors form a right hand system.
Lagrangesche Identity:
$\left(\stackrel{\to }{m}⨯\stackrel{\to }{u}\right)\cdot \left(\stackrel{\to }{v}⨯\stackrel{\to }{w}\right)$$=\left(\stackrel{\to }{m}\cdot \stackrel{\to }{v}\right)\left(\stackrel{\to }{u}\cdot \stackrel{\to }{w}\right)-\left(\stackrel{\to }{u}\cdot \stackrel{\to }{v}\right)\left(\stackrel{\to }{m}\cdot \stackrel{\to }{w}\right)$
## Geometrical approches
Equation of the line through the two points P0 und P1 given by the vectors r0 und r1:
$\stackrel{\to }{r}=\stackrel{\to }{{r}_{0}}+\lambda \left(\stackrel{\to }{{r}_{1}}-\stackrel{\to }{{r}_{0}}\right)$
Distance of two points P0 and P1:
$\left|\stackrel{\to }{{r}_{1}}-\stackrel{\to }{{r}_{0}}\right|$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 40, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8892501592636108, "perplexity": 896.2351412093548}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-13/segments/1552912202688.89/warc/CC-MAIN-20190322180106-20190322202106-00357.warc.gz"} |
http://mathhelpforum.com/advanced-math-topics/134219-complex-analysis-print.html | # complex analysis
• March 16th 2010, 11:53 PM
Kat-M
complex analysis
Let $\gamma$: [0,1] $\rightarrow C$ be any $C^1$ curve. Define $f(z)=\oint_{\gamma} \frac{1}{\zeta-z}d\zeta$
Prove that $f$ is holomorphic on $C$\ $\tilde{\gamma}$, where $\tilde{\gamma}=\{ \gamma (t):0\leq t \leq 1\}.$
• September 12th 2010, 09:43 PM
Rebesques
You can compute the difference quotient $[f(z+w)-f(z)]/w$ as $w\rightarrow 0$. The limit is (what a surprise) $f'(z)=-\int_{\gamma} \frac{1}{(\zeta-z)^2}d\zeta$.
ps. what's with the \oint? I believe that notation is,
only to signify counter-clockwise integration along a closed curve (Thinking) | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 11, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9926777482032776, "perplexity": 2927.5392908232534}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-18/segments/1430455162832.10/warc/CC-MAIN-20150501043922-00067-ip-10-235-10-82.ec2.internal.warc.gz"} |
https://www.gradesaver.com/textbooks/math/applied-mathematics/elementary-technical-mathematics/chapter-10-section-10-3-finding-binomial-factors-exercise-page-350/67 | ## Elementary Technical Mathematics
$3x(y-3)^2$
$3xy^2-18xy+27x\longrightarrow$ 3x is a common factor. $3x(y^2-6y+9)$ =$3x(\ \ \ \ \ \ \ \ \ )(\ \ \ \ \ \ \ \ \ )$ =$3x(y\ \ \ \ \ \ \ )(y\ \ \ \ \ \ \ )$ =$3x(y-\ \ )(y-\ \ )$ Since the 2nd term is negative and the 3rd is positive, the factors of 9 must both be negative. =$3x(y-3)(y-3)$ -3 and -3 have a sum of -6 and a product of 9. Since $y-3$ is repeated as a factor it is written as an exponent. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.913254976272583, "perplexity": 333.9499494037471}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-39/segments/1537267160454.88/warc/CC-MAIN-20180924125835-20180924150235-00281.warc.gz"} |
https://www.groundai.com/project/super-chandrasekhar-mass-light-curve-models-for-the-highly-luminous-type-ia-supernova-2009dc/ | Super-Chandrasekhar-Mass LC Models for SN 2009dc
# Super-Chandrasekhar-Mass Light Curve Models for the Highly Luminous Type Ia Supernova 2009dc
Yasuomi Kamiya11affiliation: Department of Astronomy, Graduate School of Science, the University of Tokyo, 7-3-1 Hongo, Bunkyo-ku, Tokyo 113-0033, Japan. 22affiliation: Kavli Institute for the Physics and Mathematics of the Universe, Todai Institutes for Advanced Study, the University of Tokyo, 5-1-5 Kashiwanoha, Kashiwa, Chiba 277-8583, Japan. 77affiliation: Research Fellow of the Japan Society for the Promotion of Science. Masaomi Tanaka33affiliation: National Astronomical Observatory of Japan, 2-21-1 Osawa, Mitaka, Tokyo 181-8588, Japan. 22affiliation: Kavli Institute for the Physics and Mathematics of the Universe, Todai Institutes for Advanced Study, the University of Tokyo, 5-1-5 Kashiwanoha, Kashiwa, Chiba 277-8583, Japan. Ken’ichi Nomoto22affiliation: Kavli Institute for the Physics and Mathematics of the Universe, Todai Institutes for Advanced Study, the University of Tokyo, 5-1-5 Kashiwanoha, Kashiwa, Chiba 277-8583, Japan. 11affiliation: Department of Astronomy, Graduate School of Science, the University of Tokyo, 7-3-1 Hongo, Bunkyo-ku, Tokyo 113-0033, Japan. Sergei I. Blinnikov44affiliation: Institute for Theoretical and Experimental Physics, 117218 Moscow, Russia. 55affiliation: Sternberg Astronomical Institute, Lomonosov Moscow State University, 119992 Moscow, Russia. 22affiliation: Kavli Institute for the Physics and Mathematics of the Universe, Todai Institutes for Advanced Study, the University of Tokyo, 5-1-5 Kashiwanoha, Kashiwa, Chiba 277-8583, Japan. Elena I. Sorokina55affiliation: Sternberg Astronomical Institute, Lomonosov Moscow State University, 119992 Moscow, Russia. 22affiliation: Kavli Institute for the Physics and Mathematics of the Universe, Todai Institutes for Advanced Study, the University of Tokyo, 5-1-5 Kashiwanoha, Kashiwa, Chiba 277-8583, Japan. Tomoharu Suzuki66affiliation: College of Engineering, Chubu University, 1200 Matsumoto-cho, Kasugai, Aichi 487-8501, Japan. 11affiliation: Department of Astronomy, Graduate School of Science, the University of Tokyo, 7-3-1 Hongo, Bunkyo-ku, Tokyo 113-0033, Japan.
###### Abstract
Several highly luminous Type Ia supernovae (SNe Ia) have been discovered. Their high luminosities are difficult to explain with the thermonuclear explosions of the Chandrasekhar-mass white dwarfs (WDs). In the present study, we estimate the progenitor mass of SN 2009dc, one of the extremely luminous SNe Ia, using the hydrodynamical models as follows. Explosion models of super-Chandrasekhar-mass (super-Ch-mass) WDs are constructed, and multi-color light curves (LCs) are calculated. The comparison between our calculations and the observations of SN 2009dc suggests that the exploding WD has a super-Ch mass of 2.2–2.4 , producing 1.2–1.4 of Ni, if the extinction by its host galaxy is negligible. If the extinction is significant, the exploding WD is as massive as 2.8 , and 1.8 of Ni is necessary to account for the observations. Whether the host-galaxy extinction is significant or not, the progenitor WD must have a thick carbon-oxygen layer in the outermost zone (20–30% of the WD mass), which explains the observed low expansion velocity of the ejecta and the presence of carbon. Our estimate on the mass of the progenitor WD, especially for the extinction-corrected case, is challenging to the current scenarios of SNe Ia. Implications on the progenitor scenarios are also discussed.
supernovae: individual (SN 2009dc) — radiative transfer — white dwarfs
slugcomment: Accepted for publication in ApJ on July 18, 2012.
## 1. Introduction
It has been widely accepted that a Type Ia supernova (SN Ia) results from a thermonuclear explosion of a carbon-oxygen (C+O) white dwarf (WD) in a close binary system. The most likely model is that the explosion is triggered by carbon ignition in the central region of the WD when the WD mass () reaches the critical mass (, 1.38 for a non-rotating C+O WD; e.g., Hillebrandt & Niemeyer, 2000; Nomoto et al., 1997, 2000). Since is very close to the Chandrasekhar’s limiting mass (Chandrasekhar mass, )111 Hereafter denotes the Chandrasekhar’s limiting mass for a non-rotating C+O WD, i.e., , where is the electron mole fraction (e.g., Chanrdasekhar, 1939)., the resulting explosions are expected to have similar properties. Actually, normal SNe Ia are used as standard candles in cosmology (Riess et al., 1998; Perlmutter et al., 1999), after correcting their luminosity dispersion by using the Pskovskii-Phillips relation (e.g., Pskovskii, 1977; Phillips, 1993).
Despite their uniformity, several unusual SNe Ia have been found to be much more luminous than normal ones. They are SN 2003fg (Howell et al., 2006), SN 2006gz (Hicken et al., 2007), SN 2007if (Scalzo et al., 2010; Yuan et al., 2010), and SN 2009dc (Yamanaka et al., 2009; Tanaka et al., 2010; Silverman et al., 2011; Taubenberger et al., 2011). These SNe Ia all show slow luminosity evolutions (e.g., Scalzo et al., 2010, Table 4). Three of them, except for SN 2003fg, show the clear absorption line of C ii in their early spectra, which are rarely detected for normal SNe Ia (e.g., Marion et al., 2006; Tanaka et al., 2008). Such extremely high luminosities require 1.2 of radioactive Ni if their explosions are spherically symmetric. In order to produce such a large amount of Ni, their progenitor C+O WDs are suggested to have super-Chandrasekhar (super-Ch) mass (i.e., ), because the exploding WDs should contain more than 0.3 of the Si-rich layer and the unburned C+O layer on top of the Ni-rich core (Howell et al., 2006; Hicken et al., 2007; Scalzo et al., 2010; Yuan et al., 2010; Yamanaka et al., 2009; Silverman et al., 2011; Taubenberger et al., 2011). Alternatively, it could also be possible to explain the extremely luminous SNe Ia by asymmetric explosions of Chandrasekhar-mass C+O WDs (Hillebrandt et al., 2007). In this paper, we focus on SN 2009dc and approximate it with a spherically symmetric model, because the spectropolarimetric observations of SN 2009dc suggest that it is a globally spherical explosion (Tanaka et al., 2010).
A super-Ch-mass C+O WD model can be formed if it is supported by rapid rotation. For example, Hachisu (1986) constructed two-dimensional models of rapidly rotating WDs. Uenishi et al. (2003) calculated the structure and evolution of two-dimensional C+O WDs that rotate by getting angular momentum form accreting matter. Yoon & Langer (2005) investigated the stability of rapidly rotating C+O WDs for a wider parameter range.
Explosions and nucleosynthsis of super-Ch-mass C+O WDs were simulated by Steinmetz et al. (1992), Pfannes et al. (2010a), and Pfannes et al. (2010b). Maeda & Iwamoto (2009) studied the bolometric light curves (LCs) of super-Ch-mass WD models. They constructed the homologously expanding models of super-Ch-mass WDs with parameters of WD mass, Ni mass, abundance distribution, and so on. Scalzo et al. (2010) studied the properties of SN 2007if, assuming a shell-surrounded super-Ch-mass WD model as a result of a WD merger. They estimated that the ejecta mass is 2.4 with 1.6 of Ni.
By applying Arnett’s law to the synthesized bolometric LCs, the Ni mass of a SN can be estimated (e.g., Arnett, 1982). Yamanaka et al. (2009) and Silverman et al. (2011) have suggested that SN 2009dc has 1.2 if they neglect the extinction by its host galaxy. They have also reported that the Ni mass could be as large as 1.6–1.7 by taking the extinction into account. Taubenberger et al. (2011) has also analytically estimated that the total mass of SN 2009dc is 2.8 and that the ejected Ni mass is 1.8 . These extreme values challenge current models and scenarios for SNe Ia.
As described above, all the past works rely on the bolometric LCs, which involves some uncertainties (see Section 2.2). In most cases, the analytic method is used to estimate the masses of ejecta and ejected Ni. To derive more accurate properties of the super-Ch candidates for discussing their progenitor scenarios, more sophisticated models are needed. In this paper, we calculate multi-color LCs for homologously expanding models of super-Ch-mass WDs for the first time. Section 2 describes our super-Ch-mass WD models and LC calculations. In Section 3, we compare our results with the photometric and spectroscopic observations of SN 2009dc to estimate the masses of the ejecta and Ni. Implications of our results are discussed in Section 4. We summarize our conclusions in Section 5.
## 2. Models and Calculations
In this section, we describe the procedures for constructing our super-Ch-mass WD models and the code for calculating their LCs, to compare with the observations of SN 2009dc. Since SN 2009dc has a continuum polarization as small as the normal SNe Ia (Tanaka et al., 2010), spherical symmetry is assumed.
### 2.1. Super-Chandrasekhar-Mass White Dwarf Models
To construct the homologously expanding models of super-Ch-mass WDs, we apply the approximations similar to those adopted by Maeda & Iwamoto (2009). The models are described by the following parameters.
• : total WD mass.
• (or ): mass (or mass fraction) of Ni. The mass fraction hereafter means the ratio to ; e.g., .
• : mass fraction of electron-captured elements (ECEs; mostly Fe, Fe, Co, and Ni). Stable Fe, Co, and Ni are included, all of which are simply assumed to have the same mass fraction.
• : mass fraction of intermediate-mass elements (IMEs). Si, S, and Ca are included, whose mass fraction ratio is , similar to that in the Chandrasekhar-mass WD model, W7 (Nomoto et al., 1984; Thielemann et al., 1986).
• : mass fraction of C+O. C and O are contained equally in mass fraction.
Since the equation
fECE+f56Ni+fIME+fCO=1 (1)
holds, we eliminate in the following discussion. In total, we have four independent parameters.
The expansion model of a super-Ch-mass WD is constructed as follows. Firstly, with the above parameters, we calculate the nuclear energy release during the explosion () by
Enuc1051erg=[0.5fECE+0.32f56Ni+1.24(1−fCO)] ×MWDM\sun (2)
(e.g., Maeda & Iwamoto, 2009). Then, the binding energy of the WD () is evaluated by Equations (22) and (32)–(34) described in Yoon & Langer (2005). Here, g cm for all models as in Maeda & Iwamoto (2009), so that increases almost linearly with . For , we extrapolate the formula of (Jeffery et al., 2006). These and give the kinetic energy of the exploding WD as
Ekin=Enuc−Ebin. (3)
How much mass fraction of Ni is synthesized at the deflagration or detonation wave depends mainly on the temperature and thus on the density at the flame front. For W7, e.g., Ni is synthesized at the flame densities of g cm. Then the total synthesized Ni mass depends on the mass contained in this density range, and thus on the presupernova density structure of the WD and the flame speed. The rotating WD is more massive than the non-rotating WD with the same central density (Yoon & Langer, 2005). For the same central density, therefore, the density profile is shallower for more massive stars, and thus the mass contained in the density range for Ni synthesis is larger; i.e., a more massive WD tends to synthesize more Ni. Also, for the same central density, faster flame produces more Ni because of less pre-expansion, and the flame speed depends on the WD mass, rotation law, density structure, and possible transition to detonation. To provide constraints on these physical processes, we calculate the light curve models with various nuclear yields and for the same central density models.
Next, the structure of the explosion model of a super-Ch-mass WD is obtained by scaling W7, the canonical Chandrasekhar-mass WD one, in a self-similar way. We scale the density () and velocity () of each radial grid in the model as
ρ∝M5/2WDE−3/2kin (4)
and
v∝M−1/2WDE1/2kin. (5)
We then consider the distribution of four element groups (ECEs, Ni, IMEs, and C+O). We assume that Ni in a super-Ch-mass WD model is mixed. The inner boundary of the mixing region is set to be km s for all the models, because the observed line velocities of Si ii are 5000 km s (Yamanaka et al., 2009). And we set the outer boundary of the mixing region based on the mass coordinate; . Here the reference mass coordinate ( ) is the outer boundary of the Ni distribution in the W7 model. This outer boundary corresponds to that of the Ni-produced zone in the W7 model. By these definitions, the mixing region is uniquely set when we choose four model parameters. Note that, for our models, the velocity at the outer boundary of the mixing region differs among models even with the same , due to the scaling by Equation (5). This mixing is important to account for the observed velocities of the Si ii line (see Section 3.2).
To demonstrate the scaling and mixing, we plot the structure and abundance distribution of the two super-Ch-mass WD models in Figure 1. The model plotted on the left panels had no IMEs at km s before being mixed; the mixing has extended Ni outwards and IMEs inwards. In some models, C+O are also mixed inwards as in the model right in Figure 1.
We set the parameter range as follows to compare the models with the observations of SN 2009dc.
• , 2.0, , 2.8.
• , 1.4, , 1.8, 1.9, 2.0.
• , 0.2, 0.3.
• , 0.2, 0.3, 0.4.
Here, is obtained by using Equation (1). Any models which have are not constructed (e.g., models with , , and ). We also excluded a model where the inner boundary of the IMEs layer without mixing is located outer than the mixing region (e.g., a model with , , and ), because no IMEs extend inwards by mixing. The parameters and of our models are listed in Columns 2–5 of Table 1.
### 2.2. Light Curve Calculations
Multi-color LCs for the constructed models are calculated with STELLA code (Blinnikov et al., 1998, 2000, 2006), which solves the one-dimensional equations of radiation hydrodynamics. STELLA was first developed to calculate LCs of Type II-L supernovae (Blinnikov & Bartunov, 1993). Its applications to SNe Ia are described in Blinnikov & Sorokina (2000), Blinnikov et al. (2006), and Woosley et al. (2007). In the present study, we use homologously expanding ejecta as input (see Section 2). Then the radiation hydrodynamics calculation is performed for each model. Maeda & Iwamoto (2009) used the one-dimensional gray radiation transfer code (Iwamoto, 1997; Iwamoto et al., 2000), where they assumed simplified opacity for line scatterings. STELLA, on the other hand, considers 155000 spectral lines in LTE assumption to calculate the opacity with expansion effect more realistically, as well as free-free/bound-free transitions and Thomson scattering.
We note here that the “bolometric” luminosity reported from the observations is the luminosity (). Yamanaka et al. (2009) derived by assuming that (actually observed) integrated luminosity () is 60% of . This is commonly applied for normal SNe Ia (Wang et al., 2009a). Since it is still unknown whether this assumption is applicable to super-Ch candidates, we directly compare theoretical and observed rather than . This is advantage of our multi-color calculations. In the calculations, the bolometric, , and LCs cover 1–50000 Å, 1650–23000 Å, and 3850–8900 Å, respectively.
### 2.3. Extinction
To compare the calculated and observed LCs, the observational data must be corrected for the extinction. In order to correct the extinction, three values are needed; the excesses for the Milky Way () and the host galaxy of SN 2009dc (), and the ratio of the -band extinction to the B V excess () of the host galaxy ( is set to be 3.1 for the Milky Way). Of these values, and of the host galaxy are somewhat difficult to estimate.
The observed Na i absorption line in the host galaxy suggests that the reddening caused by it is not negligible. To estimate , one may use the observed color. However, the observed B V of SN 2009dc is significantly different from normal SNe Ia, which may suggest that the Lira-Phillips relation (Lira, 1996) should not be applied. The other method to derive is using the equivalent width of the Na i D absorption line (EW, e.g. Turatto et al., 2003). But it is also noted that the observed EW has a (relatively) large error (Silverman et al., 2011; Taubenberger et al., 2011). Also for , a non-standard, smaller value (3.1) may be preferred for normal SNe Ia (e.g. Nobili & Goodbar, 2008; Wang et al., 2009b; Folatelli et al., 2010; Yasuda & Fukugita, 2010). If the host galaxy of SN 2009dc also has , its extinction is overestimated.
To cover most of possible ranges of extinction, we consider two extreme cases, where the extinction by the host galaxy is negligible ( mag) and significant ( mag), respectively. We set mag, and for the host galaxy (same as for the Milky way). An extinction law by Cardelli et al. (1989, Table 3) is applied.
## 3. Results
### 3.1. Bvri Light Curves
The left panel in Figure 2 shows the calculated LCs for the super-Ch-mass WD models with different . The other parameters are set to be the same ( , , and ). A clear relation is seen between the LCs and from this panel; a more massive model shows a broader LC.
Such a mass dependence is consistent with the relation between the timescale of the bolometric LC () and ,
tbol∝¯κ1/2M3/4WDE−1/2kin (6)
(Arnett, 1982), where is the opacity averaged in the ejecta (although does differ among models and with time). Note that anticorrelates with for the models plotted in the left panel of Figure 2 because and are fixed (c.f. Table 1).
We consider a timescale to see quantitatively if the LCs depends on and , analogous to Equation (6). For this purpose, we take the declining timescale (), which is defined as the time for the LC to halve its luminosity after the peak. In the left panels of Figure 3, shown are the (large panel), (small, top), and (small, right) plots for the models listed in Table 1. The LCs of the massive and less energetic models tend to be wider (i.e. larger ), which is expected from the left panel in Figure 2. For the dependence of on and , we plot against in the right panel of Figure 3. An almost linear relation is seen between and . We thus confirm a relation similar to Equation (6) for the LCs.
The calculated and observed LC are shown on the right panel in Figure 2, by setting the dates of maximum in the -band of the models and observations at the same day as day. We plot the observational data provided by Yamanaka et al. (2009), neglecting the extinction by the host galaxy (open squares). The open triangle corresponds to the early -band detection by Silverman et al. (2011), where we simply assume that the -band luminosity is equivalent to 20% of , i.e., of , assuming (cf. Wang et al., 2009a, Figure 24). The LCs of the super-Ch-mass WD models on the right panel are as luminous as the observations around the maximum, while some have relatively broader LCs or shorter rising time than the observations. Especially, the two less massive models (–2.0 ) are not preferred because they are too faint at days.
In order to find the well-fitted models, we calculate the weighted root-mean-square residual (WRMSR) of the LC,
WRMSR= ⎷1NN∑i=1⎛⎜⎝L(calc)BVRI,i−L(obs)BVRI,iδL(obs)BVRI,i⎞⎟⎠2 (7)
for each model. We use the observational data for days taken from Yamanaka et al. (2009), where the total number of the observations, , is 23 for this time range. The early detection by Silverman et al. (2011) is excluded in calculating the WRMSR. Here, and respectively denote of the observation and calculation obtained at the -th epoch. is the observational error of , which is set to 20% of . Maeda & Iwamoto (2009) used the decline rate of the bolometric LC after maximum for comparison between their models and observations. We use the above WRMSR instead of the decline rate, so that brightness can also be taken into account. The WRMSR is calculated with the observational error, and shows which model fits relatively to the observed LC.
We list the WRMSR of the models for the two extinction cases in Columns 7 and 8 of Table 1. Good agreement is found between the calculated and observed LCs for whole range of in this study, if the extinction by the host galaxy is not considered. In the case where the extinction is corrected, most models have larger WRMSR (due to smaller ), while some models with large and are better fitted.
### 3.2. Photospheric Velocity
Next, we compare the photospheric velocity () of the models and the observed line velocity of Si ii. In the last column of Table 1, listed are the ranges of our models at days. This period covers the whole phases of the spectroscopic observations by Yamanaka et al. (2009). We estimate the position of the photosphere (hence, ) in the model from the optical depth at the -band, where the rest-frame wavelength of the observed Si ii line (6355 Å) is located.
The velocity of the observed Si ii line is reported as 9000 km s at the period (Yamanaka et al., 2009). Since the Si ii absorption line is formed by the Si above the photosphere, the calculated should be smaller than the observed Si ii line velocity (for further details, see Tanaka et al., 2011, Figure 1). For the models with and 2.0 , however, the calculated are 10000 km s, much larger than the observed line velocity of Si ii. Thus, these less massive models are far from consistent with SN 2009dc. On the other hand, many models with have km s during days, which are somewhat compatible with observations. These models with small commonly have such a large C+O mass fraction as –0.4. From Equation (2), and thus are affected mainly by rather than . A model with larger has smaller , thus smaller and , being preferred for SN 2009dc.
### 3.3. Plausible Models for SN 2009dc
In order to find the most plausible model for SN 2009dc, we first select models based on LC and discussed above. We use criteria of WRMSR and km s during the days. By these criteria, models with a label AO in Column 1 of Table 1 are selected. Among them, models AM are selected for the case where the host-galaxy extinction is neglected; only models N and O for the significant extinction. In Figure 4, we plot WRMSR and maximum during the period, where the dotted lines indicates the criteria.
We further analyze multi-color LCs of these models. Figures 59 show the (left), (right top) monochromatic LCs (right bottom) of these models. In these figures, the , -, -, -, and -band LCs by Yamanaka et al. (2009) and the early detection and -band LC by Silverman et al. (2011) are plotted as squares and triangles, respectively. Filled and open symbols show the observed LCs with and without the extinction correction.
With Figures 59, we make more detailed comparisons of models AO with the observational data of SN 2009dc. The comparisons are summarized as follows:
• The early detection in Silverman et al. (2011) enables us to constrain the rising time of the models. The -band LCs of models E, H, and M are too bright at days regardless of whether the host-galaxy extinction is corrected or not. These models can be excluded because their LCs evolve too slowly.
• The remaining models for the case of the neglected host-galaxy extinction (AD, F, G, and IL) have as small as 1.2 and 1.4 , which is consistent with the analytical estimates. While model J has the smallest WRMSR, its is relatively larger than the observations. Models D, G, and L also have larger around day. Among the other models (AC, F, I, and K), the models with (AC) are preferred, because their LCs reproduce the observations well around the peak. Especially, model B has the smallest of the three models, so we suggest that model B is the most plausible model for SN 2009dc without the host-galaxy extinction.
• If the extinction by the host galaxy is significant, we have only two candidate models with (N and O). They both have similar properties; slightly larger at days, but a bit smaller around day. We regard model N as the most plausible model for the extinction-corrected case because the rising time is shorter than model O.
By these results, we suggest that the mass of progenitor WD of SN 2009dc is 2.2–2.4 if the extinction by its host galaxy is negligible, and 2.8 with the extinction if the extinction is significant. The Ni mass needed for the SN is 1.2–1.4 for the former case, and 1.8–1.9 for the latter, respectively. For the latter case with extinction, the estimated masses are consistent with those suggested by Taubenberger et al. (2011), though they assume that the mean optical opacity of SN 2009dc is similar to (normal) SN 2003du.
For SN 2007if, one of the super-Ch candidates, Scalzo et al. (2010) estimate the total mass of its progenitor to be 2.4 , as massive as we estimate for SN 2009dc. They consider a shell-structured model to explain the low Si ii line velocity and its plateau-like evolution, where the massive envelope decelerates the outer layers of the ejecta. Our calculations also reproduce the lower Si ii line velocity and similar evolution of SN 2009dc (the right-top panels of Figures 59), and suggest that the low line velocity and flat evolution of Si ii can be explained by scaled super-Ch-mass WD models, as well as the shell-shrouding models.
## 4. Discussion
### 4.1. Light Curves of Super-Chandrasekhar-Mass White Dwarf Models
#### 4.1.1 Multi-Color Light Curves
We have calculated the multi-color LCs of the super-Ch-mass WD model for the first time (right-bottom panels in Figures 59). Even the model being in good agreement with and shows discrepancy for each band to some extent, especially in the band. This deviation from observed SNe Ia in the band is also seen for a Chandrasekhar-mass WD model calculated by the code STELLA (Woosley et al., 2007, Figure 9). It could be improved by taking more spectral lines into account for the opacity calculation.
The comparison with the velocity suggests that the mixing occurred in the ejecta. It is interesting to note that LCs of SN 2009dc in the band do not clearly show double-peak features. For Chandrasekhar-mass WD models, Kasen (2006) calculated the multi-color LCs, to conclude that mixing in the ejecta could produce the single-peak LCs in the band. In fact, our models are partially mixed and their -band LCs do not show two peaks (Figures 59). It could be also the case for super-Ch-mass WD models that mixing affects their -band LCs, although some uncertainties mentioned above should be taken into account.
#### 4.1.2 Bolometric, uvoir, and Bvri Light Curves
In Figure 10, the bolometric, , and LCs are plotted for models W7, B, and N. The LCs peaks earlier than the LCs, by 2 days for W7 model, and by 10 days for model B and N, respectively. As for the peak luminosity, the LC of W7 is brighter than the LCs by 0.2 dex, while those of the two super-Ch-mass WD models, by 0.3 dex. These shifts are understood by considering the ultraviolet (UV) radiation in the early phase (Blinnikov & Sorokina, 2000).
Most of LCs for the super-Ch-mass WD models in this paper seems to be fainter for day and slightly brighter after that, than the observations (Figures 510). This discrepancy could be solved by changing the Ni distribution in the model, which powers its radiation. The more Ni outside could shorten the rising time, or make more luminous before the peak, while the less Ni inside could make fainter after that. Further detailed modeling is needed since we simply assume the Ni distribution analogous to W7.
In Figure 10, we add the observation data for days plotted in Figure 7 of Taubenberger et al. (2011) as filled pentagons, which are also corrected for the host-galaxy extinction by using their values. The calculated bolometric LC fits to the observed LC tail quite well. The observed tail lie just on model N, which means that the estimate on for the model is consistent.
### 4.2. Instabilities of Rotating White Dwarfs
We calculate by just extrapolating the formula, valid only for Yoon & Langer (2005). Using their Equations (22) and (29)–(31), we can calculate the ratio of the rotational energy () to (the magnitude of) gravitational one of () a super-Ch-mass WD with our parameter . This ratio indicates the stability of the WD. If is small enough, the WD is stable against rotation. If reaches 0.14, the WD suffers the non-axisymmetric instability (e.g. Ostriker & Bodenheimer, 1973). The above equations show that a WD with has for our . A super-Ch-mass WD with might not form. However, as a function of differs with the rotation law (e.g., Hachisu, 1986), so that it is possible that even such a massive WD as has and thus form. We thus consider the models with .
Our results in this paper suggest that the progenitor WD mass of SN 2009dc exceeds 2.2 even for the case where the host-galaxy extinction is negligible. Such a large might suggest that the explosion of this super-Ch-mass WD is triggered when its mass reaches the critical mass for the above instability of the rotating WDs (Hachisu et al., 2012).
### 4.3. Possible Progenitors and Scenarios
The progenitor WD mass for SN 2009dc in our estimate largely exceeds , thus putting severe constraints on the presupernova evolution of the binary system. This also should have important implications on the progenitor scenarios of ordinary SNe Ia, both the single degenerate (SD) and double degenerate (DD) scenarios.
#### 4.3.1 Single Degenerate Scenario
The initial masses of the WD () and its companion star () should be sufficiently large at the beginning of accretion in order to increase to 2.4 . Chen & Li (2009) argued that even for the accretion from the companion of does not obtain . However, they does not take into account the effect of mass-stripping from the companion star due to the strong WD wind (Hachisu et al., 2007). Because the mass-stripping effectively reduces the mass transfer rate from the companion to the WD, the companion star can be as massive as –7 and can reach 2.4 (Hachisu et al., 2012). Still, to realize such a massive WD, is preferable (Chen & Li, 2009; Hachisu et al., 2012).
The formation of the C+O WD with is realized in the special binaries. In stars of main-sequence mass of 8 , the C+O core mass is 1.07 to avoid off-center C-ignition before the AGB phase (e.g., Umeda et al., 1999). After the dredge-up of the He layer, the C+O core increases its mass during the AGB phase, if the binary separation is wide enough to accommodate the AGB star. In such a binary system, the C+O WD with can be formed, if the C+O core of the AGB star has already grown massive when the AGB envelope is lost in a wind or by Roche-lobe overflow. Thus is larger if the mass loss rate from the AGB star is smaller. Therefore, the C+O WDs with larger is more likely to form in the lower metallicity system and in the initially wider binary (Hachisu et al., 2007). Since the binary must also be close enough for the mass-transfer to occur, the suitable binary system could be rare, which is consistent with the low occurrence frequency of the super-Ch-mass explosion.
These requirements of large enough and in the SD scenario predicts that SNe Ia from super-Ch-mass WDs are associated with the star-forming region and low metallicity environment. It is interesting to note that the hosts of the observed super-Ch candidates are faint and star-forming galaxies except for SN 2009dc (e.g. Taubenberger et al., 2011, Table 7). For SN 2009dc, the host galaxy UGC 10064 is a passive (S0) galaxy. However, at 40 kpc away from UGC 10064, there is a blue irregular galaxy UGC 10063, which could also have a star formation in the recent past by the interaction (Silverman et al., 2011).
#### 4.3.2 Double Degenerate Scenario
For the DD scenario, to form a WD of 2.4 by merging of two C+O WDs, the primary C+O WD needs to be initially as massive as because of the following reason. To form a massive C+O core in the primary AGB star, the initial binary system needs to be wide enough. The Roche lobe overflow of the primary AGB star is so rapid that a formation of a common envelope is unavoidable. After the loss of mass and angular momentum from the common envelope, a primary C+O WD and the secondary star are left in a binary with a small separation. If the separation is too small for the secondary star to become an AGB star, the mass of the secondary C+O WD is . In order to form a WD of 2.4 , the primary C+O WD should be more massive than 1.33 , which would be very rare. In the above scenario, formation of the double C+O WDs whose initial masses are both 1.2 may not be possible because of the shrink of the binary system after the first common envelope phase.
## 5. Conclusions
To constrain the properties of SN 2009dc, we have calculated multi-band LCs for the exploding super-Ch-mass WD models with a range of model parameters. We find that the mass of the WD and other model parameters are constrained as follows.
• The observed LCs of SN 2009dc are well-explained by the super-Ch-mass WD models with –2.8 and –1.8 , if the extinction by the host galaxy is negligible.
• The observed line velocity of Si ii is consistent with of several models with –2.8 , but significantly lower than of the less massive models.
• Among our models, the most plausible model is model B with (1.2 of Ni, 0.24 of ECEs, 0.24 of IMEs, and 0.72 of C+O) for the case without the host-galaxy extinction.
• If the extinction is considered, the mass of the super-Ch-mass WD needs to be as massive as (i.e. model N with 1.8 of Ni, 0.28 of ECEs, 0.16 of IMEs, and 0.56 of C+O). We find that the fit to the observation is less successful for model N than model B.
• Such a large might suggest that the explosion of the super-Ch-WD might be related to the onset of the instability of the differentially rotating WD.
Our results in this paper are based on the simplified models of the super-Ch-mass WDs. There are still several uncertainties in the models and LCs; such as the parameterization of the models, the opacity calculated by the code, aspherical effects, and effects of possible circumstellar interaction, as well as the instability of the massive WDs. However, and of the plausible models for SN 2009dc are quite consistent with the observations, suggesting that our present approach works well for this super-Ch candidate.
We are grateful to Masayuki Yamanaka for providing us the detailed observation data of SN 2009dc, to Keiichi Maeda and Nozomu Tominaga for the constructive discussion on the construction and LC calculations of super-Ch-mass WDs models. Y.K. acknowledges the Japan Society for the Promotion of Science (JSPS) for support through JSPS Research Fellowships for Young Scientists, and his work is supported by Grant-in-Aid for JSPS Fellows #227641. The work of S.I.B. and E.I.S. in Japan is supported by the Ministry of Education, Culture, Sports, Science and Technology; and in Russia by grants RFBR 10-02-00249-a and 10-02-01398-a, the Grant of the Government of the Russian Federation (No. 11.G34.31.0047), Sci. Schools-3458.2010.2 and -3899.2010.2, and a grant IZ73Z0-128180/1 of the Swiss National Science Foundation (SCOPES). This research has been supported in part by the Grant-in-Aid for Scientific Research of MEXT (22012003, 22840009, and 23105705) and JSPS (23540262) and by World Premier International Research Center Initiative, MEXT, Japan.
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The feedback must be of minimum 40 characters and the title a minimum of 5 characters | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9668043851852417, "perplexity": 2796.867594984205}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-29/segments/1593655890105.39/warc/CC-MAIN-20200706042111-20200706072111-00383.warc.gz"} |
https://projecteuclid.org/euclid.aos/1176345153 | ## The Annals of Statistics
### On Improving Convergence Rates for Nonnegative Kernel Density Estimators
#### Abstract
To improve the rate of decrease of integrated mean square error for nonparametric kernel density estimators beyond $0(n^{-\frac{4}{5}}),$ we must relax the constraint that the density estimate be a bonafide density function, that is, be nonnegative and integrate to one. All current methods for kernel (and orthogonal series) estimators relax the nonnegativity constraint. In this paper we show how to achieve similar improvement by relaxing the integral constraint only. This is important in applications involving hazard function and likelihood ratios where negative density estimates are awkward to handle.
#### Article information
Source
Ann. Statist., Volume 8, Number 5 (1980), 1160-1163.
Dates
First available in Project Euclid: 12 April 2007
https://projecteuclid.org/euclid.aos/1176345153
Digital Object Identifier
doi:10.1214/aos/1176345153
Mathematical Reviews number (MathSciNet)
MR585714
Zentralblatt MATH identifier
0459.62031
JSTOR | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9672158360481262, "perplexity": 1705.5108718775643}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-30/segments/1563195528869.90/warc/CC-MAIN-20190723043719-20190723065719-00242.warc.gz"} |
https://quant.stackexchange.com/questions/16181/swaption-model-for-forward-swap-rate/16563 | # swaption model for forward swap rate
I have another question about interest rates. In this case it is about swaption and how to come up with a pricing formula. For the rest of my question I use the notation from Brigo. The payoff of a payer swaption discountad from the maturity $T_\alpha$ to the current time $t$ is given by
$$D(t,T_\alpha)N\left(\sum^\beta_{i=\alpha +1 }P(T_\alpha,T_i)\tau_i(F(T_\alpha;T_{i-1},T_i)-K)\right)^+$$
where
• $D(t,T_i)$ the discount factor at $t$ of time $T_i$
• N some notional
• $\tau_i$, general daycount convention for between $T_{i-1}$ and $T_i$
• $F(T_\alpha;T_{i-1},T_i)$ forward rate at $T_\alpha$ between $T_{i-1}$ and $T_i$
• strike rate $K$
• $P(T_\alpha,T_i)$ zero coupon bond at $T_\alpha$ with maturity $T_i$.
denoting with $S:=S_{\alpha,\beta}(0)$ the forward swap rate, i.e. that $K$ which makes the contract fair $(=0)$ we can come up with models for $S$. Assuming a log normal model we derive a Black like formula.
However, I'm interested in the case where $dS=\sigma dW$, i.e. $S$ is normally distributed (Bachelier model). How does a pricing forumla for a swaption look like? I just can find Black formula on the web. Many thanks for the reference / answer.
One can write for the payoff of an swaption $$\sum_i\tau_i P_{i+1}(S_{\alpha,\beta}(T_\alpha)-K)^+$$ and therefore the pricing equation follows Joshi's explainations. To derive the above equation use that the swap rate is given by $$S_{\alpha,\beta} = \sum_i \frac{\tau_iP_{i+1}}{\sum_i\tau_iP_{i+1}}F^i,$$ where $F^i$ are the corresponding forward rates.
well just take the Bachelier formula with $r=d=0$ $S_0 = S_{\alpha,\beta}$ and then multiply by the annuity.
The annuity will be $$\sum \limits_i \tau_i P_{i+1}.$$ where $P_{i+1}$ is the df for $t_{i+1}.$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9569377899169922, "perplexity": 676.7853913684477}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-35/segments/1566027318894.83/warc/CC-MAIN-20190823150804-20190823172804-00122.warc.gz"} |
http://mathhelpforum.com/calculus/115605-integral-result-help.html | Math Help - Integral Result help!
1. Integral Result help!
I am confused how to integrate the equation below the result
should be /6 ???
$\int\limits_{0}^{1}\frac{1}{\sqrt{4-x^2}}\, dx$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 1, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9998130202293396, "perplexity": 3313.271942806224}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-49/segments/1416931007150.95/warc/CC-MAIN-20141125155647-00250-ip-10-235-23-156.ec2.internal.warc.gz"} |
https://www.zora.uzh.ch/id/eprint/135005/ | # On the computation of finite bottom-quark mass effects in Higgs boson production
Mueller, Romain; Öztürk, Deniz Gizem (2016). On the computation of finite bottom-quark mass effects in Higgs boson production. Journal of High Energy Physics, 2016(8):55.
## Abstract
We present analytic results for the partonic cross-sections contributing to the top-bottom interference in Higgs production via gluon fusion at hadron colliders at NLO accuracy in QCD. We develop a method of expansion in small bottom-mass for master integrals and combine it with the usual infinite top-mass effective theory. Our method of expansion admits a simple algorithmic description and can be easily generalized to any small parameter. These results for the integrated cross-sections will be needed in the computation of the renormalization counter-terms entering the computation of finite bottom-quark mass effects at NNLO.
## Abstract
We present analytic results for the partonic cross-sections contributing to the top-bottom interference in Higgs production via gluon fusion at hadron colliders at NLO accuracy in QCD. We develop a method of expansion in small bottom-mass for master integrals and combine it with the usual infinite top-mass effective theory. Our method of expansion admits a simple algorithmic description and can be easily generalized to any small parameter. These results for the integrated cross-sections will be needed in the computation of the renormalization counter-terms entering the computation of finite bottom-quark mass effects at NNLO.
## Statistics
### Citations
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21 citations in Scopus®
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Item Type: Journal Article, refereed, original work 07 Faculty of Science > Physics Institute 530 Physics Physical Sciences > Nuclear and High Energy Physics English 2016 13 Feb 2017 15:21 01 Apr 2020 22:39 Springer 1029-8479 Gold Publisher DOI. An embargo period may apply. https://doi.org/10.1007/JHEP08(2016)055 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9256156086921692, "perplexity": 2474.153899832016}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320300010.26/warc/CC-MAIN-20220116180715-20220116210715-00510.warc.gz"} |
https://brilliant.org/problems/shortest-way-out/ | # Shortest way out
Geometry Level 2
Above is a cube $$ABCDEFGH$$. $$ABCD$$ is the ceiling, $$EFGH$$ is the floor, and the rest are the walls. Additionally, $$AB = 2$$. A lizard is on point $$F$$ and is wanting to go to point $$A$$. If the lizard can only travel via walls, what is the measure of the shortest distance the lizard can take in order to reach its destination?
The answer can be expressed in the form of $$A + B\sqrt{C}$$ such that $$A$$, $$B$$ and $$C$$ are prime numbers or $$0$$. Find $$A + \frac{C}{B}$$.
× | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8391340374946594, "perplexity": 350.5777686409374}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-39/segments/1537267156423.25/warc/CC-MAIN-20180920081624-20180920101624-00297.warc.gz"} |
http://mathhelpforum.com/differential-equations/195493-reoccurance-relation-differential-equation.html | ## from reoccurance relation to a differential equation
Hi,
say I have a reoccurance relation:
$a_{t+1} = f(a_t)$
and I want to try and solve it as a differential eq:
$\frac { da_t }{ dt }\cong \frac { a_{ t+1 }-a_{ t } }{ \Delta t } =\frac { f(a_{ t })-a_{ t } }{ \Delta t } =...$
What $\Delta t$ do I take? can I just use $\Delta t = 1 ..?$
thank you | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 4, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9164837598800659, "perplexity": 1082.1213172514465}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-40/segments/1443737994631.83/warc/CC-MAIN-20151001221954-00216-ip-10-137-6-227.ec2.internal.warc.gz"} |
https://tex.stackexchange.com/questions/324160/make-bottom-margin-smaller-while-using-div-calc?answertab=oldest | # Make bottom margin smaller while using div=calc
Problem: the bottom margin is too large, i.e. I want the text to run further towards the footer, i.e. closer to the footer rule in the example below, but still keep div=calc such that the rest fits as it does now. Minimum Working Example showing the problem:
\documentclass[DIV=calc, paper=a4, fontsize=11pt, twocolumn]{scrartcl}
\usepackage{lipsum}
\usepackage{fancyhdr}
\pagestyle{fancy}
\usepackage{lastpage}
\lfoot{}
\cfoot{}
\rfoot{\footnotesize Page \thepage\ of \pageref{LastPage}}
\renewcommand{\footrulewidth}{0.4pt}
\begin{document}
\lipsum[1-8]
\end{document}
One solution is to add footheight=0mm to the options of \documentclass{scrartcl}: | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8961918354034424, "perplexity": 4397.606999322831}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320301737.47/warc/CC-MAIN-20220120100127-20220120130127-00088.warc.gz"} |
https://math.stackexchange.com/questions/304314/the-set-of-all-functions-from-mathbbn-to-0-1-is-uncountable | # The set of all functions from $\mathbb{N} \to \{0, 1\}$ is uncountable?
How can I prove that the set of all functions from $\mathbb{N} \to \{0, 1\}$ is uncountable?
Edit: This answer came to mind. Is it correct?
This answer just came to mind. By contradiction suppose the set is $\{f_n\}_{n \in \mathbb{N}}$. Define the function $f: \mathbb{N} \to \{0,1\}$ by $f(n) \ne f_n(n)$. Then $f \notin\{f_n\}_{n \in \mathbb{N}}$.
• Think about binary expansion of real numbers. Or look for the duplicate. – Julien Feb 14 '13 at 21:11
• @user62268 yeah that works, i posted the same – Dominic Michaelis Feb 14 '13 at 21:24
• Your answer that came to mind is correct, and in fact is exactly Cantor's diagonalization argument for the power set of natural numbers being larger than the set of natural numbers itself. – Joe Z. Feb 14 '13 at 21:28
You can proof it by contraposition. I will identify the functions with sequence so $a_{n}$=$a(n)$. Now lets say it's countable, now let $a_{nk}=a_k(n)$ be the $k$-th function. Now we construct the function $$b(k)=\left\{ \begin{array}{rl} 1 & a_{kk}=0\\ 0& a_{kk}=1\end{array}\right.$$ You should be able to do the rest.
• You don't need contraposition. You can do a direct proof that goes like this: Let $f$ be a mapping from $\mathbb N$ to $2^{\mathbb N}$. We can construct an element $b$ of $2^{\mathbb N}$ just as you did: $b(k) = 1-f(k)(k)$, and show that $b$ is not in the range of $f$. Therefore $f$ is not a bijection. But $f$ was completely general, so we have just proved that no mapping from $\mathbb N$ to $2^{\mathbb N}$ is a bijection, QED. – MJD Feb 14 '13 at 22:25
Hint: Show that $\{0,1\}^\mathbb N$ is equinumerous with $\mathcal P(\mathbb N)$ and use Cantor's theorem to conclude there is no bijection between $\mathbb N$ and $\mathcal P(\mathbb N)$.
Hint : use the diadic developpement of elements of $[0,1]$.
Hint: Each function from $\mathbb{N} \to \{0, 1\}$ is isomorphic to a subset of $\mathbb{N}$. Simply count $n \in \mathbb{N}$ to be part of the subset of $f(n) = 1$. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9393473267555237, "perplexity": 148.21484397753812}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-35/segments/1566027315329.55/warc/CC-MAIN-20190820113425-20190820135425-00100.warc.gz"} |
http://www.download-now.net/Minnesota/differentiation-of-error-function.html | Address Rogers, MN 55374 (763) 428-7677
# differentiation of error function Minnetonka Beach, Minnesota
This directly results from the fact that the integrand e − t 2 {\displaystyle e^{-t^ − 2}} is an even function. This usage is similar to the Q-function, which in fact can be written in terms of the error function. Handbook of Differential Equations, 3rd ed. Google search: Google's search also acts as a calculator and will evaluate "erf(...)" and "erfc(...)" for real arguments.
Why are so many metros underground? Dit beleid geldt voor alle services van Google. Zwillinger, D. Weisstein. "Bürmann's Theorem" from Wolfram MathWorld—A Wolfram Web Resource./ E.
Springer-Verlag. IEEE Transactions on Communications. 59 (11): 2939–2944. PARI/GP: provides erfc for real and complex arguments, via tanh-sinh quadrature plus special cases. comm., May 9, 2004).
Laden... New York: Gordon and Breach, 1990. Sloane, N.J.A. W.
Integrals and Series, Vol.2: Special Functions. W. Note that some authors (e.g., Whittaker and Watson 1990, p.341) define without the leading factor of . derivatives error-function share|cite|improve this question edited Apr 23 at 9:02 kamil09875 4,3592729 asked Apr 23 at 7:44 Rakesh 11 The error function erf($x$) is just $\frac{2}{\sqrt\pi}\int_0^xe^{-t^2}\ dt$, so its
Engineering and Design Solutions 10.954 weergaven 23:29 Gaussian - Duur: 4:28. Washington, DC: Hemisphere, pp.385-393, 1987. Boston, MA: Academic Press, p.122, 1997. Orlando, FL: Academic Press, pp.568-569, 1985.
Steve Grambow 22.895 weergaven 9:49 Approximation of Error in Hindi - Duur: 42:24. The integrand ƒ=exp(−z2) and ƒ=erf(z) are shown in the complex z-plane in figures 2 and 3. New York: Chelsea, 1999. Some authors discuss the more general functions:[citation needed] E n ( x ) = n ! π ∫ 0 x e − t n d t = n ! π ∑
Step-by-step Solutions» Walk through homework problems step-by-step from beginning to end. Asymptotic expansion A useful asymptotic expansion of the complementary error function (and therefore also of the error function) for large real x is erfc ( x ) = e − Wolfram Education Portal» Collection of teaching and learning tools built by Wolfram education experts: dynamic textbook, lesson plans, widgets, interactive Demonstrations, and more. Conf., vol. 2, pp. 571–575. ^ Van Zeghbroeck, Bart; Principles of Semiconductor Devices, University of Colorado, 2011. [1] ^ Wolfram MathWorld ^ H.
What Was "A Lot of Money" In 1971? Another approximation is given by erf ( x ) ≈ sgn ( x ) 1 − exp ( − x 2 4 π + a x 2 1 You can only upload photos smaller than 5 MB. See also Related functions Gaussian integral, over the whole real line Gaussian function, derivative Dawson function, renormalized imaginary error function Goodwin–Staton integral In probability Normal distribution Normal cumulative distribution function, a
Je moet dit vandaag nog doen. Hardy, G.H. Beoordelingen zijn beschikbaar wanneer de video is verhuurd. Because these numbers are not symbolic objects, you get the floating-point results:A = [erf(1/2), erf(1.41), erf(sqrt(2))]A = 0.5205 0.9539 0.9545Compute the error function for the same numbers converted to symbolic objects.
A complex generalization of is defined as (39) (40) Integral representations valid only in the upper half-plane are given by (41) (42) SEE ALSO: Dawson's Integral, Erfc, Erfi, Fresnel Integrals, Gaussian For integer , (16) (17) (18) (19) (Abramowitz and Stegun 1972, p.299), where is a confluent hypergeometric function of the first kind and is a gamma function. The defining integral cannot be evaluated in closed form in terms of elementary functions, but by expanding the integrand e−z2 into its Maclaurin series and integrating term by term, one obtains ISBN 978-0-486-61272-0.
The denominator terms are sequence A007680 in the OEIS. MathWorks does not warrant, and disclaims all liability for, the accuracy, suitability, or fitness for purpose of the translation. Help answer pleaseeeeeeee? It is an entire function defined by (1) Note that some authors (e.g., Whittaker and Watson 1990, p.341) define without the leading factor of .
doi:10.1090/S0025-5718-1969-0247736-4. ^ Error Function and Fresnel Integrals, SciPy v0.13.0 Reference Guide. ^ R Development Core Team (25 February 2011), R: The Normal Distribution Further reading Abramowitz, Milton; Stegun, Irene Ann, eds. Retrieved 2011-10-03. ^ Chiani, M., Dardari, D., Simon, M.K. (2003). Wolfram|Alpha» Explore anything with the first computational knowledge engine. Prudnikov, A.P.; Brychkov, Yu.A.; and Marichev, O.I.
so the integrand is exp(-v^2). The error function has special values for these parameters:[erf(sym(0)), erf(sym(Inf)), erf(sym(-Inf))]ans = [ 0, 1, -1]Compute the error function for complex infinities. Washington, DC: Hemisphere, pp.385-393 and 395-403, 1987. For previous versions or for complex arguments, SciPy includes implementations of erf, erfc, erfi, and related functions for complex arguments in scipy.special.[21] A complex-argument erf is also in the arbitrary-precision arithmetic | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8966580629348755, "perplexity": 4162.835971981371}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-22/segments/1558232256997.79/warc/CC-MAIN-20190523003453-20190523025453-00434.warc.gz"} |
https://infoscience.epfl.ch/record/178555 | Infoscience
Journal article
# VLT adaptive optics search for luminous substructures in the lens galaxy towards SDSS J0924+0219
The anomalous flux ratios between quasar images are suspected of being caused by substructures in lens galaxies. We present new deep and high-resolution H and Ks imaging of the strongly lensed quasar SDSS J0924+0219 obtained using the ESO VLT with adaptive optics and the laser guide star system. SDSS J0924+0219 is particularly interesting because the observed flux ratio between the quasar images vastly disagree with the predictions from smooth mass models. With our adaptive optics observations we find a luminous object, Object L, located similar to 0.3 '' to the north of the lens galaxy, but we show that it cannot be responsible for the anomalous flux ratios. Object L as well as a luminous extension of the lens galaxy to the south are seen in the archival HST/ACS image in the F814W filter. This suggests that Object L is part of a bar in the lens galaxy, as also supported by the presence of a significant disk component in the light profile of the lens galaxy. Finally, we find no evidence of any other luminous substructure that may explain the quasar images flux ratios. However, owing to the persistence of the flux ratio anomaly over time (similar to 7 years), a combination of microlensing and millilensing is the favorite explanation for the observations. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8171958327293396, "perplexity": 1924.8595269966884}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-13/segments/1490218189031.88/warc/CC-MAIN-20170322212949-00476-ip-10-233-31-227.ec2.internal.warc.gz"} |
http://kea-monad.blogspot.com/2009/03/m-theory-lesson-265.html | occasional meanderings in physics' brave new world
Name:
Location: New Zealand
Marni D. Sheppeard
## Friday, March 20, 2009
### M Theory Lesson 265
There is a nice series of papers by Godsil et al on combinatorics associated to MUBs. Consider the simple qubit Combescure matrix This is the only choice for the eigenvectors of the Pauli matrix $\sigma_{Y}$ that satisfies the following properties. The Schur multiplication of two matrices simply defines entries by the product of matching entries from the two components $A$ and $B$. That is $M_{ij} = A_{ij} B_{ij}$. Under this product, an inverse for $R_2$ is found relative to the democratic matrix (the Schur identity): An invertible matrix in this sense is type II if $M (M^{-1})^{T} = n I$, where $I$ is the ordinary identity matrix. This works for $R_{2}$, although only $R_{2}^{8} = I$.
A type II matrix is a spin model, in the sense of Jones, if all vectors of the form
$Me_{i} \circ M^{-1}e_{j}$
(for $e_{i}$ the standard basis vectors) are eigenvectors for $M$. One checks that this holds for $R_2$. Note that the Fourier (Hadamard) operator $F_{2}$, although a type II matrix, is not a spin model matrix.
CarlBrannen said...
Nice to see the importance of the Democratic matrix showing up here.
Regarding cosmology, the modification of GR that I believe we have to reach is nicely described in the new PI lecture: Towards the end of the cosmological constant problem! : Niayesh Afshordi
This is a modification of GR. There are two important points of compatibility with the gravity I've been working on from the quantum side: (a) Assumes preferred reference frame, and (b) the added material (which he called aether) propagates faster than light.
March 20, 2009 8:01 PM
CarlBrannen said...
There is a related arXiv paper, 0807.2639
March 20, 2009 8:09 PM | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9535279273986816, "perplexity": 689.517711166679}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-26/segments/1529267864546.30/warc/CC-MAIN-20180622143142-20180622163142-00123.warc.gz"} |
https://www.omnicalculator.com/physics/resultant-force | # Net Force Calculator
Created by Purnima Singh, PhD
Reviewed by Steven Wooding
Last updated: Oct 19, 2022
Omni's net force calculator allows you to determine the resultant force on a body when several forces act simultaneously on it.
Continue reading this article to know what net force is and how to find the net force on an object. You will also see some examples of net force calculations.
If you need help calculating force from Newton's second law of motion, check out the force calculator.
## What is net force? – Net force definition
When a number of forces act simultaneously on an object or a system, the net force or resultant force is the vector sum of all these forces.
We know that the effect of a force $F$ acting on an object of mass $m$ is to accelerate the object according to the equation:
$\quad a = \frac{F}{m}$
Our magnitude of acceleration calculator is a handy tool in case you want to explore more about acceleration.
The effect of net force would accelerate the object by the same amount as all the actual forces acting on the object. So we can say that the net force is a single force that would produce the same effect as all the forces working together.
Now that we know what net force is, let us understand how to find the net force on an object.
## How to find net force – Net force equation
We know that force is a vector quantity, i.e., we need to specify both the magnitude and direction of a force to give its complete description. This means that we should add the individual forces to find the net force just like we add other vectors (check out the vector addition calculator to learn all about vector addition).
To understand this, let us consider a simple scenario where two forces $F_1$ and $F_2$ are acting on a body from two different directions. We can represent these two forces as two vectors $\overrightarrow{F_1}$ and $\overrightarrow{F_2}$ acting at angles $\theta_1$ and $\theta_2$ (see figure 1).
Using the triangle law of vector addition, we know that if two vectors acting simultaneously on a body can be represented in magnitude and direction by the two sides of a triangle taken in one order, the resultant of these two vectors can be represented in magnitude and direction by the third side of the triangle taken in the opposite order (see the net force diagram in figure 1).
Hence, we get the resultant vector:
$\ \ \overrightarrow{F} = \overrightarrow{F_1} + \overrightarrow{F_2}$
If several forces are acting together (see figure 2), we can apply the polygon rule and write a more general net force formula as:
\begin{aligned} \overrightarrow{F} &= \overrightarrow{F_1} + \overrightarrow{F_2} + \overrightarrow{F_3} + ... + \overrightarrow{F_n}\\ \text{ or:}\\ \overrightarrow{F} &= \sum_{i=1}^\infty \overrightarrow{F_i} \end{aligned}
In the next section, we will see a step-by-step guide to how to find the magnitude and direction of the resultant force?
## How to calculate net force
To get the magnitude of the resultant force $F$, we will use the net force equation. We will first resolve each of the forces $F_1$ and $F_2$ into their respective rectangular components:
$\quad F_{1x} = F_1\cos \theta_1 \\\ \\ \quad F_{1y} = F_1\sin \theta_1$
Similarly, for the force $F_2$, we can write:
$\quad F_{2x} = F_2\cos \theta_2\\\ \\ \quad F_{2y} = F_2\sin \theta_2$
Since the components of a vector are scalar quantities, we can now add them. To find the horizontal component $F_x$ of the resultant force, we will sum all the horizontal components of the individual forces:
$\quad F_x = F_{1x} + F_{2x}$
In the same way, the vertical component will be:
$\quad F_y = F_{1y} + F_{2y}$
Finally, we will calculate the magnitude of the resultant force using:
$\quad F = \sqrt{F_x^2 + F_y^2}$
And the angle of the resultant force with respect to the horizontal axis with:
$\quad \theta = \tan^{-1}\frac{F_y}{F_x}$
For $n$ number of forces, we can write a general formula as:
$\quad F_x = \sum_{i=1}^\infty F_{ix}\\\ \\ \quad F_y = \sum_{i=1}^\infty F_{iy}$
Using the net force formula, we can calculate the magnitude of the resultant force using:
$\quad F = \sqrt{F_x^2 + F_y^2}$
And the angle of the resultant force with respect to the horizontal axis using:
$\quad \theta = \tan^{-1}\frac{F_y}{F_x}$
## Examples of net force calculation
To further understand how to find the resultant force, we will consider two simple examples.
First we will consider a simple case where two forces $\overrightarrow{F_1}$ and $\overrightarrow{F_2}$ are applied to an object in opposite directions such that;
• $|F_1| = 10\ N$ and $\theta_1 = 0 \degree$; and
• $|F_2| = 15\ N$ and $\theta_2 = 180 \degree$.
An example of this case would be when you and your friend are sitting in front of each other and trying to push a book towards each other.
Let us see how to calculate the net force.
1. First, we will find the vertical and horizontal components for both the forces:
• $F_{1x} = 10 \cdot \cos 0 \degree$ and $F_{1y} =10 \cdot \sin 0 \degree$.
$\implies F_{1x} = 10$ and $F_{1y} =0$
• $F_{2x} = 15 \cdot \cos 180 \degree$ and $F_{2y} =15 \cdot \sin 180 \degree$.
$\implies F_{1x} = -15$ and $F_{1y} =0$
2. As the vertical components, $F_{1y}$ and $F_{2y}$ are zero, the vertical component of the resultant force will be zero. We can calculate the horizontal component as:
• $F_x = 10 + (-15)$, or
$F_x = -5\ N$
3. According to the definition of the net force, the magnitude of the resultant force will be:
• $F = \sqrt{F_x^2} = 5\ N$
and the direction will be along the direction of the larger force, i.e., $180 \degree$.
Now let us consider another case where the magnitude of the forces is the same as in the previous example, but now both the forces act in the same direction, i.e., $\theta_1 = \theta_2 = 180 \degree$. For example, when you and your friend try to push a heavy box together. In this case, the net force would be $25\ N$ along the direction of both the forces, i.e., $180 \degree$.
In the next section, we will see how to solve the same problem using our resultant force calculator.
## How to use the net force calculator
Let us see how to use the net force calculator:
1. Enter the magnitude ($|F_1| = 10\ N$ and $|F_2| = 15\ N$) and direction ($\theta_1 = 0 \degree$ and $\theta_2 = 180 \degree$) of different forces acting on the body. The directions are measured with respect to the positive x-axis.
2. You can add data for up to 10 forces; fields will appear as you need them.
3. The resultant force calculator will display the magnitude ($F = 5\ N$) and direction ($\theta = 180 \degree$) of the net force. It will also show the values of the horizontal and vertical components of the resultant force.
To convert between different units of force, head on to Omni's force converter.
## FAQ
### How do I find the resultant force acting on an object?
To find the resultant force or net force acting on an object, follow the given instructions:
1. Determine the horizontal and vertical components of all the individual forces by using the formula;
• Horizontal component –
Fₓ = F cos θ
• Vertical component –
Fᵧ = F sin θ
2. Add the individual horizontal and vertical components to get the horizontal and vertical components of the resultant force.
3. Sum the square of the horizontal and vertical components of the resultant force and take the square root of the result. You will get the magnitude of the resultant force.
### When the net force on an object is zero?
According to Newton's first law of motion, if an object is either at rest or moving in a straight line with constant velocity (no acceleration), the net force on the object is zero.
### What are the units of net force?
The units for net force are the same as the unit of force. The SI unit of the net force is the newton (N), and the cgs unit of the net force is dyne.
### How does net force affect the speed of an object?
From Newton's second law of motion, we know that the acceleration or rate of change of speed of an object is directly proportional to the net force acting on the object.
Purnima Singh, PhD
Force 1 (F₁)
N
Angle 1 (θ₁)
deg
Force 2 (F₂)
N
Angle 2 (θ₂)
deg
You can add up to 10 different forces. Each angle is measured with respect to the positive x-axis.
Resultant force
Horizontal component (Fx)
0.97
N
Vertical component (Fy)
0.26
N
Magnitude of resultant force (F)
1
N
Direction of resultant force (θ)
15
deg
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https://brilliant.org/discussions/thread/paying-paradox-help/ | Let's say you want to buy a $97 shirt. You borrow$50 from Friend 1 and $50 from Friend 2. You buy the shirt and get$3 as change. You keep $1 for yourself, one for Friend 1 and the other for Friend 2. Thus, making your debt to Friend 1$49 and Friend 2 $49 If you add the debts, 49+49, it is 98, and then add your$1 which makes $99. Where is the missing dollar? Note by Abijah Bautista 5 years, 9 months ago MarkdownAppears as *italics* or _italics_ italics **bold** or __bold__ bold - bulleted- list • bulleted • list 1. numbered2. list 1. numbered 2. list Note: you must add a full line of space before and after lists for them to show up correctly paragraph 1paragraph 2 paragraph 1 paragraph 2 [example link](https://brilliant.org)example link > This is a quote This is a quote # I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world" # I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world" MathAppears as Remember to wrap math in $ ... $ or $ ... $ to ensure proper formatting. 2 \times 3 $2 \times 3$ 2^{34} $2^{34}$ a_{i-1} $a_{i-1}$ \frac{2}{3} $\frac{2}{3}$ \sqrt{2} $\sqrt{2}$ \sum_{i=1}^3 $\sum_{i=1}^3$ \sin \theta $\sin \theta$ \boxed{123} $\boxed{123}$ ## Comments Sort by: Top Newest See this as: By adding the debts, you are calculating how much you owe the 2 friends, which is $\98$. This tallies with the amount of money that you have spent or have: $98=97$ (from the shirt you bought) $+1$ (that you kept. ) - 5 years, 9 months ago Log in to reply Ok, so lets think about each person separatedly: (You:0 F1:50 F2:50 shop:0) You take the money: (You:100 F1:0 F2:0 shop:0) You buy the shirt (You:3 F1:0 F2:0 shop:97) You give each one 1 dollar (You:1 F1:1 F2:1 shop:97) You pay the debts (You:1-(49+49) F1:1+49 F:1+49 shop:97) (You:-97 F1:50 F:50 shop:97) I dont see any missing dollar.. In your text you dont take the final payment in account to sum all the money :p Actually the sum is always 100 dollars and you'll end up with a -97 debt just as expected. Anyway why are you adding debts? It doesnt have any meaning :P - 5 years, 9 months ago Log in to reply You should subtract it.. of course! Instead you are adding it.. :) - 5 years, 9 months ago Log in to reply Very good question, all though, I do not believe it works that way sadly. Lets think of it another way: Debt is a separate thing from the money itself, it is how much you owe the person. I don't know exactly why it does this with two people, but it seems to not happen with one person only. This is very interesting. - 5 years, 9 months ago Log in to reply By what Logic are you adding the 1$ to the debt of 98\$ ???
- 5 years, 9 months ago
This problem is very much like this one.
http://en.wikipedia.org/wiki/Missingdollarriddle
- 5 years, 9 months ago
it is with me only...because , from the beginning i don't have any money but bought a shirt with all frnds money. at last i have one dollar with me..thats the missing dollar
- 5 years, 9 months ago
you are supposed to subtract one, not add.
- 5 years, 8 months ago | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 14, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.979242205619812, "perplexity": 3446.9135911548015}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-35/segments/1566027313987.32/warc/CC-MAIN-20190818165510-20190818191510-00522.warc.gz"} |
http://www.orcunkoraliseri.com/2017/09/06092017-operative-temperature-mean.html | ### 06.09.2017 - Operative Temperature / Mean Radian Temperature
Figure.Thermometer
# Operative temperature (air temperature + mean radiant temperature + air velocity)
Operative temperature is indictaion of thermal comfort comes from air temperature, mean radiant temperature and air speed. It is helpful to define or predict thermal comfort for occupants of a building. Basically, thermal comfort is related with environmental elements, which are air temperature, air velocity, relative humidity and the consolidation of conditions, also individual properties such as clothing, metabolic heat, acclimatisation, state of health. In many spaces, with low air velocity and where air temperature and mean radiant temperaturemay be similar, air temperature alone can be a reasonable indicator of thermal comfort. However, in spaces where surfaces may be heated or cooled, where there is significant thermal mass, or where solar radiation is present, air and radiant temperatures may be very different and so it is necessary to take account of radiant temperatures in assessing thermal comfort...
Calculation:
Operative temperature = (tr + (ta x √10v)) / (1+√10v)
Where ta = air temperature / tr = mean radiant temperature / v = air speed (m/s)
Or
Operative temperature = (( hr x tr) + (hc x ta )) / ( hr + hc )
Where hc = convective heat transfer coefficient / hr = radiative heat transfer coefficient
Or
Operative temperature = (ta + tr)/2
Where the air speed is less than 0.1m/s, (as is typical in buildings) radiative and convective heat transfers may be similar, and so the equation can be simplified to:
Mean radiant temperature (MRT) is a measure of the average temperature of the surfaces that surround a particular point, with which it will exchange thermal radiation. If the point is exposed to the outside, this may include the sky temperature and solar radiation
Calculation:
Mean radiant temperature could be calculated by globe thermometer. This is a hollow copper sphere painted matt black (to give it a high emissivity) with a temperature sensor at its centre. From the temperature recorded, along with air velocity and air temperature. (needed to account for convective heat exchange)
MRT = globe temperature + 2.42 x air velocity in m/s (globe temperature air temperature)
Or
The mean radiant temperature (MRT) is a means of expressing the influence of surface temperatures on occupant comfort. It can be calculated many ways. In its simplest and least accurate form it is a homogenous steady state area weighted average of the uncontrolled or unconditioned surface temperatures (AUST) written as
Tmr = T1A1 + T2A2 + …+ TNA / ( A1 + A2 + …+ AN )
where,
Tmr = mean radiant temperature, °R
TN = surface temperature of surface N, °R (calculated or measured)
AN = area of surface | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9208168387413025, "perplexity": 3462.044430044098}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662517245.1/warc/CC-MAIN-20220517095022-20220517125022-00472.warc.gz"} |
http://mathoverflow.net/questions/126339/generators-for-a-certain-congruence-subgroup-of-sln-z | # Generators for a certain congruence subgroup of SL(n,Z)
I'm looking for a reference (or quick proof) of the following fact. Fix some $n \geq 3$ and some $\ell \geq 2$. Set $\Gamma_n(\ell) = \text{ker}(\text{SL}_n(\mathbb{Z}) \rightarrow \text{SL}_n(\mathbb{Z}/\ell \mathbb{Z}))$. Next, set
$$\Gamma_n'(\ell) = \{\text{A \in \Gamma_n(\ell) | for all diagonal entries a_{ii} of A, we have \ell^2 | (a_{ii}-1)}\}.$$
It is an easy exercise to show that $\Gamma_n'(\ell)$ is a subgroup of $\Gamma_n(\ell)$. Finally, for all $1 \leq i,j \leq n$ with $i \neq j$, let $e_{ij}$ be the elementary matrix obtained by taking the identity matrix and placing a $1$ at position $(i,j)$.
Observe that $e_{ij}^{\ell} \in \Gamma_n'(\ell)$. The fact I'm interested in is the fact that $\Gamma_n'(\ell)$ is generated by the set of all the $e_{ij}^{\ell}$. I've seen this asserted without proof in print before (though I don't recall exactly where), but I've been unable to work out a proof myself.
-
One can probably break up the proof into two steps: show it for $n=3$, and then show that $\Gamma_{n+1}'(\ell)$ is generated by the $n+1$ copies of $\Gamma_n'(\ell)$ sitting inside it in the natural way. I'm not sure how much easier the case $n=3$ is than the general case though (and the statement is false for $n=2$). – Ian Agol Apr 8 '13 at 5:20
There is a Theorem of Tits which says that the group generated by $e_{ij}^l$ has finite index in $SL_n({\mathbb Z})$ if $n\geq 3$. The paper is a Comptes rendus announcement (generating systems of congruence subgroups, CR. ACad. Sci 283 (1976), no. 9 A693-A695 (see the following review http://www.ams.org/mathscinet/search/publdoc.html?amp=&loc=revcit&revcit=244257&vfpref=html&r=16&mx-pid=424966). By a result of Mennicke, finite index subgroups of $SL_n({\mathbb Z})$ are congruence subgroups ($n\geq 3$). So, you need only check that the congruence closure of the group generated by $e_{ij}^l$ in $SL_n({\mathbb Z})$ is of the desired type. This is a purely local checking (prime by prime, for primes dividing $l$) and can be done easily.
ADDED: what you are asking for is (seemingly) stronger than Tits' theorem: not only do the $e_{ij}^l$ generate a finite index subgroup, they generate a specific congruence subgroup. But, as the previous paragraph shows, this is $equivalent$ to the result of Tits.
-
I guess the folllowing is a short explicit proof of Tits' result for this very special case, assuming CSP for $\rm{SL}_n(\bf{Z})$ (so this should be seen as a lenghty comment on the above answer). The key step is to show that the group $\Lambda=\langle e_{ij}^\ell\rangle$ contains $\Gamma_n(\ell^2)$: the commutator $[e_{ij}^\ell,e_{km}^\ell]$ is equal to $1+\ell^2( [n_{ij},n_{km}] - n_{ij}^2 - n_{km}^2)+ O(\ell^3)$ where $n_{ab}=e_{ab}-1$ is a nilpotent matrix. On the other hand you have $(e_{ij}^\ell)^\ell=1+\ell^2 n_{ij}$, and as $n_{ij}^2=n_{i,j+1}$ it follows that $\Lambda$ contains in fact also the matrices $1+\ell^2 [n_{ij},n_{km}]$ modulo $\ell^3$. So you finally get that $\Lambda\Gamma_n(\ell^3)$ contains all matrices $1+\ell^2a$ where $a$ is a sum of $n_{ij}$s and their Lie brackets, and these generate the $\bf{Z}/\ell$-module $\mathfrak{sl}_n(\bf{Z}/\ell)$. As remarked by Aakumadula below (and apparently the fact that $\Gamma_n'(\ell)$ contains $\Gamma_n(\ell^2)$ also follows from a theorem of Tits, see her/his answer to this question), it remains to show that for a prime $p$ not dividing $\ell$ the $e_{ij}^\ell$ generate $\Gamma_n$ modulo $\Gamma_n(p)$: in this case we have $\Lambda\Gamma_n(p)=\langle e_{ij}\rangle\Gamma_n(p)$ and it is well-known that the images of the $e_{ij}$ generate $\rm{SL}_n(\bf{F}_p)$. So we have that the congruence closure of $\Lambda$ contains $\Gamma_n(\ell^2)$ and by CSP it follows that $\Lambda\supset\Gamma_n(\ell^2)$.
To conclude: now you have only to show that the $e_{ij}^l$ generate $\Gamma_n'(\ell)$ modulo $\ell^2$. But this is obvious as $\Gamma_n'(\ell)/\Gamma_n(\ell^2)$ is just the subspace of the abelian group $\Gamma_n(\ell)/\Gamma_n(\ell^2)\cong \mathfrak{sl}_n(\bf{Z}/\ell)$ of matrices with zeroes on the diagonal.
-
I agree that the key is to show that the group contains $\Gamma_n(\ell^2)$. But I don't understand your argument for this. Can you expand it a bit? Thanks! – Edward Cooper Apr 6 '13 at 2:43
I do not understand Cooper's remark. The theorem of Tits quoted above shows that $e_{ij}^l$ generate a congruence subgroup, which can only be $\Gamma _n(l^2)$. – Venkataramana Apr 6 '13 at 4:08
@Cooper: I edited to give more details. – Jean Raimbault Apr 6 '13 at 12:09
@jean: your argument only proves the local statement that in $SL_n$ of the $l$-adic integers (suppose $l$ is a prime), the $e_{ij}^l$ generate the appropriate congruence closure. It does not prove that at the global level (i.e. the integral points of $SL_n$, ), the $e_{ij}^l$ generate the appropriate congruence subgroup. You do need the theorem of Tits (and that is a non-trivial result). – Venkataramana Apr 6 '13 at 13:44
@Aakumadula: you're right, there is also need to check that at primes $p$ not dividing $\ell$ the $e_{ij}^\ell$ generate all of $\rm{SL}_n(\bf{Z}_p)$. However this can be done in an elementary way since the $e_{ij}$ generate $\rm{SL}_n(\bf{F}_p)$. Edited to add that. – Jean Raimbault Apr 6 '13 at 14:20
FIrst I'd clarify that your notation $e_{ij}^\ell$ actually refers to the matrix with diagonal entries 1, the off-diagonal $(i,j)$ entry equal to $\ell$, and other entries 0.
I don't know what you've read, but since these matrix calculations are quite old and also deeply embedded in the study of the Congruence Subgroup Problem, it's a good idea to look into some of the relevant older literature. On a concrete level, the emphasis is on the group $\Gamma: = \mathrm{SL}_n(\mathbb{Z})$ and its subgroups of finite index, when $n \geq 3$ (the case $n=2$ being much more complicated). Here the key players are the (normal) principal congruence subgroups you've denoted by $\Gamma_n(\ell)$ and the interleaved elementary subgroups: inverse images of finite groups generated by elementary/unipotent matrices.
Key references available online include the 1964 announcement by Bass-Lazard-Serre here and the detailed follow-up by Bass-Milnor-Serre here.
Some of the concrete calculations you are looking for are also written down in section 17 of my (typewritten) 1980 Springer Lecture Notes Arithmetic Groups. The older lecture notes Algebraic K-Theory by Bass (1968) contain a vast amount of detail, and of course there are newer treatises including those by Hahn and O'Meara along with a new book by Weibel.
ADDED: The short paper by Tits cited by Aakumadula is definitely helpful for your question, though it's dependent on the earlier work and is not readily available online (nor is the ancient review I wrote). The literature on congruence subgroups is extensive and often far more general than what you need, but I don't see a direct computational proof of the result you read somewhere. (Advice: Keep track of those sources.) Also, notation varies in the subject, but your choice of $e_{ij}$ is unfortunate since that symbol usually means the matrix with a single nonzero entry $1$. A more usual convention is to write something like $x_{ij}(\ell)$ for your unipotent matrix.
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You my also need to use Lemma 3.1 of arXiv:1301.6082 to deal with the distinction between the subgroup generated by the $e^\ell_{ij}$ and the normal subgroup generated by them, meaning the smallest normal subgroup of the elementary group that contains them. In the congruence subgroup problem one tends to use normal subgroups. – Wilberd van der Kallen Apr 3 '13 at 13:26
As van der Kallen alluded to, Bass-Milnor-Serre (and its antecedents) only proved that the $e_{ij}^{\ell}$ (which, by the way, seems pretty unambiguous to me; it means the $\ell^{\text{th}}$ power of $e_{ij}$) normally generate the whole group $\Gamma_n(\ell)$. This isn't quite what I want, which is a somewhat finer result. – Edward Cooper Apr 3 '13 at 14:39
@Wilberd van der Kallen : Are you sure you got the arXiv identifier right? When I went there, I got a paper entitled "Reaction-diffusion model Monte Carlo simulations on the GPU", which doesn't sound right at all. – Edward Cooper Apr 3 '13 at 14:40
@Edward : Bass-Milnor-Serre and its "antecedents" is not quite right. There s this paper of Tits (which uses Bass-Milnor-Serre) where he proves that the group generated by $e_{ij}^l$ is a finite index subgroup for $n\geq 3$. – Venkataramana Apr 4 '13 at 1:16
@Edward Cooper. Sorry. Try arXiv:1303.60882 by Stepanov. – Wilberd van der Kallen Apr 10 '13 at 6:57 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8929564952850342, "perplexity": 214.78429744396007}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-26/segments/1466783398516.82/warc/CC-MAIN-20160624154958-00018-ip-10-164-35-72.ec2.internal.warc.gz"} |
http://mathhelpforum.com/advanced-algebra/38310-characteristic-value-vector.html | # Math Help - Characteristic Value and Vector
1. ## Characteristic Value and Vector
If A is nonsingular, show that the characteristic values of A^-1 are the reciprocals of A, and the A and A^-1 have the same characteristic vectors.
I need help getting started on this. Any help would be greatly appreciated. Thanks,
Jim
2. Fix an eigenvalue $\lambda$ of A and let v be an eigenvector w.r.t. the eigenvalue, write down $Av=\lambda v$. Since A is invertible, $\lambda$ is nonzero and you can multiply both sides of the equation by the inverse of A. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 3, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9843419790267944, "perplexity": 253.4439948814863}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-23/segments/1406510271862.24/warc/CC-MAIN-20140728011751-00408-ip-10-146-231-18.ec2.internal.warc.gz"} |
https://brilliant.org/practice/taylor-series-approximation/ | ×
## Taylor Series
The Taylor series is a polynomial of infinite degree used to represent functions like sine, cube roots, and the exponential function. They're how some calculators (and Physicists) make approximations.
# Taylor Series Approximation
Let $$f(x)$$ be a function such that $$f(0)=3, f'(0)=1,$$ and $$f(3)=1.$$ Using a quadratic Taylor polynomial of $$f(x),$$ we can approximate the value of $$f''(0)$$ as $f''(0) \approx -\frac{A}{B},$ where $$A$$ and $$B$$ are coprime integers. Find the value of $$AB.$$
Using the quadratic Taylor polynomial of $$f(x) = \frac{1}{x}$$ at $$x=5,$$ find the approximate value of $$\frac{1}{7}$$ multiplied by $$5^3.$$
For the linear Taylor polynomial $$g(x)=ax+b$$ of $$f(x)=-\frac{1}{x-1}$$ at $$x=2,$$ find the error $$e$$ defined by $e=\int_{2}^{3}{(g(x)-f(x))^2 dx}.$
Using the quadratic Taylor polynomial of $$f(x) = \ln{x}$$ at $$x=4,$$ we can approximate the value of $$\ln{9}$$ in the form $\ln{A}+B,$ where $$A$$ and $$B$$ are rational numbers. Find the value of $$2A^2B.$$
Let $$f(x)$$ be a function such that $$f(0)=1, f'(0)=2,$$ and $$f(3)=1.$$ Using a quadratic Taylor polynomial of $$f(x),$$ we can approximate the value of $$f''(0)$$ as $f''(0) \approx -\frac{A}{B},$ where $$A$$ and $$B$$ are coprime integers. Find the value of $$AB.$$
× | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9479524493217468, "perplexity": 60.00997376911694}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-04/segments/1484560284429.99/warc/CC-MAIN-20170116095124-00020-ip-10-171-10-70.ec2.internal.warc.gz"} |
http://mathhelpforum.com/algebra/90163-need-help-word-problem.html | # Math Help - Need help on word problem.
1. ## Need help on word problem.
There are 396 persons in a theater. If the ratio of women to men is 2:3, and the ratio of men to children is 1:2, how many men are in the theater?
2. Hi,
Are you sure your numbers are correct? I'm getting 396*6/13 men, which I'm pretty sure isn't physically possible...
If you are having problems with the words, simply get rid of them! Make equations, which are usually much easier to work with. Maths notation is actually quite a modern thing - previously people would do maths in words and sentences, which is hard! For instance, "=" was not introduce until 1557(MacTutor). Also, in some ways maths merely takes away any external noise and gets to the crux of the problem, and what you are dealing with here is quite a good example of this. As sentences, it is hard to solve/understand, but as equations it is less so. Another good example is a thing called Group Theory - essentially it is the abstract study of symmetry, but it has developed into a very powerful tool. Anyway, I'm rambling. However, my point is this - get rid of the words! Make it into something you can deal with.
3. ## Here is my answer.
I think the answer should be this.
women:men =2:3
men:children=1:2
therefore women:men:children=2:3:6
men=396x3/11=108
There are 108 men. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9005951285362244, "perplexity": 667.398849950325}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-15/segments/1397609526102.3/warc/CC-MAIN-20140416005206-00366-ip-10-147-4-33.ec2.internal.warc.gz"} |
https://perminc.com/resources/publications/coal-bed-characterization-studies-x-ray-computerized-tomography-ct-micro-ct-techniques/ | ## Coal Bed Characterization Studies with X-ray Computerized Tomography (CT) and Micro-CT Techniques
Saites, F., Wang, G., Guo, R., Mannhardt, K. and Kantzas, A.
DOI: 10.2118/2006-027
CIM 2006-027 presented at the 57th Annual Technical Meeting of the Petroleum Society held in Calgary, June 13-15, 2006.
## ABSTRACT
Much of the focus of CBM reservoir assessment in Canada is based on understanding cleats and natural fractures, both in outcrop and in core taken from well bores. Coal characterization studies using imaging devices are presented. X-ray computerized tomography (CT) and micro CT are used on coal samples to provide a better understanding of fracture morphology and apertures. Images are collected in samples of approximately 10 cm in diameter with resolution of (0.4 mm) 2 using X-ray CT, and samples of 1 cm in diameter with resolution of (5_m) 2 using micro CT. Visualization at both resolutions allows for discussion and comparison of structural characteristics at both scales. X-ray CT images are processed to obtain density “logs” and maps under different overburden pressures. The pore space contained helium, and, in one set of scans, argon. Bulk densities increase with increasing overburden pressure. Fracture patterns in both cores were reconstructed from the images using a fracture identification algorithm. Manipulation of the density maps can also provide local strains as a function of increasing overburden pressure.
A full version of this paper is available on OnePetro Online. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.83595871925354, "perplexity": 4556.4923048615055}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-16/segments/1585370506477.26/warc/CC-MAIN-20200401223807-20200402013807-00532.warc.gz"} |
http://link.springer.com/article/10.1140/epjb/e2013-40726-6 | , 86:419
# Investigation of a universal behavior between Néel temperature and staggered magnetization density for a three-dimensional quantum antiferromagnet
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## Abstract
We simulate the three-dimensional quantum Heisenberg model with a spatially anisotropic ladder pattern using the first principles Monte Carlo method. Our motivation is to investigate quantitatively the newly established universal relation T N /√c 3 ∝ ℳ s near the quantum critical point (QCP) associated with dimerization. Here T N , c, and ℳ s are the Néel temperature, the spinwave velocity, and the staggered magnetization density, respectively. For all the physical quantities considered here, such as T N and ℳ s , our Monte Carlo results agree nicely with the corresponding results determined by the series expansion method. In addition, we find it is likely that the effect of a logarithmic correction, which should be present in (3 + 1)-dimensions, to the relation T N /√c 3 ∝ ℳ s near the investigated QCP only sets in significantly in the region with strong spatial anisotropy. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8815093040466309, "perplexity": 1986.9760908932235}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-10/segments/1393999642518/warc/CC-MAIN-20140305060722-00076-ip-10-183-142-35.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/is-this-the-correct-approach.568793/ | # Is this the correct approach?
1. Jan 18, 2012
### Jamin2112
1. The problem statement, all variables and given/known data
2. Relevant equations
Fourier transform, Gaussian filter
3. The attempt at a solution
First of all, someone in the class did it correctly and here's what they said:
Second of all, let me make sure I have a correct understanding of this.
We have a 20 x 262144 matrix (note that that 262144 = 643) each row i, column j is the amplitude of the signal. We're supposed to fft and sum each row before dividing by 20, to get the average power at each frequency. Further, if you look at the code at the bottom you see that the 262144 columns are actually frequencies meant to be in a 64 x 64 x 64 matrix. So we have a 64 x 64 x 64 matrix, each index being a frequency and each containing an average power value. Find the location of the maximum power value. Do an inverse fft. Put a Gaussian filter around the frequency we found. Then look at the plot of the average signal with the filter.
Basically correct?
Can you offer guidance or do you also need help?
Similar Discussions: Is this the correct approach? | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9311078786849976, "perplexity": 394.02993842840834}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-39/segments/1505818687324.6/warc/CC-MAIN-20170920142244-20170920162244-00442.warc.gz"} |
https://math.hecker.org/2014/03/30/linear-algebra-and-its-applications-exercise-3-1-18/ | ## Linear Algebra and Its Applications, Exercise 3.1.18
Exercise 3.1.18. Suppose that $S = \{0\}$ is the subspace of $\mathbb{R}^4$ containing only the origin. What is the orthogonal complement of $S$ ($S^\perp$)? What is $S^\perp$ if $S$ is the subspace of $\mathbb{R}^4$ spanned by the vector $(0, 0, 0, 1)$?
Answer: Every vector is orthogonal to the zero vector. (In other words, $v^T0 = 0$ for all $v$.) So the orthogonal complement of $S = \{0\}$ is the entire vector space, or in this case $S^\perp = \mathbb{R}^4$.
If $S$ is spanned by the vector $(0, 0, 0, 1)$ then all vectors in $S$ are of the form $(0, 0, 0, d)$. Any vector whose last entry is zero is orthogonal to vectors in $S$. In other words, for vectors of the form $(a, b, c, 0)$ the inner product with a vector in $S$ is
$a \cdot 0 + b \cdot 0 + c \cdot 0 + 0 \cdot d = 0+0+0+0 = 0$
The space of vectors of the form $(a, b, c, 0)$ is spanned by the vectors $(1, 0, 0, 0)$, $(0, 1, 0, 0)$, and $(0, 0, 1, 0)$, which are linearly independent and form a basis for the subspace. Thus $S^\perp$ is the subspace of $\mathbb{R}^4$ with basis vectors $(1, 0, 0, 0)$, $(0, 1, 0, 0)$, and $(0, 0, 1, 0)$.
NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.
If you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition, Dr Strang’s introductory textbook Introduction to Linear Algebra, Fourth Edition and the accompanying free online course, and Dr Strang’s other books.
This entry was posted in linear algebra and tagged . Bookmark the permalink. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 29, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9207940697669983, "perplexity": 98.65904065915865}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-39/segments/1537267159359.58/warc/CC-MAIN-20180923114712-20180923135112-00322.warc.gz"} |
https://tug.org/pipermail/texhax/2008-December/011404.html | # [texhax] apacite and url
Alan Litchfield alan at alphabyte.co.nz
Thu Dec 4 20:18:55 CET 2008
```Hi Christian,
I can't answer your second question, but as to the first the use of url's is
extensively handled in the document apacite.pdf. If you have TeXLive you
already have this, or you can get it from your local ctan mirror.
Basically you can use either url = {...} as a bib field or you can use
\url{...} within a field, say howpublished = {...\url{...}...} for example.
HIH
Alan
Christian Deindl wrote:
> hi,
>
> I have two questions.
>
> I'm using apacite as bibliography style.
> Two of my bibtex entries need to be cited with an url.
> unfourtunately this produces an error message "missing \$ inserted".
> Is there a way to cite an url?
> perhaps I need an additional package?
>
> the second question is somewaht related to the first.
> since my paper is in german, is there a good german style, who is also | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9487059712409973, "perplexity": 4536.802028306767}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-30/segments/1563195526799.4/warc/CC-MAIN-20190720235054-20190721021054-00478.warc.gz"} |
http://www.chegg.com/homework-help/questions-and-answers/proton-projected-positive-x-direction-region-uniform-electric-field-e-600-105-i-n-c-t-0-pr-q2678805 | ## proton in a feild
A proton is projected in the positive x direction in to a region of a uniform electric field E = - 6.00 × 105 i N/C at t=0. The proton travels 7.00 cm as it comes to rest. (9 points)
Determine
the acceleration of the proton,
its initial speed, and
the time interval over which the proton comes to rest. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8754882216453552, "perplexity": 870.5943291169283}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2013-20/segments/1368705956734/warc/CC-MAIN-20130516120556-00060-ip-10-60-113-184.ec2.internal.warc.gz"} |
http://www.msad49moodle.org/course/index.php?categoryid=54 | ### Technology Curriculum
Write a concise and interesting paragraph here that explains what this course is about | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9741511344909668, "perplexity": 3776.492220853597}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-43/segments/1508187825308.77/warc/CC-MAIN-20171022150946-20171022170946-00823.warc.gz"} |
https://www.physicsforums.com/threads/help-setting-up-equation-to-find-curl-of-navier-stokes-equation.625948/ | # Help Setting Up Equation To Find Curl of Navier-Stokes Equation
1. Aug 6, 2012
### AKBob
1.
The problem statement, all variables and given/known data
I'm having trouble using equation 2.1 or 2.2 in the article to find the curl of the navier-stokes equation. I understand how to find curl, but can't make sense of the explanation/steps in the document provided by the professor.
2. Relevant equations
All relavent equations are included in the two attachments.
3. The attempt at a solution
I'm really having trouble getting started. The document provided by my professor says to "First evaluate the '(Beta)yk X v' term, substitute that in (v is the vector discussed at the top)," but the equation at the top looks like a general equation, and I'm starting to get frustrated.
Any help/ideas/suggestions would really be appreciated.
File size:
4.3 KB
Views:
53
File size:
47.6 KB
Views:
81
2. Aug 6, 2012
### voko
All the equations in the attachments are broken beyond repair.
3. Aug 6, 2012
### Muphrid
Yeah, sorry, use some LaTeX please. It'll be easier on everyone (well, except maybe you). Example:
$$\rho \left(\frac{\partial v}{\partial t} + v \cdot \nabla v \right) = - \nabla p + f + \overline T(\nabla)$$
Is given by
Code (Text):
$$\rho \left(\frac{\partial v}{\partial t} + v \cdot \nabla v \right) = - \nabla p + f + \overline T(\nabla)$$
Similar Discussions: Help Setting Up Equation To Find Curl of Navier-Stokes Equation | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.802988588809967, "perplexity": 2409.272924544255}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-09/segments/1518891814857.77/warc/CC-MAIN-20180223213947-20180223233947-00080.warc.gz"} |
http://mathhelpforum.com/calculus/15356-laplace-transform.html | Math Help - Laplace transform
1. Laplace transform
Could someone show me how to do the following Laplace transform?
f(t) =
e^(2t) if 0 < t < 8
1 if t > 8
F(s) = ?
I'm pretty lost. Thanks for any help.
2. Originally Posted by PvtBillPilgrim
Could someone show me how to do the following Laplace transform?
f(t) =
e^(2t) if 0 < t < 8
1 if t > 8
F(s) = ?
I'm pretty lost. Thanks for any help.
Well we can do thus using just the definition of the Laplace Transform:
$
(\mathcal{L} f) (s) = \int_0^{\infty} f(t) e^{-st} dt
= \int_0^8 e^{2t}e^{-st} dt + \int_8^{\infty} e^{-st}dt
= \int_0^8 e^{t(2-s)} dt + \int_8^{\infty} e^{-st}dt
$
........ $=\frac{1}{s-2}[1-e^{8(2-s)}] + \frac{e^{-8s}}{s}$
RonL
3. Makes sense.
Thank you very much. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 2, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9870262145996094, "perplexity": 1094.0793464909727}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-35/segments/1440645176794.50/warc/CC-MAIN-20150827031256-00098-ip-10-171-96-226.ec2.internal.warc.gz"} |
https://www.zora.uzh.ch/id/eprint/143117/ | # Jupiter’s formation and its primordial internal structure
Lozovsky, Michael; Helled, Ravit; Rosenberg, Eric D; Bodenheimer, Peter (2017). Jupiter’s formation and its primordial internal structure. The Astrophysical Journal, 836(2):227.
## Abstract
The composition of Jupiter and the primordial distribution of the heavy elements are determined by its formation history. As a result, in order to constrain the primordial internal structure of Jupiter, the growth of the core and the deposition and settling of accreted planetesimals must be followed in detail. In this paper we determine the distribution of the heavy elements in proto-Jupiter and determine the mass and composition of the core. We find that while the outer envelope of proto-Jupiter is typically convective and has a homogeneous composition, the innermost regions have compositional gradients. In addition, the existence of heavy elements in the envelope leads to much higher internal temperatures (several times 104 K) than in the case of a hydrogen–helium envelope. The derived core mass depends on the actual definition of the core: if the core is defined as the region in which the heavy-element mass fraction is above some limit (say, 0.5), then it can be much more massive (~15 ${M}_{\oplus }$) and more extended (10% of the planet's radius) than in the case where the core is just the region with 100% heavy elements. In the former case Jupiter's core also consists of hydrogen and helium. Our results should be taken into account when constructing internal structure models of Jupiter and when interpreting the upcoming data from the Juno (NASA) mission.
## Abstract
The composition of Jupiter and the primordial distribution of the heavy elements are determined by its formation history. As a result, in order to constrain the primordial internal structure of Jupiter, the growth of the core and the deposition and settling of accreted planetesimals must be followed in detail. In this paper we determine the distribution of the heavy elements in proto-Jupiter and determine the mass and composition of the core. We find that while the outer envelope of proto-Jupiter is typically convective and has a homogeneous composition, the innermost regions have compositional gradients. In addition, the existence of heavy elements in the envelope leads to much higher internal temperatures (several times 104 K) than in the case of a hydrogen–helium envelope. The derived core mass depends on the actual definition of the core: if the core is defined as the region in which the heavy-element mass fraction is above some limit (say, 0.5), then it can be much more massive (~15 ${M}_{\oplus }$) and more extended (10% of the planet's radius) than in the case where the core is just the region with 100% heavy elements. In the former case Jupiter's core also consists of hydrogen and helium. Our results should be taken into account when constructing internal structure models of Jupiter and when interpreting the upcoming data from the Juno (NASA) mission.
## Statistics
### Citations
Dimensions.ai Metrics
19 citations in Web of Science®
24 citations in Scopus®
### Altmetrics
Detailed statistics
Item Type: Journal Article, refereed, original work 07 Faculty of Science > Institute for Computational Science 530 Physics Physical Sciences > Astronomy and Astrophysics Physical Sciences > Space and Planetary Science English February 2017 09 Jan 2018 21:48 28 Jul 2020 12:05 IOP Publishing 1538-4357 Green Publisher DOI. An embargo period may apply. https://doi.org/10.3847/1538-4357/836/2/227
Preview
Content: Published Version
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View at publisher | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.840344250202179, "perplexity": 1113.868920556465}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-40/segments/1600400191780.21/warc/CC-MAIN-20200919110805-20200919140805-00153.warc.gz"} |
https://edurev.in/course/quiz/attempt/7813_Test-Signal-Systems/6126487f-c86a-4f53-af90-01cd9394778b | # Test: Signal & Systems
## 25 Questions MCQ Test GATE ECE (Electronics) 2023 Mock Test Series | Test: Signal & Systems
Description
Attempt Test: Signal & Systems | 25 questions in 75 minutes | Mock test for GATE preparation | Free important questions MCQ to study GATE ECE (Electronics) 2023 Mock Test Series for GATE Exam | Download free PDF with solutions
QUESTION: 1
### The even part of the signal x(n) = u(n)
Solution:
x(n) = u(n)
x(-n) = u(-n)
Even part of x(n),
QUESTION: 2
### The value of discrete time signal at non-integer is :
Solution:
The value of discrete time signal and non-integers is zero.
QUESTION: 3
### The ROC of the signal x(t) = 4
Solution:
Since there is no common ROC is ROC and laplace transform of x(t) does not exist.
QUESTION: 4
The input and output relationship of a system is described as y(t) = ax + b, this system is linear
Solution:
For a linear system equation 3 and 4 must be equal, so b = 0, whereas for any value of a.
QUESTION: 5
Which of the following signal is an example of an anti causal signal
Solution:
For an anti causal signal, the value of a signal must be zero for t > 0 so signal is e-at u(-t)
QUESTION: 6
The maximum phase-shift provided by the given system is
Solution:
QUESTION: 7
Determine the fundamental period of x(n) =
Solution:
QUESTION: 8
Consider a signal x(t) = 4 rect (t/6) and its Fourier - transform is x(ω). Determine the area under the curve in the ω-domain.
Solution:
QUESTION: 9
If the Fourier transform of x(t) is x(ω), then determine the Fourier transform of x(at-b)
Solution:
QUESTION: 10
A pole zero pattern of a certain filter is shown in the figure. The filter must be
Solution:
Since locations of poles and zeros are mirror image to each other about vertical axis, so it is an all pass filter
QUESTION: 11
Determine the laplace-transform of signal x(t) shown in figure
Solution:
QUESTION: 12
Determine the total energy of x(t) = 12 Sin (6t)
Solution:
x(t) = 12 Sinc (6t)
So, x(t) = 2 rect (f/6)
According to Parsvell's Theorem,
QUESTION: 13
Determine the bandwidth of the signal x(t) = e-at u(t), so that it contains 90% of its total energy.
Solution:
QUESTION: 14
Determine the output y(t) for an input x(t) = e-2t u(t), if the step response of the system is t u(t).
Solution:
QUESTION: 15
Simplify the following expression:
δ(-t) * u(t) * e-t/2 u(t)
Solution:
QUESTION: 16
Consider the analog signal x(t) = 3 Cos 100πt. This signal is sampled with a frequency of 75 samples per second and at the reconstruction side an ideal LPF having cut-off frequency is 30Hz is used. Determine the frequency component at the output of filter.
Solution:
QUESTION: 17
The input and output relationship of a system is given as,
Solution:
It is the input-output relationship of an accumulator.
For any value of n, the output y(n) is depends only on the previous and present input
So it is causal.
It is also liner It is unstable system. Because the output is not bounded for the bounded input.
QUESTION: 18
A signal x(t) = 2 (1 - Cos2πt) is sampled with a sampling frequency of 10 Hz. Determine the z-transform of sampled signal.
Solution:
QUESTION: 19
Given,
Determine the ROC of its z-transform.
Solution:
ROC : |Z| > 1/3
QUESTION: 20
If a signal f(t) has energy E, the energy of the signal f(2t) is equal to
Solution:
QUESTION: 21
The correlation between two signals x1(t) and x2(t) is 6. If average power of x1(t) is 10 and x2(t) is 8.
Then determine the power of x1(t) + x2(t)
Solution:
Total energy is given as,
E = E1 + E2 + 2 Rx (τ)
= 10 + 8 + 2 × 6
= 30
QUESTION: 22
, the co-efficient of term e-t in f(t) will be :
Solution:
QUESTION: 23
S1 : δ(n) is an energy signal having energy is 1.
S2 : u(n) is a power signal having power is 1/2.
Which of the above is/are correct?
Solution:
Energy of the signal is given as
QUESTION: 24
The input and output relationship of a system is given as,
The system is :
Solution:
QUESTION: 25
Determine the impulse response of the inverse system, if the impulse response of the system is u(n).
Solution:
Use Code STAYHOME200 and get INR 200 additional OFF Use Coupon Code | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8693585395812988, "perplexity": 4843.193669360612}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662525507.54/warc/CC-MAIN-20220519042059-20220519072059-00183.warc.gz"} |
https://socratic.org/questions/what-is-pascal-s-triangle#138674 | Precalculus
Topics
What is Pascal's triangle?
Apr 17, 2015
One of the most interesting Number Patterns is Pascal's Triangle. It is named after Blaise Pascal.
To build the triangle, always start with "1" at the top, then continue placing numbers below it in a triangular pattern.
Each number is the two numbers above it added together (except for the edges, which are all "1").
Interesting part is this:
The first diagonal is just "1"s, and the next diagonal has the counting numbers. The third diagonal has the triangular numbers. The fourth diagonal has the tetrahedral numbers. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9191370010375977, "perplexity": 1266.6216714942193}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320303956.14/warc/CC-MAIN-20220123015212-20220123045212-00442.warc.gz"} |
https://ratnuu.wordpress.com/tag/technical/ | You are currently browsing the tag archive for the ‘Technical’ tag.
While trying to understand the Luby transform (LT) code, I stumbled upon the well known coupon collector’s problem. It took a while for me to figure out the connection, but as it turned out, there is a stronger connection between these two. In LT parlance, if we were to use only degree one packets (that is, packets sent and collected as it is) what is the expected number of packets to be collected (when collected randomly, one at a time) such that all the required packets are collected atleast once. For illustration let us say we have $n$ information packets at the transmitter. The receiver collectes these one by one at random. How many packets on the average, we need to collect until we have collected all of the $n$ different information packets. Remember we are collecting the packets randomly (On the other hand, if we were to collect things deterministically, we just need to collect $n$ packets to get all these $n$, when done without replacement).
Assume that there are $n$ distinct coupon types. We have, a large pool of these coupons at disposal. Every time you go to the shop you collect a coupon picked uniformly at random from the pool. The picked coupon has equal probability of being any of the $n$ types. Naturally, some of the collected coupons (over multiple visits to the shop) may be of the same type. The question asked is this: Suppose the coupon collector aims to have coupons of all types. How many (number of visits) coupons he has to collect till he possess all the $n$ distinct types of coupons?
In expectation, the coupon collector should make $n \log(n) + O(1)$ visits to the shop in order to have atleast one copy of all $n$ distinct types of coupons . This coupon collector problem can sound a little confusing to a fresh reader. For simplicity sake we can assume that, there are $n$ differently coloured coupons at the shop. The question then is, on average (i.e., expectation) how many times one needs to visit (each visit fetch a coupon) the shop so that all coloured coupons are fetched atleast once.
There are $n$ different type of coupons. The coupon collector collects a coupon upon each visit. The collected coupon is among the $n$ types, picked uniformly at random (from a set of possibly large pool of coupons) . Since the coupon is drawn uniformly at random, there is a non zero probability that some of the collected coupons over multiple visits may be of the same type. Suppose that at some stage, the coupon collector has $r$ different type of coupons collected. The probability that his next visit fetch a new coupon type (not of the $r$ types he already have in the kitty) is $p_r=\frac{n-r}{n}$. So, the expected number of coupons to be collected to fetch a new coupon type is $\frac{n}{n-r}$. Let us denote this number by $E\left[N_r\right]$.
The expected value $E\left[N_i\right]=\frac{1}{p_i}=\frac{n}{n-i}$. From this we can compute the expected value of $N$. In other words, $E[N]$, the expected number of coupons to be collected (i.e, number of visits to the shop!) so that, the he would have all the different $n$ types of coupons is:
$E[N]=\displaystyle \sum_{i=1}^{n-1} {\frac{n}{n-i}}=n\sum_{i=1}^{n-1}{\frac{1}{i}}=nH(n)=n\log(n)+O(1)$
So, what is the trouble? This number $n\log(n)$ is prohibitively high a number to be acceptable (as decoding time of $n\log (n)$ is significantly higher than the wishful linear time $n$!). So, simply using degree $1$ is not a good idea. This is why Luby went ahead and identified some smarter distribution like Soliton (and its variants proposed later on, such as robust soliton and then the recent raptor codes by Amin).
leeyoongu on LT codes decoding using Wiedem… vinod kumar on Yesudas and Rafi singing same… Ramesh on The split lot corner stor… sreelesh. on Yesudas and Rafi singing same… The new Prime gap |… on Improved bounds on Prime numbe… | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 26, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8638216853141785, "perplexity": 767.4026370762635}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-30/segments/1500549423842.79/warc/CC-MAIN-20170722022441-20170722042441-00543.warc.gz"} |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Statistical_Mechanics/Advanced_Statistical_Mechanics/The_Grand_Canonical_Ensemble/Particle_number_fluctuations | Particle number fluctuations
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In the grand canonical ensemble, the particle number $$N$$ is not constant. It is, therefore, instructive to calculate the fluctuation in this quantity. As usual, this is defined to be
$\Delta N = \sqrt{\langle N^2 \rangle - \langle N \rangle^2}$
Note that
$$\zeta{\partial \over \partial \zeta}\zeta {\partial \over \partial \zeta}\ln {\cal Z}(\zeta,V,T)$$ $${1 \over {\cal Z}}\sum_{N=0}^{\infty}N^2 \zeta^N Q(N,V,T) -{1 \over {\cal Z}^2} \left[\sum_{N=0}^{\infty} N \zeta^N Q(N,V,T)\right]^2$$ $$\langle N^2 \rangle - \langle N \rangle^2$$
Thus,
$\left(\Delta N\right)^2 =\zeta{\partial \over \partial \zeta} \zeta {\partial \over \partial \zeta} \ln {\cal Z} (\zeta, V, T) = ({KT}^2){\partial^2 \over \partial \mu^2}\ln {\cal Z}(\mu,V,T) = kTV{\partial^2 P \over \partial \mu^2}$
In order to calculate this derivative, it is useful to introduce the Helmholtz free energy per particle defined as follows:
$a(v,T) = {1 \over N}A(N,V,T)$
where $$v={V \over N} = {1 \over \rho}$$ is the volume per particle.
The chemical potential is defined by
$$\mu$$ $${\partial A \over \partial N} =a(v,T) + N{\partial a \over \partial v}{\partial v \over \partial N}$$ $$a(v,T) - v{\partial a \over \partial v}$$
Similarly, the pressure is given by
$P = -{\partial A \over \partial V} = -N{\partial a \over \partial v}{\partial v \over \partial V} = -{\partial a \over \partial v}$
Also,
${\partial \mu \over \partial v} = -v{\partial^2 a \over \partial v^2}$
Therefore,
${\partial P \over \partial \mu} = {\partial P \over \partial v}{\partial v \over \partial \mu} = {\partial^2 a \over \partial v^2} \left[v{\partial^2 a \over \partial v^2}\right]^{-1} = {1 \over v}$
and
${\partial^2 P \over \partial \mu^2} = {\partial \over \partial v}{\partial P \over \partial \mu}{\partial v \over \partial \mu} = {1 \over v^2} \left[ v {\partial^2 a \over \partial v^2}\right]^{-1}= -{1 \over v^3 \partial P/\partial v}$
But recall the definition of the isothermal compressibility:
$\kappa_T = -{1 \over V}{\partial V \over \partial P}=-{1 \over v \partial p/\partial v}$
Thus,
${\partial^2 P \over \partial \mu^2} = {1 \over v^2}\kappa_T$
and
$\Delta N = \sqrt{\frac{\langle N \rangle kT \kappa_T}{v}}$
and the relative fluctuation is given by
${\Delta N \over N} = {1 \over \langle N \rangle}\sqrt{\frac{\langle N \rangle kT \kappa_T}{v}} \sim {1 \over \sqrt {\langle N \rangle }}\rightarrow 0\;\;{as}\langle N \rangle \rightarrow \infty$
Therefore, in the thermodynamic limit, the particle number fluctuations vanish, and the grand canonical ensemble is equivalent to the canonical ensemble. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9891257882118225, "perplexity": 128.36545920233834}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662625600.87/warc/CC-MAIN-20220526193923-20220526223923-00556.warc.gz"} |
http://math.stackexchange.com/questions/105978/can-an-element-be-a-quadratic-residue-and-a-generator-mod-p | # Can an element be a quadratic residue and a generator (mod p)?
i.e. is is possible for
• $g$ to be a generator$\mod{p}$, and
• $g \equiv x^2 \mod{p}$ for some $x$
I'm guessing not, as I think $x$ can't be expressed as a power of $g$, contradicting g being a generator?
-
Dear malikyo_o: Your argument is almost complete! You should try harder! – Pierre-Yves Gaillard Feb 5 '12 at 14:15
If $p$ is an odd prime, then no. $\varphi(p)=p-1$ is even and the multiplicative group $\!\!\!\pmod{p}$ has order $\varphi(p)$. If $g=x^2$, then $g^{(p-1)/2}=x^{p-1}=1\pmod{p}$. But if $g$ is a generator of the multiplicative group $\!\!\!\pmod{p}$, $g^k\not=1\pmod{p}$ for $0<k<p-1$.
In fact, the only $p$ for which $\varphi(p)$ is odd is $p=2$, and $1=1^2$ is a generator for the multiplicative group $\!\!\!\pmod{2}$. For all other $p$, $g=x^2$ cannot be a generator of the multiplicative group $\!\!\!\pmod{p}$.
If you consider $p=2$ to be a prime number (and I wouldn't know how you could defend not doing so) then you will see that $1$ is both a quadratic residue and a generator mod $2$ (admittedly not very spectacular, but true nontheless). This is the only case, as explained the answer by robjohn.
I have added that $p$ is an odd prime in the first paragraph. Thanks for keeping me on my toes. – robjohn Feb 5 '12 at 14:53 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9613246917724609, "perplexity": 167.32608462635002}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-10/segments/1394010451932/warc/CC-MAIN-20140305090731-00081-ip-10-183-142-35.ec2.internal.warc.gz"} |
https://math.libretexts.org/TextMaps/Analysis/Book%3A_Real_Analysis_(Boman_and_Rogers)/7%3A_Intermediate_and_Extreme_Values/7.4%3A_The_Supremum_and_the_Extreme_Value_Theorem | $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$
# 7.4: The Supremum and the Extreme Value Theorem
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Skills to Develop
• Explain supremum and the extreme value theorem
Theorem 7.3.1 says that a continuous function on a closed, bounded interval must be bounded. Boundedness, in and of itself, does not ensure the existence of a maximum or minimum. We must also have a closed, bounded interval. To illustrate this, consider the continuous function $$f(x) =tan^{-1}x$$ defined on the (unbounded) interval $$(-∞,∞)$$.
Figure $$\PageIndex{1}$$: Graph of $$f(x) =tan^{-1}x$$.
This function is bounded between $$-\frac{π}{2}$$ and $$\frac{π}{2}$$, but it does not attain a maximum or minimum as the lines $$y = ±\frac{π}{2}$$ are horizontal asymptotes. Notice that if we restricted the domain to a closed, bounded interval then it would attain its extreme values on that interval (as guaranteed by the EVT).
To find a maximum we need to find the smallest possible upper bound for the range of the function. This prompts the following definitions.
Definition: $$\PageIndex{1}$$
Let $$S ⊆ R$$ and let $$b$$ be a real number. We say that $$b$$ is an upper bound of $$S$$ provided $$b ≥ x$$ for all $$x ∈ S$$.
For example, if $$S = (0,1)$$, then any $$b$$ with $$b ≥ 1$$ would be an upper bound of $$S$$. Furthermore, the fact that $$b$$ is not an element of the set $$S$$ is immaterial. Indeed, if $$T = [0,1]$$, then any $$b$$ with $$b ≥ 1$$ would still be an upper bound of $$T$$. Notice that, in general, if a set has an upper bound, then it has infinitely many since any number larger than that upper bound would also be an upper bound. However, there is something special about the smallest upper bound.
Definition: $$\PageIndex{2}$$
Let $$S ⊆ R$$ and let $$b$$ be a real number. We say that $$b$$ is the least upper bound of $$S$$ provided
1. $$b ≥ x$$ for all $$x ∈ S$$. ($$b$$ is an upper bound of $$S$$)
2. If $$c ≥ x$$ for all $$x ∈ S$$, then $$c ≥ b$$. (Any upper bound of $$S$$ is at least as big as $$b$$)
In this case, we also say that $$b$$ is the supremum of $$S$$ and we write
$b = \sup(S)$
Notice that the definition really says that $$b$$ is the smallest upper bound of $$S$$. Also notice that the second condition can be replaced by its contrapositive so we can say that $$b = \sup S$$ if and only if
1. $$b ≥ x$$ for all $$x ∈ S$$
2. If $$c < b$$ then there exists $$x ∈ S$$ such that $$c < x$$
The second condition says that if a number $$c$$ is less than $$b$$, then it can’t be an upper bound, so that $$b$$ really is the smallest upper bound.
Also notice that the supremum of the set may or may not be in the set itself. This is illustrated by the examples above as in both cases, $$1 = \sup (0,1)$$ and $$1 = \sup [0,1]$$. Obviously, a set which is not bounded above such as $$N = {1, 2, 3, ...}$$ cannot have a supremum. However, for non-empty sets which are bounded above, we have the following.
Theorem $$\PageIndex{1}$$: The Least Upper Bound Property (LUBP)
Let $$S$$ be a non-empty subset of $$R$$ which is bounded above. Then $$S$$ has a supremum.
Sketch of Proof
Since $$S \neq \varnothing$$, then there exists $$s ∈ S$$. Since $$S$$ is bounded above then it has an upper bound, say $$b$$. We will set ourselves up to use the Nested Interval Property. With this in mind, let $$x_1 = s$$ and $$y_1 = b$$ and notice that $$∃ x ∈ S$$ such that $$x ≥ x_1$$ (namely, $$x_1$$ itself) and $$∀ x ∈ S$$, $$y_1 ≥ x$$. You probably guessed what’s coming next: let $$m_1$$ be the midpoint of $$[x_1,y_1]$$. Notice that either $$m_1 ≥ x$$, $$∀x ∈ S$$ or $$∃ x ∈ S$$ such that $$x ≥ m_1$$. In the former case, we relabel, letting $$x_2 = x_1$$ and $$y_2 = m_1$$. In the latter case, we let $$x_2 = m_1$$ and $$y_2 = y_1$$. In either case, we end up with $$x_1 ≤ x_2 ≤ y_2 ≤ y_1, y_2 - x_2 = \frac{1}{2} (y_1 - x_1)$$, and $$∃ x ∈ S$$ such that $$x ≥ x_2$$ and $$∀x ∈ S$$, $$y_2 ≥ x$$. If we continue this process, we end up with two sequences, ($$x_n$$) and ($$y_n$$), satisfying the following conditions:
1. $$x_1 ≤ x_2 ≤ x_3 ≤ ...$$
2. $$y_1 ≥ y_2 ≥ y_3 ≥ ...$$
3. $$∀ n, x_n ≤ y_n$$
4. $$\lim_{n \to \infty } (y_n - x_n) = \lim_{n \to \infty }\frac{1}{2^{n-1}} (y_1 - x_1) = 0$$
5. $$∀ n,∃ x ∈ S$$ such that $$x ≥ x_n$$ and $$∀x ∈ S, y_n ≥ x$$
By properties 1-5 and the NIP there exists $$c$$ such that $$x_n ≤ c ≤ y_n, ∀ n$$. We will leave it to you to use property 5 to show that $$c = \sup S$$.
Exercise $$\PageIndex{1}$$
Complete the above ideas to provide a formal proof of Theorem $$\PageIndex{1}$$.
Notice that we really used the fact that $$S$$ was non-empty and bounded above in the proof of Theorem $$\PageIndex{1}$$. This makes sense, since a set which is not bounded above cannot possibly have a least upper bound. In fact, any real number is an upper bound of the empty set so that the empty set would not have a least upper bound.
The following corollary to Theorem $$\PageIndex{1}$$ can be very useful.
Corollary $$\PageIndex{1}$$
Let ($$x_n$$) be a bounded, increasing sequence of real numbers. That is, $$x_1 ≤ x_2 ≤ x_3 ≤···$$. Then ($$x_n$$) converges to some real number $$c$$.
Exercise $$\PageIndex{2}$$
Prove Corollary $$\PageIndex{1}$$.
Hint
Let $$c = \sup {x_n|n = 1,2,3,...}$$. To show that $$\lim_{n \to \infty } x_n = c$$, let $$\epsilon > 0$$. Note that $$c - \epsilon$$ is not an upper bound. You take it from here!
Exercise $$\PageIndex{3}$$
Consider the following curious expression $$\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{\cdots }}}}$$. We will use Corollary $$\PageIndex{1}$$ to show that this actually converges to some real number. After we know it converges we can actually compute what it is. Of course to do so, we need to define things a bit more precisely. With this in mind consider the following sequence ($$x_n$$) defined as follows:
$x_1 = \sqrt{2}$
$x_{n+1} = \sqrt{2+x_n}$
1. Use induction to show that $$x_n < 2$$ for $$n = 1, 2, 3, ....$$
2. Use the result from part (a) to show that $$x_n < x_{n+1}$$ for $$n = 1, 2, 3, ...$$
3. From Corollary $$\PageIndex{1}$$, we have that ($$x_n$$) must converge to some number $$c$$. Use the fact that ($$x_{n+1}$$) must converge to $$c$$ as well to compute what $$c$$ must be.
We now have all the tools we need to tackle the Extreme Value Theorem.
Theorem $$\PageIndex{2}$$: Extreme Value Theorem (EVT)
Suppose $$f$$ is continuous on $$[a,b]$$. Then there exists $$c$$, $$d ∈ [a,b]$$ such that $$f(d) ≤ f(x) ≤ f(c)$$, for all $$x ∈ [a,b]$$.
Sketch of Proof
We will first show that $$f$$ attains its maximum. To this end, recall that Theorem Theorem 7.3.1 tells us that $$f[a,b] = {f(x)|x ∈ [a,b]}$$ is a bounded set. By the LUBP, $$f[a,b]$$ must have a least upper bound which we will label $$s$$, so that $$s = \sup f[a,b]$$. This says that $$s ≥ f(x)$$,for all $$x ∈ [a,b]$$. All we need to do now is find a $$c ∈ [a,b]$$ with $$f(c) = s$$. With this in mind, notice that since $$s = \sup f[a,b]$$, then for any positive integer $$n$$, $$s - \frac{1}{n}$$ is not an upper bound of $$f[a,b]$$. Thus there exists $$x_n ∈ [a,b]$$ with $$s - \frac{1}{n} < f(x_n) \leq s$$. Now, by the Bolzano-Weierstrass Theorem, ($$x_n$$) has a convergent subsequence($$x_{n_k}$$) converging to some $$c ∈ [a,b]$$. Using the continuity of $$f$$ at $$c$$, you should be able to show that $$f(c) = s$$. To find the minimum of $$f$$, find the maximum of $$-f$$.
Exercise $$\PageIndex{4}$$
Formalize the above ideas into a proof of Theorem $$\PageIndex{2}$$.
Notice that we used the NIP to prove both the Bolzano-Weierstrass Theorem and the LUBP. This is really unavoidable, as it turns out that all of those statements are equivalent in the sense that any one of them can be taken as the completeness axiom for the real number system and the others proved as theorems. This is not uncommon in mathematics, as people tend to gravitate toward ideas that suit the particular problem they are working on. In this case, people realized at some point that they needed some sort of completeness property for the real number system to prove various theorems. Each individual’s formulation of completeness fit in with his understanding of the problem at hand. Only in hindsight do we see that they were really talking about the same concept: the completeness of the real number system. In point of fact, most modern textbooks use the LUBP as the axiom of completeness and prove all other formulations as theorems. We will finish this section by showing that either the Bolzano-Weierstrass Theorem or the LUBP can be used to prove the NIP. This says that they are all equivalent and that any one of them could be taken as the completeness axiom.
Exercise $$\PageIndex{5}$$
Use the Bolzano-Weierstrass Theorem to prove the NIP. That is, assume that the Bolzano-Weierstrass Theorem holds and suppose we have two sequences of real numbers, ($$x_n$$) and ($$y_n$$), satisfying:
1. $$x_1 ≤ x_2 ≤ x_3 ≤ ...$$
2. $$y_1 ≥ y_2 ≥ y_3 ≥ ...$$
3. $$∀ n, x_n ≤ y_n$$
4. $$\lim_{n \to \infty } (y_n - x_n) = 0$$
Prove that there is a real number $$c$$ such that $$x_n ≤ c ≤ y_n$$, for all $$n$$.
Since the Bolzano-Weierstrass Theorem and the Nested Interval Property are equivalent, it follows that the Bolzano-Weierstrass Theorem will not work for the rational number system.
Exercise $$\PageIndex{6}$$
Find a bounded sequence of rational numbers such that no subsequence of it converges to a rational number.
Exercise $$\PageIndex{7}$$
Use the Least Upper Bound Property to prove the Nested Interval Property. That is, assume that every non-empty subset of the real numbers which is bounded above has a least upper bound; and suppose that we have two sequences of real numbers ($$x_n$$) and ($$y_n$$), satisfying:
1. $$x_1 ≤ x_2 ≤ x_3 ≤ ...$$
2. $$y_1 ≥ y_2 ≥ y_3 ≥ ...$$
3. $$∀ n, x_n ≤ y_n$$
4. $$\lim_{n \to \infty } (y_n - x_n) = 0$$
Prove that there exists a real number $$c$$ such that $$x_n ≤ c ≤ y_n$$, for all $$n$$. (Again, the $$c$$ will, of necessity, be unique, but don’t worry about that.)
Hint
Corollary $$\PageIndex{1}$$ might work well here.
Exercise $$\PageIndex{8}$$
Since the LUBP is equivalent to the NIP it does not hold for the rational number system. Demonstrate this by finding a non-empty set of rational numbers which is bounded above, but whose supremum is an irrational number.
We have the machinery in place to clean up a matter that was introduced in Chapter 1. If you recall (or look back) we introduced the Archimedean Property of the real number system. This property says that given any two positive real numbers $$a,b$$, there exists a positive integer $$n$$ with $$na > b$$. As we mentioned in Chapter 1, this was taken to be intuitively obvious. The analogy we used there was to emptying an ocean $$b$$ with a teaspoon $$a$$ provided we are willing to use it enough times $$n$$. The completeness of the real number system allows us to prove it as a formal theorem.
Theorem $$\PageIndex{3}$$: Archimedean Property of $$\mathbb{R}$$
Given any positive real numbers $$a$$ and $$b$$, there exists a positive integer $$n$$, such that $$na > b$$.
Exercise $$\PageIndex{9}$$
Prove Theorem $$\PageIndex{3}$$.
Hint
Assume that there are positive real numbers $$a$$ and $$b$$, such that $$na ≤ b, ∀ n ∈ N$$. Then $$N$$ would be bounded above by $$b/a$$. Let $$s = \sup (N)$$ and consider $$s - 1$$.
Given what we’ve been doing, one might ask if the Archimedean Property is equivalent to the LUBP (and thus could be taken as an axiom). The answer lies in the following problem.
Exercise $$\PageIndex{10}$$
Does $$\mathbb{Q}$$ satisfy the Archimedean Property and what does this have to do with the question of taking the Archimedean Property as an axiom of completeness?
### Contributor
• Eugene Boman (Pennsylvania State University) and Robert Rogers (SUNY Fredonia) | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9886834025382996, "perplexity": 89.56363463220312}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-34/segments/1534221210735.11/warc/CC-MAIN-20180816113217-20180816133217-00290.warc.gz"} |
https://www.physicsforums.com/threads/2-questions-one-wave-one-delta-function.89725/ | # 2 questions one wave one delta function
1. Sep 19, 2005
### Phymath
1st question
what the heck does a "minimum" mean when talking about interference in waves, i got a question of the like y = 1.19(1 + 2 cos p)sin(kx - wt + p) is the superpostion function of three waves one which is p out of phase of the first and another which is p out of phase of the second wave. What value of p gives the minimum, i have no idea what that means I'm guessin when the amplitude is 0 or when pi/2 - kx + wt = p but how do i find that?
2nd question
i have the function
$$\int^{\infty}_{-\infty} (6-5x^5)\delta(x) dx$$
now by defintion of the delta function because 0 is contained with-in (as is all numbers) between the limits should it not = 0? thanks anyone
2. Sep 19, 2005
### StNowhere
2nd question:
Delta function:
$$\int^{\infty}_{-\infty} f(x)\delta(x-a) dx = f(a)$$
Using that, it looks to me like your value is 6
I'll look at the first question a little more before I hazard a guess on it.
3. Sep 19, 2005
### Phymath
how is it 6 when $$f(x) = 6-5x^4$$, and $$\delta(x) = \delta(x-0)$$?
4. Sep 19, 2005
### HallsofIvy
Because the "definition" of the delta function that you refer to requires that
$$\int_{-\infty}^{\infty}f(x)\delta(x)dx= f(0)$$! | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8514612317085266, "perplexity": 1351.5237844030023}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-04/segments/1547583659654.11/warc/CC-MAIN-20190118005216-20190118031216-00590.warc.gz"} |
https://planetmath.org/InequalityOfLogarithmicAndAsymptoticDensity | # inequality of logarithmic and asymptotic density
For any $A\subseteq\mathbb{N}$ we denote $A(n):=|A\cap\{1,2,\ldots,n\}|$ and $S(n):=\sum\limits_{k=1}^{n}\frac{1}{k}$.
Recall that the values
$\underline{d}(A)=\liminf_{n\to\infty}\frac{A(n)}{n}\qquad\overline{d}(A)=% \limsup_{n\to\infty}\frac{A(n)}{n}$
are called lower and upper asymptotic density of $A$.
The values
$\underline{\delta}(A)=\liminf_{n\to\infty}\frac{\sum\limits_{k\in A;k\leq n}% \frac{1}{k}}{S(n)}\qquad\overline{\delta}(A)=\limsup_{n\to\infty}\frac{\sum% \limits_{k\in A;k\leq n}\frac{1}{k}}{S(n)}$
are called lower and upper logarithmic density of $A$.
We have $S(n)\sim\ln n$ (we use the Landau notation). This follows from the fact that $\lim\limits_{n\to\infty}S(n)-\ln n=\gamma$ is Euler’s constant. Therefore we can use $\ln n$ instead of $S(n)$ in the definition of logarithmic density as well.
The sum in the definition of logarithmic density can be rewritten using Iverson’s convention as $\sum_{k=1}^{n}\frac{1}{k}[k\in A]$. (This means that we only add elements fulfilling the condition $k\in A$. This notation is introduced in [1, p.24].)
###### Theorem 1.
For any subset $A\subseteq\mathbb{N}$
$\underline{d}(A)\leq\underline{\delta}(A)\leq\overline{\delta}(A)\leq\overline% {d}(A)$
holds.
###### Proof.
We first observe that
$\displaystyle\frac{1}{k}[k\in A]=\frac{A(k)-A(k-1)}{k},$ $\displaystyle D(n):=\sum_{k=1}^{n}\frac{1}{k}[k\in A]=\frac{A(n)}{n}+\sum_{k=1% }^{n-1}\frac{A(k)}{k(k+1)}$
There exists an $n_{0}\in\mathbb{N}$ such that for each $n\geq n_{0}$ it holds $\underline{d}(A)-\varepsilon\leq\frac{A(n)}{n}\leq\overline{d}(A)+\varepsilon$.
We denote $C:=1+S(n_{0})$. For $n\geq n_{0}$ we get
$\displaystyle D(n)\leq C+\sum_{k=n_{0}}^{n-1}\frac{A(k)}{k}\cdot\frac{1}{k+1}% \leq C+(\overline{d}(A)+\varepsilon)\sum_{k=n_{0}}^{n-1}\frac{1}{k+1}\sim(% \overline{d}(A)+\varepsilon)\ln n,$ $\displaystyle\overline{\delta}(A)=\limsup_{n\to\infty}\frac{D(n)}{\ln n}\leq% \overline{d}(A)+\varepsilon.$
This inequality holds for any $\varepsilon>0$, thus $\overline{\delta}(A)\leq\overline{d}(A)$.
For the proof of the inequality for lower densities we put $C^{\prime}:=\sum_{k=1}^{n_{0}-1}\frac{A(k)}{k(k+1)}-(\underline{d}(A)-% \varepsilon)S(n_{0})$. We get
$\displaystyle D(n)\geq C^{\prime}+(\underline{d}(A)-\varepsilon)S(n_{0})+(% \underline{d}(A)-\varepsilon)\sum_{k=n_{0}}^{n}\frac{1}{k+1}=\\ \displaystyle C^{\prime}+(\underline{d}(A)-\varepsilon)S(n)\sim(\underline{d}(% A)-\varepsilon)\ln n$
and this implies $\underline{\delta}(A)\geq\underline{d}(A)$. ∎
For the proof using Abel’s partial summation see [4] or [5].
###### Corollary 1.
If a set has asymptotic density, then it has logarithmic density, too.
A well-known example of a set having logarithmic density but not having asymptotic density is the set of all numbers with the first digit equal to 1.
It can be moreover proved, that for any real numbers $0\leq\underline{\alpha}\leq\underline{\beta}\leq\overline{\beta}\leq\overline{% \alpha}\leq 1$ there exists a set $A\subseteq\mathbb{N}$ such that $\underline{d}(A)=\underline{\alpha}$, $\underline{\delta}(A)=\underline{\beta}$, $\overline{\delta}(A)=\overline{\beta}$ and $\overline{d}(A)=\overline{\alpha}$ (see [2]).
## References
• 1 R. L. Graham, D. E. Knuth, and O. Patashnik. Concrete mathematics. A foundation for computer science. Addison-Wesley, 1989.
• 2 L. Mišík. Sets of positive integers with prescribed values of densities. Mathematica Slovaca, 52(3):289–296, 2002.
• 3 H. H. Ostmann. Additive Zahlentheorie I. Springer-Verlag, Berlin-Göttingen-Heidelberg, 1956.
• 4 J. Steuding. http://www.math.uni-frankfurt.de/~steuding/steuding/prob.pdfProbabilistic number theory.
• 5 G. Tenenbaum. Introduction to analytic and probabilistic number theory. Cambridge Univ. Press, Cambridge, 1995.
Title inequality of logarithmic and asymptotic density InequalityOfLogarithmicAndAsymptoticDensity 2014-03-24 9:16:11 2014-03-24 9:16:11 kompik (10588) kompik (10588) 8 kompik (10588) Theorem msc 11B05 AsymptoticDensity LogarithmicDensity | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 35, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9937368631362915, "perplexity": 557.0113696943137}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-17/segments/1618038119532.50/warc/CC-MAIN-20210417102129-20210417132129-00104.warc.gz"} |
https://chemistry.stackexchange.com/questions/21860/a-misunderstanding-about-the-energy-profile-of-reactions-with-a-catalyst-involve/42391 | A misunderstanding about the energy profile of reactions with a catalyst involved
All of us are aware of the importance of the catalysts in bio-chemistry. For a high school learner like me, catalysts ,and therefore, enzymes play a bridge-like role that connect high school bio to high school chemistry.
Yet, I got baffled about the two common energy profiles drawn for an anonymous reaction with a presence of a catalyst. The question may seem rudimentary, I agree, but the possible answers to my question were too technical for me to understand.
One of the energy profiles had several curves in the pathway for $E$, and the other one has only one curve in the pathway where the catalyst should have affected.
My guess is that the first energy profile is about a net change in $E_\mathrm{a}$ but the second one demonstrates the reality when an exothermic reaction occurs.
Now, here are the questions:
• Why are the two energy profiles different? Please be as if you're teaching thermo-chemistry to a little kid!
• Why do those curves occur in the latter energy profile and is there a way to know how many of those curves will occur in the presence of a catalyst?
Image credits: chemwiki.ucdavis.edu (page here) and wiki (page here).
The upper graph is just the most simple way to visualize the effect of a catalyst on a reaction $\ce{S -> P}$:
The activation energy is lowered.
The activation energy for the reaction in that direction is the difference of the energies of the starting material $S$ and a transition state $TS^\#$.
Since it is the same starting marterial in the presence or absence of the catalyst, the energy of the transition state is different.
Can the same transition state have two different energies - just through the remote distance magical action of a catalyst located somewhere? Probably not!
It is much more plausible that - in the absence and presence of a catalyst - two completely different (different in structure and different in energy) transition states are involved.
Exactly this situation is described in the second graph!
The catalyst "reacts" with the starting material, either by forming a covalent bond, by hydrogen bonding, etc. and thus opens a different reaction channel with different intermediates and a transition state not found in in the non-catalyzed reaction.
In the end, the same product $P$ is formed and the catalyst is regenerated, but this doesn't mean that the catalyst wasn't heavily involved in the reaction going on.
1# Curve 1 is just for a simple case where it's showing the comparison between a theoretical pathway for a catalyzed and non-catalyzed reaction but it has an another significant which I will discuss in part two.
2# How many of these curves can occur: Infinite (Theoretically) but in practical case you have to consider all types of possible elementary reaction in your model. An elementary reaction is a reaction which can occur in one step or the reactant, product and TS has well defined position on potential energy surface. For example if you have a simple ketone hydrogenation reaction, you can get an overall energy profile for gas phase reaction but on a catalytic surface there are lots of possibilities. At first the ketone and hydrogen will adsorb on the surface and then 1st hydrogen addition can occur either on C of the ketone group or on the hydrogen of the ketone group. They are intuitively main reaction pathway but there can be much more.
In the above image I was working on LA hydrogenation and you can see there are two different pathway for the hydrogenation and there might be some other more pathway. And if you want to capture the real chemistry you have to make sure you are including all important intermediates. So your energy profile may look like this Here the picture becomes quite complicated with lots of elementary reaction. To determine the overall activation energy of a catalyzed reaction scientists follow two methods
method-1: Energetic span theory: Given by Shaik et al. I am adding the link to that paper here, http://pubs.acs.org/doi/ipdf/10.1021/ar1000956
2) Apparent activation barrier: This is an age old method. Here the sensitivity of turn over frequency (TOF) with the change of temperature is measured and defined as apparent activation barrier. $$E_a=-R\frac{\delta~ln(TOF)}{\delta(\frac{1}{T})}$$
Now you can plot the first E profile with your gas phase activation energy and apparent activation energy of the catalyzed reaction.
A reaction involving more than one elementary step has one or more intermediates being formed which, in turn, means there is more than one energy barrier to overcome. In other words, there is more than one transition state lying on the reaction pathway. As it is intuitive that pushing over an energy barrier or passing through a transition state peak would entail the highest energy, it becomes clear that it would be the slowest step in a reaction pathway. However, when more than one such barrier is to be crossed, it becomes important to recognize the highest barrier which will determine the rate of the reaction. This step of the reaction whose rate determines the overall rate of reaction is known as rate determining step or rate limiting step. The height of energy barrier is always measured relative to the energy of the reactant or starting material. Different possibilities have been shown in figure 6
From the link you gave on energy profile i found this. Basically there is no "net change" in $E_{a}$ . In the first graph in question the curve which is at the top is the representing how the reaction proceeds without a catalyst and the one below is of the reaction with catalyst. In the second graph there are 4 intermediate reactions/transition states before the product is formed. Hence you have 4 peaks of different activation energy as shown.
A catalyst basically provides alternate routes to a reaction by forming the "transition/metastable/intermediate state at a lower energy than the original reaction. You will study later that this has to do with basic Collision theory and molecular dynamics.
In the given graphs a catalyst can make the R->P through just one route where only one transition happens (as in the graph 1) or for 4(or even more) transition states (as in the graph 2). For one set of conditions(P,T,V etc) I am guessing the number of these peaks will be unique. It is also possible that for a certain catalyst and reactant the number of peaks / intermediate states is constant. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8471270799636841, "perplexity": 641.1781576397732}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-10/segments/1581875145859.65/warc/CC-MAIN-20200223215635-20200224005635-00355.warc.gz"} |
https://www.physicsforums.com/threads/buoancy-volume-question.893658/ | # Homework Help: Buoancy volume question
1. Nov 16, 2016
### NY152
1. The problem statement, all variables and given/known data
A 69.5 kg man floats in freshwater with 2.95% of his volume above water when his lungs are empty, and 4.75% of his volume above water when his lungs are full. Calculate the volume of air, in liters, that he inhales (this is called his lung capacity). Neglect the weight of air in his lungs.
2. Relevant equations
d=m/v
freshwater: d=1000 kg/m^3
3. The attempt at a solution
From the given information, there's a 1.8% increase in volume. I'm just not sure where to start given the information above.
2. Nov 16, 2016
### Staff: Mentor
You should be able to determine the man's volume for both cases. Start by considering the volume of water he displaces in order to float.
3. Nov 16, 2016
### NY152
Okay so I did ((69.5 kg/1000 kg/m^3)*(4.75/100))-((69.5/1000 kg/m^3)(2.95/100))= .00125 m^3 then converted that to liters which is 1.25 Liters. The hw system says "it looks like you may have confused the denominator and the numerator, check your algebra" Not sure where I'm going wrong here though...
4. Nov 16, 2016
### Staff: Mentor
You've found the difference between 4.75% of the displaced water and 2.95% of the displaced water. That's not what you're looking for. The displaced water volume is constant and smaller than the man's volume. You need to find the two volumes of the man.
By the way, save yourself a heap of typing and just assign a variable name to repeated quantities. Call the volume of water displaced vw, for example.
5. Nov 16, 2016
### NY152
Oh okay, so I'd do the same sort of math setup but use 95.25% and 97.05% instead?
6. Nov 16, 2016
### Staff: Mentor
No. Let's concentrate on one of the volumes for the man first.
You've correctly determined that he displaces a volume of water $V_w = M/ρ$, where M is his mass and ρ the density of water. That is also the amount of his volume that is below water (since it's displacing the water). Let's call the man's total volume for the first case (the 2.95% above water case) $V_o$. What would be the volume above water (in symbols, no numbers yet)? | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8877604007720947, "perplexity": 1508.1938613619034}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-30/segments/1531676589557.39/warc/CC-MAIN-20180717031623-20180717051623-00341.warc.gz"} |
https://quantiki.org/wiki/Superfidelity | Superfidelity
Superfidelity is a measure of similarity between density operators. It is defined as
$G(\rho,\sigma) = \mathrm{tr}\rho\sigma + \sqrt{1-\mathrm{tr}(\rho^2)}\sqrt{1-\mathrm{tr}(\sigma^2)},$
where σ and ρ are density matrices.
Superfidelity was introduced inmiszczak09sub as an upper bound for fidelity.
Properties
Super-fidelity has also properties which make it useful for quantifying distance between quantum states. In particular we have:
• Bounds: 0 ≤ G(ρ1, ρ2) ≤ 1.
• Symmetry: G(ρ1, ρ2) = G(ρ2, ρ1).
• Unitary invariance: for any unitary operator U, we have G(ρ1, ρ2) = G(Uρ1U † , Uρ2U † ).
• Concavity:
G(ρ1, αρ2 + (1 − α)ρ3) ≥ αG(ρ1, ρ2) + (1 − α)G(ρ1, ρ3)
for any ρ1, ρ2, ρ3 ∈ ΩN and α ∈ [0, 1].
• Supermultiplicativity: for ρ1, ρ2, ρ3, ρ4 ∈ ΩN we have
G(ρ1 ⊗ ρ2, ρ3 ⊗ ρ4) ≥ G(ρ1, ρ3)G(ρ2, ρ4). | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9983946084976196, "perplexity": 145.1643232569495}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-17/segments/1524125947690.43/warc/CC-MAIN-20180425022414-20180425042414-00391.warc.gz"} |
https://physics.stackexchange.com/questions/746057/free-boson-twisted-boundary-condition-and-t2-partition-function | Free boson twisted boundary condition and $T^2$ partition function
Many CFT textbooks discuss free boson theory and free fermion theories on the torus.
The partition function for the boson theory (without compactification and orbifold) is obtained by summing over the Verma modules from all possible highest weight states $$|\alpha \rangle$$ coming from the vertex operators $$e^{i \alpha X}$$. The result reads
$$\sqrt{\frac{1}{\operatorname{Im}\tau}} \frac{1}{\eta(\tau) \overline{\eta(\tau})} \ ,$$ where the $$\operatorname{Im}\tau$$ factor comes from integrating over all possible "momenta" $$\alpha$$, while the $$\eta(\tau)$$ comes from summing over Virasoro descendants.
For free fermion theory, the torus partition function takes in all four sectors, (R, R), (R, NS), (NS, R), (NS, NS), corresponding to different boundary conditions of the field $$\psi$$.
The uncompactified free boson theory also has a twisted sector as well, with an anti-periodic boundary condition $$X(e^{2\pi i}z) = - X(z)$$. In the big yellow book, this situation is discussed, where the two-point function of $$X$$ and the stress tensor is computed. But what is left untouched is the representation theory in this sector.
I wonder what is the $$T^2$$ partition function of the uncompactified is we also consider anti-periodic boundary condition?
• Perhaps I'm misunderstanding the question, but we do take into account the twisted sector while compactifying on an orbifold, whereupon the "real" Hilbert space only consists of $\mathbb Z_2$-invariant states (you can do this with a projection operator in the trace) Jan 18 at 16:07
• @NiharKarve You are right, when $\mathbb{Z}_2$ orbifolding, people do consider the twisted sector. But what about before orbifolding and compactification, just the vanilla free boson theory? Jan 18 at 17:10
• @NiharKarve Ok I see. In the textbooks at hand, people always impose the twisted condition together with the compactification $X \sim X + 2\pi R$. I guess what I want to ask is if the compactification is optional, just do orbifold, and if so, what is the Hilbert space. I'm guessing one simply removes the $\mathbb{Z}_2$ non-invariant states from the original boson theory? Jan 19 at 1:22
• @NiharKarve Asked differently, what is the $T^2$ partition function if we also consider the anti-periodic boundary condition? For periodic condition, the partition function is as written in the post. Jan 19 at 1:50
• 1) Only the condition $X\sim X + 2\pi R$ is compactification on the circle, orbifold compactification needs the additional $\mathbb Z_2$ symmetry. 2. Naïvely all you have to do is start with the circle-compactified theory and then remove the non-$\mathbb Z_2$-invariant states, but modular invariance of the partition function forces inclusion of states generated by half-integer moded oscillators: that's the twisted sector. So the actual partition function is $\frac12 Z_\text{circ}+\left|\frac\eta{\vartheta_2}\right|+\left|\frac\eta{\vartheta_3}\right|+\left|\frac\eta{\vartheta_4}\right|$. Jan 19 at 9:31
$$\renewcommand{\Im}{\operatorname{Im}}$$Let's dissect the question a bit. First, the periodic BCs on the free boson on a torus correspond to (R,R). But of course, no one stops you from imposing (R,NS), (NS,R) and (NS,NS) BCs and compute the partition function. The physical meaning of this is having turned on a background $$\mathbb{Z}_2$$ gauge field along either or along both cycles of the torus. So to rephrase your question, you essentially ask the following:
We know that $$Z_{\text{(R,R)}}[\tau] = \frac{1}{\sqrt{\Im(\tau)}}\frac{1}{\left|\eta(\tau)\right|^2}.\tag{0.1}\label{1}$$ What is $$Z_{(\bullet,\circ)}[\tau]$$ with $$(\bullet,\circ)\in\{\text{(R,NS), (NS,R), (NS,NS)}\}?$$
There are two ways to answer, and both of them use the following fact of life:
The partition function of the non-compact scalar on a torus $$\mathbb{T}^2_\tau:=\mathbb{C}/\left(\mathbb{Z}\oplus\tau\mathbb{Z}\right)$$, with $$\tau\in\mathbb{H}$$ (the upper half plane), can be read from the partition function of the compact scalar by sending the compactification radius to infinity, $$R\to\infty$$.
Way 1
The simplest way to obtain the answer is as follows.
Well, it suffices to go look at the compactified case and stare at the orbifold computation. For example stare at equation (8.24) in Ginsparg's notes. You will see that it is only the (R,R) sector that contributes to the $$R$$ dependence. Therefore, the (R,NS), (NS,R) and (NS,NS) are identical in the uncompactified case (when you sent $$R\to\infty$$). So we have \begin{align} Z_\text{(R,NS)}[\tau] &= \left|\frac{2\eta(\tau)}{\vartheta_2(\tau)}\right| \tag{1.1}\label{1.1} \\ Z_\text{(NS,R)}[\tau] &= \left|\frac{\eta(\tau)}{\vartheta_4(\tau)}\right| \tag{1.2}\label{1.2}\\ Z_\text{(NS,NS)}[\tau] &= \left|\frac{\eta(\tau)}{\vartheta_3(\tau)}\right|. \tag{1.3}\label{1.3} \end{align}
Way 2
Another way, is to do the computation from scratch. Namely, go back to the path integral and compute, say in the (R,R) case $$Z_\text{(R,R)}[\tau] = \frac{\operatorname{vol}(\text{zero-modes})}{\sqrt{\operatorname{det}_\text{(R,R)}'\!\big(\partial\bar\partial\big)}}.\tag{2.1}\label{2.1}$$ Up to factors of $$2$$ and $$\pi$$, you can then see the following: $$\operatorname{vol}(\text{zero-modes}) = \sqrt{\Im(\tau)}\tag{2.2}\label{2.2}$$ and the non-zero eigenvalues of $$\partial\bar\partial$$ on a torus with (R,R) BCs are simply $$\lambda^\text{(R,R)}_{n,m} = \frac{1}{\Im(\tau)^2}\left|n+\tau m\right|^2, \qquad (n,m)\in\mathbb{Z}^2\setminus(0,0),$$ giving $$\operatorname{det}'_\text{(R,R)}(\partial\bar\partial) = \Im(\tau)^2\left|\eta(\tau)\right|^4.\tag{2.3}\label{2.3}$$ Altogether plugging \eqref{2.2} and \eqref{2.3} in \eqref{2.1}, gives \eqref{1}.
Now, for the other boundary conditions, all you have to do is observe that an NS BC along either cycle shifts either (or both) $$n$$ or $$m$$ by $$\frac{1}{2}$$, therefore, e.g. for (R,NS) BCs you have $$\lambda^\text{(R,NS)}_{n,m} = \frac{1}{\Im(\tau)}\left| n+\tau\left(m+\frac{1}{2}\right) \right|^2, \qquad n,m\in\mathbb{Z}^2.$$ Note that now you don't have a zero-mode anymore. So all you have to do now is compute the determinant of $$\partial\bar\partial$$ with these boundary conditions and find $$Z_{(\bullet,\circ)}[\tau] = \frac{1}{\sqrt{\det_{(\bullet,\circ)}(\partial\bar\partial)}}.$$ Doing this should land you on \eqref{1.1}-\eqref{1.3}.
• Thanks for your answer! I didn't realize the $R$ independence of $R$ in the other sectors previously. Nice observation. So from your answer, somehow the Hilbert space of the "twisted sector" in the $R \to +\infty$ limit is still like some direct sum of $|m, n\rangle$ subsectors? How to understand this at the level of allowed primaries? Jan 19 at 13:44
• As a comparison, for the uncompactified boson theory, the untwisted sectors would consider $|\alpha\rangle$ sectors with continuous $\alpha$. Jan 19 at 13:46
• Actually, the RR sector partition function of the $\mathbb{Z}_2$ orbifold theory is computed in Ginsparg, eq (8.6). It's not obvious to me that the $R \to +\infty$ limit of the double infinite sum reproduces the expected factor $1/\sqrt{\operatorname{Im\tau}}$. Could you clarify a bit? Jan 20 at 5:58
• In (8.6) you can't directly take a limit $R\to\infty$ because the exponent contains both a $\propto R$ piece and a $\propto 1/R$ piece. To take the limit you must first Poisson resum one of the two sums so that you end up with something only $\propto R$ and then only the $(0,0)$ term contributes. The Poisson resummation also spits out a factor of $1/\sqrt{\mathrm{Im}(\tau)}$ from the Fourier transform of the exponential. Jan 20 at 12:45
• I see, I guess you are referring to Ginsparg's eq (8.3), where the sum is the Poisson resummation of eq (8.6). Indeed, each term (except when $n, n' = 0$) goes to zero individually when $r \to \infty$. Thanks! Jan 21 at 3:23 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 34, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9687023162841797, "perplexity": 555.0924783341181}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296943562.70/warc/CC-MAIN-20230320211022-20230321001022-00067.warc.gz"} |
https://chemistry.stackexchange.com/questions/18967/validity-of-troutons-rule?noredirect=1 | Validity of Trouton's Rule
I would like to know under what circumstances is Trouton's rule obeyed by liquids and why certain systems i.e. substances may deviate from Trouton's rule.
By the way I know what Trouton's rule is and I have a vague idea that sometimes Trouton's rule is not obeyed due to the presence of hydrogen bonding in liquids such as water but I would like to know more about other reasons why certain situations the rule is invalid.
The rule says that for many liquids the entropy of vaporization is almost the same, around ${\rm 85\,J K^{-1}}$.
It's true that the rule states that entropy of vapourization is $85 – 88~\mathrm{J/K}$.However, the rule hardly works for high ordered substances exhibiting hydrogen bonding. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8006572127342224, "perplexity": 752.81580274957}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-45/segments/1603107882103.34/warc/CC-MAIN-20201024080855-20201024110855-00495.warc.gz"} |
https://socratic.org/questions/how-do-you-solve-6-x-4-8-x-2#257878 | Algebra
Topics
# How do you solve 6(x+4)=8(x-2)?
Apr 23, 2016
$6 \left(x + 4\right) = 8 \left(x - 2\right)$
$6 x + 24 = 8 x - 16$
$40 = 2 x$
$20 = x$
#### Explanation:
To solve this equation, we first must expand both sets of brackets. To do this, multiply the number or letter on the outside of the brackets by the numbers or letter on the inside.
$6 \left(x + 4\right)$
$6 x + 24$
$6$ multiplied by $x$ gives $6 x$
$6$ multiplied by $4$ gives $24$
So the sum currently looks like this:
$6 x + 24 = 8 \left(x + 2\right)$
Now we must expand the brackets on the right side of the equals sign.
$8 \left(x - 2\right)$
$8 x - 16$
$8$ multiplied by $x$ gives $8 x$
$8$ multiplied by $- 2$ gives $- 16$
So now the sum looks a little more approachable, like this:
$6 x + 24 = 8 x - 16$
To finish the sum, we have to have the like terms on the same side of the equal sign. In other words, we have to get the $x$'s on one side of the equals sign and the numbers on the other.
First, we have to cancel out the $- 16$ by adding $16$ to both sides of the equals sign.
This gives $6 x + 40 = 8 x$
Then, cancel out the $6 x$ by taking $6 x$ off both sides of the equals sign.
This gives $40 = 2 x$
Finally, to fully simplify the answer, divide both sides of the equals sign by two to get a single $x$
Giving: $20 = x$
Hope this helped!
##### Impact of this question
746 views around the world | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 31, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9303158521652222, "perplexity": 227.15524812970577}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320304572.73/warc/CC-MAIN-20220124155118-20220124185118-00675.warc.gz"} |
https://tivadardanka.com/blog/how-the-dot-product-measures-similarity/ | # How the dot product measures similarity
In machine learning, we use the dot product every day.
Its definition is far from revealing. For instance, what does the sum of coordinate products have to do with similarity?
There is a beautiful geometric explanation behind.
The dot product is one of the most fundamental concepts in machine learning, making appearances almost everywhere. By definition, the dot product (or inner product) is defined between two vectors as the sum of coordinate products.
## The fundamental properties of the dot product
To peek behind the curtain, there are three key properties that we have to understand.
First, the dot product is linear in both variables. This property is called bilinearity.
Second, the dot product is zero if the vectors are orthogonal. (In fact, the dot product generalizes the concept of orthogonality beyond Euclidean spaces. But that's for another day :) )
Third, the dot product of a vector with itself equals the square of its magnitude.
## The geometric interpretation of the dot product
Now comes the interesting part. Given a vector $y$, we can decompose $x$ into the two components $x_o$ and $x_p$. One is parallel to $y$, while the other is orthogonal to it.
In physics, we apply the same decomposition to various forces all the time.
The vectors $x_o$ and $x_p$ are characterized by two properties:
1. $x_p$ is a scalar multiple of $y$,
2. and $x_0$ is orthogonal to $x_p$ (and thus to $y$).
We are going to use these properties to find an explicit formula for $x_p$. Spoiler alert: it is related to the dot product.
Due to $x_0$ being orthogonal to $y$, we can use the bilinearity of the dot product to express the $c$ in $x_p = c y$.
By solving for $c$, we get that it is the ratio of the dot product and the magnitude of $y$.
If both $x$ and $y$ are unit vectors, the dot product simply expresses the magnitude of the orthogonal projection!
## Dot product as similarity
Do you recall how the famous trigonometric functions sine and cosine are defined?
Let's say that the hypotenuse of our right triangle is a unit vector and one of the legs is on the $x$-axis. Then the trigonometric functions equal the magnitudes of the projections to the axes.
Using trigonometric functions, we see that the dot product of two unit vectors is the cosine of their enclosed angle $\alpha$! This is how the dot product relates to cosine.
If $x$ and $y$ are not unit vectors, we can scale them and use our previous discovery to get the cosine of $\alpha$.
The closer to its value to $1$, the more similar $x$ and $y$ are. (In a sense.) In machine learning, we call this quantity the cosine similarity.
Now you understand why. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 29, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.995685875415802, "perplexity": 156.58549568040013}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446710764.12/warc/CC-MAIN-20221130124353-20221130154353-00594.warc.gz"} |
http://math.stackexchange.com/questions/62754/converting-from-one-representation-of-a-field-element-to-another | # Converting From One Representation of a Field Element to Another
Let $A \subseteq B \subseteq C$ be fields and let $\alpha$, $\beta$, $\gamma$ be such that $A(\alpha) = B$, $B(\beta) = C$, $A(\gamma) = C$. Assume $B$ and $C$ have finite degree over $A$.
Let $m(\alpha,A)$ be the minimal polynomial of $\alpha$ over $A$, let $m(\beta,B)$ be the minimal polynomial of $\beta$ over $B$, let $m(\beta,A)$ be the minimal polynomial of $\beta$ over $A$, and let $m(\gamma,A)$ be the minimal polynomial of $\gamma$ over $A$.
Let $c \in C$. We have, on the one hand, $$c=\sum_{i = 1}^{[C:B]} b_i \beta^{i-1}, \quad b_i \in B$$ $$b_i=\sum_{j = 1}^{[B:A]} a_{ij} \alpha^{j-1}, \quad a_{ij} \in A$$ and on the other hand $$c=\sum_{k = 1}^{[C:A]} c_k \gamma^{k-1}, \quad c_k \in A$$ We also have $$\gamma = \sum_{i = 1}^{[C:B]} d_i \beta^{i-1}, \quad d_i \in B$$ $$d_i=\sum_{j = 1}^{[B:A]} e_{ij} \alpha^{j-1}, \quad e_{ij} \in A$$ If the polynomials $m(\alpha,A)$, $m(\beta,B)$, $m(\beta,A)$, $m(\gamma,A)$ and the numbers $a_{ij}$, $e_{ij}$ are known explicitly, how can I calculate the $c_i$?
ADDED: Now that Joriki and Jyriki have helped me to formulate the problem correctly, I see the solution is not so hard. Since $$\gamma = \sum_{i = 1}^{[C:B]} \sum_{j = 1}^{[B:A]} e_{ij} \alpha^{j-1}\beta^{i-1}$$ we can find numbers $f_{ijk}$ such that $$\gamma^{k-1} = \sum_{i = 1}^{[C:B]} \sum_{j = 1}^{[B:A]} f_{ijk} \alpha^{j-1}\beta^{i-1}$$ by using the polynomials $m(\alpha,A)$, $m(\beta,B)$ to express large powers of $\alpha$ and $\beta$ in terms of smaller powers. Then $$c = \sum_{i = 1}^{[C:B]} \sum_{j = 1}^{[B:A]} a_{ij} \alpha^{j-1}\beta^{i-1}$$ and also $$c = \sum_{k=1}^{[C:A]} c_k \left( \sum_{i = 1}^{[C:B]} \sum_{j = 1}^{[B:A]} f_{ijk} \alpha^{j-1}\beta^{i-1} \right) = \sum_{i = 1}^{[C:B]} \sum_{j = 1}^{[B:A]} \left( \sum_{k=1}^{[C:A]} c_k f_{ijk} \right) \alpha^{j-1}\beta^{i-1}$$ Therefore $$a_{ij} = \sum_{k=1}^{[C:A]} c_k f_{ijk}$$ So we are reduced to solving this system of system $[C:A] = [C:B] \cdot [B:A]$ linear equations for the $c_k$.
-
With some difficulty, I think. Take a fairly simple example. Let $A$ be the rationals, $\alpha=\sqrt2$, $\beta=\sqrt3$, $\gamma=\sqrt2+\sqrt3$. You have $c=a_{11}+a_{12}\sqrt2+a_{21}\sqrt3+a_{22}\sqrt6$, and you want $c=c_1+c_2(\sqrt2+\sqrt3)+c_3(\sqrt2+\sqrt3)^2+c_4(\sqrt2+\sqrt3)^3$. Expressing the $c_i$ in terms of the $a_{ij}$ (and the minimal polynomials) looks mildly unpleasant. – Gerry Myerson Sep 8 '11 at 6:30
I would have thought that you can't, since there may be several values of $\gamma$ with the same $A(\gamma)$ and $m(\gamma,A)$ that will generally have different $c_i$? For example, consider $A=\mathbb Q$, $\alpha=1$, $\beta=\sqrt2$ and $\gamma=\pm\sqrt2$. Then $A(\gamma)=\mathbb Q(\sqrt2)$ and $m(\gamma,A)=x^2-2$ for both signs, but the sign of $c_2$ is flipped. – joriki Sep 8 '11 at 6:46
@Gerry Unpleasant or not, I need to find an algorithm do it. – maxpower Sep 8 '11 at 9:20
@maxpower: I don't understand. How do the $a_{ij}$ help? They're the same for both signs in my example. – joriki Sep 8 '11 at 9:29
@maxpower: I agree with joriki. The problem specification must include a way of identifying the element $\gamma$ in terms of $\alpha$ and $\beta$. Another example would be $\alpha=\sqrt2$, $\beta=i$, $\gamma$ any primitive eighth root of unity. All four sign combination are possible in $\gamma=\pm\alpha(1\pm\beta)/2$ in that all those numbers share the same minimial polynomial $x^4+1$. Unless you know which combination is $\gamma$, the problem cannot be solved. – Jyrki Lahtonen Sep 8 '11 at 9:44
The $c_i$ are not determined by the given data. Different values of $\gamma$ can lead to identical $A(\gamma)$ and $m(\gamma,A)$ but different values of $c_i$, as in the examples given in comments by Jyrki and me. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9802478551864624, "perplexity": 102.74673284490034}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-35/segments/1440645151768.51/warc/CC-MAIN-20150827031231-00048-ip-10-171-96-226.ec2.internal.warc.gz"} |
https://orbital-mechanics.space/orbital-maneuvers/nonimpulsive-orbital-maneuvers.html | # Nonimpulsive Orbital Maneuvers#
Up to now, all the maneuvers we have considered have been impulsive. This means that they happen extremely quickly relative to the time scale of the overall maneuver or trajectory. Practically, this means that the position vector is constant during the impulse.
However, there are very useful maneuvers that can be performed by providing nonimpulsive thrust. For instance, providing a very low thrust over a sufficiently long time period may be more efficient, depending on the propellant source, than impulsive maneuvers. Nonimpulsive maneuvers typically include propulsion devices such as solar sails and ion engines.
Since the position is changing during the impulse, we must return to the equation of motion and add an additional force term to the right hand side:
$\ddot{\vector{r}} = -\mu\frac{\vector{r}}{r^3} + \frac{\vector{F}}{m}$
Assuming that force is a thrust is provided in the same direction as the velocity, then:
$\vector{F} = T\frac{\vector{v}}{v}$
where $$T$$ is the magnitude of the thrust force and $$\vector{v} = \dot{\vector{r}}$$. If the thrust is provided in the opposite direction of the velocity, then a negative sign should be added to the previous equation. In any case, this results in three scalar equations of motion:
\begin{aligned}\ddot{x} &= -\mu\frac{x}{r^3} + \frac{T}{m}\frac{\dot{x}}{v} & \ddot{y} &= -\mu\frac{y}{r^3} + \frac{T}{m}\frac{\dot{y}}{v} & \ddot{z} &= -\mu\frac{z}{r^3} + \frac{T}{m}\frac{\dot{z}}{v}\end{aligned}
In addition, to provide the thrust $$T$$, the rocket motors must eject propellant overboard. This causes the mass of the spacecraft to decrease according to:
$\frac{dm}{dt} = -\frac{T}{I_{sp}g_0}$
where $$m$$ is the instantaneous mass of the spacecraft, $$I_{sp}$$ is the specific impulse of the engine/propellant combination, and $$g_0$$ is the sea-level acceleration of gravity.
This set of differential equations does not have an analytical solution in general. However, we can construct a numerical solution by writing the system of ODEs as the six components of the state vector (3 position and 3 velocity) plus the equation for the mass. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9616224765777588, "perplexity": 498.1666457456828}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296949642.35/warc/CC-MAIN-20230331113819-20230331143819-00524.warc.gz"} |
https://tex.stackexchange.com/questions/297334/lyx-document-compiled-in-osx | # LyX document compiled in OSX?
LyX is a wysiwyg LaTeX editor that I find useful to write exercises fast instead of TextMate that I use for long-term projects. So the below shows an exercise that I would like to print as PDF: however I am getting the following errors Undefined control sequence that are very vague. I cannot see any \begin{document} or other LaTeX commands in LyX in the view which makes debugging harder than in text-based editors. So
How can I compile a document in LyX?
• the fact that \mathbb is what's unidentified means that amsfonts (or if you need more symbols, amssymb, which loads amsfonts) isn't loaded. so try adding that to your job. (but i don't know how, not being a lyx user.) – barbara beeton Mar 4 '16 at 17:51
• @barbarabeeton thank you for the observation: I realized from that to search for packages and I found a solution, provided below for new LyX users, nice software :) – hhh Mar 4 '16 at 18:06
• What options do you have set in Document > Settings > Math Options? – scottkosty Mar 4 '16 at 18:35
• @scottkosty thank you for the observation, added the alternative method here. – hhh Mar 5 '16 at 16:20
• @hhh looks good. I would add that there is an advantage of doing it the LyX way because that way LyX knows about the packages that are being used. This way LyX can load them in the best order (sometimes this makes a difference in LaTeX, although in this case I doubt it). – scottkosty Mar 5 '16 at 16:57
and add your preamble such as \usepackage{amsmath,amsfonts,amssymb,amsthm} so
and CMD+R to get the document compiled like in TextMate -- it works without writing any \begin{document}.
Document > Settings > Math Options
More graphical method as pointed out by the comment is to tongle the settings in Math Options like the below and in this case you don't need to add the mathematical packages to LaTeX Preamble. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9380436539649963, "perplexity": 1781.0977786582907}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570986657949.34/warc/CC-MAIN-20191015082202-20191015105702-00201.warc.gz"} |
https://math.stackexchange.com/questions/386766/power-series-and-integral-test | # Power Series and Integral test
so I'm studying for my exams and there are a few questions that I don't completely understand. I need help with questions (b) and (d). For (b), I ended up getting diverging, because the limit is infinity?. Also I have no clue how to do (d)
So I also spent all night trying to learn power series and I need help with (d), (e) and (f). I ended up obtaining the right answer for (d), but I'm not sure if my method of working it out is correct. So basically what I did first was do the ratio test and I ended up getting |x^2|limit k->infinity (log(1+k))/log(2+k). Am I suppose to use L'hopital's rule to find the limit? Because I used it and the interval ended up being from -1infinity (k^k)/(l+1)^k ,what do I do from there? ,
Btw the answers to (b) and (d) for the integral test is converging for both of them. And the answers to (d), (e) and (f) for the power series is 1, 2e and 0, respectively.
• Help is very much appreciated :) – George Randall May 9 '13 at 16:46
Under "Integral Test":
A useful result all students should work out at least once in their "calculus lives" is $\int_1^{\infty} \frac{1}{x^p} \ dx \ ,$ to see for what values of $p$ this integral converges or diverges (you may have already seen this when you covered Type I improper integrals). It will help in spotting which "p-series" $\Sigma_{n=1}^{\infty} \frac{1}{n^p}$ converge.
In your set, you'll need to use u-substitution in examining $\int_0^{\infty} \frac{1}{(n+1)^{\gamma}} \ dx \ ,$ and partial fraction decomposition for $\int_0^{\infty} \frac{1}{(x+1) \ \cdot \ (x+2)} \ dx \$ (or use the hint). (And having re-read your last sentences, yes, (b) and (d) converge.) $$\\$$
Under "Power Series" :
I believe you are correct for (d), (e) , and (f). The Ratio Test for (d) produces
$$\lim_{k \rightarrow \infty} | \ \frac{x^{2k+2}}{x^{2k}} \ \cdot \ \frac{\log (k+1)}{\log (k+2)} \ | \ = \ \lim_{k \rightarrow \infty} | \ x^2 \ \cdot \ \frac{\log (k+1)}{\log (k+2)} \ | ,$$
and I don't think you need a lot of justification to declare that the limit for the ratio of logarithms is $1$ . (The ratio is equivalent to $\log_{k + 2} (k+1)$ ).
As for (e) and (f), these both hinge on dealing with $k^k$ in some manner. For (e), the ratio includes the factors
$$\frac{(k+1)! \cdot k^k}{k! \cdot (k+1)^{(k+1)}} \ ,$$
which reduce to
$$(k+1) \ \cdot \ \frac{ k^k}{ (k+1)^{(k+1)}} \ = \ ( \frac{k}{ k+1})^k \ ,$$
for which the limit at infinity can be found by appropriate use of l'Hopital's Rule on "indeterminate powers", or familiarity with the behavior of this function (as many members of this forum handle it).
Here is one example, your first one. When you integrate that square root term, you get 2*SQRT[x+1] and so this integral is divergent (WHY?) Therefore the Series is divergent. Third example, the anti derivative is an arctan, so can you show that this one will be convergent?
First example of Radius of Convergence, with ratio test (check your book for definition!) you can find that interval of convergence is between -3 and 3. So what's the Radius? Now you can try some more... | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9436496496200562, "perplexity": 177.40614224999294}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-22/segments/1558232256546.11/warc/CC-MAIN-20190521182616-20190521204616-00463.warc.gz"} |
https://planetmath.org/MatrixPnorm | # matrix p-norm
A class of matrix norms, denoted $\|\cdot\|_{p}$, is defined as
$\|\,A\,\|_{p}=\sup_{x\neq 0}\frac{\|\,Ax\,\|_{p}}{\|\,x\,\|_{p}}\qquad{}x\in% \mathbb{R}^{n},A\in\mathbb{R}^{m\times n}.$
The matrix $p$-norms are defined in terms of the vector $p$-norms (http://planetmath.org/VectorPNorm).
An alternate definition is
$\|\,A\,\|_{p}=\max_{\|\,x\,\|_{p}=1}\|\,Ax\,\|_{p}.$
As with vector $p$-norms, the most important are the 1, 2, and $\infty$ norms. The 1 and $\infty$ norms are very easy to calculate for an arbitrary matrix:
$\begin{array}[]{ll}\|\,A\,\|_{1}&=\displaystyle\max_{1\leq j\leq n}\sum_{i=1}^% {m}|a_{ij}|\\ \|\,A\,\|_{\infty}&=\displaystyle\max_{1\leq i\leq m}\sum_{j=1}^{n}|a_{ij}|.% \end{array}$
It directly follows from this that $\|\,A\,\|_{1}=\|\,A^{T}\,\|_{\infty}$.
The calculation of the $2$-norm is more complicated. However, it can be shown that the 2-norm of $A$ is the square root of the largest eigenvalue of $A^{T}A$. There are also various inequalities that allow one to make estimates on the value of $\|\,A\,\|_{2}$:
$\frac{1}{\sqrt{n}}\|\,A\,\|_{\infty}\leq\|\,A\,\|_{2}\leq\sqrt{m}\|\,A\,\|_{% \infty}.$
$\frac{1}{\sqrt{m}}\|\,A\,\|_{1}\leq\|\,A\,\|_{2}\leq\sqrt{n}\|\,A\,\|_{1}.$
$\|\,A\,\|_{2}^{2}\leq\|\,A\,\|_{\infty}\cdot\|\,A\,\|_{1}.$
$\|\,A\,\|_{2}\leq\|\,A\,\|_{F}\leq\sqrt{n}\|\,A\,\|_{2}.$
($\|\,A\,\|_{F}$ is the Frobenius matrix norm)
Title matrix p-norm MatrixPnorm 2013-03-22 11:43:22 2013-03-22 11:43:22 mathcam (2727) mathcam (2727) 20 mathcam (2727) Definition msc 15A60 msc 00A69 MatrixNorm VectorNorm FrobeniusMatrixNorm | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 19, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9756673574447632, "perplexity": 682.419563762664}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-13/segments/1552912202474.26/warc/CC-MAIN-20190320230554-20190321012554-00258.warc.gz"} |
http://scitation.aip.org/content/aip/journal/pof2/20/7/10.1063/1.2958319 | • journal/journal.article
• aip/pof2
• /content/aip/journal/pof2/20/7/10.1063/1.2958319
• pof.aip.org
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Model of a truncated fast rotating flow at infinite Reynolds number
USD
10.1063/1.2958319
By L. Bourouiba1,a)
View Affiliations Hide Affiliations
Affiliations:
1 McGill University, 805 Sherbrooke W., Room 945, Montréal, Québec H3A 2K6, Canada
a) Electronic mail: [email protected].
Phys. Fluids 20, 075112 (2008)
/content/aip/journal/pof2/20/7/10.1063/1.2958319
http://aip.metastore.ingenta.com/content/aip/journal/pof2/20/7/10.1063/1.2958319
## Figures
FIG. 1.
Initial horizontal spectra of ICs, IC: I and IC: II.
FIG. 2.
Time series of the energy contributions and for , 0.2, and 0.01, initiated with IC: I (a) and IC: II (b). The time axis is the dimensional time.
FIG. 3.
Time series of the 2D enstrophy (a) and the energy of the vertical component of the 2D field (b) for , 0.2, and 0.01 for both IC: I and IC: II. The time axis is the dimensional time.
FIG. 4.
Time series of the and for the simulation initialized with IC: I (a) and IC: II (b). The time axis is the nondimensional time .
FIG. 5.
Horizontal wavenumber spectra of (upper panel) and (lower panel) for and for IC: I (left column) and IC: II (right column). The theoretical spectra have been offset for clarity. The initial numerical spectra are denoted and multiple lines are for different times.
FIG. 6.
Time series of [centroid of 2D energy spectra defined by Eq. (39)], with nondimensional time and for . Both simulations started with IC: I and IC: II are shown.
FIG. 7.
Horizontal wavenumber spectra for and for IC: I (a) and IC: II (b). The theoretical spectra have been offset for clarity. The initial numerical spectra are denoted and the multiple lines are for different times.
FIG. 8.
Vertical spectra of 3D energy, , for simulations initiated with IC: I (a) and IC: II (b). The initial numerical spectra are denoted and the multiple lines are for different times. The theoretical spectra have been offset for clarity only for the simulation IC: II (a).
FIG. 9.
Time averaged 3D energy spectrum in log-log scale at an initial time [(a) and (b)], an intermediate time range below such that [(c) and (d)], an intermediate time [(e) and (f)], and the end of the simulations with [(g) and (h)]. Both the and time ranges correspond to times larger than , i.e., beyond the decoupled phase. and the ICs are IC: I (left column) and IC: II (right column). The colors are normalized for each graph such that the maximum (minimum) value of the modal spectrum is represented by the brightest (darkest) color.
## Tables
Table I.
The timestep , the rotation rate , the final output time , the 2D Rossby number , and the Robert filter parameter for each of the selected simulations.
/content/aip/journal/pof2/20/7/10.1063/1.2958319
2008-07-30
2014-04-24
Article
content/aip/journal/pof2
Journal
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This is a required field | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8208197951316833, "perplexity": 4408.467698320504}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-15/segments/1398223206147.1/warc/CC-MAIN-20140423032006-00048-ip-10-147-4-33.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/boundary-value-problem.152879/ | # Boundary value problem
1. Jan 24, 2007
### John O' Meara
Schodinger's equation for one-dimensional motion of a particle whose potential energy is zero is
$$\frac{d^2}{dx^2}\psi +(2mE/h^2)^\frac{1}{2}\psi = 0$$
where $$\psi$$ is the wave function, m the mass of the particle, E its kinetic energy and h is Planck's constant. Show that
$$\psi = Asin(kx) + Bcos(kx)$$ ( where A and B are constants) and $$k =(2mE/h^2)^\frac{1}{2}$$ is a solution of the equation.
Using the boundary conditions $$\psi=0$$ when x=0 and when x=a, show that
(i) the kinetic energy $$E=h^2n^2/8ma^2$$
(ii) the wave function $$\psi = A sin(n\pi\times x/a)$$ where n is any integer. (Note if $$sin(\theta) = 0 then \theta=n\pi$$)
My attempt:
A*sin(0) + B*cos(0) = 0, => 0 +B =0 => B = 0.
Therefore
A*sin(k*a)=0, Therefore $$(2mE/h^2)^\frac{1}{2}a = n\pi => E=n^2\pi^2h^2/2ma^2$$
1. The problem statement, all variables and given/known data
2. Relevant equations
3. The attempt at a solution
Last edited: Jan 24, 2007
2. Jan 24, 2007
### John O' Meara
(ii) should read A*sin(n*pi*x/a), I'm just notgood at boundary value problems.
3. Jan 24, 2007
### HallsofIvy
Well, first, you haven't shown that that $\psi$ does, in fact, satisfy the differential equation. After that, yes, B= 0. Now, assuming A is not 0, that is that $\psi$ is not itself identically 0, then yes, we must have sin(ka)= 0 so that $ka= (2mE/h^2)^{\frac{1}{2}}= n\pi$. E follows eactly as you say.
1. The problem statement, all variables and given/known data
2. Relevant equations
3. The attempt at a solution[/QUOTE]
4. Jan 25, 2007
### John O' Meara
I was able to show that $$\psi$$ is a solution of the equation, it is (i) and (ii) that I had trouble doing espically (ii) i.e., $$\psi = A sin(n\pi\times x/a)$$. Where did he get the argument "$$n\pi\times x/a$$", or more importantly how does he expect me to get that argument of the sine.
Also remember in (i) the answer he has for E $$=h^2n^2/8ma^2$$, not what I got for E. Thanks for the help.
5. Jan 26, 2007
### John O' Meara
I hope someone can tell me how $$\psi$$ can go from $$Asin((2mE/h^2)^\frac{1}{2}x) to Asin(n\pi\times x/a)$$. Thanks for the help.
6. Jan 26, 2007
### HallsofIvy
Well, when x= a, the argument is $n\pi$ what is $sin(n\pi)[/tex]? Remember that you were told that [itex]\psi(0)= 0$ and $\psi(a)= 0$. Knowing that cos(0)= 1 tells us that the second constant, B, must be 0. That leaves Asin(kx). We must have Asin(ka)= 0 and we don't want A= 0 (that would mean our function is always 0) so we must have sin(ka)= 0. For what x is sin(x)= 0? Multiples of $\pi$ of course:
$ka= n\pi$. For that to be true, k must be equal to $n\pi/a$
You were also told that $k= \sqrt{2ME/h^2}$ and you now know $k= n\pi/a[\itex] so [itex]n\pi/a= \sqrt{2ME/h^2}$. Solve that for E. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9840323328971863, "perplexity": 853.8932731634618}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-13/segments/1521257648205.76/warc/CC-MAIN-20180323083246-20180323103246-00389.warc.gz"} |
https://en.m.wikipedia.org/wiki/Vector_(mathematics) | # Vector (mathematics and physics)
(Redirected from Vector (mathematics))
In mathematics and physics, a vector is an element of a vector space.
For many specific vector spaces, the vectors have received specific names, which are listed below.
Historically, vectors were introduced in geometry and physics (typically in mechanics) before the formalization of the concept of vector space. Therefore, one often talks about vectors without specifying the vector space to which they belong. Specifically, in a Euclidean space, one considers spatial vectors, also called Euclidean vectors which are used to represent quantities that have both magnitude and direction, and may be added, subtracted and scaled (i.e. multiplied by a real number) for forming a vector space.[1]
## Vectors in Euclidean geometry
In classical Euclidean geometry (i.e., synthetic geometry), vectors were introduced (during the 19th century) as equivalence classes under equipollence, of ordered pairs of points; two pairs (A, B) and (C, D) being equipollent if the points A, B, D, C, in this order, form a parallelogram. Such an equivalence class is called a vector, more precisely, a Euclidean vector.[2] The equivalence class of (A, B) is often denoted ${\displaystyle {\overrightarrow {AB}}.}$
A Euclidean vector is thus an equivalence class of directed segments with the same magnitude (e.g., the length of the line segment (A, B)) and same direction (e.g., the direction from A to B).[3] In physics, Euclidean vectors are used to represent physical quantities that have both magnitude and direction, but are not located at a specific place, in contrast to scalars, which have no direction.[4] For example, velocity, forces and acceleration are represented by vectors.
In modern geometry, Euclidean spaces are often defined from linear algebra. More precisely, a Euclidean space E is defined as a set to which is associated an inner product space of finite dimension over the reals ${\displaystyle {\overrightarrow {E}},}$ and a group action of the additive group of ${\displaystyle {\overrightarrow {E}},}$ which is free and transitive (See Affine space for details of this construction). The elements of ${\displaystyle {\overrightarrow {E}}}$ are called translations.
It has been proven that the two definitions of Euclidean spaces are equivalent, and that the equivalence classes under equipollence may be identified with translations.
Sometimes, Euclidean vectors are considered without reference to a Euclidean space. In this case, a Euclidean vector is an element of a normed vector space of finite dimension over the reals, or, typically, an element of ${\displaystyle \mathbb {R} ^{n}}$ equipped with the dot product. This makes sense, as the addition in such a vector space acts freely and transitively on the vector space itself. That is, ${\displaystyle \mathbb {R} ^{n}}$ is a Euclidean space, with itself as an associated vector space, and the dot product as an inner product.
The Euclidean space ${\displaystyle \mathbb {R} ^{n}}$ is often presented as the Euclidean space of dimension n. This is motivated by the fact that every Euclidean space of dimension n is isomorphic to the Euclidean space ${\displaystyle \mathbb {R} ^{n}.}$ More precisely, given such a Euclidean space, one may choose any point O as an origin. By Gram–Schmidt process, one may also find an orthonormal basis of the associated vector space (a basis such that the inner product of two basis vectors is 0 if they are different and 1 if they are equal). This defines Cartesian coordinates of any point P of the space, as the coordinates on this basis of the vector ${\displaystyle {\overrightarrow {OP}}.}$ These choices define an isomorphism of the given Euclidean space onto ${\displaystyle \mathbb {R} ^{n},}$ by mapping any point to the n-tuple of its Cartesian coordinates, and every vector to its coordinate vector.
## Vectors in specific vector spaces
• Column vector, a matrix with only one column. The column vectors with a fixed number of rows form a vector space.
• Row vector, a matrix with only one row. The row vectors with a fixed number of columns form a vector space.
• Coordinate vector, the n-tuple of the coordinates of a vector on a basis of n elements. For a vector space over a field F, these n-tuples form the vector space ${\displaystyle F^{n}}$ (where the operation are pointwise addition and scalar multiplication).
• Displacement vector, a vector that specifies the change in position of a point relative to a previous position. Displacement vectors belong to the vector space of translations.
• Position vector of a point, the displacement vector from a reference point (called the origin) to the point. A position vector represents the position of a point in a Euclidean space or an affine space.
• Velocity vector, the derivative, with respect to time, of the position vector. It does not depend of the choice of the origin, and, thus belongs to the vector space of translations.
• Pseudovector, also called axial vector, an element of the dual of a vector space. In an inner product space, the inner product defines an isomorphism between the space and its dual, which may make difficult to distinguish a pseudo vector from a vector. The distinction becomes apparent when one changes coordinates: the matrix used for a change of coordinates of pseudovectors is the transpose of that of vectors.
• Tangent vector, an element of the tangent space of a curve, a surface or, more generally, a differential manifold at a given point (these tangent spaces are naturally endowed with a structure of vector space)
• Normal vector or simply normal, in a Euclidean space or, more generally, in an inner product space, a vector that is perpendicular to a tangent space at a point. Normals are pseudovectors that belong to the dual of the tangent space.
• Gradient, the coordinates vector of the partial derivatives of a function of several real variables. In a Euclidean space the gradient gives the magnitude and direction of maximum increase of a scalar field. The gradient is a pseudo vector that is normal to a level curve.
• Four-vector, in the theory of relativity, a vector in a four-dimensional real vector space called Minkowski space
## Tuples that are not really vectors
The set ${\displaystyle \mathbb {R} ^{n}}$ of tuples of n real numbers has a natural structure of vector space defined by component-wise addition and scalar multiplication. When such tuples are used for representing some data, it is common to call them vectors, even if the vector addition does not mean anything for these data, which may make the terminology confusing. Similarly, some physical phenomena involve a direction and a magnitude. They are often represented by vectors, even if operations of vector spaces do not apply to them.
## Vectors in algebras
Every algebra over a field is a vector space, but elements of an algebra are generally not called vectors. However, in some cases, they are called vectors, mainly due to historical reasons.
### Vector fields
A vector field is a vector-valued function that, generally, has a domain of the same dimension (as a manifold) as its codomain,
### Miscellaneous
• Ricci calculus
• Vector Analysis, a textbook on vector calculus by Wilson, first published in 1901, which did much to standardize the notation and vocabulary of three-dimensional linear algebra and vector calculus
• Vector bundle, a topological construction that makes precise the idea of a family of vector spaces parameterized by another space
• Vector calculus, a branch of mathematics concerned with differentiation and integration of vector fields
• Vector differential, or del, a vector differential operator represented by the nabla symbol ${\displaystyle \nabla }$
• Vector Laplacian, the vector Laplace operator, denoted by ${\displaystyle \nabla ^{2}}$ , is a differential operator defined over a vector field
• Vector notation, common notation used when working with vectors
• Vector operator, a type of differential operator used in vector calculus
• Vector product, or cross product, an operation on two vectors in a three-dimensional Euclidean space, producing a third three-dimensional Euclidean vector
• Vector projection, also known as vector resolute or vector component, a linear mapping producing a vector parallel to a second vector
• Vector-valued function, a function that has a vector space as a codomain
• Vectorization (mathematics), a linear transformation that converts a matrix into a column vector
• Vector autoregression, an econometric model used to capture the evolution and the interdependencies between multiple time series
• Vector boson, a boson with the spin quantum number equal to 1
• Vector measure, a function defined on a family of sets and taking vector values satisfying certain properties
• Vector meson, a meson with total spin 1 and odd parity
• Vector quantization, a quantization technique used in signal processing
• Vector soliton, a solitary wave with multiple components coupled together that maintains its shape during propagation
• Vector synthesis, a type of audio synthesis
## Notes
1. ^ "vector | Definition & Facts". Encyclopedia Britannica. Retrieved 2020-08-19.
2. ^ In some old texts, the pair (A, B) is called a bound vector, and its equivalence class is called a free vector.
3. ^ "1.1: Vectors". Mathematics LibreTexts. 2013-11-07. Retrieved 2020-08-19.
4. ^ "Vectors". www.mathsisfun.com. Retrieved 2020-08-19.
5. ^ "Compendium of Mathematical Symbols". Math Vault. 2020-03-01. Retrieved 2020-08-19.
6. ^ a b Weisstein, Eric W. "Vector". mathworld.wolfram.com. Retrieved 2020-08-19. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 15, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9519175291061401, "perplexity": 357.23068706848636}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-04/segments/1610703524743.61/warc/CC-MAIN-20210121101406-20210121131406-00774.warc.gz"} |
https://fr.maplesoft.com/support/help/maple/view.aspx?path=Student%2FStatistics%2FChiSquareSuitableModelTest | ChiSquareSuitableModelTest - Maple Help
Student[Statistics]
ChiSquareSuitableModelTest
apply the chi-square suitable model test
Calling Sequence ChiSquareSuitableModelTest(X, F, options)
Parameters
X - observed data sample F - function, algebraic; probability distribution or random variable to match data against options - (optional) equation(s) of the form option=value where option is one of bins, level, output, or range; specify options for the ChiSquareSuitableModelTest function
Description
• The ChiSquareSuitableModelTest function performs the chi-square suitable model test upon an observed data sample against a known random variable or probability distribution. It works by determining bins for a histogram from the probability distribution, then classifying the entries of X into these bins, and finally testing whether the resulting histogram matches the histogram for the probability distribution.
• The first parameter X is a data sample of observed data to use in the analysis.
• The second parameter F is a random variable or probability distribution that is compared to the observed data sample.
• This test is only appropriate if there is prior knowledge of any parameters in the distribution. If any of the parameters in the distribution have been fitted to the data sample in question, then an adjustment of the degrees-of-freedom parameter is necessary. This adjustment is not available in the current implementation.
Options
The options argument can contain one or more of the options shown below.
• bins='deduce' or posint
This option indicates the number of bins to use when categorizing data from X and probabilities from F. If set to 'deduce' (default), the function attempts to determine a reasonable value for this option. This parameter is ignored if the distribution is discrete.
• range='deduce' or range
This option indicates the range to use when considering data values - data outside of the range is discarded during processing. If set to 'deduce' (default), the function attempts to determine a suitable range.
• level=float
This option is used to specify the level of the analysis (minimum criteria for the observed data to be considered well-fit to the expected data). By default, this value is 0.05.
• output=report or plot or both
If the option output is not included or is specified to be output=report, then the function will return a report. If output=plot is specified, then the function will return a plot of the sample test. If output=both is specified, then both the report and the plot will be returned.
Examples
> $\mathrm{with}\left(\mathrm{Student}\left[\mathrm{Statistics}\right]\right):$
Initialize an array of data
> $X≔\mathrm{Sample}\left(\mathrm{NormalRandomVariable}\left(0,1\right),100\right):$
Perform the suitable model test upon this sample.
> $\mathrm{ChiSquareSuitableModelTest}\left(X,\mathrm{UniformRandomVariable}\left(0,1\right),\mathrm{bins}=10\right)$
Chi-Square Test for Suitable Probability Model ---------------------------------------------- Null Hypothesis: Sample was drawn from specified probability distribution Alt. Hypothesis: Sample was not drawn from specified probability distribution Bins: 10 Degrees of Freedom: 9 Distribution: ChiSquare(9) Computed Statistic: 301.6000000 Computed p-value: 0. Critical Values: 16.9189774487099 Result: [Rejected] This statistical test provides evidence that the null hypothesis is false.
$\left[{\mathrm{hypothesis}}{=}{\mathrm{false}}{,}{\mathrm{criticalvalue}}{=}{16.9189774487099}{,}{\mathrm{distribution}}{=}{\mathrm{ChiSquare}}{}\left({9}\right){,}{\mathrm{pvalue}}{=}{0.}{,}{\mathrm{statistic}}{=}{301.6000000}\right]$ (1)
> $\mathrm{ChiSquareSuitableModelTest}\left(X,\mathrm{NormalRandomVariable}\left(0,1\right),\mathrm{bins}=10\right)$
Chi-Square Test for Suitable Probability Model ---------------------------------------------- Null Hypothesis: Sample was drawn from specified probability distribution Alt. Hypothesis: Sample was not drawn from specified probability distribution Bins: 10 Degrees of Freedom: 9 Distribution: ChiSquare(9) Computed Statistic: 14.80000000 Computed p-value: .0965781731648307 Critical Values: 16.9189774487099 Result: [Accepted] This statistical test does not provide enough evidence to conclude that the null hypothesis is false.
$\left[{\mathrm{hypothesis}}{=}{\mathrm{true}}{,}{\mathrm{criticalvalue}}{=}{16.9189774487099}{,}{\mathrm{distribution}}{=}{\mathrm{ChiSquare}}{}\left({9}\right){,}{\mathrm{pvalue}}{=}{0.0965781731648307}{,}{\mathrm{statistic}}{=}{14.80000000}\right]$ (2)
If the output=plot option is included, then a report will be returned.
> $\mathrm{ChiSquareSuitableModelTest}\left(X,\mathrm{NormalRandomVariable}\left(0,1\right),\mathrm{bins}=10,\mathrm{output}=\mathrm{plot}\right)$
If the output=both option is included, then both a report and a plot will be returned.
> $\mathrm{report},\mathrm{graph}≔\mathrm{ChiSquareSuitableModelTest}\left(X,\mathrm{NormalRandomVariable}\left(0,1\right),\mathrm{bins}=10,\mathrm{output}=\mathrm{both}\right):$
Chi-Square Test for Suitable Probability Model ---------------------------------------------- Null Hypothesis: Sample was drawn from specified probability distribution Alt. Hypothesis: Sample was not drawn from specified probability distribution Bins: 10 Degrees of Freedom: 9 Distribution: ChiSquare(9) Computed Statistic: 14.80000000 Computed p-value: .0965781731648307 Critical Values: 16.9189774487099 Result: [Accepted] This statistical test does not provide enough evidence to conclude that the null hypothesis is false. Histogram Type: default Data Range: -1.6348567543439 .. 2.21337958939036 Bin Width: .128274544791142 Number of Bins: 30 Frequency Scale: relative
> $\mathrm{report}$
$\left[{\mathrm{hypothesis}}{=}{\mathrm{true}}{,}{\mathrm{criticalvalue}}{=}{16.9189774487099}{,}{\mathrm{distribution}}{=}{\mathrm{ChiSquare}}{}\left({9}\right){,}{\mathrm{pvalue}}{=}{0.0965781731648307}{,}{\mathrm{statistic}}{=}{14.80000000}\right]$ (3)
> $\mathrm{graph}$
References
Kanju, Gopal K. 100 Statistical Tests. London: SAGE Publications Ltd., 1994.
Sheskin, David J. Handbook of Parametric and Nonparametric Statistical Procedures. London: CRC Press, 1997.
Compatibility
• The Student[Statistics][ChiSquareSuitableModelTest] command was introduced in Maple 18. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 11, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9674355983734131, "perplexity": 2969.5185974766255}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320304217.55/warc/CC-MAIN-20220123081226-20220123111226-00280.warc.gz"} |
https://hal-cea.archives-ouvertes.fr/cea-00828207 | # Convection and differential rotation properties of G and K stars computed with the ASH code
* Corresponding author
Abstract : The stellar luminosity and depth of the convective envelope vary rapidly with mass for G-and K-type main sequence stars. In order to understand how these properties influence the convective turbulence, differential rotation, and meridional circulation , we have carried out 3D dynamical simulations of the interiors of rotating main sequence stars, using the anelastic spherical harmonic (ASH) code. The stars in our simulations have masses of 0.5, 0.7, 0.9, and 1.1 M , corresponding to spectral types K7 through G0, and rotate at the same angular speed as the sun. We identify several trends of convection zone properties with stellar mass, exhibited by the simulations. The convective velocities, temperature contrast between up-and downflows, and meridional circulation velocities all increase with stellar luminosity. As a consequence of the trend in convective velocity, the Rossby number (at a fixed rotation rate) increases and the convective turnover timescales decrease significantly with increasing stellar mass. The 3 lowest mass cases exhibit solar-like differential rotation, in a sense that they show a maximum rotation at the equator and minimum at higher latitudes, but the 1.1 M case exhibits anti-solar rotation. At low mass, the meridional circulation is multi-cellular and aligned with the rotation axis; as the mass increases, the circulation pattern tends toward a unicellular structure covering each hemisphere in the convection zone.
Keywords :
Document type :
Journal articles
Complete list of metadatas
Cited literature [32 references]
https://hal-cea.archives-ouvertes.fr/cea-00828207
Contributor : Marianne Leriche <>
Submitted on : Wednesday, September 30, 2020 - 5:15:42 PM
Last modification on : Wednesday, September 30, 2020 - 5:17:51 PM
Long-term archiving on: : Monday, January 4, 2021 - 8:43:21 AM
### File
Brow1.pdf
Files produced by the author(s)
### Citation
S.P. Matt1, O. Do Cao, B.P. Brown, A.S. Brun. Convection and differential rotation properties of G and K stars computed with the ASH code. Astronomical Notes / Astronomische Nachrichten, Wiley-VCH Verlag, 2011, 332 (9-10), pp.897-906. ⟨10.1002/asna.201111624⟩. ⟨cea-00828207⟩
Record views | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8834047913551331, "perplexity": 4457.04578818443}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-04/segments/1610703511903.11/warc/CC-MAIN-20210117081748-20210117111748-00358.warc.gz"} |
https://standards.globalspec.com/std/798178/api-mpms-8-4 | ### This is embarrasing...
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# API MPMS 8.4
## Manual of Petroleum Measurement Standards Chapter 8 - Sampling Section 4 - Standard Practice for Sampling and Handling of Fuels for Volatility Measurement
inactive
Organization: API Publication Date: 1 December 2004 Status: inactive Page Count: 18
##### scope:
This practice covers procedures and equipment for obtaining, mixing, and handling representative samples of volatile fuels for the purpose of testing for compliance with the standards set forth for volatility related measurements applicable to light fuels. The applicable dry vapor pressure equivalent range of this practice is 13 to 105 kPa (2 to 16 psia).
This practice is applicable to the sampling, mixing, and handling of reformulated fuels including those containing oxygenates.
The values stated in SI units are to be regarded as the standard except in some cases where drawings may show inch-pound measurements which are customary for that equipment.
This standard does not purport to address all of the safety concerns, if any, associated with its use. It is the responsibility of the user of this standard to establish appropriate safety and health practices and determine the applicability of regulatory limitations prior to use.
*A Summary of Changes section appears at the end of this standard.
### Document History
May 1, 2020
Manual of Petroleum Measurement Standards Chapter 8.4 Standard Practice for Sampling and Handling of Fuels for Volatility Measurement
This practice covers procedures and equipment for obtaining, mixing, and handling representative samples of volatile fuels for the purpose of testing for compliance with the standards set forth for...
December 1, 2017
Manual of Petroleum Measurement Standards Chapter 8.4 Standard Practice for Sampling and Handling of Fuels for Volatility Measurement
This practice covers procedures and equipment for obtaining, mixing, and handling representative samples of volatile fuels for the purpose of testing for compliance with the standards set forth for...
March 1, 2014
Manual of Petroleum Measurement Standards Chapter 8.4 Standard Practice for Sampling and Handling of Fuels for Volatility Measurement
This practice covers procedures and equipment for obtaining, mixing, and handling representative samples of volatile fuels for the purpose of testing for compliance with the standards set forth for...
December 1, 2004
Manual of Petroleum Measurement Standards Chapter 8 - Sampling Section 4 - Standard Practice for Sampling and Handling of Fuels for Volatility Measurement
This practice covers procedures and equipment for obtaining, mixing, and handling representative samples of volatile fuels for the purpose of testing for compliance with the standards set forth for...
API MPMS 8.4
December 1, 2004
Manual of Petroleum Measurement Standards Chapter 8 - Sampling Section 4 - Standard Practice for Sampling and Handling of Fuels for Volatility Measurement
This practice covers procedures and equipment for obtaining, mixing, and handling representative samples of volatile fuels for the purpose of testing for compliance with the standards set forth for...
January 1, 1995
Manual of Petroleum Measurement Standards Chapter 8 - Sampling Section 4 - Standard Practice for Manual Sampling and Handling of Fuels for Volatility Measurement
1 Scope 1.1 The applicable dry vapor pressure equivalent range of this standard is 13-105 kilopascals (2-16 pounds per square inch absolute). 1.2 This standard is applicable to the sampling, mixing,... | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.93471759557724, "perplexity": 4123.015727371569}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320303845.33/warc/CC-MAIN-20220122103819-20220122133819-00543.warc.gz"} |
https://phys.libretexts.org/Courses/University_of_California_Davis/UCD%3A_Physics_7C/10%3A_Electromagnetism/10.1%3A_Fields/10.1.3%3A_Fields_in_Physics | $$\require{cancel}$$
# 10.1.3: Fields in Physics
There are three fields in which we will be interested for physics 7C:
1. the Gravitational Field
2. the Electric Field
3. the Magnetic Field
Currently, the most familiar of these is the gravitational field, so the motivation for using fields will start here.
## The Gravitational Field of Earth
Let us start by making a simple statement, which is very imprecise: the Earth’s gravity is stronger than the Moon’s gravity. Justification for this statement comes from watching videos of the astronauts on the Moon and who fall more slowly and leap higher than on Earth. In science we must be more precise. If we calculate the force of the Moon on the Apollo lander, this is much greater than the force of Earth on an apple. We see that we cannot make a blanket statement that the force of gravity on Earth is always greater than the force of gravity on the Moon.
The solution to comparing the Earth's and Moon's gravities is rather simple: we need to compare apples to apples. If we ask what $$\mathbf{F}_{\text{Earth on apple}}$$ is and what $$\mathbf{F}_{\text{Moon on apple}}$$ is then we find $$\mathbf{F}_{\text{Earth on apple}}> \mathbf{F}_{\text{Moon on apple}}$$. More generally, it is true for any object $$X$$: $| \mathbf{F}_{\text{Earth on X}} \text{(at surface of Earth)}| > | \mathbf{F}_{\text{Moon on X}} \text{(at surface of Moon)}|$ This is a more precise version of what we mean when we say that the Earth’s gravity is stronger than the Moon’s gravity.
We can actually do a little bit better than this. The force of gravity does not distinguish between apples, oranges or skyscrapers. If we could build a skyscraper with the mass of an apple, $$\mathbf{F}_{\text{Earth on skyscraper}}$$ would be the same as $$\mathbf{F}_{\text{Earth on apple}}$$; to compare the strength of the gravitational field we don’t need to use exactly the same object, just two objects with the same mass.
Now let's come back to the gravitational field. Recall from Physics 7B that the force of gravity between two masses is $| \mathbf{F}_{\text{mass 1 on mass 2}}| = \dfrac{G M_1 M_2}{r^2} = M_2 \left( \dfrac{G M_1}{r^2} \right)$ where $$r$$ is the distance between the centers of mass, and the direction of the force pulls the masses together. $$G$$ is known as the universal gravitational constant, and is equal to $$6.67 \times 10^{−11} \text{ Nm}^2 \text{kg}^{−2}$$. $$G$$ is a universal constant meaning that it takes the same value regardless of the problem we are doing. Because $$G$$ is so small, we do not notice the gravitational attraction of objects around us unless the object has an enormous mass.
Now let's ask the question “what would the force of the Earth be on an object of mass 1 kg located a distance $$r$$ away?" We can calculate this:$| \mathbf{F}_{\text{Earth on 1 kg object}} | = (1 kg) \times \left( \dfrac{G M_{Earth}}{r^2} \right)$If we agree to always compare the gravity of an object by referring to what force it would exert on a second 1 kg mass then we can do this for any second mass. We find that for any object: $| \mathbf{F}_{\text{Earth on object}} = M_{object} \left( \dfrac{G M_{Earth}}{r^2} \right) = M_{object} |\mathbf{g}_{Earth} |$The quantity on the right, which refers to the Earth and distance, is the gravitation field of the Earth. We denote this $$\mathbf{g}_{Earth}$$:$| \mathbf{g}_{Earth} | \equiv \dfrac{G}{M_{Earth}}{r^2}$As this definition might suggest, $$\mathbf{g}_{Earth}$$ is a vector field with units of acceleration that points towards the Earth. Once we know $$\mathbf{g}_{Earth}$$ we can easily calculate the force on any other mass:$|\mathbf{F}_{\text{Earth on object}} | = M_{object} | \mathbf{g}_{Earth} |$We have seen this relationship many times before (this is why we chose to call the gravitational field $$\mathbf{g}_{Earth}$$ rather than some other letter). Like all fields, the value of $$\mathbf{g}_{Earth}$$ depends on position; it decreases with increased $$r$$, which is the distance from the center of the Earth to the position of interest.
## The Direct and Field Model of Forces
In the way that we have introduced the gravitational field the field is simply a shortcut. Instead of saying “the force a 1 kg object would feel, if placed here, is 5 N” we can simply say “the gravitational field here is 5 N/kg”. The field is not necessary to determine the gravitational force between two objects, it is simply convenient. We will see later that we actually need to talk about fields if we want energy and momentum to be conserved, but for now we will simply treat them as a shortcut.
With this in mind, we have two separate ways of discussing how a gravitational force acts between two objects. The first is where we calculate the force by putting numbers into Newton’s gravitational law without considering the field: the Direct Model: $\text{Object #1} \xrightarrow{\text{creates force on}} \text{Object #2}$ The other way we could think about this is in our new language of fields; one mass creates the field and the other feels the effects: Field Model: $\text{Object #1} \xrightarrow{\text{creates field}} \mathbf{g}_{\text{obj #1}} \xrightarrow{\text{exerts force on}} \text{Object #2}$
Instead of thinking of one object directly exerting a force on another we think of one object (referred to as the “source”) creating a field and then that field creates a force on the second object (sometimes referred to as the “test object”). Of course, both methods are calculations of the same thing and yield the same answer, as the next Example will show:
Example #1
A. What gravitation force does the Earth exert on a 2 kg book on its surface?
B. What gravitation force does the Earth exert on the same book 2 km above its surface?
Use both the direct and field methods (The mass and radius of the Earth can be found online)
Solution
A. Direct Method
After looking up the mass and radius of the Earth we can use Newton's law: $\mathbf{F}_{\text{Earth on book}} = \dfrac{GM_{Earth}M_{book}}{r^2_{Earth}}$ $=\dfrac{(6.67 \times 10^{−11} \text{ N m}^2 \text{ kg}^{−2})(5.98 \times 10^{24} \text{ kg})(2 \text{ kg})}{(6380000 \text{ m})^2}$ $= 19.6 \text{ N}$
A. Field Method
We know that $$\mathbf{g}_{Earth} = 9.8 \text{ N/kg}$$ at the surface of the Earth. Normally we approximate this to 10 N/kg, but let us be more precise for this example. $\mathbf{F}_{\text{Earth on book}} = M_{book} \mathbf{g}_{Earth} = (2 \text{ kg})(9.8 \text{ N/kg}) = 19.6 \text{ N}$
Notice how this calculation was much easier, since we already knew $$\mathbf{g}_{Earth}$$.
B. Direct Method
This proceeds almost exactly the same as before. The two tricky points here are that we have to recall that $$r$$ is the distance from the center of the Earth, and to change 10,000 km into meters.
\begin{align} \mathbf{F}_{\text{Earth on book}} &= \dfrac{GM_{Earth}M_{book}}{r_{Earth}+ 10,000 \text{ km})^2} \\[5pt] &=\dfrac{(6.67 \times 10^{−11} \text{ N m}^2 \text{ kg}^{−2})(5.98 \times 10^{24} \text{ kg})(2 \text{ kg})}{(6,380,000\text{ m} + 10,000,000 \text{ m})^2} \\[5pt] &= 3.0 \text{ N} \end{align}
B Field method
We don't know $$\mathbf{g}_{Earth}$$ at a distance of 10,000 km from the surface of the Earth off the top of our head, we calculate it first. $\mathbf{g}_{Eart} \text{(at 10,000 km from the surface)} = \dfrac{GM_{Earth}}{(r_{Earth} + 10,000 \text{ km})^2}$ $=\dfrac{(6.67 \times 10^{−11} \text{ N m}^2 \text{ kg}^{−2})(5.98 \times 10^{24} \text{ kg})(2 \text{ kg})}{(6,380,000\text{ m} + 10,000,000 \text{ m})^2}$ $= 1.5 \text{ N/kg}$ We can calculate the force the Earth exerts on the book: $\mathbf{F}_{\text{Earth on book}} = M_{book} \mathbf{g}_{Earth} = (2 \text{ kg})(1.5 \text{ N/kg}) = 3.0 N$ Because we did not know $$\mathbf{g}_{Earth}$$ before starting the problem the field method was longer. But if we were asked to do the same calculation for a different mass 10,000 km above the Earth’s surface we now have $$\mathbf{g}_{Earth}$$ and could do it much quicker.
## Which Mass Creates the Field? (Newton's Third Law)
In the example above, when using the field method we decided that the Earth would create the field and the book would respond to it. This seems quite acceptable, as we are used to the Earth exerting a gravitational force. But what if we decided to use a book and a chair in our example? Which would be the “source” for the gravitational field, and which would be pulled by the field?
To try and work out the answer to this question, let us think about the same problem using the direct model of forces. Calculating the magnitude of the force of the book on the chair gives $|\mathbf{F}_{\text{Book on chair}}| = \dfrac{GM_{book}M_{chair}}{r^2}$ where $$r$$ is the distance between them. Now let us calculate the force of the chair on the book: $|\mathbf{F}_{\text{Chair on book}}| = \dfrac{GM_{book}M_{chair}}{r^2} = |\mathbf{F}_{\text{Book on chair}}|$ These forces have the same magnitude, but pull in opposite directions. This is not a coincidence, but is a consequence of Newton’s third law that we learned in 7B: $\mathbf{F}_{\text{A on B}} = - \mathbf{F}_{\text{B on A}}$
In the language of the field model we see that the answer is both the chair and the book create a field. To do the complete problem in the field model we would have to look at both the book's field acting on the chair and the chair's field acting on the book $\text{Book} \xrightarrow{\text{creates field}} \mathbf{g}_{Book} \xrightarrow{\text{exerts force on}} \text{Chair}$ $\text{Chair} \xrightarrow{\text{creates field}} \mathbf{g}_{Chair} \xrightarrow{\text{exerts force on}} \text{Book}$
An important consequence of this is that to be affected by a field, an object must also create a field of the same type. Note that an object does not feel its own field, only the field of all external objects. In other words, the object does not self-interact, but to feel an external gravitational field, for example, it must also create its own gravitational field.
While an object that feels a field must also create the same field, when we are emphasizing an object's ability to create a field we refer to it as the "source" of the field. When we emphasize an object's response to a field we call it the "test" object. We can usually adopt this convention when we can ignore the response of the source object to the field of the test object. However, as we just learned, both objects create a field and are affected by the other's field. For gravity, any object with mass creates and feels a gravitational field. For the electric field, any object with (electric) charge is a source and feels the field. For the magnetic field, as discussed later, the source is moving electric charges. Likewise, charges feel magnetic fields only when they are in motion.
Exercise
In Example #1 above, would we have to worry about the force of the book on the Earth? If not, why not?
## Field Lines
We learned above that the gravitational field of earth is given by $$|\mathbf{g}_{Earth}| = \dfrac{GM_{Earth}}{r^2}$$ where $$r$$ is the distance of the test object from our source, Earth. A similar argument as the one above can be made for any spherical object with mass. We can see using this method hat the gravitational field created by a spherical mass at a distance $$r$$ away is $\mathbf{g}_{spherical} = \dfrac{GM_{spherical}}{r^2}$This equation also applies to point masses. The direction of the gravitational field is always pulling inward. To represent the field, let's create a field map by picking a set of points and drawing vectors to indicate the direction and magnitude of the gravitational field:
The vectors only refer to the value of the fields at the location that they start. The actual length of the vectors is arbitrary, because we could always define some scale to convert the vectors to the right lengths and units. The ratio of lengths between two different vectors is not arbitrary – to accurately represent the vector field, all vectors should be scaled down for the field map exactly the same way.
If we look at the previous field map of the Earth’s gravitational field, we see that far from the Earth it is almost impossible to read the direction of those arrows. We could enlarge them, but then we'd need to scale up every arrow just as much. Some of the arrows near Earth are quite big, so this can become very messy. Furthermore, the size of the arrows limits the amount of information we could put on that part of the map.
To address these shortcomings, we introduce a different representation of a vector field by using field lines. To construct field lines, we draw a continuous lines starting at a point, always going along the direction of the field. An example for the gravitational field of the Earth is shown below:
Notice that the length of the arrows no longer corresponds to the strength of the field, so the strength cannot be read directly from looking at this picture. However this picture contains all the information as the field map. While we cannot look at the length of the arrows to get the strength of the field, we can instead look at how closely packed the field lines are. The closer together we see the lines (near the Earth, in this example) the stronger the field. The further apart the field lines are (far from the Earth), the weaker the field.
If we start with the field line diagram, we can reconstruct the field at any given point the following way:
• Direction: Take a tangent to the field line at that point. This is the direction of the vector field at that point.
• Magnitude: Given by the density of the surrounding lines (denser lines mean longer vectors).
While the vector map is the most direct representation of the field, we will return to using field lines when it's convenient.
Vector Map Field Lines
Pros
• Scales well for vectors that differ in magnitude
Cons
• Hard scaling and readability issues for areas of small field
• Must work to reconstruct the magnitude of the field.
## Gauss's Law
One thing that has not been made explicit in the discussion of field lines so far is that they cannot just start or stop in any location. Where the field lines start or stop depends on the field.
For the gravitational field, field lines start at far distances away and can only stop when they encounter a mass. If there are no masses in a particular region then field lines cannot be created or destroyed. The number of field lines that stop at a mass is proportional to the mass of the object encountered. Thus not all the field lines in our Earth example will “stop” at the Earth (they might hit a satellite, for example). Don’t worry about field lines passing through physical objects; remember that field lines they are only a representation of the field.
For the electric field, field lines start on positive charges and end on negative charges. This makes electric examples slightly more subtle, because if the number of field lines entering is the same as the number leaving it could indicate either that the region has no charge in it or it has an equal number of positive and negative charges. For electric fields looking at the number of field lines entering and exiting a region only tells us about the net charge in the region, not how many individual charges are in that region.
In either case, in regions with no mass (for gravity) or no net charge (for electric field lines), field lines cannot be created or destroyed. Therefore if the number of field lines entering a region is different from the number leaving we must have a field source (i.e. a mass or electric charge) in that region. By knowing the difference, we can figure out exactly how much of that source is in the region. This is the essence of Gauss's Law, which we now make more precise.
Think of a region inside an imaginary closed surface. In other words, imagine a shape with an inside that you cannot get out of unless you go through the surface. A box with a lid is a closed surface, but a cup is not. We can use any imaginary region we desire for Gauss's Law, provided that it is a closed surface.
Once we pick our region, we inspect how many field lines enter it, and how many leave. The only way a field line can get in (or out) is by going through this surface. If a different number of field lines come in than go out, then field lines are being created or destroyed inside. For gravitational field lines, this would indicate a mass; for electric field lines, this would indicate a net charge. If there is no difference between the field lines entering and leaving there is no mass inside or no net charge, and overall no field lines are being created or destroyed.
The amount of field lines going through the surface is referred to as the flux. An increase in flux, for example, can either be represented with more field lines, or by field lines that are drawn closer together. Gauss’s law is simply the statement that $\text{Net flux} \propto \text{(# field lines entering)} - \text{(# field lines leaving)}$ $\text{Net flux} \propto \begin{cases} \text{total mass inside surface} & \quad \text{(gravity)} \\ \text{total charge inside surface} & \quad \text{(electric)} \\ \end{cases}$
A useful analogy to a closed surface and field lines is a leaky bucket filled with water. The bucket is not closed, but you can imagine a closed surface that immediately surrounds the bucket. If water is leaking out of the bucket, it is also leaking out from the inside of this imaginary surface. Therefore the water must be passing through the surface. If we measure the flux of water through our surface, we can measure the amount of water that has leaked out of our bucket.
Of course, you did not need consider imaginary surfaces to realize that the bucket is leaking. But this familiar example displays what we do with fields; if the field lines are coming out of a surface, then something is creating them. If we are losing field lines in a region, something is destroying them.
###### Gauss's Law and Magnetism?
So far Gauss’s law has been discussed for gravitational fields and electric fields, but no mention of magnetic fields has been made. That is because, to the best of our knowledge magnetic monopoles (pieces of magnetic "charge") do not exist. Instead all known magnetic fields are created by moving electric fields. As a consequence magnetic fields do not “start” or “end” but instead make complete loops. Because magnetic field lines never start or end the number of magnetic field lines entering a closed surface is always equal to the number of magnetic field lines leaving that surface. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.891442596912384, "perplexity": 459.4064775852259}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-13/segments/1552912203168.70/warc/CC-MAIN-20190324022143-20190324044143-00528.warc.gz"} |
http://yadda.icm.edu.pl/yadda/element/bwmeta1.element.12a32edb-6aea-3267-9c2b-9a7a8c7f3caf | PL EN
Preferencje
Język
Widoczny [Schowaj] Abstrakt
Liczba wyników
Czasopismo
## Filozofia (Philosophy)
2007 | 62 | 4 | 324-328
Tytuł artykułu
### ANAPHORA AND RESTRICTED QUANTIFICATION
Autorzy
Warianty tytułu
Języki publikacji
SK
Abstrakty
EN
The aim of the paper is to analyze the different approaches to anaphora with the restricted quantifiers. An important point is distinguishing the anaphoric process, which is in fact structured, from the outcomes of that process. The requirements which we put on anaphora (referential dependence and extensional identity of semantic values of antecedent and anaphoric expressions, together with preserving the meaning of analyzed sentences) cannot be met by equipping the classical semantic theories of anaphora (e.g. the analyses of Keenan or Neale). Anaphora is then explained as an algorithmic process in which the semantic value of the anaphoric expression is a higher-order structured function. This function can also be represented as an algorithm consisting of two main sub-algorithms: a calling procedure of picking up the semantic value of a restricted quantifier (also interpreted as a special kind of algorithm) and an execution procedure containing the semantic value already selected. The result is that the semantic values/structured functions of the anaphoric expressions depend on semantic values of the antecedent expressions without violating the principle of preserving the meaning. Given the identity of the extensional relevant algorithmic parts, also the extensions of these functions are the same.
Słowa kluczowe
EN
Czasopismo
Rocznik
Tom
Numer
Strony
324-328
Opis fizyczny
Rodzaj publikacji
ARTICLE
Twórcy
autor
• J. Podrouzek, Filozoficky ustav SAV, Klemensova 19, 813 64 Bratislava, Slovak Republic
Bibliografia
Typ dokumentu
Bibliografia
Identyfikatory
CEJSH db identifier
07SKAAAA02465117 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8647796511650085, "perplexity": 2738.101153110865}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-47/segments/1510934806620.63/warc/CC-MAIN-20171122175744-20171122195744-00016.warc.gz"} |
https://chemistry.stackexchange.com/questions/63974/finding-out-pka-of-acid-from-molar-conductivity | Finding out pKa of acid from molar conductivity
I'm reading about the electrical properties of solution where there is a problem like that:
The molar conductivity of $0.0250\ \mathrm M$ $\ce{HCOOH(aq)}$ is $4.61\ \mathrm{mS\ m^2\ mol^{-1}}$. Determine the $\mathrm pK_\mathrm a=-\log K_\mathrm a$ of the acid. (limiting ionic conductivity of $\ce{H+}=34.96\ \mathrm{mS\ m^2\ mol^{-1}}$ and limiting ionic conductivity of $\ce{OH-}=19.91\ \mathrm{mS\ m^2\ mol^{-1}}$)
I have to solve the problem using the equation below:
$$\frac1{\Lambda_\mathrm m}=\frac1{\Lambda_\mathrm m^0}+\frac{\Lambda_\mathrm mc}{K_\mathrm a\left(\Lambda_\mathrm m^0\right)^2}$$
Here which limiting ionic conductivity should I use to solve the problem? $\ce{H+}$ or $\ce{OH-}$?
You need to add the limiting ionic conductivities for $\ce{H+}$ and $\ce{OH-}$ together to get the limiting ionic conductivity for all the ions in solution ($\Lambda_{0}$, which will replace $\Lambda^{0}_{\mathrm m}$ in your equation).
This arises from a simplification for calculating $\Lambda_{0}$ in weak electrolyte solutions (such as yours) according to Kohlrausch's Law in which it is stated:
Each ionic species makes a contribution to the conductivity of the solution that depends only on the nature of that particular ion, and is independent of the other ions present.
from which we can then estimate $\Lambda_{0}$ as:
$$\Lambda_{0} = \sum_{i}\lambda_{i,+}^{0} + \sum_{i}\lambda_{i,-}^{0}$$
$$\Lambda_{0} = \underbrace{(34.96 + 19.91)}_{54.87}\ \mathrm{mS\cdot m^{2}\cdot mol^{-1}}$$
$${1\over 4.61} = {1\over 54.87} + {4.61\times 0.025\over K_{\mathrm a}\times (54.87)^{2}}$$
$$K_{\mathrm a} = 1.926\times 10^{-4} \implies \mathrm{p}K_{\mathrm a} = 3.72$$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8376170992851257, "perplexity": 487.39716148923077}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-35/segments/1566027322160.92/warc/CC-MAIN-20190825000550-20190825022550-00235.warc.gz"} |
https://simple.wikipedia.org/wiki/Solar_mass | # Solar mass
Solar mass is a unit of measurement of mass. It is equal to the mass of the Sun, about 332,950 times the mass of the Earth, or 1,048 times the mass of Jupiter. Masses of other stars and groups of stars are listed in terms of solar masses. Its mathematical symbol and value are:
${\displaystyle M_{\odot }=1.98892\times 10^{30}{\hbox{ kg}}}$ (kg being kilograms) | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 1, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9837262630462646, "perplexity": 454.6880607820162}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764500215.91/warc/CC-MAIN-20230205032040-20230205062040-00019.warc.gz"} |
https://stats.stackexchange.com/questions/173991/does-x-a-random-variable-usually-refer-to-the-population-or-the-sample?noredirect=1 | # Does $X$ (a random variable) usually refer to the population or the sample? [duplicate]
Like the title says, does $X$ (a random variable) usually refer to the population or the sample?
A random variable, by definition, is a misurable function. If you estract a sample from a population, they follow the same distribution, so the random variable can describe both. When you use the term "Random Variable" you refer to a function from a set to another, the first being called "sample space" and the second $\mathbb{R}^n$. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8239141702651978, "perplexity": 446.0807992518465}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-31/segments/1627046153531.10/warc/CC-MAIN-20210728060744-20210728090744-00512.warc.gz"} |
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