text: C have continuous partial derivatives and they satisfy the Cauchy Riemann equations then Z @U f(z)dz= 0: Proof. f(z) is analytic on and inside the curve C. That is, the roots of z2 + 8 are outside the curve. Do the same integral as the previous example with the curve shown. An illustration is Hadamard's example: The Cauchy problem for the Laplace equation $$\Delta u = \ \frac{\partial ^ {2} u }{\partial x ^ {2} } + \frac{\partial ^ {2} u }{\partial y ^ {2} } + \frac{\partial ^ {2} u }{\partial z ^ {2} } = 0$$ Cauchy Theorem when internal efforts are bounded, and for fixed normal n (at point M), the linear mapping n ↦ t ( M ; n ) is continuous, then t ( M ; n ) is a linear function of n , so that there exists a second order spatial tensor called Cauchy stress σ such that Theorem 14.3 (Cauchy’s Theorem). FÀX¥Q.Pu -PAFhÔ(¥  Then, (5.2.2) I = ∫ C f ( z) z 4 d z = 2 π i 3! Prove that if $f$ is analytic at then $\displaystyle{\mid f'(z) \mid^2 = \left (\frac{\partial u}{\partial x} \right )^2 + \left ( \frac{\partial v}{\partial x} \right )^2}$ and $\displaystyle{\mid f'(z) \mid^2 = \left (\frac{\partial u}{\partial y} \right )^2 + \left ( \frac{\partial v}{\partial y} \right )^2}$. However note that $\displaystyle{1 = \frac{\partial u}{\partial x} \neq \frac{\partial v}{\partial y} = -1}$ ANYWHERE. The Riemann Mapping Theorem; Complex Integration; Complex Integration: Examples and First Facts; The Fundamental Theorem of Calculus for Analytic Functions; Cauchy's Theorem and Integral Formula; Consequences of Cauchy's Theorem and Integral Formula; Infinite Series of Complex Numbers; Power Series; The Radius of Convergence of a Power Series THE CAUCHY MEAN VALUE THEOREM JAMES KEESLING In this post we give a proof of the Cauchy Mean Value Theorem. The contour integral becomes I C 1 z − z0 dz = Z2π 0 1 z(t) − z0 dz(t) dt dt = Z2π 0 ireit reit Theorem (Some Consequences of MVT): Example (Approximating square roots): Mean value theorem finds use in proving inequalities. However note that $\displaystyle{1 = \frac{\partial u}{\partial x} \neq \frac{\partial v}{\partial y} = -1}$ ANYWHERE. What is an intuitive way to think of Cauchy's theorem? Find out what you can do. 4.3.2 More examples Example 4.8. Stã|þtÇÁ²vfÀ& Iæó>@dÛ8.ËÕ2?hm]ÞùJõ:³@ØFÃ¦ÄÔç¯3³$W°¤hxÔIÇç/ úÕØØ¥¢££`ÿ3 Wikidot.com Terms of Service - what you can, what you should not etc. Solution: This one is trickier. Physics 2400 Cauchy’s integral theorem: examples Spring 2017 and consider the integral: J= I C [z(1 z)] 1 dz= 0; >1; (4) where the integration is over closed contour shown in Fig.1. Group of order$105$has a subgroup of order$21$5. Example Evaluate the integral I C 1 z − z0 dz, where C is a circle centered at z0 and of any radius. Recall from The Cauchy-Riemann Theorem page that if$A \subseteq \mathbb{C}$is open,$f : A \to \mathbb{C}$with$f = u + iv$, and$z_0 \in A$then$f$is analytic at$z_0$if and only if there exists a neighbourhood$\mathcal N$of$z_0$with the following properties: We also stated an important result that can be proved using the Cauchy-Riemann theorem called the complex Inverse Function theorem which says that if$f'(z_0) \neq 0$then there exists open neighbourhoods$U$of$z_0$and$V$of$f(z_0)$such that$f : U \to V$is a bijection and such that$\displaystyle{\frac{d}{dw} f^{-1}(w) = \frac{1}{f'(z)}}$where$w = f(z)$. HBsuch In particular, has an element of order exactly . Q.E.D. 2. dz, where. Cauchy’s theorem Simply-connected regions A region is said to be simply-connected if any closed curve in that region can be shrunk to a point without any part of it leaving a region. Example 5.2. ∫ C ( z − 2) 2 z + i d z, \displaystyle \int_ {C} \frac { (z-2)^2} {z+i} \, dz, ∫ C. . Thus by the Cauchy-Riemann theorem,$f(z) = e^{z^2}$is analytic everywhere. Logarithms and complex powers 10. Theorem 23.3 we know that all of the derivatives of f are also analytic in D.Inparticular, this implies that all the partials of u and v of all orders are continuous. Let Cbe the unit circle. In cases where it is not, we can extend it in a useful way. 3)¸%ÀÄ¡*Å2:à)Ã2 Let f ( z) = e 2 z. Corollary of Cauchy's theorem … Now, having found suitable substitutions for the notions in Theorem 2.2, we are prepared to state the Generalized Cauchy’s Theorem. One can use the Cauchy integral formula to compute contour integrals which take the form given in the integrand of the formula. Liouville’s theorem: bounded entire functions are constant 7. ANALYSIS I 9 The Cauchy Criterion 9.1 Cauchy’s insight Our diﬃculty in proving “a n → ‘” is this: What is ‘? Then$u(x, y) = e^{x^2 - y^2} \cos (2xy)$and$v(x, y) = e^{x^2 - y^2} \sin (2xy)$. We have, by the mean value theorem, , for some such that . Recall from the Cauchy's Integral Theorem page the following two results: The Cauchy-Goursat Integral Theorem for Open Disks: If$f$is analytic on an open disk$D(z_0, r)$then for any closed, piecewise smooth curve$\gamma$in$D(z_0, r)$we have that: (1) This means that we can replace Example 13.9 and Proposition 16.2 with the following. View and manage file attachments for this page. Examples. Re(z) Im(z) C. 2. If you want to discuss contents of this page - this is the easiest way to do it. Suppose that$f$is analytic. Addeddate 2006-11-11 01:04:08 Call number 29801 Digitalpublicationdate 2005/06/21 Identifier complexintegrati029801mbp Identifier-ark ark:/13960/t0rr1q351 Theorem 7.4.If Dis a simply connected domain, f 2A(D) and is any loop in D;then Z f(z)dz= 0: Proof: The proof follows immediately from the fact that each closed curve in Dcan be shrunk to a point. Suppose $$R$$ is the region between the two simple closed curves $$C_1$$ and $$C_2$$. The first order partial derivatives of$u$and$v$clearly exist and are continuous. Something does not work as expected? If is a finite group, and is a prime number dividing the order of , then has a subgroup of order exactly . In Figure 11 (a) and (b) the shaded grey area is the region and a typical closed 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. Except for the proof of the normal form theorem, the material is contained in standard text books on complex analysis. Residues and evaluation of integrals 9. Let V be a region and let Ube a bounded open subset whose boundary is the nite union of continuous piecewise smooth paths such that U[@UˆV. Click here to toggle editing of individual sections of the page (if possible). (4) is analytic inside C, J= 0: (5) On the other hand, J= JI +JII; (6) where JI is the integral along the segment of the positive real axis, 0 x 1; JII is the Example 4.3. Let$f(z) = f(x + yi) = x - yi = \overline{z}$. A practically immediate consequence of Cauchy's theorem is a useful characterization of finite p-groups, where p is a prime. Then as before we use the parametrization of the unit circle Determine whether the function$f(z) = \overline{z}$is analytic or not. If a function f is analytic at all points interior to and on a simple closed contour C (i.e., f is analytic on some simply connected domain D containing C), then Z C f(z)dz = 0: Note. Cauchy’s formula 4. Compute Z C cos(z) z(z2 + 8) dz over the contour shown. If the real and imaginary parts of the function f: V ! Cauchy’s theorem requires that the function $$f(z)$$ be analytic on a simply connected region. Watch headings for an "edit" link when available. Cauchy’s theorem 3. Im(z) Im(z) 2i 2i C Solution: Let f(z) = cos(z)=(z2 + 8). Theorem 23.7. Cauchy’s integral theorem An easy consequence of Theorem 7.3. is the following, familiarly known as Cauchy’s integral theorem. Solution The circle can be parameterized by z(t) = z0 + reit, 0 ≤ t ≤ 2π, where r is any positive real number. Related. Change the name (also URL address, possibly the category) of the page. Laurent expansions around isolated singularities 8. We will now look at some example problems in applying the Cauchy-Riemann theorem. Compute Z C 1 (z2 + 4)2 Example 4.4. Argument principle 11. See pages that link to and include this page. Notify administrators if there is objectionable content in this page. The notes assume familiarity with partial derivatives and line integrals. The first order partial derivatives of$u$and$v$clearly exist and are continuous. )©@¤Ä@T\A!sbM°1q¼GY*|z¹ô\mT¨sd. With Cauchy’s formula for derivatives this is easy. The path is traced out once in the anticlockwise direction. A remarkable fact, which will become a theorem in Chapter 4, is that complex analytic functions automatically possess all Append content without editing the whole page source. Determine whether the function$f(z) = e^{z^2}$is analytic or not using the Cauchy-Riemann theorem. Theorem (Cauchy’s integral theorem 2): Let Dbe a simply connected region in C and let Cbe a closed curve (not necessarily simple) contained in D. Let f(z) be analytic in D. Then Z C f(z)dz= 0: Example: let D= C and let f(z) be the function z2 + z+ 1. The stronger (better) version of Cauchy's Extension of the MVT eliminates this condition. Cauchy’s Theorem Cauchy’s theorem actually analogue of the second statement of the fundamental theorem of calculus and integration of familiar functions is facilitated by this result In this example, it is observed that is nowhere analytic and so need not be independent of choice of the curve connecting the points 0 and . Determine whether the function$f(z) = \overline{z}$is analytic or not. Then$u(x, y) = x$and$v(x, y) = -y$. New content will be added above the current area of focus upon selection$\displaystyle{\frac{\partial u}{\partial x}, \frac{\partial u}{\partial y}, \frac{\partial v}{\partial x}, \frac{\partial v}{\partial y}}$,$\displaystyle{\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}}$,$\displaystyle{\frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}}$,$\displaystyle{\frac{d}{dw} f^{-1}(w) = \frac{1}{f'(z)}}$,$f(z) = f(x + yi) = x - yi = \overline{z}$,$\displaystyle{1 = \frac{\partial u}{\partial x} \neq \frac{\partial v}{\partial y} = -1}$,$\displaystyle{\mid f'(z) \mid^2 = \left (\frac{\partial u}{\partial x} \right )^2 + \left ( \frac{\partial v}{\partial x} \right )^2}$,$\displaystyle{\mid f'(z) \mid^2 = \left (\frac{\partial u}{\partial y} \right )^2 + \left ( \frac{\partial v}{\partial y} \right )^2}$, Creative Commons Attribution-ShareAlike 3.0 License. Click here to edit contents of this page. If we assume that f0 is continuous (and therefore the partial derivatives of u … This should intuitively be clear since$f$is a composition of two analytic functions. Re(z) Im(z) C. 2. Example 1 The function $$f\left( x \right)$$ is differentiable on the interval $$\left[ {a,b} \right],$$ where $$ab \gt 0.$$ Show that the following equality ${\frac{1}{{a – b}}\left| {\begin{array}{*{20}{c}} a&b\\ {f\left( a \right)}&{f\left( b \right)} \end{array}} \right|} = {f\left( c \right) – c f’\left( c \right)}$ holds for this function, where $$c \in \left( {a,b} \right).$$ The following theorem says that, provided the ﬁrst order partial derivatives of u and v are continuous, the converse is also true — if u(x,y) and v(x,y) obey the Cauchy–Riemann equations then Solution: Since ( ) = e 2 ∕( − 2) is analytic on and inside , Cauchy’s theorem says that the integral is 0. Check out how this page has evolved in the past. Then there is … Cauchy Mean Value Theorem Let f(x) and g(x) be continuous on [a;b] and di eren-tiable on (a;b). The interior of a square or a circle are examples of simply connected regions. In particular, a finite group G is a p-group (i.e. The mean value theorem says that there exists a time point in between and when the speed of the body is actually . Theorem 2says thatitisnecessary for u(x,y)and v(x,y)toobey the Cauchy–Riemann equations in order for f(x+iy) = u(x+iy)+v(x+iy) to be diﬀerentiable. z +i(z −2)2. . How to use Cayley's theorem to prove the following? They are: So the first condition to the Cauchy-Riemann theorem is satisfied. Since the integrand in Eq. Theorem 2.9 Let Mbe an oriented smooth manifold with corners and Bbe an n-dimensional body in M. Suppose that and are bounded n-forms on B and ˝is a continuous function on the bundle of oriented hyperplanes! 3. Then$u(x, y) = x$and$v(x, y) = -y$. Let$f(z) = f(x + yi) = x - yi = \overline{z}$. For example, for consider the function . So one of the Cauchy-Riemann equations is not satisfied anywhere and so$f(z) = \overline{z}$is analytic nowhere. Power series expansions, Morera’s theorem 5. If f(z)=u(z)+iv(z)=u(x,y)+iv(x,y) is analytic in a … So, we rewrite the integral as Z C cos(z)=(z2 + 8) z dz= Z C f(z) z dz= 2ˇif(0) = 2ˇi 1 8 = ˇi 4: Example 4.9. 0. They are given by: So$\displaystyle{\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}}$everywhere. Cauchy's Integral theorem concept with solved examples Subject: Engineering Mathematics /GATE maths. f ‴ ( 0) = 8 3 π i. I use Trubowitz approach to use Greens theorem to examples, which examples showing how residue calculus can help to calculate some deﬁnite integrals. 1. I have deleted my non-Latex post on this theorem. Cauchy saw that it was enough to show that if the terms of the sequence got suﬃciently close to each other. The partial derivatives of these functions exist and are continuous. Identity principle 6. 1. The Cauchy-Goursat Theorem Cauchy-Goursat Theorem. General Wikidot.com documentation and help section. Cauchy's Integral Theorem Examples 1. Compute. They are: So the first condition to the Cauchy-Riemann theorem is satisfied. View/set parent page (used for creating breadcrumbs and structured layout). So one of the Cauchy-Riemann equations is not satisfied anywhere and so$f(z) = \o… It is a very simple proof and only assumes Rolle’s Theorem. For example, a function of one or more real variables is real-analytic if it is diﬀerentiable to all orders on an open interval or connected open set and is locally the sum of its own convergent Taylor series. To do so, we have to adjust the equation in the theorem just a bit, but the meaning of the theorem is still the same. Do the same integral as the previous examples with the curve shown. Then from the proof of the Cauchy-Riemann theorem we have that: The other formula can be derived by using the Cauchy-Riemann equations or by the fact that in the proof of the Cauchy-Riemann theorem we also have that: \begin{align} \quad \frac{\partial u}{\partial x} = 1 \quad , \quad \frac{\partial u}{\partial y} = 0 \quad , \quad \frac{\partial v}{\partial x} = 0 \quad , \quad \frac{\partial v}{\partial y} = -1 \end{align}, \begin{align} \quad f(z) = f(x + yi) = e^{(x + yi)^2} = e^{(x^2 - y^2) + 2xyi} = e^{x^2 - y^2} e^{2xyi} = e^{x^2 - y^2} \cos (2xy) + e^{x^2 - y^2} \sin (2xy) i \end{align}, \begin{align} \quad \frac{\partial u}{\partial x} = 2x e^{x^2 - y^2} \cos (2xy) - 2y e^{x^2 - y^2} \sin (2xy) = e^{x^2 - y^2} [2x \cos (2xy) - 2y \sin (2xy)] \end{align}, \begin{align} \quad \frac{\partial v}{\partial y} = -2ye^{x^2 - y^2} \sin(2xy) + 2x e^{x^2 - y^2} \cos (2xy) = e^{x^2 - y^2}[2x \cos (2xy) - 2y \sin (2xy)] \end{align}, \begin{align} \quad \frac{\partial u}{\partial y} =-2ye^{x^2 - y^2} \cos (2xy) - 2x e^{x^2 - y^2} \sin (2xy) = -e^{x^2 - y^2}[2x \sin (2xy) + 2y \cos (2xy)] \end{align}, \begin{align} \quad \frac{\partial v}{\partial x} = 2xe^{x^2 - y^2}\sin(2xy) + 2ye^{x^2 - y^2}\cos(2xy) = e^{x^2 - y^2}[2x \sin (2xy) + 2y \cos(2xy)] \end{align}, \begin{align} \quad f'(z) = \frac{\partial u}{\partial x} + i \frac{\partial v}{\partial x} \end{align}, \begin{align} \quad \mid f'(z) \mid = \sqrt{ \left( \frac{\partial u}{\partial x} \right )^2 + \left ( \frac{\partial v}{\partial x} \right )^2} \end{align}, \begin{align} \quad \mid f'(z) \mid^2 = \left( \frac{\partial u}{\partial x} \right )^2 + \left ( \frac{\partial v}{\partial x} \right )^2 \end{align}, \begin{align} \quad f'(z) = \frac{\partial v}{\partial y} -i\frac{\partial u}{\partial y} \end{align}, Unless otherwise stated, the content of this page is licensed under. all of its elements have order p for some natural number k) if and only if G has order p for some natural number n. 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Or not using the Cauchy-Riemann equations is not satisfied anywhere and So $f$ is analytic not! Edit '' link when available text books on complex analysis intuitively be clear since $f z... Integral as the previous example do the same integral as the previous example other... Order exactly breadcrumbs and structured layout ) { z }$ the path is traced out once in anticlockwise! X + yi ) = x $and$ v ( x + yi =! How this page power series expansions, Morera ’ s theorem: bounded functions... Proving inequalities re ( z ) = \o… 1 Service - what you can, what you not. And they satisfy the Cauchy integral formula to compute contour integrals which take the given! = f ( z ) dz= 0: proof are constant 7 that we can extend it a... Z^2 } $is analytic or not MVT eliminates this condition ( C_1\ ) and \ ( C_2\ ) not! If the real and imaginary parts of the page ( used for creating breadcrumbs structured! Has a subgroup of order$ 105 $has a subgroup of order exactly the anticlockwise direction 8... Cauchy Riemann equations then z @ u f ( x + yi ) = e 2.. Or a circle are examples of simply connected regions terms of Service - what you can, what you not! This means that we can extend it in a useful way, what you should not etc vs! Some Consequences of MVT ): mean value theorem says that there exists a time point in between and the! Connected regions useful way with the curve shown a finite group G is useful. Theorem finds use in proving inequalities equations then z @ u f ( z ) = \o… 1 corollary cauchy's theorem examples! Of any radius should intuitively be clear since$ f ( z ) = x ! ) = x - yi = \overline { z } $is a useful characterization of finite,! Also URL address, possibly the category ) of the formula example with the shown... Clear since$ f ( z ) Im ( z ) dz= 0: proof a centered. It is not satisfied anywhere and So $f ( x, y ) =$... Link to and include this page example 13.9 and Proposition 16.2 with the curve shown intuitive. Normal form theorem, the material is contained in standard text books on complex analysis to editing! 4 ) 2 Cauchy ’ s theorem: bounded entire functions are constant 7 only Rolle... Contour shown below and Evaluate the integral i C 1 ( z2 + 4 ) 2 Cauchy ’ theorem! 16.2 with the following shown below and Evaluate the same integral as in the integrand of the page )... $is analytic everywhere content in this page has evolved in the of... Between the two simple closed curves \ ( C_1\ ) and \ ( C_2\ ) z z0! And include this page$ 105 $has a subgroup of order 105... If there is objectionable content in this page has evolved in the.. Some Consequences of MVT ): mean value theorem,, for some such that 105$ a! I have deleted my non-Latex post on this theorem liouville ’ s.! Are examples of simply connected regions ( 5.2.2 ) i = ∫ C f z.


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# Homework Help: Why is the R resistance equal to the internal resistance

1. Apr 3, 2005

### apchemstudent

The answer to the attachment is if the R resistance was equal to the internal resistance. How was this determined? Please explain this to me because i have no idea how they can solve this. Thanks

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2. Apr 5, 2005

### apchemstudent

anyone have any idea as to how the answer was reached?

3. Apr 5, 2005

### asrodan

I haven't taken any courses on circuits yet and only vaguely remember what I covered in intro. But here's what I think.

The potential of the battery and the current produced are both constant.

Since the only components are the battery and the resistor, the potential drop across the resistor must equal the potential provided by the battery. So the maximum value of R for which the circuit would "function" is when it is equal to the internal resistance.

It = IR where t = internal resistance.

4. Apr 6, 2005

### GCT

latex

From what I recall recently

$$V_{battery} = emf-IR_{internal}$$

assuming that the "load" resistor is $$R_{eq}$$, (I forgot some of the terminologies), than

$$P=I^{2}R$$, where the $I=EMF/(R+R_{internal})$

thus

$$P= \frac {(EMF^{2}(R))}{(R+R_{internal})^{2}}$$
Try observing the trend for an infinitely large value of R and infinitely small value.

I'm pretty new to this also, so I'm not absolutely certain

5. Apr 6, 2005

### Delta

I once derived this as an out of class exercise:

$$P = \frac{E^2R}{(R+r)^2}$$ as shown above.

where P is the power, R is the load, r is the internal resistance and E is the battery EMF.

We are trying to find the point at which P is a maximum by varying the load R, therefore we involve a bit of calculus with respect to R.

We need the quotient rule.

$$\frac{d}{dx}(\frac{u}{v}) = \frac{V\frac{du}{dx}-U\frac{dv}{dx}}{V^2}$$

letting $$u = E^2R$$ and $$v = (R+r)^2$$

So ....

$$\frac{dP}{dR}=\frac{E^2(R+r)^2 - 2(R+r)(E^2R)}{(R+r)^{4}}$$

at maximum power $$\frac{dP}{dR}=0$$
$$(R+r)^{4}$$ can not equal zero, therefore $$E^2(R+r)^2 - 2(R+r)(E^2R)$$ must equal zero.

Rearranging this eventually gets to r = R, thus showing at maximum power the load resistance equals the internal resistance

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# GATE2017-26

$$\frac{dx}{dt}=sin\left ( t \right ),\:\:\:\:x\left ( 0 \right )=0$$
the value of $x$ at $t=\pi /3$ , is _____________.


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# Journal of Operator Theory

Volume 65, Issue 2, Spring 2011  pp. 403-417.

Some remarks on Haagerup's approximation property

Authors Jon P. Bannon (1) and Junsheng Fang (2)
Author institution: (1) Department of Mathematics, Siena College, Loudonville NY, 12065, U.S.A.
(2) Department of Mathematics, Texas A\& M University, College Station TX, 77843-3368, U.S.A.

Summary:  A finite von Neumann algebra $\mathcal{M}$ with a faithful normal trace $\tau$ has Haagerup's approximation property if there exists a pointwise deformation of the identity in $2$-norm by subtracial compact completely positive maps. In this paper we prove that the subtraciality condition can be removed. This enables us to provide a description of Haagerup's approximation property in terms of correspondences. We also show that if $\mathcal{N}\subset \mathcal{M}$ is an amenable inclusion of finite von Neumann algebras and $\mathcal{N}$ has Haagerup's approximation property, then $\mathcal{M}$ also has Haagerup's approximation property.

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# A 1.22-g spider oscillates on its web, which has a damping constant of b = 3.10×10−5...

###### Question:

A 1.22-g spider oscillates on its web, which has a damping constant of b = 3.10×10−5 kg/s .

How long does it take for the spider's amplitude of oscillation to decrease by 18.0 % ?

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##### :MB[ QUata[duo) opqnop 244 osn ION O Mv[ 4HM dazs ypea Ajusn [ V=%[8 quawarduoj a41 4IM V JO JUJUapduo) %1 JO UOQUn 341 SL ILja} ISIY J1 su Aposop YOoT) "SMe] XwItap! J3s Sutsn Aq V =Vj (anv IBT? & puv V S138 [[B JOJ ohoxd (squrod 01)
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text:
dirichletEta( z )

The Dirichlet eta function of z in Math. Defined by

$\eta(z) = \sum_{k=1}^\infty \frac{ (-1)^{k-1} }{ k^z }$

Related to the Riemann zeta function by

$\eta(z) = \left( 1 - 2^{1-z} \right) \zeta(z)$

Real part on the real axis:

Imaginary part on the real axis is zero.

Real part on the imaginary axis:

Imaginary part on the imaginary axis:

Real part on the complex plane:

Imaginary part on the complex plane:

Absolute value on the complex plane:

Related functions:   zeta

Function category: zeta functions


text:
# [lltx] Problem with slanted fonts

Ulrike Fischer news3 at nililand.de
Sun Sep 19 16:24:52 CEST 2010

Am Sun, 19 Sep 2010 14:37:31 +0300 schrieb Khaled Hosny:

> On Sun, Sep 19, 2010 at 01:27:38PM +0200, Ulrike Fischer wrote:
>> Am Sat, 18 Sep 2010 15:58:37 +0930 schrieb Will Robertson:
>>
>>
>>>>> I have installed TexLive 2010 and am trying out Lualatex.
>>>>> When typesetting the following example:
>>>>>
>>>>> \documentclass{article}
>>>>> \usepackage{fontspec}
>>>>> \setmainfont{Gentium Basic}
>>>>> \begin{document}
>>>>> \textsl{hello}
>>>>> \end{document}
>>>>>
>>>>> the resulting pdf file contains the following line:
>>>>>
>>>>> Basic/I:mode=node;script=latn;language=DFLT;hello
>>>>>
>>>>> I wouldn't expect to find the font features in there...
>>>>> What am I doing wrong here?
>>>>
>>>> Nothing. It is a bug, due to the space in the font name.
>>>>
>>>> Until the bug is repaired you can remove the space from the font
>>>> name (GentiumBasic), or you can use \textit instead of \textsl. This
>>>> will give the same result anyway as long as you don't use the
>>>> SlantedFont-option to declare a specific slanted font.
>>
>>> I've been trying to track down the origin of this bug and so far
>>> I'm out of luck.
>>
>> I have tracked it a bit down. The source is 1. that \slshape use sub
>> in \DeclareFontShape and 2. that the "font declaration" is stored in
>> a command. The result depends also on the actual font. In the
>> following example Times New Roman gives "only" some unwanted text in
>> the pdf (because "Times" is the name of an existing font too), while
>> Lucida gives directly an error (! Font
>> \EU2/testfam/m/n/10=name:Lucida not loadable:). Attention: If you
>> use both method (with and without command \fontdesc) for the same
>> font they can be side effect.
>
> So, how XeTeX is dealing with this? If its behaviour can be emulated in
> luaotfload then we should do it, else it might be something for LuaTeX
> to handle.

I have no idea how xetex is going around. But I found the place
where it fails: in \do at subst@correction. A failing example gives
this with tracingmacros =1 :

EU2/testfam/m/n -><->"name:Lucida Sans"

<->"name:Lucida Sans" ->

\subst at correction ->\EU2/testfam/m/it/10 \global \font
\EU2/testfam/m/n/10 \fontname \font \relax
luaotfload | Updating the font names database:
luaotfload | Scanning TEXMF and OS fonts...
... and then error

A non failing shows this (notice the missing space!):

EU2/testfam/m/n -><->"name:LucidaSans"

<->"name:LucidaSans" ->

\subst at correction ->\EU2/testfam/m/it/10 \global \font
\EU2/testfam/m/n/10 \fontname \font \relax


text:
# Fourier Transform of a sin(2pi*x)

1. Oct 1, 2015

### BobP

I have been very briefly introduced to Fourier transformations but the topic was not explained especially well (or I just didn't understand it!)

We were shown the graphs with equations below and then their fourier transformation (RHS). I understand the one for cos(2pist) but NOT the sin(2pist).

Here is my reasoning: The sin wave shown has a constant frequency (that is larger than the cosine function; hence its frequency values are larger). It is possible to have a "-ve" frequency because waves can travel in both directions. I do not understand why the amplitude on the sin graph is negative though? If it was negative on both it would make sense but it isn't!
I know that both FT boil down to a cosine expression (as the sin expression is lost when integrating between infinities) and that makes it even more confusing for me! Please can someone help me understand. thanks

2. Oct 1, 2015

### blue_leaf77

Just a wording correction though, amplitude by definition is a positive quantity. As for your question, two functions connected through Fourier transform are in general complex, that is, they have modulus and phase. The reason why the negative frequency component in the FT of sine points down is that because this component are 180 degree out of phase from the positive frequency component.
You don't integrate the delta function alone, do you. There is also $e^{i\omega t}$ in the integrand.

3. Oct 1, 2015

### BobP

But surely the cos points are also 180 degrees out of phase? I am not too sure I understand what you mean by this

4. Oct 1, 2015

### BobP

Oh Wait. Do you mean because sin is an odd function whereas cos is an even function?

5. Oct 1, 2015

### blue_leaf77

No, the Fourier transform of cosine is real.
A complex number $z$ can be expresed as $z=|z|e^{i\theta}$ where $|z|$ the modulus and $\theta$ the phase. You can write the Fourier transform of sine as
$$\frac{1}{2}(\delta(\nu+s) - \delta(\nu-s)) = \frac{1}{2}(\delta(\nu+s) + e^{i\pi} \delta(\nu-s))$$
EDIT: the Fourier transform of cosine is real and symmetric.

Last edited: Oct 1, 2015
6. Oct 1, 2015

### BobP

so are you saying that the "negative" ampltidue on the sin curve reflects the fact that it is the imaginary component of the sin expression (as per Euler formula)

7. Oct 1, 2015

### blue_leaf77

No, the negative sign appears because of the real component of $e^{i\pi}$ instead.
$$e^{i\pi}=-1$$

8. Oct 1, 2015

### BobP

I see...thanks for the help.
Truth be told, I understood none of the maths initially but your help has pointed me to videos / math books that have explained the issue to me so I now understand it (I hope!)

Really grateful!


text:
PL EN

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Journal

## Acta Physica Polonica A

1997 | 92 | 6 | 1101-1108
Article title

### Stochastic Resonance in System with On-Off Intermittency

Authors
Content
Title variants
Languages of publication
EN
Abstracts
EN
Stochastic resonance in a chaotic threshold-crossing system exhibiting on-off intermittency and attractor bubbling: the logistic map with the control parameter varying randomly or chaotically in time is studied in the case of weak additive and multiplicative periodic forcing. In both cases signal-to-noise ratio shows dependence on the forcing frequency; in the case of multiplicative forcing this dependence appears even for very small frequencies. It is shown that this is a result of a very long characteristic time scale, typical of systems with on-off intermittency.
Keywords
EN
Publisher

Journal
Year
Volume
Issue
Pages
1101-1108
Physical description
Dates
published
1997-12


text:
# best electronics schematic program for noobe

#### VernonLS

Joined Oct 9, 2019
42
What is the best/easiest to learn/low cost/free electronics cad program for beginners to use. I have been using Visio just because I own it, but it often gets lost simply trying to connect wires and it has no component library. I have tried Eagle, but it is way, way more than I need and the guy who does the tutorial should be shot because he talks so slow that I fall asleep. I have also tried EasyEDA and it has some really nice features, but it can't handle changing the drawing size or printing it larger or smaller mostly because it all about physical components, not electronics. I also tried ProfiCAD, but found it severely limiting. So far the best and easiest to learn one I have found is CircuitLab, but the trial period ends quickly and I just can't justify spending $16 -$30 a month for my hobby. I don't care about PCB layout as I am only building hobby boards, not manufacturing a product. I would like to have a library of common IC parts, relays, common symbols, etc. Simulation is nice, but a must-have.

Easy to learn is high on my list of desires, able to handle complex designs is low. Hope others will share their favorites.

#### dl324

Joined Mar 30, 2015
9,331
I tried a couple of programs before I settled on Eagle. I don't know why people claim Eagle is difficult to learn. I didn't find that to be the case.

That being said, I don't like the new versions of Eagle that require an account. I use a very old version and don't care for most of the "improvements" they've made. There's one that I'd like (more flexible overbar in annotations), but I can live without the rest.

KiCad shows up often in these "best" queries. I thought about checking it out, but something (I think the download size) discouraged me.

Joined Jul 18, 2013
19,421
Maybe not that easy, but anything/program etc that is comprehensive also usually has a learning curve.
I originally used an older OrCad but have since moved over to something newer, namely the Free/open-source KiCad.
It presently has seven professional contributors to it, I recently posted a link to some introductory Utube videos.
Max.

#### MrChips

Joined Oct 2, 2009
19,750
I don't know what is "best" but you can add CircuitMaker 2000 to the list.

#### ebeowulf17

Joined Aug 12, 2014
2,960
I'm a huge fan of DipTrace. I think it's roughly the same capabilities as Eagle, it's free, and it does everything I've ever needed it to. I'm one of the people who thought Eagle was difficult and horribly counter intuitive. Learning DipTrace was really very easy by comparison. Of course, we all have different learning styles and workflow preferences, but that was my experience.


text:
• ms@ms.lt
• +370 607 27 665
• My work is in the Public Domain for all to share freely.

Introduction E9F5FC

Understandable FFFFFF

Questions FFFFC0

Notes EEEEEE

Software

## Book.Topos istorija

2019 balandžio 16 d., 12:40 atliko AndriusKulikauskas -
Pridėtos 72-74 eilutės:

* [[http://www.math.harvard.edu/~lurie/papers/HTT.pdf | Higher Topos Theory]] Jacob Lurie
2019 balandžio 06 d., 11:27 atliko AndriusKulikauskas -
Pridėta 75 eilutė:
* [[https://www.youtube.com/watch?v=vmcbm5FxRJE | Colin McLarty: Nonetheless one should learn the language of topos]]
2018 rugsėjo 30 d., 20:59 atliko AndriusKulikauskas -
Ištrinta 71 eilutė:
* [[https://en.wikipedia.org/wiki/Yoneda_lemma | Wikipedia: Yoneda lemma]]
2018 rugsėjo 30 d., 20:57 atliko AndriusKulikauskas -
Ištrintos 45-71 eilutės:

%center%Attach:YonedaLemma.png

'''Yoneda Lemma'''

Key points: Every object has a morphism to itself. Every set function has at least one element in its range.

What: Set of properties. Why: Set of relationships as dictaded by its relationships but especially its relationship with itself (its essence).

What is left unspoken: Sets are labelled - elements are labels. Categories are unlabelled only structure. Is this related to model theory?

The properties of an entity correspond to the analogous stances: "This is the essence of the entity, the property that makes it what it is."

Hom(A,–) gives the perspectives from A, including perspectives on perspectives.

If F is the empty functor, or if F(A) is empty, then there are no natural transformations.

F(A) is a set of labels.

* The Yoneda Lemma asserts that {$C^{op}$} embeds in {${\textbf{Set}}^C$} as a full subcategory.
* The functor category {$D^C$} has as objects the functors from ''C'' to ''D'' and as morphisms the natural transformations of such functors. The Yoneda lemma describes representable functors in functor categories.

Let F be an arbitrary functor from C to Set. Then Yoneda's lemma says that: For each object A of C, the natural transformations from {$h^A$} to F are in one-to-one correspondence with the elements of F(A). That is,

{${Nat} (h^{A},F)\cong F(A)$}

Moreover this isomorphism is natural in A and F when both sides are regarded as functors from SetC x C to Set.
2018 rugsėjo 30 d., 20:56 atliko AndriusKulikauskas -
Pakeistos 1-2 eilutės iš
See: [[Category theory]]
į:
See: [[Category theory]], [[Yoneda Lemma]]
Ištrintos 3-4 eilutės:

* Explain how substitution of eigenvalues into symmetric functions can add information. And relate this to the Yoneda lemma where associativity adds information when we go from objects (node-variables) to morphisms (matrix edges).
2018 birželio 22 d., 12:17 atliko AndriusKulikauskas -
Pridėtos 32-33 eilutės:

"Sketches of an Elephant: A Topos Theory Compendium 2 Volume Set"
2018 birželio 22 d., 12:07 atliko AndriusKulikauskas -
Pridėtos 30-31 eilutės:

Combinatorics - create objects "for children" - then count them - the formulas appear in other contexts - gives clues for related toposes.
2018 birželio 22 d., 11:20 atliko AndriusKulikauskas -
Pridėtos 28-29 eilutės:

Sheaves are not interesting but rather, what we see from our external point of view, and what needs to be removed to see the interesting internal point of view.
2018 birželio 22 d., 10:58 atliko AndriusKulikauskas -
Pridėtos 26-27 eilutės:

Can define objects and math before logic.
2018 birželio 22 d., 10:49 atliko AndriusKulikauskas -
Pridėtos 24-25 eilutės:

Example: internal view (of complicated curves) are embedded in an external space (and the latter is removed in the topos).
2018 birželio 22 d., 10:46 atliko AndriusKulikauskas -
Pridėtos 22-23 eilutės:

Double negation in the simpler, topos world, almost holds - but when you translate into the richer world - you get a highly nontrivial statement.
2018 birželio 20 d., 19:15 atliko AndriusKulikauskas -
Pridėtos 20-21 eilutės:

Morita equivalence.
2018 birželio 20 d., 18:21 atliko AndriusKulikauskas -
Pakeista 19 eilutė iš:
The logic of geometry is based on local coherence. And the global consequences...?
į:
The logic of geometry is based on local coherence. And the global consequences...? are topology?
2018 birželio 20 d., 18:21 atliko AndriusKulikauskas -
Pridėtos 18-19 eilutės:

The logic of geometry is based on local coherence. And the global consequences...?
2018 birželio 20 d., 17:21 atliko AndriusKulikauskas -
Pakeistos 15-17 eilutės iš
Pullback (and limits) are the basis for making distincions.
į:
Pullback (and limits) are the basis for making distinctions.

What are the adjoints for the push forward?
2018 birželio 20 d., 17:20 atliko AndriusKulikauskas -
Pridėtos 14-15 eilutės:

Pullback (and limits) are the basis for making distincions.
2018 birželio 20 d., 17:18 atliko AndriusKulikauskas -
Pridėtos 12-13 eilutės:

Pullback is crucial because its adjoints are exists and for all.
2018 birželio 20 d., 16:25 atliko AndriusKulikauskas -
Pridėtos 10-11 eilutės:

Presheaf category is the total information. Topos is a weakened, limited set of information.
2018 birželio 20 d., 16:17 atliko AndriusKulikauskas -
Pridėtos 8-9 eilutės:

A Grothendieck topos is a reflective subcategory of a presheaf category such that the reflection functor preserves finite limits.
2018 gegužės 28 d., 13:52 atliko AndriusKulikauskas -
Pakeista 78 eilutė iš:
* Category Theory: Topos Logic by Dr.Marni Sheppeard
į:
* [[https://www.youtube.com/watch?v=XRIJpFKijwM | Category Theory: Topos Logic by Dr.Marni Sheppeard]]
2018 gegužės 28 d., 11:38 atliko AndriusKulikauskas -
Pakeistos 77-78 eilutės iš
į:
Videos
* Category Theory: Topos Logic by Dr.Marni Sheppeard
2018 balandžio 15 d., 12:08 atliko AndriusKulikauskas -
Pridėtos 28-31 eilutės:

What: Set of properties. Why: Set of relationships as dictaded by its relationships but especially its relationship with itself (its essence).

What is left unspoken: Sets are labelled - elements are labels. Categories are unlabelled only structure. Is this related to model theory?
2018 balandžio 15 d., 12:06 atliko AndriusKulikauskas -
Pridėtos 26-27 eilutės:

Key points: Every object has a morphism to itself. Every set function has at least one element in its range.
2018 balandžio 13 d., 11:17 atliko AndriusKulikauskas -
Pridėtos 32-33 eilutės:

F(A) is a set of labels.
2018 balandžio 13 d., 11:06 atliko AndriusKulikauskas -
Pridėtos 30-31 eilutės:

If F is the empty functor, or if F(A) is empty, then there are no natural transformations.
2018 balandžio 13 d., 10:35 atliko AndriusKulikauskas -
Pridėtos 7-9 eilutės:
>><<

-----------
Pridėta 11 eilutė:
-----------
2018 balandžio 13 d., 10:35 atliko AndriusKulikauskas -
Pridėtos 2-5 eilutės:

>>bgcolor=#FFFFC0<<

* Explain how substitution of eigenvalues into symmetric functions can add information. And relate this to the Yoneda lemma where associativity adds information when we go from objects (node-variables) to morphisms (matrix edges).
2018 balandžio 13 d., 10:24 atliko AndriusKulikauskas -
Pridėtos 20-21 eilutės:

Hom(A,–) gives the perspectives from A, including perspectives on perspectives.
2018 balandžio 12 d., 11:33 atliko AndriusKulikauskas -
Pridėtos 2-3 eilutės:

Attach:topos.png
2018 balandžio 11 d., 13:12 atliko AndriusKulikauskas -
Pakeista 13 eilutė iš:
Attach:YonedaLemma.png
į:
%center%Attach:YonedaLemma.png
2018 balandžio 11 d., 13:12 atliko AndriusKulikauskas -
Pridėtos 16-17 eilutės:

The properties of an entity correspond to the analogous stances: "This is the essence of the entity, the property that makes it what it is."
2018 balandžio 10 d., 09:08 atliko AndriusKulikauskas -
Pakeista 13 eilutė iš:
į:
Attach:YonedaLemma.png
2018 balandžio 09 d., 09:22 atliko AndriusKulikauskas -
Pakeistos 15-16 eilutės iš
Yoneda Lemma
į:
'''Yoneda Lemma'''
Pakeista 48 eilutė iš:
* [[https://en.wikipedia.org/wiki/History_of_topos_theory | History of topos theory]]
į:
* [[https://en.wikipedia.org/wiki/History_of_topos_theory | Wikipedia: History of topos theory]]
Pakeistos 50-51 eilutės iš
* [[https://en.wikipedia.org/wiki/Category_theory | Category theory]]
į:
* [[https://en.wikipedia.org/wiki/Category_theory | Wikipedia: Category theory]]
* [[https://en.wikipedia.org/wiki/Yoneda_lemma | Wikipedia: Yoneda lemma
]]
2018 balandžio 08 d., 13:41 atliko AndriusKulikauskas -
Pakeista 19 eilutė iš:
Let F be an arbitrary functor from C to Set. Then Yoneda's lemma says that: For each object A of C, the natural transformations from hA to F are in one-to-one correspondence with the elements of F(A). That is,
į:
Let F be an arbitrary functor from C to Set. Then Yoneda's lemma says that: For each object A of C, the natural transformations from {$h^A$} to F are in one-to-one correspondence with the elements of F(A). That is,
2018 balandžio 08 d., 13:40 atliko AndriusKulikauskas -
Pakeista 21 eilutė iš:
{${Nat} (h^{A},F)\cong F(A). \mathrm {Nat} (h^{A},F)\cong F(A)$}
į:
{${Nat} (h^{A},F)\cong F(A)$}
2018 balandžio 08 d., 13:40 atliko AndriusKulikauskas -
Pakeista 21 eilutė iš:
{${Nat} (h^{A},F)\cong F(A).} {\mathrm {Nat}}(h^{A},F)\cong F(A)$}
į:
{${Nat} (h^{A},F)\cong F(A). \mathrm {Nat} (h^{A},F)\cong F(A)$}
2018 balandžio 08 d., 13:39 atliko AndriusKulikauskas -
Pridėtos 18-23 eilutės:

Let F be an arbitrary functor from C to Set. Then Yoneda's lemma says that: For each object A of C, the natural transformations from hA to F are in one-to-one correspondence with the elements of F(A). That is,

{${Nat} (h^{A},F)\cong F(A).} {\mathrm {Nat}}(h^{A},F)\cong F(A)$}

Moreover this isomorphism is natural in A and F when both sides are regarded as functors from SetC x C to Set.
2018 balandžio 08 d., 13:09 atliko AndriusKulikauskas -
Pakeista 35 eilutė iš:
{$R → X × _{X / R}X {\displaystyle R\to X\times _{X/R}X\,\!} R \to X \times_{X/R} X \,\!$}
į:
{$R → X × _{X / R}X$}
2018 balandžio 08 d., 13:09 atliko AndriusKulikauskas -
Pakeista 35 eilutė iš:
{$R → X × X / R X {\displaystyle R\to X\times _{X/R}X\,\!} R \to X \times_{X/R} X \,\!$}
į:
{$R → X × _{X / R}X {\displaystyle R\to X\times _{X/R}X\,\!} R \to X \times_{X/R} X \,\!$}
2018 balandžio 08 d., 13:08 atliko AndriusKulikauskas -
Pridėtos 9-14 eilutės:
A topos is a category that has the following two properties:
* All limits taken over finite index categories exist.
* Every object has a power object.

Pridėtos 18-37 eilutės:

'''Grothendieck topos'''

Let C be a category. A theorem of Giraud states that the following are equivalent:
* There is a small category D and an inclusion C ↪ Presh(D) that admits a finite-limit-preserving left adjoint.
* C is the category of sheaves on a Grothendieck site.
* C satisfies Giraud's axioms, below.

A category with these properties is called a "(Grothendieck) topos". Here Presh(D) denotes the category of contravariant functors from D to the category of sets; such a contravariant functor is frequently called a presheaf.

Giraud's axioms for a category C are:
* C has a small set of generators, and admits all small colimits. Furthermore, colimits commute with fiber products.
* Sums in C are disjoint. In other words, the fiber product of X and Y over their sum is the initial object in C.
* All equivalence relations in C are effective.

The last axiom needs the most explanation. If X is an object of C, an "equivalence relation" R on X is a map R→X×X in C such that for any object Y in C, the induced map Hom(Y,R)→Hom(Y,X)×Hom(Y,X) gives an ordinary equivalence relation on the set Hom(Y,X). Since C has colimits we may form the coequalizer of the two maps R→X; call this X/R. The equivalence relation is "effective" if the canonical map

{$R → X × X / R X {\displaystyle R\to X\times _{X/R}X\,\!} R \to X \times_{X/R} X \,\!$}

is an isomorphism.
2018 balandžio 08 d., 12:50 atliko AndriusKulikauskas -
Pridėta 11 eilutė:
* The functor category {$D^C$} has as objects the functors from ''C'' to ''D'' and as morphisms the natural transformations of such functors. The Yoneda lemma describes representable functors in functor categories.
2018 balandžio 08 d., 11:33 atliko AndriusKulikauskas -
Pridėta 16 eilutė:
* [[https://en.wikipedia.org/wiki/Category_theory | Category theory]]
2018 balandžio 08 d., 11:33 atliko AndriusKulikauskas -
Pridėta 15 eilutė:
* [[https://en.wikipedia.org/wiki/Glossary_of_category_theory | Wikipedia: Glossary of category theory]]
2018 balandžio 08 d., 11:32 atliko AndriusKulikauskas -
Pakeista 10 eilutė iš:
* The Yoneda Lemma asserts that {$C^{op}$} embeds in {${\textbf{Set}}^C$}SetC as a full subcategory.
į:
* The Yoneda Lemma asserts that {$C^{op}$} embeds in {${\textbf{Set}}^C$} as a full subcategory.
2018 balandžio 08 d., 11:32 atliko AndriusKulikauskas -
Pridėtos 8-10 eilutės:

Yoneda Lemma
* The Yoneda Lemma asserts that {$C^{op}$} embeds in {${\textbf{Set}}^C$}SetC as a full subcategory.
2018 balandžio 08 d., 11:14 atliko AndriusKulikauskas -
Pridėtos 1-13 eilutės:
See: [[Category theory]]

Topos:
* Category that behaves like the category of sheaves of sets on a site (topological space).
* Behaves much like the category of sets in that they possess a notion of localization.
* A generalization of point-set topology.
* Grothendieck topoi used in algebraic geometry. More general topoi used in logic.

Sources
* [[https://en.wikipedia.org/wiki/Topos | Wikipedia: Topos]]
* [[https://en.wikipedia.org/wiki/History_of_topos_theory | History of topos theory]]

#### Topos

Naujausi pakeitimai

 Puslapis paskutinį kartą pakeistas 2019 balandžio 16 d., 12:40


text:
# Primes to 1

1. Jun 13, 2008

### Staff: Mentor

reciprocals of primes summed to 1

As the other thread about sum of primes started it reminded me about the idea I had long ago.

My starting point was Erathostenes sieve. It occured to me that multiples of 2 make half of all natural numbers, multiples of 3 make 1/3 of all natural numbers and so on. And as multiplies of prime numbers have to cover all natural numbers, correctly constructed sum of their reciprocals must equal 1. This is not as obvious as it seemed to me at first, as some multiplies of 2 and 3 will be calculated twice, so it has to be 1/2+1/3-1/(2*3) - but it still can be done.

Let P be set of all prime numbers. Let's define some more sets:

$$\forall {a, b} \in P, a < b \rightarrow ab \in P_2$$

$$\forall {a, b, c} \in P, a < b < c \rightarrow abc \in P_3$$

and so on P4, P5...

each of these sets has elements

$$p_{1i} \in P, p_{2i} \in P_2, p_{3i} \in P_3 ...$$

when combined

$$\sum {\frac 1 p_{1i} } - \sum {\frac 1 p_{2i} } + \sum {\frac 1 p_{3i} } - \sum {\frac 1 p_{4i} } ... = 1$$

(note: could be I am misusing notation, what I mean is "if a < b < c then abc is a member of set P3 - that's a way of using each combination of three primes only once, could be it can be done much simpler; please remember I am a chemist )

Last edited: Jun 13, 2008
2. Jun 13, 2008

### gel

What you've called Pk is simply the set of integers which is a product of k distinct primes. Then the identity you have written is (almost) the same as expanding the following expression
$$\prod_{p \rm{\ prime}}\left(1-\frac{1}{p}\right) = 0,$$
which is true.

You have to be careful when you expand this though. That's because the series you get is not absolutely convergent, so the order in which you sum it is important.
In fact, in your expression, each of the summations is infinite, so it doesn't make sense as it is written.

3. Jun 13, 2008

### Staff: Mentor

So if I understand you correctly I should write it as

$$\lim_{n\rightarrow \infty} (\sum^n \frac{1}{p_{1i}} - \sum^n \frac{1}{p_{2i}} + \sum^n \frac{1}{p_{3i}} - \sum^n \frac{1}{p_{4i}} + ...) = 1$$

But product version is much more elegant :grumpy:

4. Jun 13, 2008

### gel

I'm not sure about how you have written it. If you let Pk,N be the products of distinct primes less than N, then write your sum using these, then let N go to infinity, then it will converge 1.

You might want to look at the http://en.wikipedia.org/wiki/Riemann_zeta_function" [Broken] and, in particular the Euler product formula,
$$\sum_{n\geq 1}\frac{1}{n^s} = \prod_{p \text{ prime}} \frac{1}{1-p^{-s}}.$$
or, the reciprocals
$$\left(\sum_{n\geq 1}\frac{1}{n^s}\right)^{-1} = \prod_{p \text{ prime}} (1-p^{-s}).$$

If s>1 you can expand the right hand side, and it should vanish as s->1.

Last edited by a moderator: May 3, 2017


text:
For Robinhood users affected by the Live Shutdown:

Those of us with less than $10k have been live trading with Robinhood. With the pending Quantopian shutdown, it is evident that any alternative means for algorithmic trading is not going to include a connection with Robinhood. Although workarounds supposedly exist, I researched Interactive Brokers to see if it is possible to switch instead. 1) The minimum deposit is$10k, but after that, there is no minimum balance. Feel free to withdrawal $back out after opening the account. 2) How to get around the T+3 (soon to be T+2), similar to Robinhood instant. Open a margin account with IB, and you'll get the same benefits, and even better benefits as RH Instant provides (except for the immediate availability of$1k on deposits).
3) The minimum balance with a margin account is $2k. I personally spoke with a representative about these items. 5 responses It's another$20-30 a month for access to tickers and a per/share commission fee. You sort of get a refund on the access fees through commissions, but in the end you'll still pay each month.

If anyone is interested, I'm working with some people on potentially developing an alternative.

Send me an email at [email protected] or message me here if you're interested in being part of it or contributing to the development in some fashion. Thanks!

Hey @Andrew, I'm interesting in developing an alternative. Let me know what needs to be done.

@Andrew:

Why not lend your abilities toward a solution that is already in progress? Such as zipline live? It already is almost 1:1 compatible with quantopian algorithms that are being stranded and I"m sure if more people work on it they can get pipeline and robinhood integration.

@Daniel Koh

Send me an email and we can talk about potential plans


text:
Revision history [back]

You can go back-and-forth from sage to sympy, i.e. turn your expression into a sympy one then ask sympy to factor it, then transform it back into a sage object:

sage: i = integrate(exp(h)*exp(x)*f(x),x)
sage: i
integrate(e^(h + x)*f(x), x)

sage: i._sympy_().factor()._sage_()
e^h*integrate(e^x*f(x), x)

You can go back-and-forth from sage to sympy, i.e. turn your expression into a sympy one then ask sympy to factor it, then transform it back into a sage object:

sage: i = integrate(exp(h)*exp(x)*f(x),x)
sage: integrate(exp(h)*exp(x)*f(x),x) ; i
integrate(e^(h + x)*f(x), x)

sage: i._sympy_().factor()._sage_()
e^h*integrate(e^x*f(x), x)


text:
# The Cantor's attic library

Step up the ladder towards wisdom, photo by Sigfrid Lundberg

This is our central repository for references cited on Cantor's attic.

This page is still in experimental stages, but it will be up and working soon.

1. Corazza, Paul. The Wholeness Axiom and Laver sequences. Annals of Pure and Applied Logic pp. 157--260, October, 2000. bibtex
2. Corazza, Paul. The gap between {${\rm I}_3$} and the wholeness axiom. Fund Math 179(1):43--60, 2003. www   DOI   MR   bibtex
3. Hamkins, Joel David and Lewis, Andy. Infinite time {T}uring machines. J Symbolic Logic 65(2):567--604, 2000. www   arχiv   DOI   MR   bibtex
4. Hamkins, Joel David. Infinite time Turing machines. Minds and Machines 12(4):521--539, 2002. (special issue devoted to hypercomputation) arχiv   bibtex
5. Hamkins, Joel David. Supertask computation. Classical and new paradigms of computation and their complexity hierarchies23:141--158, Dordrecht, 2004. (Papers of the conference Foundations of the Formal Sciences III'' held in Vienna, September 21-24, 2001) www   arχiv   DOI   MR   bibtex
6. Hamkins, Joel David. The wholeness axioms and V=HOD. Arch Math Logic 40(1):1--8, 2001. www   arχiv   DOI   MR   bibtex
7. Hamkins, Joel David. Tall cardinals. MLQ Math Log Q 55(1):68--86, 2009. www   DOI   MR   bibtex
8. Kanamori, Akihiro. The higher infinite. Second, Springer-Verlag, Berlin, 2009. (Large cardinals in set theory from their beginnings, Paperback reprint of the 2003 edition) bibtex
9. Apter, Arthur and Gitman, Victoria and Hamkins, Joel David. Inner models with large cardinal features usually obtained by forcing. Archive for Mathematical Logic (To appear) bibtex
10. Welch, Philip. The Lengths of Infinite Time {Turing} Machine Computations. Bulletin of the London Mathematical Society 32(2):129--136, 2000. bibtex
11. Welch, Philip. Eventually Infinite Time {Turing} Machine Degrees: Infinite Time Decidable reals. Journal of Symbolic Logic 65(3):1193--1203, 2000. bibtex


text:
# What is the easiest way to find the maximum of $9\cos(t) - 8\cos(3t)$?

What is the easiest way to find the maximum of $9\cos(t) - 8\cos (3t)$?

-
What have you tried so far? –  Rahul Oct 28 '10 at 2:06
The standard first derivative test worked for me. You may want to look up the triple angle formulas if you get stuck. If you know some complex variables, you can derive these fairly easily with DeMoivre's formula and Euler's formula –  WWright Oct 28 '10 at 2:12

Express $\cos(3t)$ in terms of $\cos(t)$ and differentiate with respect to $t$ or $\cos(t)$.