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# Differentiation on Smooth Manifolds without Metric

1. Nov 27, 2008

### dx

Hi,

I'm confused about what differentiation on smooth manifolds means. I know that a vector field $$v$$ on a manifold $$M$$ is a function from $$C^{\infty}(M)$$ to $$C^{\infty}(M)$$ which is linear over $$R$$ and satisfies the Leibniz law. This should be thought of, I'm told, as a 'derivation' on smooth functions on the manifold, i.e. differentiation. Intuitively, a vector field is a field of little arrows on the manifold, and its action on a function is just the directional derivative of the function in the direction of the tangent vector at that point.

What I don't quite understand is what it means to differentiate on a space without any notion of distance. To take a simpler example, consider a vector space $$R^2$$ (not the euclidean vector space with metric, $$E^2$$). What is a smooth function on a vector space? What is differentiation of smooth functions on a vector space without any other structure?

2. Nov 27, 2008

### morphism

But there is a notion of distance, at least locally. After all, an n-manifold is locally like R^n. This is the point of defining a smooth structure on a manifold: it gives us a notion of differentiability. Think about this as you would a topology: without a topology, it doesn't make sense to talk about continuity.

3. Nov 27, 2008

### dx

What is distance on $$R^n$$? There are many possible ways to define a metric on $$R^n$$; it doesn't in the mathematical sense come equipped with one.

4. Nov 27, 2008

### quasar987

When no metric on R^n is specified, it it implicit that we mean the euclidean one...

In particular, when we say a manifold is a locally euclidean topological space, we mean is it locally homeomorphic to R^n, given the topology of the euclidean norm.

5. Nov 28, 2008

### OrderOfThings

Assume you have a two-dimensional smooth manifold $M$, a smooth function $f$ and a vector (field) $\mathbf{v}$. And you would like to calculate the directional derivative of $f$ at some point $p$.

Since it is a manifold you have a coordinate system $(x,y)$ around the point and since the function is smooth you also have the partial derivates $\partial f/\partial x,\, \partial f/\partial y$ in this coordinate system. A directional derivative $L_vf$ is then a linear combination of these partial derivatives. If the vector has coordinates $(dx,dy)$ in the coordinate system (strictly: in the induced tangent space coordinate system), then the directional derivative is
$$L_vf=\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy[/itex]. 6. Dec 3, 2008 ### dx As far as I can tell, the definition of a manifold doesn't involve any metric, implicit or explicit. It is only required that it is locally like the vector space $R^n$. Anyway, I figured out what my problem was. I was thinking of the directional derivative as [tex] \nabla f(a) \cdot v$$ instead of as

$$\nabla_v f(a) = \lim_{t \rightarrow 0} \frac{1}{t}[ f(a + tv) - f(a) ]$$.

The latter definition is covaraint with respect to linear transformations (homeomorphisms are locally linear transformations) and requires only vector space structure, while the former needs a notion of dot product.

Last edited: Dec 3, 2008
7. Dec 3, 2008

### quasar987

A manifold is required to be locally like R^n as a topological space, not as a vector space! (What would that mean anyway?)

And the topology implicit in this definition is the standard one, in which distances are measured according to the law of Pythagoras.

8. Dec 3, 2008

### Hurkyl

Staff Emeritus
And a differentiable manifold is required to be locally like R^n as a differentiable manifold. And a smooth manifold is required to be locally like R^n as a smooth manifold.

The standard topology doesn't really say much about distances.

9. Dec 3, 2008

### dx

The topology of a topological space X is defined by the set of open sets T, no notion of distance is needed.

10. Dec 3, 2008

### quasar987

True, but I find it important to keep in mind that a manifold is a topological space locally like R^n with the topology produced by the familiar euclidean metric. So a manifold is a possibly alien being, but locally familiar.

11. Dec 3, 2008

### morphism

I'm not sure I buy that the standard topology on R^n doesn't have much to do with distances. It does come from a metric after all (from a norm even, and any norm at that).

12. Dec 3, 2008

### mathwonk

why not just read the definition in any book on manifolds?

13. Dec 4, 2008

### dx

You can use the Euclidean metric to define the open sets, but once you have the open sets you forget about the metric. The set of open sets T is what defines the topology of the space. There are many metrics which produce the same topology, so the standard topology on R^n has nothing to do with the Euclidean metric in particular. The point is that we only need an idea of whether two points are "nearby", not how far apart they are.

14. Dec 4, 2008

### mathwonk

even on the real line derivatives have nothing to do with a metric, since you divide deltay by delta x. i.e. changing the metric by a scalar multiple does not affect it. it does relate to the affine structure.

but thats because there you want the derivative to be a number. on a manifold a derivative is a linear map between to abstract vector spaces of equivalence classes of curves, and no metric is needed.

or of you prefer, one does refer to a metric on R^n but the specific metric does not matter, only the result of the process, which can be done with a different metric, equivalent up to diffeomorhism.

really you should read a standard book on this.

15. Dec 4, 2008

### Doodle Bob

...what he said. In fact, you can get a free copy of one of those standard books here: http://www.math.harvard.edu/~shlomo/ (Advanced Calculus) Check out p. 373.

16. Dec 5, 2008

### mathwonk

play with this: a manifold is a topological ,space equipped with a family of real valued functions on each open set that behave like diffble functions (closed under lin ear combinations..). Moreover assume that each point p has a nbhd U and a homeomorphism of that nbhd with a ball B around the origin in R^n making the family of diffble functions on U correspond to the family of all smooth functions on B.

then define a map f from one manifold to another to be diffble if for each point p in the domain, the pullback of every function diffble near f(p) is diffble near p.

for such a function we will define the derivative as follows: first we define a tangent vector at p as an equivalence class of parametrized curves through p, where two curves are equivalent iff composing both of the m with every smooth function at p gives two functions defined near t=0 and having the same derivative at 0.

then check that diffble functions between manifolds preserve equivalence of curves, and define the derivative at p of f, as the map induced by f on equivalence classes of curves at p.

hows that? is the metric sufficiently obscured in that definition for your taste?

Last edited: Dec 5, 2008
17. Dec 6, 2008

### OrderOfThings

Well, yes, but the notation $a+tv$ is a bit problematic. You are adding a vector to a point, so strictly speaking you are going off on a tangent. Instead you can restate the definition by first choosing any curve $\gamma(t)$ such that $\gamma(0)=a,\, \gamma'(0)=v$ and then defining the derivative to be

$$\nabla_v f(a) = \lim_{t \rightarrow 0} \frac{f(\gamma(t)) - f(\gamma(0))}{t}.$$

That said, you are in good company using this kind of notation; Arnold speaks of "bent vectors" in some place and even Cartan are doing similar things. So maybe there are some hidden ideas to be understood here.

But note that although the concept of directional derivative is perfectly valid without a metric, you need a metric to make it really useful. This is since then you can normalize the direction vector which makes it possible to compare the derivative value in different directions. You might even calculate the direction of, say, the steepest descent.

Without a metric the only interesting direction is the one for which $\nabla_v f(a)=0$, since this will always evaluate to zero even if v is rescaled.

Last edited: Dec 6, 2008
18. Dec 6, 2008

### dx

Actually, I was working on a chart, so i was adding a vector to a vector. In fact, this definition doesn't depend on the which chart you use, since the transition functions are smooth, and therefore locally linear transformations.

19. Dec 6, 2008

### dx

That's not correct if you're using the standard definition of a derivative as the limit of a difference quotient. $$\Delta y$$ doesn't depend on the metric on the domain of the function, but $$\Delta x$$ does, so changing the metric changes the derivative. If you use the covariant definition of derivative, then the metric doesn't matter. See post #6.

The meaning of the part in bold is what I was confused about, and it was resolved in post #6.

Last edited: Dec 6, 2008
20. Dec 6, 2008

### Hurkyl

Staff Emeritus
As a notational aside....

Suppose you want to specify in an expression a tangent vector v along with the point P upon which it's based. I think the most common way to do it is something like (P, v), but I've found that coopting + for this purpose is extremely useful.

Definition: If P is a point on a manifold, and v is a tangent vector at P, then the expression P+v is the corresponding point of the tangent bundle.

This notation has various useful features. My favorite is, given a differentiable map f of smooth manifolds, you have this relationship between the pushforward on the tangent bundle and the derivative:
$$f_*(P + v) = f(P) + f'(P) v$$
and, generally speaking, notationifies the intuition that tangent vectors capture the idea of an infinitessimal.

I vague recall getting some use out of this for lie groups/algebras, but I don't remember exactly how it worked out.


text: - Interior-point solver for convex separable block-angular problems Jordi Castro (jordi.castroupc.edu) Abstract: Constraints matrices with block-angular structures are pervasive in Optimization. Interior-point methods have shown to be competitive for these structured problems by exploiting the linear algebra. One of these approaches solved the normal equations using sparse Cholesky factorizations for the block constraints, and a preconditioned conjugate gradient (PCG) for the linking constraints. The preconditioner is based on a power series expansion which approximates the inverse of the matrix of the linking constraints system. In this work we present an efficient solver based on this algorithm. Some of its features are: it solves linearly constrained convex separable problems (linear, quadratic or nonlinear); both Newton and second-order predictor-corrector directions can be used, either with the Cholesky+PCG scheme or with a Cholesky factorization of normal equations; the preconditioner may include any number of terms of the power series; for any number of these terms, it estimates the spectral radius of the matrix in the power series (which is instrumental for the quality of the preconditioner). The solver has been hooked to SML, a structure-conveying modelling language based on the popular AMPL modeling language. Computational results are reported for some large and/or difficult instances in the literature: (1) multicommodity flow problems; (2) minimum congestion problems; (3) statistical data protection problems using $\ell_1$ and $\ell_2$ distances (which are linear and quadratic problems, respectively), and the pseudo-Huber function, a nonlinear approximation to $\ell_1$ which improves the preconditioner. In the largest instances, of up to 25 millions of variables and 300000 constraints, this approach is from two to three orders of magnitude faster than state-of-the-art linear and quadratic optimization solvers. Keywords: interior-point methods; structured problems; normal equations; preconditioned conjugate gradient; large-scale optimization; optimization software Category 1: Optimization Software and Modeling Systems Category 2: Convex and Nonsmooth Optimization (Convex Optimization ) Citation: J. Castro, Interior-point solver for convex separable block-angular problems, Research Report DR 2014/03, Dept. of Statistics and Operations Research, Universitat Politècnica de Catalunya, Barcelona, 2014. Download: [PDF]Entry Submitted: 10/08/2014Entry Accepted: 10/08/2014Entry Last Modified: 11/04/2014Modify/Update this entry Visitors Authors More about us Links Subscribe, Unsubscribe Digest Archive Search, Browse the Repository Submit Update Policies Coordinator's Board Classification Scheme Credits Give us feedback Optimization Journals, Sites, Societies Optimization Online is supported by the Mathematical Optmization Society.

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# Is there a data structure that can find the kth smallest in constant time with logarithmic add and delete operations?

I'm looking for a single or a conjunction of data structures that can find the kth smallest element in constant time, delete the kth smallest element in logarithmic time, and add a new element in logarithmic time. The closest I've found to this is the order statistic tree which achieves logarithmic add and delete operations but has logarithmic time complexity for accessing the kth smallest element.

Any ideas along these lines would be much appreciated.

Clarification: kth smallest means that for any variable k value (non-constant), the data structure should be able to retrieve that value

• Is a hash map not acceptable? It is O(1) for all the things you mentioned: en.wikipedia.org/wiki/Hash_table – hLk Jun 7 '19 at 0:37
• @KhanPower The access time I'm referring to in the question is to get the value at an index for a list that is in sorted order. Because a hash table mixes up the ordering, getting the ith element aka the ith smallest/largest element would require resorting – MathGeek Jun 7 '19 at 0:45
• @MathGeek What is that current problem you are trying to solve? Can you add a reference to that problem or explain that problem? – John L. Jun 7 '19 at 1:47
• Would you settle for amortized $O(1)$ complexity? – Andrej Bauer Aug 7 '19 at 22:10
• en.m.wikipedia.org/wiki/Skip_list#Indexable_skiplist is pretty close to what you want. – ryan Apr 4 at 1:49

An array is the only thing that will get you consistent lookup in constant time.

Removal & Insertion time $$\in O(n)$$

If you can fit your data to a model (I.e. a polynomial) and the size of your elements remains somewhat constant you can have almost constant lookup with log addition/removal

• I understand where you are coming from. But to me, even if an array is the only possible way to get constant lookup, I can't really say definitively that we wouldn't be able to lower the removal and insertion down to logarithmic time with some trickery (some pretty clever data structures can be implemented in an array alone like an array based heap for example which has many logarithmic operations). This would be different if you could somehow prove that you couldn't. As for the model fitting, I can't really make assumptions about the data. Also isn't the insertion into an array linear time? – MathGeek Jun 7 '19 at 3:55

Kth smallest number in a dynamic stream of numbers. Idea from the above source.

Have two heaps, one a max heap with $$k$$ elements (all the elements less than the $$k$$th largest element and the $$k$$th largest element itself are here) and the rest of the elements are in the other min heap.

1. Inserting a new element: If the new element is less than the top of the max heap (which is the current $$k$$th largest element), then insert it into the max heap, after that (I mean to say after the heapification following the insertion) pop the top element and insert it into the min heap. If the new element is greater than or equal to the current $$k$$th largest insert it into the min heap.

2. Deletion: If the element is in the min heap (ie., greater than equal to the $$k$$th largest) then simply delete from the min heap*. Else delete the element from the max heap and pop the top of the min heap and insert it into the max heap.

3. Look up: simply output the top of the min heap. You'll have to suitably handle the corner case where there aren't enough elements ie. the total number of elements are less than $$k$$.

Please look into deletion of element in $$O(\log n)$$ time. It's a little complicated – you've gotta keep track of the positions of the elements in the heap so that you have $$O(\log n)$$ if you use a set (in C++) or $$O(1)$$ amortized if you use hash tables/unordered set. After finding the element in heap you've gotta replace it with the last element in the heap and let it sift down or up, which is $$O(\log n)$$.

• This answer assumes $k$ is fixed for the lifetime of the data structure, which the OP has clarified to not be the case. – Reinstate Monica Aug 7 '19 at 18:00


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### Still have math questions?

Suppose that $$\ 2000$$ is placed in a savings account at an annual rate of $$72 \%$$ , compounded monthly. Assuming that no withdrawals are made, how long will it take for the account to grow to $$\ 3058$$ ? Do not round any intermediate computations, and round your answer to the nearest hundredth.
2000$$\times$$(1.072)$$^{x}$$=3058


text:
# 8.7 Logarithmic Functions

Logarithms are the inverse of exponential functions – they allow us to undo exponential functions and solve for the exponent. They are also commonly used to express quantities that vary widely in size.

Logarithm Equivalent to an Exponential

The logarithm (base $b$) function, written $\log_b (x)$, is the inverse of the exponential function (base $b$), $b^x$.

This means the statement $b^a=c$ is equivalent to the statement $\log_b (c)=a$.

#### Example 1

Write these exponential equations as logarithmic equations:

1. $2^3=8$
2. $5^2=25$
3. $10^{-4}=\frac{1}{10000}$

1. $2^3=8$ is equivalent to $\log_2(8)=3$.
2. $5^2=25$ is equivalent to $\log_5(25)=2$.
3. $10^{-4}=\frac{1}{10000}$ is equivalent to $\log_{10}\left(\frac{1}{10000}\right)=-4$.

#### Example 2

Solve $2^x=10$ for $x$.

By rewriting this equation using a logarithm, we get $x=\log_2(10)$.

While this does define a solution, and an exact solution at that, you may find it somewhat unsatisfying since it is difficult to compare this expression to the decimal estimate we made earlier. Also, giving an exact expression for a solution is not always useful – often we really need a decimal approximation to the solution. Luckily, this is a task calculators and computers are quite adept at. Unluckily for us, most calculators and computers will only evaluate logarithms of two bases, base 10 and base $e$. Happily, this ends up not being a problem, as we’ll see briefly.

Common and Natural Logarithms

The common log is the logarithm with base 10, and is typically written $\log(x)$.
The natural log is the logarithm with base $e$, and is typically written $\ln(x)$.

#### Example 3

Evaluate $\log(1000)$ using the definition of the common log.

To evaluate $\log(1000)$, we can say $x=\log(1000)$, then rewrite into exponential form using the common log base of 10:$10^x=1000.$
From this, we might recognize that 1000 is the cube of 10, i.e., $10^3=1000$, so $x = 3$.
We also can use the inverse property of logs to write $\log_{10}\left(10^3\right) =3$.

 Number Number as exponential log(number) 1000 $10^3$ 3 100 $10^2$ 2 10 $10^1$ 1 1 $10^0$ 0 0.1 $10^{-1}$ -1 0.01 $10^{-2}$ -2 0.001 $10^{-3}$ -3

#### Example 4

Evaluate $\log(500)$ using you calculator or computer.

Using a computer or calculator, we can evaluate and find that $\log(500)\approx 2.69897$.

Because logarithms are inverse functions of exponentials, we can, in particular, use them to solve exponential equations, where the unknown is in the exponent, in much the same way as we use the inverse properties of addition vs. subtraction or multiplication vs. division when rearranging equations to solve for an unknown value.

Properties of Logarithms

Inverse Properties:

(1) $\log_b(b^x)=x$

(2) $b^{\log_b(x)}=x$

Exponent Property

(3) $\log_b\left(A^r\right)=r\,\log_b(A)$

Here is an example how these properties are useful:

Suppose we have the following equation and had to determine the value of $r$:

$2\cdot 1.8^x=17.2$

To solve for $x$, we first have to determine which operations, and in which order, are acting on $x$, and then undo them in the opposite order. Because $x$ is first put into the exponent of 1.8, and then the new value is multiplied by 2, we first have to undo the last operation, i.e., we must divide by 2:

$1.8^x=\frac{17.2}{2}$

Looking at the equation now, we can see that the only thing happening to $x$ is that $x$ is put into the exponent of 1.8. The opposite operation would be the inverse property (1), so we could apply $log_{1.8}$ to get just $x$. However, the typical scientific calculator only calculates the common and the natural log, so $log_{1.8}$ won’t be helpful when applied to the other side of the equation. So instead we will apply either $\log$ or $ln$ to both sides of the equation, and then bring the exponent down by using the exponent property (3):

\begin{align*} \ln(1.8)^x&=\ln\left(\frac{17.2}{2}\right)\Rightarrow x\ln(1.8)=\ln\left(\frac{17.2}{2}\right)\\ &\Rightarrow x=\frac{\ln\left(\frac{17.2}{2}\right)}{\ln(1.8)}\approx 3.660788 \end{align*}

General process for solving exponential equations:

1. Isolate the exponential expressions when possible.
2. Take the logarithm of both sides.
3. Utilize the exponent property for logarithms to pull the variable down from the exponent.
4. Continue to use algebra to solve for the variable.

#### Example 5 – solving for a variable in the exponent

In the last section, we predicted the population (in billions) of India $t$ years after 2008 by using the function $f(t)=1.14(1+0.0134)^t$. If the population continues following this trend, when will the population reach 2 billion?

Answer: $t =?$, $f(t) = 2$, $f(t)=1.14(1+0.0134)^t$

 Initial equation. $1.14(1.0134)^t=2$ Divide by 1.14 to isolate the exponential expression. $1.0134^t=\dfrac{2}{1.14}$ Take the logarithm to both sides of the equation. $\ln\left(1.0134^t\right)=\ln\left(\dfrac{2}{1.14}\right)$ Apply the exponent property on the right side. $t\cdot \ln(1.0134)=\ln\left(\dfrac{2}{1.14}\right)$ Divide both sides by $\ln(1.0134)$ $t = \dfrac{\ln\left(\dfrac{2}{1.14}\right)}{\ln(1.0134)}$ Calculate. $t\approx 42.23 \text{ years}$

If this growth rate continues, the model predicts the population of India will reach 2 billion about 42 years after 2008, or approximately in the year 2050.

#### Example 6 – solving for a variable in the exponent

Solve $5e^{-0.3t}=2$ for $t$.

 State the initial equation. $5e^{-0.3t}=2$ Divide by 5 to isolate the exponential. $e^{-0.3t}=\frac{2}{5}$ Take the natural log of both sides. $\ln\left(e^{-0.3t}\right)=\ln\left(\dfrac{2}{5}\right)$ Use the inverse property for logs (2). $-0.3t=\ln\left(\dfrac{2}{5}\right)$ Divide by -0.3. $t = \dfrac{\ln\left(\dfrac{2}{5}\right)}{-0.3}$ Calculate. $t\approx 3.054$

In addition to solving exponential equations, logarithmic expressions are very useful in compound interest questions.

#### Example 7 – compound interest

Let’s consider an example from the previous section: Suppose an investment of $2,000 at the annual rate of 1.89% can be compounded quarterly and continuously. Determine when the total amount, including interest, will reach$2,500 in each of the two cases of compounding.

Answer: $t=?$

Compounding quarterly: $A_q(t)=P\left(1+\frac{r}{m}\right)^{mt}=2000\left(1+\frac{0.0189}{4}\right)^{4t}$

\begin{align*} A_q(t)&=2500\Rightarrow 2000\left(1+\frac{0.0189}{4}\right)^{4t}=2500\\ &\Rightarrow\left(1+\frac{0.0189}{4}\right)^{4t}=\frac{2500}{2000}\\ &\Rightarrow\ln\left(\left(1+\frac{0.0189}{4}\right)^{4t}\right)=\ln\left(\frac{2500}{2000}\right)\\ &\Rightarrow 4t\cdot\ln\left(1+\frac{0.0189}{4}\right)=\ln\left(\frac{2500}{2000}\right)\\ &\Rightarrow t=\frac{\ln\left(\frac{2500}{2000}\right)}{4\cdot \ln\left(1+\frac{0.0189}{4}\right)}\\ &\Rightarrow t\approx 11.834 \end{align*}

Since 11.834 years is between 11 years and three quarters and 12 years, the time it will for the investment to reach 2500 by compounding quarterly is 12 years. Compounding continuously: $A_c(t)=Pe^{rt}=2000e^{0.0189t}$ \begin{align*} A_c(t)&=2500\Rightarrow 2000e^{0.0189t}=2500\\ &\Rightarrow e^{0.0189t}=\frac{2500}{2000}\\ &\Rightarrow\ln\left(e^{0.0189t}\right)=\ln\left(\frac{2500}{2000}\right)\\ &\Rightarrow 0.0189t\ln(e)=\ln\left(\frac{2500}{2000}\right)\\ &\Rightarrow 0.0189t=\ln\left(\frac{2500}{2000}\right)\\ &\Rightarrow t=\frac{\ln\left(\frac{2500}{2000}\right)}{0.0189}\\ &\Rightarrow t\approx 11.807 \end{align*} Since 11.807 years is approximately 11 years and $0.807(12)$ months, the time it will for the investment to reach2500 by compounding continuously is 11 years and 10 months.

While we don’t often need to sketch the graph of a logarithm, it is helpful to understand the basic shape.

Graphical Features of the Logarithm

Graphically, given the function $g(x)=\log_b(x)$.

• The graph has a horizontal intercept at (1, 0).
• The graph has a vertical asymptote at $x = 0$.
• The graph is increasing and concave down.
• The domain of the function is $x \gt 0$, or $(0, \infty)$ in interval notation.
• The range of the function is all real numbers, or $(-\infty, \infty)$ in interval notation.

When sketching a general logarithm with base $b$, it can be helpful to remember that the graph will pass through the points $(1, 0)$ and $(b, 1)$.

To get a feeling for how the base affects the shape of the graph, examine the graphs below:

Another important observation made was the domain of the logarithm: $x \gt 0$. Like the reciprocal and square root functions, the logarithm has a restricted domain which must be considered when finding the domain of a composition involving a log.

#### Example 8 – domain of a logarithmic function

Find the domain of the function $f(x)=\log(5-2x)$.

The logarithm is only defined when the input is positive, so this function will only be defined when $5-2x \gt 0$.Solving this inequality, $-2x \gt -5$, so $x\lt \frac{5}{2}$.The domain of this function is $x\lt \frac{5}{2}$, or, in interval notation, $\left(-\infty, \frac{5}{2} \right)$.


text:
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# Powers: Basic Rules - Raising a Power to Another Power

$$(3 \times 10^{-1}) \times (3 \times 10^{-1}) =$$
Incorrect. [[snippet]]
Incorrect. [[snippet]]
Correct. [[snippet]] >$$(3 \times 10^{-1}) \times (3 \times 10^{-1}) = (3 \times 10^{-1})^2$$ Split the power to individual bases, creating a power in scientific form. >$$(3 \times 10^{-1})^2=3^2 \times 10^{-2}$$ >$$3^2 \times 10^{-2}=9 \times 10^{-2}$$ Since the 10 part has an exponent of -2, shift the decimal point two places to the **left**. >$$9 \times 10^{-2}=0.09$$
Incorrect. [[snippet]]
Incorrect. [[snippet]]
$$0.001$$
$$0.009$$
$$0.09$$
$$0.1$$
$$0.9$$


text:
Margin Explainer

# What is Margin

Margin is the collateral that you need to post when entering into a leveraged derivatives contract. The amount required to enter into a new position is referred to as Initial Margin. The Initial Margin for a position depends up on both the position size and the selected leverage.
If the trade against you, the unrealised loss in your position is adjusted against the initially posted margin. When the remaining margin (Initial Margin - Unrealised Loss) goes below Maintenance Margin, liquidation process is triggered.

# Risk Limits: Margin Requirement vs. Position Size

The Initial Margin and Maintenance margin requirements for any contract are increased as the position size increases. Increasing the margin requirement helps to ensure orderly liquidation of large positions and reduces the incidences of auto-deleveraging.
Margin requirement remains flat till a certain position size (Position Threshold). Once position size crosses this threshold, margin requirement increases linearly with position size.
When Position size is less than or equal to Position Threshold:
$Initial\ Margin\% = Initial\ Margin\%_{MIN}$
$Maintenance\ Margin\% = Maintenance\ Margin\%_{MIN}$
and, when Position size is greater than Position Threshold:
$Initial\ Margin\% = Initial\ Margin\%_{MIN} + Slope_{IM} * (Position\ Size - Position\ Threshold )$
$Maintenance\ Margin\% = Maintenance\ Margin\%_{MIN} + Slope_{MM} * (Position\ Size - Position\ Threshold )$
The values of
$Initial\ Margin\%_{MIN}$
,
$Maintenance\ Margin\%_{MIN}$
, Position Threshold and Slope vary from contract to contract and are available in the contract specifications.

# Isolated Margin

Delta Exchange follows an Isolated Margin approach in which margin is explicitly assigned to each position and is not shared across positions. However, cross-margin like functionalities can be achieved by enabling auto margin top up.
A position is specific to a particular contract. Thus, for each contract, two margin sub-accounts are maintained:
• Position Margin: Margin allocated to all existing (may include multiple long/ short positions) in given derivatives contract
• Order Margin: Margin allocated to all open orders in a given derivatives contract
It is worth noting that Unrealized PnL is not factored into Position Margin or Order Margin.
Position Margins and Order Margins are allocated from your Wallet Balance. Thus, at any time, the amount that remains unallocated is what is available for placing a new order. This is referred to as Available Balance.
$Available\ Balance = Wallet\ Balance - (Position\ Margin + Order\ Margin)$
Every time a new order is placed, the system does three things:
1. 1.
computes the Reservation margin for this new order,
2. 2.
checks whether
$Reservation \ Margin <= Available \ Balance$
, and
3. 3.
if
$2$
holds, then transfer Reservation Margin amount from Available Balance to the Order Margin of the contract.

# Reserved Margin Computation

Initial Margin (IM) requirements for a standalone order are as follows:
$IM = (Initial\ Margin\% * \#Contracts * Multiplier * Limit\_Bid\_Price)$
$IM = (InitialMargin\% * \#Contracts * Multiplier * MarkPrice)$
• Sell limit order
$IM = (InitialMargin\% * \#Contracts * Multiplier * Max (Limit\ Offer\ Price, BestBid)$
• Sell market order
$IM = (InitialMargin\% * \#Contracts * Multiplier * Max (MarkPrice, BestBid)$
Now, if there are existing positions/ open orders in the contract, then the Initial Margin requirement for the new combination of position and open orders is recomputed. For this computation, positions or orders on opposite side are netted in such a manner that for two offsetting orders, margin is reserved only once.
Reserved Margin is then the difference of the Initial Margin requirement for the combined position (existing + new order) and the Position Margin and Order Margin currently allocated to the derivative contract.

# Impact of order cancellations on margin

In a given contract, Order Margin account contains margin blocked for all the current open orders. If one or more of these open orders is cancelled, the Initial Margin requirement for the remaining open orders and existing orders is recomputed. The new Initial Margin requirement will either be same as earlier or lower. In the latter case, excess margin is released.


text:
# 1  Creating Networks

From the simplest two host network to complex enterprise or carrier networks spanning the globe a network still comes down to nodes (switches and hosts) connected by links. One of the most fundamental network design and deployment decisions comes down to what to build/own and what to rent/lease. Typically we don't get much choice on whether we get to own or rent links or nodes, but we can have a wide choice in what technologies to use.

Once we need our network to extend beyond a single building we generally have to use other peoples networks to create the links we need. For example carriers offer many different WAN services as this example network map the the carrier Centurylink shows.

## 1.1  Issues with using other peoples networks

### 1.1.1 Incompatibility of address types

1.  Example: A company uses private IPv4 addresses but wants to interconnect sites across the public internet.   For more detail: RFC1918 specifies private IPv4 address ranges.

2. Related Example: Multiple users within a private address space need to share a singe public IP address → NAT (Network address translation).  See Wikipedia and IETF RFC2663, section 4.1.2 "Network Address Port Translation (NAPT)" is what we usually just call NAT.

“NAPT extends the notion of translation one step further by
also translating transport identifier (e.g., TCP and UDP
port numbers, ICMP query identifiers). This allows the
transport identifiers of a number of private hosts to be
multiplexed into the transport identifiers of a single
external address. NAPT allows a set of hosts to share a
single external address.”

### 1.1.2 Compatibility of signal types

Example: I want to support a legacy telephone system and the only network infrastructure available is IP based. A standardized technique for doing this over IP or MPLS is called psuedowires

• Pseudowire architecture RFC3985

• Structure-Aware Time Division Multiplexed (TDM) Circuit Emulation Service over Packet Switched Network CESoPSN. See section 5.2 as an illustration of a straight forward way of packing DS0 signals into a packet. Also use this to illustrate compromises between latency and efficiency.

### 1.1.3 Incompatibility of layers

Example: I want to extend an Ethernet LAN between two sites and the only network available between this is an IP network.

# 2 Adaptation, Layering, and Tunneling

As part of normal network operations we expect to add layers of extra functionality above the link and network layer protocols we use. In fact both Ethernet Frames and IP datagrams contain explicit areas in their headers to tell the reciever what type of payload they are carrying. See Figures 1 and 2. In addition the next layer protocol may need to do some extra work to aid in the transmission over the lower level. We generally call this process adaptation.

From RFC7042 we get a sampling of Ethertypes:

   0x0800  Internet Protocol Version 4 (IPv4)
0x0808  Frame Relay ARP
0x22F3  TRILL
0x22F4  L2-IS-IS
0x8035  Reverse Address Resolution Protocol (RARP)
0x86DD  Internet Protocol Version 6 (IPv6)
0x880B  Point-to-Point Protocol (PPP)
0x880C  General Switch Management Protocol (GSMP)
0x8847  MPLS
0x8848  MPLS with upstream-assigned label
0x8861  Multicast Channel Allocation Protocol (MCAP)
0x8863  PPP over Ethernet (PPPoE) Discovery Stage
0x8864  PPP over Ethernet (PPPoE) Session Stage
0x893B  TRILL Fine Grained Labeling (FGL)
0x8946  TRILL RBridge Channel

From IANA here are a sampling of IPv4 protocol numbers:

1   ICMP    Internet Control Message
2   IGMP    Internet Group Management
3   GGP     Gateway-to-Gateway
4   IPv4    IPv4 encapsulation
5   ST      Stream
6   TCP     Transmission Control
7   CBT     CBT
8   EGP     Exterior Gateway Protocol
9   IGP     any private interior gateway
10  BBN-RCC-MON     BBN RCC Monitoring
11  NVP-II  Network Voice Protocol
12  PUP     PUP
13  ARGUS   ARGUS
14  EMCON   EMCON
15  XNET    Cross Net Debugger
16  CHAOS   Chaos
17  UDP     User Datagram

With PPP we saw the type of effort needed to put packets over bit or byte streams, i.e., we needed to introduce some framing to delimit the start and end of the packet and such. What about if we want to carry old fashioned voice DS0 circuits over a packet network? Couldn't we just gather a bunch of bytes from the DS0s and put them into a packet? Actually that is kind of what is done (Figures 3 and 4).

One tricky part of putting "circuits" over "packets" is timing recovery, i.e., making sure that the bytes get played out at the right speed at the other end of the network. From the above diagram we see that the CES service uses the Real-time Transport Protocol to help with this function.

From looking at the the previous headers for CES one should wonder about the efficiency of such an approach. When we talk about the efficiency of a digital communication scheme we generally mean the ratio of payload bytes (or bits) transported to the total bits transported.

Example 1: Consider an old fashioned T1 frame that carries 24 DS0 bytes along with 1 bit of synchronization overhead. The efficiency of such a scheme is:

$Efficiency = \frac{24 \times 8}{24 \times 8 + 1} = 99.482\%$

Example 2: Consider an Ethernet frame with 1200 bytes of payload. Assuming no padding we have 7 octets of preamble, 1 octet for SFD, 6 octets for destination address, 6 octets for source address, 2 octets for type, and 4 octets for the frame check sequence. This gives a total of 26 bytes of overhead. Hence the efficiency for this packet transmission is:

$Efficiency = \frac{1200 bytes}{1200 bytes + 26 bytes} = 97.88\%$

What if the payload was only 100 bytes?

$Efficiency = \frac{100 bytes}{100 bytes + 26 bytes} = 79.37\%$

For better efficiency we want to have large payloads for our packets. It looks like our old fashioned T1 line is more "efficient" than modern Ethernet transmission? Or is more analysis and context needed to make the assesment?

### 2.2.1 Latency in Real Time Communications

For real time communications we have an additional issue that can get in the way of efficiency and that is latency.

Example: For a 64 Kbps voice channel how long does it take to buffer up 100 bytes for transmission? What about 1000 bytes? Answer:

$time = \frac{100 bytes \times 8 bits/byte}{64Kbps} = 12.5msec$

$time = \frac{1000 bytes \times 8 bits/byte}{64Kbps} = 125.0msec$

For comparison the approximate distance from the US to India is 13600km and the speed of light is approximately 300,000km/s which would give a delay of approximately 45.3msec. Hence to achieve the highest efficiency in filling an Ethernet packet with DS0 voice samples would result in a delay equivalent to about three times the fiber optic transmission delay from the US to India. Hence we see that for real time communications there is a trade off between efficiency and latency.

## 2.3 What's Tunneling?

As we've seen with "normal" protocol layering, we frequently encapsulate payloads into additional "containers" to help "adapt" that payload to the link or network layer packet format. For example shown in Figures 5 and 6 are the UDP and TCP transport layer headers used with the IP network layer.

So how is tunneling different from the usual encapsulation and adaptation that we see in our layer protocol stacks? In general we use the term tunneling when the payload being encapsulated is a packet (or signal) from a layer equal to or lower than that the one we will be using to carry that packet or signal. We'll see or have seen examples of the following:

• Physical layer — DS0s — over IP
• Link Layer — Ethernet — over IP
• IP — say with private address — over IP — with public addresses

Typical network architecture with tunneling providing interconnection:

Network Address Translation (NAT), Half a Tunnel?: Somewhat similar to tunneling is NAT, Figure 8, which from a network diagram looks like half a tunnel.

### 2.3.1 Middleboxes

In the tunnel network diagram, Figure 7, we added special boxes, labeled TE, to denote the tunnel end points. Such processing may take place in a separate piece of network equipment or may be combined with other network equipment such as a router. In either case the tunneling functionality is very different from routing. In a home router typically NAT, Firewall, IP routing, Ethernet Switching, and wireless access point are all combined in the same physical box.

Things like tunnel endpoints, NAT, and Firewalls may exist in the IP "path" between source and destination and have different functionality than IP routers. At the IETF such entities are called middleboxes. RFC3234 discusses why these have become so important to the architecture of the internet and describes approximately twenty different types of middleboxes.

# 4 Virtual Private Networks

A nice set of definitions for VPNs and the problem they solve comes from RFC3809 1, June 2004:

   Corporations and other organizations have become increasingly
dependent on their networks for telecommunications and data
communication.  The data communication networks were originally built
as Local Area Networks (LAN).  Over time the possibility to
interconnect the networks on different sites has become more and more
important.  The connectivity for corporate networks has been supplied
by service providers, mainly as Frame Relay (FR) or Asynchronous
Transfer Mode (ATM) connections, and more recently as Ethernet and
IP-based tunnels.  This type of network, interconnecting a number of
sites over a shared network infrastructure is called Virtual Private
Network (VPN).  If the sites belong to the same organization, the VPN
is called an Intranet.  If the sites belong to different
organizations that share a common interest, the VPN is called an
Extranet.

## 4.1 Network Contexts

The network context — who is connecting to whom, for what reason, etc... — greatly influences the types of problems encountered in creating a VPN and the technologies utilized.

One decomposition of the VPN networking problem from 2:

• Remote Access VPN:
• Site-to-Site VPN:
• Intranet VPNs connecting sites within an organization
• Extranet VPNs connecting two different organizations

Another decomposition could be based on who provisions the VPNs:

• Provider Provisioned VPNs (PPVPN)
• The bulk of recent standard activitions at the IETF and IEEE have be focused on PPVPNs.
• Customer or Enterprise Provisioned VPNs
• Based on user configured tunnels
• Based on enterprise configured IEEE 802.1Q (Ethernet) VLANs

Still another decomposition can be based on the layer that the VPN provides service. At one point in time the IETF had three different working groups for layer 1, 2, and 3 VPNs respectively.

Finally when the VPN is offered by a service provider (carrier) they may offer different types of services such as:

• Virtual Private Wire Service (VPWS) — like the point to point tunnels we've looked at.
• Virtual Private LAN Service (VPLS) — which takes care of getting an Ethernet frame to the appropriate location as if all locations were on a single LAN.

## 4.2 Ethernet VLANs

Fundamental ideas of Ethernet VPNs:

1. Allow multiple Virtual “LANs” (really bridged networks) to share the same physical infrastructure. With each VLAN given a unique Identifier, known as a VID (VLAN identifier)

2. Restrict forwarding and broadcasting to individual VLANs, i.e., provides traffic isolation within a common network.

3. Enhances learning bridges to understand VLAN tags and such.

Examples: separate departments within a company, multiple tenant office spaces, multiple client applications within a data center.

### 4.2.1 VLAN Implementation Outline

The first thing we need to enable VLANs is a place in the Ethernet frame header to put the VLAN ID (VID). Recall the Ethernet frame structure shown in Figure 12.

It seems that there isn't any space for a VID, i.e., all the fields are used. The trick that that IEEE adopted was to define new Ethertypes to support VLANs and then appending the "regular" Ethertype, i.e., the one that indicates the upper layer protocol as part of the payload as shown in Figure 13.

After the destination and source address a “tag header” is inserted and the FCS is recomputed. The header now consists of the Tag Protocol identifier (TPID) — essentially a new EtherType — and Tag Control Information (TCI).

Figure 14 shows a list of the VLAN EtherTypes (from IEEE 802.1Q-2011, this is 1365 page document):

In addition to the VID = VLAN Identifier the Tag Control Information (TCI) field contains congrestion and QoS related subfields: DEI = Drop Eligible Indicator and PCP=Priority Code Point as shown in Figure 15.

Formally 802.Q describes VLAN functionality via an additional layer called the “Enhanced Internal Sublayer Service”.

### 4.2.2 How does Frame "tagging" work?

Who inserts VLAN tags onto packets?

Switches do!  Switches can be configured to assign VIDs based on port or protocol.

How are VLANs configured in a typical case?

1. Two fundamental VLAN switch port types: “tagged” and “untagged”.

2. Some ports will be “access” ports

3. Some ports will be “trunk” (switch-to-switch ports)

4. Configure “access” ports as “untagged” and set the PVID (port VLAN ID) to the VID of the VLAN that the host should belong to.  Untagged frames from the host will get the PVID added when entering the bridged network. Tagged frames for the host will have their tags removed.

5. Configure “trunk” ports as “tagged”.  Such ports can belong to multiple VLANs and the tags are needed to restrict forwarding.

6. How can we get traffic to pass between VLANs?  (IP) Route it!

1. Create a “tagged” port on one of the switches and configure it to belong to all VLANs that you want to route between.

2. Hook up a “lollipop” (aka “one-armed”) router to this multi-VLAN port.

3. Note that most Linux distributions have support for IEEE 802.Q tagging. In current Ubuntu this configured with the vconfig command that comes with the vlan package. Hence you could create an inexpensive lollipop router out of Linux.

Example low cost VLAN capable switch:

http://www.newegg.com/Product/Product.aspx?Item=N82E16833704093

### 4.2.3 Bandwidth Efficiency via MSTP

Ethernet VLANs help us achieve traffic isolation and restrict broadcast domains but what about better link utilization? The problem with bridged networks from a bandwidth/link utilization perspective is that they are tree topology networks! Consider the network shown in Figure 16 where the circles denote hosts and the color represents the logical grouping of the hosts, i.e., we want to put the blue and orange colored hosts into separate VLANs.

Ethernet spanning tree protocol prunes links that would result in loops, i.e., they are not used, and could give us a logical connectivity for this network as shown in Figure 17. A key property of tree topologies is that there is only one path between any two nodes. Looking at Figure 17 we see that for some of the blue hosts to communicate with their peers will require paths traversing up to four links and this traffic will cross links used by the orange nodes to communicate.

The multiple spanning tree protocol (MSTP) feature in 802.1Q allows us to assign different spanning trees to different VLANs. In our example network we could come up with different trees for the orange and blue VLANs as shown in Figure 18 where the different link colors are used to denote the two different trees in use and the the VLANs using them.

The advantage of this VLAN to tree assignment is even more obvious when we prune off links in each VLAN tree that will not be carrying traffic for that VLAN as shown in 19 where we can see that each link in this network is carrying traffic for only one VLAN. As we will see a bit more when we study SDNs that general mesh networks possess a great many trees and that trees play an important role in different traffic forwarding paradigms. In addition, SDNs will give us the most freedom to choose forwarding trees.


text:
# Using of matrices in simulink block diagrams

All simulations I've done in simulink as yet have not matrix in their block diagram like the following picture :

and they have block diagram structure like :

for simulating a system that has matrices with variable parameters in matrice ,in its structrue, how can use matrices as block diagram in it? for example for following system:

I can't set and adapt the dimesnsion of blocks.

If you want to build a block that has matrix operations there are to blocks that will let you do this, the Fcn block and the MATLAB function block, you can

You can carry through matrices with signals as shown below. The example shows a 1x3 vector multiplied by itself and displayed in the scope as a 3x3 matrix.

Care must be taken to ensure size is preserved for the matrix operation

But the math you have shown above does not make sense, as you should either be creating two states (for example $x_4$ should be split into two states) OR you should be carrying the $x_4$ through to simplify $\dot{x_4}$ to a scalar in your analysis of the system. Usually states are scalars at least with my experience with nonlinear control.

$$\left( \begin{bmatrix} u_2 \\ u_3 \end{bmatrix}+ \begin{bmatrix} \frac{-l K_4}{l_1} & 0 \\ 0 & \frac{-l K_4}{l_1} \end{bmatrix} \right) x_4$$

• thanks for your help although I don't understand solution for conditions that parameters of matrices are state variables. can you explain more? – Rwy5 Jan 25 '18 at 19:56
• What do you want me to explain? You haven't defined your problem properly and you haven't taken the time to upvote. Where did you get the system you described above? – Voltage Spike Jan 25 '18 at 19:58

To perform matrix multiplication in simulation is as simple as using a gain block.

A gain block doesn't have to be a scalar, it can be a matrix.

• Thanks for your help, but my problem is the conditions, when elements of matrix are not scalar and they're variables such as the degree of pitching or rolling. what is your suggestion about this? – Rwy5 Jan 27 '18 at 16:22


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## Introduction

Breast cancer remains the most commonly diagnosed type of cancer and the second most common cause of cancer-related death in women1,2. Improved patient survival requires a multidisciplinary approach, with specialists in pathology, medical, surgical, and radiation oncology3. Clinical management and accurate diagnosis of breast cancer type, staging, and grade require tissue biopsies. Biopsy specimens must be processed onto slides and stained using immunohistochemical (IHC) staining procedures4. Hematoxylin and Eosin (H&E) is a common stain4 used to identify nuclei and cytoplasm; Hematoxylin stains nuclei blue; Eosin stains the cytoplasm pink4,5,6. Examination of tissue biopsy specimens, prepared onto slides, is currently considered the gold-standard clinical diagnosis of cancer6. A recent update to the American Joint Committee on Cancer (AJCC) breast cancer staging manual7 now also incorporates histological features, such as grade and receptor status, into the prognostic staging process. However, the manual process of histopathological analysis is laborious, time-consuming, and limited by the quality of the specimen and the experience of the pathologist8,9. As a result, there has been growing interest in automating and digitizing pathological workflows, in addition to using computer-aided diagnosis (CAD) to streamline and optimize the process of tissue analysis8.

The emerging field of digital pathology has been made possible with the advent of high-resolution whole slide image (WSI) scanners10 and has further been driven by advancements in imaging technologies, data storage, and computer vision algorithms11. The high-resolution WSIs, obtained by scanning histopathological specimens, allow for in silico analysis. The application of artificial intelligence (AI) – in particular, its subfields of machine learning (ML) and deep learning (DL) – for in silico analysis are proving to be promising tools10. Researchers and data scientists in AI have applied convolutional neural networks (CNN) and machine classifiers to segment and classify objects, predict disease diagnosis, and treatment response; to tailor medical treatment, develop diagnostic assays, and determine patient response to treatment therapies10. For example, Wang et al.,12 demonstrated that in combining a CNN and random forest classifiers, mitotic nuclei could accurately be detected in H&E stained breast tissue. Digital pathology aims to develop clinically relevant computational methods to enhance specimen characterization and automate the detection of conventional features such as nuclear pleomorphism (e.g. nuclear shape and size) and tissue-spatial characteristics (e.g. cellular distribution)13.

Recent developments in CNN architectures represent the rapidly growing field of DL14. CNNs, trained on sizeable supervised image datasets, have been used to develop state-of-the-art learning algorithms15, capable of carrying out classification and segmentation tasks for medical image analysis. However, a significant challenge with advancing these algorithms includes the limited availability of supervised histopathological datasets. A dataset review is presented in Table 1 and outlines publicly available supervised histopathological datasets16,17,18,19,20,21,22,23,24,25. These datasets feature expert annotations of histopathological structure from a variety of tissues and staining techniques. A number of these datasets contain breast tissue samples; however, the method in which the histopathological structures were annotated is not consistent. Of the seven datasets that contain breast tissue, two datasets17,21 were not exhaustively annotated, and two other datasets18,23 contained annotations where the borders of the nuclei were not outlined. In general, training CNNs for semantic or instance segmentation requires exhaustive and full annotations of specific histopathological structures19.

U-Net26 is an example of a deep CNN (DCNN) that provides semantic segmentations. The network was first developed by Ronneberger et al., building on the fully convolutional network27 (FCN). U-Net improved on the FCN by reducing training time and enhancing localization with the addition of long skip connections and up-sampling layers to the encoder-decoder architecture. Since the introduction of U-Net, it is now recognized that the network can yield state-of-the-art segmentation results28,29,30,31.

Innovative segmentation algorithms that have gained widespread attention further include Mask regional convolutional neural network32 (Mask R-CNN), a two-stage DCNN. He et al.,32 introduced Mask R-CNN to provide instance segmentation, expanding on faster regional convolutional neural network33 (Faster R-CNN). At a high-level Mask R-CNN localizes each object instance to a bounding box then segments each instance.

Many recent state-of-the-art DCNN architectures, together with network weights pre-trained on large supervised image repositories, have been made publicly available for researchers and data scientists to implement. However, one of the restrictive factors in breast pathomics is the limited availability of supervised histopathological datasets. Previous studies34,35 and computer vision competitions22,36 have tested the generalizability of novel DCNN on multi-organ and multi-organism datasets; however, no study has evaluated if DCNNs trained on a variety of organs can accurately segment breast tumor nuclei. Therefore, this study aimed to determine if DCNNs, trained on histology images independent of breast tissue, can accurately segment tumor nuclei of the breast. Two DCNN architectures, one providing semantic segmentation (U-Net) and one providing instance segmentation (Mask R-CNN), will be evaluated. Additionally, classical segmentation methods will segment tumor nuclei of the breast, and the results compared to those of the DCNNs.

## Methods

### Training datasets

The training dataset implemented in this study featured open-source images from the Multi-Organ Nucleus Segmentation (MoNuSeg) challenge16,37. The MoNuSeg dataset, obtained from National Cancer Institute’s cancer genome atlas38 (TCGA), included high-resolution WSI of H&E stained slides from nine tissue types, digitized in eighteen different hospitals at 40 × magnification16. The types of tissue included breast, liver, kidney, prostate, bladder, colon, stomach, lung, and brain. Sub-regions (1,000 × 1,000 pixel), densely populated with nuclei, were extracted from the WSI. This study used the ground truth annotations released during the challenge. These ground truth annotations included all epithelial and stromal cells, which were annotated by students, then approved by an expert pathologist, with less than a 1% error in detection16. There were approximately 28,000 nuclei annotated across the entire dataset. The training set for this study consisted of colour normalized39 H&E images from all tissue types, excluding breast.

### Testing datasets

The hold-out (test) set exclusively contained H&E-stained images of breast tissue. There were fifty-eight images in total from two separate datasets. Eight images were from the MoNuSeg dataset and fifty from the Triple Negative Breast Cancer (TNBC) dataset19. The ground truth annotations associated with these images were curated and released with the respective dataset. The TNBC dataset contained fifty H&E stained images taken at 40 × magnification from eleven TNBC patients19. Three to eight sub-regions (512 × 512 pixel), with varying cellularity, were extracted per patient. The annotations were performed by three expert pathologists. Annotated cells included: normal epithelial cells, myoepithelial breast cells, invasive carcinoma cells, fibroblasts, endothelial cells, adipocytes, macrophages, and inflammatory cells. As the cell class was unavailable, with respect to both datasets, performance measures were based on segmenting all annotated cells within the test images.

All images were colour normalized39, then tiled to a size of 256 × 256 pixels with a 50% overlap. Post-processing the probability map, from any given U-Net like DCNN, involved calculating a threshold that maximized precision at a given intersections over union (IoU) threshold. Furthermore, we implemented morphological operations that filled missing pixels and removed small predicted artifacts.

### Software and hardware

All software related to this study was written in Python programming language version 3.7.6., using Anaconda (https://www.anaconda.com). Each DCNN was trained and implemented with Keras version 2.3.140 and Tensorflow version 2.1.041. U-Net like architectures were implemented with the segmentation models package (https://github.com/qubvel/segmentation_models) and Mask R-CNN was implemented with the matterport package (https://github.com/matterport/Mask_RCNN). The ImageJ2-Fiji42,43 (Fiji) package (https://imagej.nih.gov/ij/download.html) was applied for nuclei segmentation. Otsu threshold44 and watershed transform45,46 were applied with the scikit-image package (https://scikit-image.org) version 0.17.2. All experiments were performed on a workstation equipped with an AMD (Advanced Micro Devices, Inc., Santa Clara, USA) Ryzen Threadripper 1920X 12-Core Processor, 64 GB of RAM, and a single NVIDIA (NVIDIA Corporation, Santa Clara, USA) GeForce RTX 2080 Ti graphics processing unit (GPU).

### Classical segmentation techniques

Fiji is an open-source image processing and analysis software package that features a built-in nuclei segmentation pipeline. To prepare the images for segmentation, they were first colour normalized39, then converted to gray-scale. The pipeline involved implementing a Gaussian filter, thresholding the images, followed by applying the watershed transform. The classical segmentation techniques (Otsu threshold, Watershed transform, Fiji) applied to the images of this study, were used to compare the performance of DCNNs to such techniques.

### U-net

Eight DCNNs with U-Net like architecture were trained for this study; furthermore, the encoders of the DCNNs were replaced with the following networks: visual geometry group47 (VGG)-16, VGG-1947, Residual Networks48 (ResNet)-50, ResNet-10148, ResNet-15248, Dense convolutional Network49 (DenseNet)-121, DenseNet-20149, and Inception-v350. These DCNN encoders were chosen because of their success in image classification14,51 and their wide use in a range of instance and semantic segmentation tasks52,53,54. Weights pre-trained on the ImageNet55 dataset initialized each DCNN. All DCNN decoders consisted of up-sampling, concatenation of the respective feature map from the encoding path, followed by two blocks of 3 × 3 convolutions, batch normalization56, and rectified linear units57 (ReLU) activation. The final layer of all DCNNs consisted of a 1 × 1 convolution with SoftMax activation.

All U-Net like DCNNs implemented training images with randomly applied combinations of rotation, distortion, skew, shear, and flip augmentations. The normalized training images was passed through an augmentation pipeline in groupings of H&E images, annotations, and weighted maps; ensuring that all corresponding images remained spatially aligned. All images were further tiled to a size of 256 × 256 pixels, with a 10% overlap.

There were 6,720 images used during training, split into a ratio of 80:20 between the training and validation sets, respectively. Adaptive moment estimation58 (Adam) optimizer with a learning rate of 1e-4 was used to fine-tune all U-Net like DCNNs. All DCNNs were trained until convergence. Each DCNN had a combined loss function of weighted cross-entropy26 and soft Dice (Eq. 1):

$${\mathcal{L}}_{combined}= {\mathcal{L}}_{WCE}+ {(1-\mathcal{L}}_{Dice})$$
(1)

Equation (2), which represents weighted cross-entropy26 was formulated as:

$${\mathcal{L}}_{WCE}= -\sum_{\mathcal{X}\in {\mathbb{X}}}{\mathcal{W}}_{map}\left(x\right) \mathrm{log}({p}_{a(x)}(\mathcal{X}))$$
(2)

where $$\mathcal{X}$$ was the image used during training, $$\alpha$$ was the ground truth annotation of each image, across all classes, and $$\mathcal{W}$$ was the weighted map introduced to the loss function. Dice loss (Eq. 3) was formulated as:

$${\mathcal{L}}_{Dice}= \frac{2\sum_{\mathcal{X}\in {\mathbb{X}}}p\left(\mathcal{X}\right)gt\left(\mathcal{X}\right)}{\sum_{\mathcal{X}\in {\mathbb{X}}}({p(x)}^{2}+ {gt\left(\mathcal{X}\right)}^{2})}$$
(3)

where $$\mathrm{\rm P}$$ was the output of the SoftMax activation function, and $$gt$$ was the ground truth annotation.

Prior to training Mask R-CNN, all training images were tiled to a resolution of 256 × 256 pixels with a 10% overlap, for a total of 560 images. The images were further randomly split into training and validation sets at a ratio of 80:20, respectively. The CNN backbone of Mask R-CNN was ResNet-101 and through optimization, Gradient Descent59 was chosen as the optimizer for the network. Furthermore, ResNet-101 was initialized with weights pre-trained on the common objects in context60 (COCO) dataset. Gradient Descent was set to a momentum to 0.9, a learning rate of 1e-3, a weight regularizer of 1e-4, and trained until convergence.

### Ensemble networks

U-Net ensemble calculated the pixel-wise mean of the probability maps from the U-Net like DCNNs with VGG-19, ResNet-101, and DenseNet-121 encoders.

A gradient boosting network (GB U-Net) was developed to improve localization. GB U-Net is a two-stage network where at a high-level, concatenates feature maps from U-Net like DCNNs with VGG-19, ResNet-101, and DenseNet-121 encoders with the colour normalized H&E image, then uses the concatenated images as input to an additional U-Net (Fig. 1). The feature maps concatenated with the H&E image were taken from the last convolutional layer, just before the SoftMax activation. The second stage of GB U-Net was trained with a slightly modified encoder, which featured repeated blocks of 3 × 3 convolutions, batch normalization56, ReLU57 activation, and Dropout61 layers. The decoder followed the same structure as all previous U-Net like DCNNs. GB U-Net was trained until convergence using the 6,720 augmented H&E images, masks, and weighted maps previously mentioned. The network featured a combined loss function of weighted cross-entropy26 and soft Dice, previously outlined.

### Evaluation metrics

This study's evaluation metrics were Aggregated Jaccard Index16 (AJI) and mean average precision (mAP). Kumar et al.,16 proposed AJI, which penalized pixel and object level detection. AJI penalized missed objects, falsely detected objects, along with under and over segmented objects. It did this by computing the ratio of aggregated cardinality of intersection and union of predicted nuclei matched with annotated nuclei. The mAP is a popular metric used in object detection and segmentation tasks. The metric yields average precision over multiple IoU thresholds. Specifically, ten IoU thresholds starting at 0.5 and increasing linearly by 0.05 to 0.95, were used for this study.

## Results

Table 2a,b summarize the results of the DL and classical segmentation methods for the MoNuSeg and TNBC datasets, respectively. Table 2a is composed of three categories, 1) classical segmentation methods (Otsu threshold, watershed transform, Fiji), 2) U-Net like DCNNs with varying encoders, 3) Mask R-CNN and ensemble networks. With an AJI of 0.3396 and mAP of 0.237, Fiji considerably outperformed the Otsu thresholding and watershed transform. However, the DL methods outperformed Fiji across all metrics. In comparing all U-Net like DCNNs, the Densenet-201 encoder scored highest in AJI with 0.5083, while the ResNet-101 encoder scored highest in mAP with 0.3318. With the highest AJI, the Densenet encoder surpassed all U-Net like DCNNs in both object level and pixel-wise segmentation accuracy. In contrast, the ResNet encoder demonstrated that it excelled in accurately classifying nuclei from other histopathological structures. Furthermore, the DenseNet-201 encoder scored highest in F1 at an IoU threshold of 0.5 and recall at IoU thresholds of 0.5 and 0.7; demonstrating that the network excelled in correctly identifying all nuclei, while also provided the best balance in correctly identifying and separating nuclei from other structures. With regards to the TNBC dataset (Table 2b) the ResNet-101 encoder scored highest in AJI with 0.5080, mAP with 0.3306, and AP at both IoU thresholds.

The overall top-scoring network in both datasets was GB U-Net. For the MoNuSeg dataset GB U-Net achieved an AJI of 0.5331 and a mAP of 0.3909, while for the. TNBC dataset the network achieved an AJI of 0.5403 and mAP of 0.3772.

Figure 2 displays AP plotted against the evaluated IoU thresholds for several segmentation methods using the MoNuSeg test dataset. GB U-Net and Mask R-CNN scored highest in this metric. Both networks displayed similar AP values throughout the IoU thresholds; however, GB U-Net displayed superior performance in the lower thresholds, while Mask R-CNN performed better in thresholds greater than 0.65. Both the ResNet-101 encoder and U-Net Ensemble displayed similar mAP scores, and Fig. 2 also demonstrated that both AP values remained similar across all IoU thresholds. A noticeable difference between the networks was that U-Net ensemble scored slightly higher in AP throughout the mid-range of IoU thresholds (0.55–0.8). Additionally, Fig. 2 demonstrated that the DCNNs outperformed the top-scoring classical segmentation method (Fiji) across all IoU thresholds.

Figure 3 displays the predicted maps of the MoNuSeg test images with the lowest and highest combined AJI. The predicted maps were colour coded with green indicating correctly classified pixels (true positive), red indicating pixels not classified as nuclei (false negative), and blue indicating pixels that were miss classified as nuclei (false positive). Row 1 (Fig. 3) provided examples of predicted maps associated with the lowest scoring image. The AJI values of these images are summarized in Table 3. The image was difficult to segment accurately, as large areas of the image displayed densely crowded regions of tumor nuclei. Fiji struggled to correctly separate nuclei within these areas, which resulted in over-segmentation. However, Fiji also under-segmented the surrounding areas. The DCNNs were better able to separate nuclei within the crowded areas; however, the networks tended to over-segment the outer areas. The top-scoring networks, GB U-Net and Mask R-CNN performed similarly on both images; however, under qualitative analysis, the main difference between the networks was that Mask R-CNN tended to over-segment images compared to GB U-Net.

## Discussion

This was the first study that investigated DCNN’s accuracy in segmenting invasive carcinoma of the breast when trained on H&E-stained digital images of liver, kidney, prostate, bladder, colon, stomach, lung, and brain cancers. The eleven DCNNs trained for this study included: eight U-Net like architectures with various encoders, two gradient boosting networks, and Mask R-CNN. Furthermore, the study's three classical segmentation methods included: Otsu threshold, watershed transform, and Fiji. Overall, the DCNNs outperformed all classical segmentation methods. The top-scoring DCNN in AJI and mAP was GB U-Net. Mask R-CNN scored slightly lower; however, it outperformed all U-Net like DCNNs in the MoNuSeg test dataset. Segmentation performance of the DCNNs demonstrated that, in using transfer learning, the networks were able to effectively implement features learned independently of breast tissue to segment tumor nuclei within breast tissue.

The work presented in this study reflects the emerging field of pathomics, which combines digital pathology and novel AI-based software. Overall, pathomics-based approaches provide novel techniques to augment clinical workflows by automating WSI feature extraction for diagnostic, treatment, and prognostic applications. For example, breakthroughs in AI-based technologies have automated histological grading of breast cancer specimens62, mitosis detection12, nuclei pleomorphism segmentation63, and tubule nuclei classification of estrogen receptor (ER) + breast cancers64. Furthermore, DL networks have automated important diagnostic and prognostic factors such as the classification of human epidermal growth factor receptor 2 (HER2) score65, along with ER and progesterone receptor (PR) status66. These molecular markers are fundamental to clinical decision making both at the time of diagnosis and recurrence, if recurrence occurs, to select the most effective therapies. Recent work has also identified certain tumor nuclei morphological features to be independent prognostic factors for tumor size and tubular formation67, alongside features that are significantly associated with eight-year disease-free survival68. Additionally, recent work by Ali et al.,69,70 introduced a prognostic assay that evaluated lymphocyte density as a predictor of pathological complete response (pCR) in breast cancer patients with neoadjuvant chemotherapy regimens. As the field of oncology moves into an era of precision medicine71, novel insights and innovations in tumor nuclei segmentation and feature extraction may lead to more precise breast cancer treatment selection. One such way is AI driven diagnostic and theragnostic tools that leverage DL architectures.

U-Net excels in biomedical semantic segmentation29; however, the network struggles with correctly classifying close or touching objects. In semantic segmentation, each pixel is identified with a class label specific to the object. Object instances within the same class are not separated; therefore, all foreground objects will possess the same class label within binary segmentation. Objects that are close together or overlapping, i.e., areas of crowded nuclei, may be incorrectly classified as a single object. Weighted cross-entropy26 is a loss function used to improve pixel classification of close or touching objects. Pre-calculated weighted maps26, are passed to the network and used to weight the cross-entropy loss. Weighted cross-entropy severely penalizes incorrect pixel classification of close or touching nuclei.

Symmetrical networks such as U-Net’s encoder-decoder architecture further possess an opportunity to modify the network’s structure and improve performance. The original U-Net encoder followed a typical CNN structure, where successive convolution, activation, and pooling layers calculated feature maps. However, the complexity of training DCNNs has resulted in the advancement of network architectures15 and contributed to the development of novel networks such as: VGG47, ResNet48, DenseNet49, and GoogLeNet50. These relatively newer networks have achieved superior state-of-the-art object detection results14 with improvements in feature extraction, minimizing the vanishing gradient, and aiding network convergence. By modifying the network architecture or replacing the standard encoder with a deeper CNN and using U-Nets long skip connections, researchers have improved object and segmentation level performance52,53,72.

Mask R-CNN is a novel network that was proposed to provide instance level annotations. The first stage of Mask R-CNN is identical to region proposal networks33 (RPN) of Faster-RCNN, which proposes class probability maps. The CNN backbone of Mask R-CNN provides the feature map as input for the convolutional layer specific to RPN; the convolutional layer's output then feeds both regression and classification layers. The regression layer predicts region proposals, while the classification layer predicts the probability of an object bound within the proposal. The second stage of Mask R-CNN uses the region proposals to calculate the object class, bounding box offset, and segments the object using the FCN27. Region of interest (RoI) align was introduced with Mask R-CNN to correct the RoI pool's roundoff error, aligning the segmented image with the object. RoI pool, introduced with fast regional convolutional neural network73 (Fast R-CNN), was developed to reduce computational time by taking feature maps, defined by a RoI, and scaling them to a fixed size; this prompted misalignment between the feature map and input image. RoI align incorporated bilinear interpolation to calculate floating-point values of the sampling points, avoiding quantization. Mask R-CNN outputs bounding box coordinates, object class probability score, and a segmented map of each object instance.

ML-based approaches work exceptionally well when the training and testing data are drawn from the same feature space74. In circumstances where data are sparse, transfer learning can transfer knowledge from one domain to a task within an independent domain74. However, it is imperative to ensure that both domains are related to avoid negative transfer. Transfer learning has been extensively implemented by data scientists for tasks such as text classifiers75,76,77. However, this was the first study investigating the use of transfer learning to segment tumor nuclei of breast tissue exclusively, to the author's knowledge. Currently there are minimal open-source annotated breast tissue datasets (Table 1), therefore it is essential to identify if DCNNs can leverage transfer learning to accurately segment breast nuclei from open-source datasets. Previous studies34,35,78 have used the MoNuSeg and TNBC datasets to evaluate the segmentation accuracy of novel DCNNs. The MoNuSeg test dataset of these studies included fourteen tissue samples from seven organs, three (bladder, colon, stomach) of which were withheld from the training set. Specifically, work by Graham et al78., compared a novel DCNN (HoVer-Net) to various other DCNNs. In the first experiment Graham and colleagues trained the DCNNs using the MoNuSeg dataset. In comparing the results of the current study to those of Graham and colleagues, GB U-Net outperformed Cell Profiler79 (AJI: 0.366), QuPath80 (AJI: 0.432), FCN827 (AJI: 0.281), FCN8 + WS (AJI: 0.429), SegNet81 (AJI: 0.377), SegNet + WS (AJI: 0.508), deep contour-aware network (DCAN)82 (AJI: 0.525), and CNN316 (AJI: 0.508). In the second experiment Graham and colleagues evaluated the DCNNs using the TNBC dataset. In comparing their results to the current study GB U-Net outperformed FCN8 + WS (AJI: 0.506), FCN8 (AJI: 0.281), Mask R-CNN (AJI: 0.529), DCAN (AJI: 0.537), Micro-Net83 (AJI: 0.531), and DIST19 (AJI: 0.523). Furthermore, a study by Wang et al34., evaluated various DCNNs trained using the MoNuSeg dataset. In comparing their results to the current study GB U-Net, again, outperformed FCN (AJI: 0.452), U-Net26 (AJI: 0.513), SegNet (AJI: 0.505), and DCAN (AJI: 0.518). Additionally, a study by Liu et al35., used thirteen images from the TNBC dataset to evaluate multiple DCNNs. In comparing their results to the current study GB U-Net outperformed GCN84 (AJI: 0.1907), Mask R-CNN (AJI: 0.5297), and pixel2pixel85 (AJI: 0.4760). However, novel DCNNs including HoVer-Net (AJI: 0.618 & 0.59), Bending Loss34 (AJI: 0.641), and panoptic segmentation35 (AJI: 0.5865) achieved a higher AJI compared to GB U-Net.

Many of the studies mentioned above featured DCNNs developed to provide precise segmentation of tumor nuclei, explicitly improving on segmenting areas of densely populated nuclei. Differentiation of nuclei is essential for spatial features; however, a study by Boucheron et al86., identified that, specific to tumor nuclei classification, perfectly segmented nuclei do not guarantee optimum classification accuracy. Although GB U-Net did not outperform all DCNNs the results demonstrate that transfer learning can be implemented while training DCNNs to segment breast tumor nuclei if precise segmentations are not explicitly required.

A limitation of this study is the relatively small dataset used for training and testing, based on the limited open-source datasets. Future work will involve expanding the number of images and types of tissues included in the dataset. Generative adversarial networks (GANs) are an innovative approach to creating synthetic histopathological images. GANs are a promising approach to data synthesis, in which synthetic H&E stained histopathology images and ground truth images can be generated from features learned from the training data87,88,89. GANs have the potential to significantly increase histopathological training sets without requiring additional annotations from expert pathologists.

## Conclusion

The study's objective was to determine if transfer learning could be implemented with DCNNs to segment tumor nuclei of breast tissue accurately. This study introduced a novel ensemble network (GB U-Net), which scored highest in AJI and mAP. Mask R-CNN scored slightly lower in AJI and mAP, compared to GB U-Net; however, one of the main differences between the networks was that Mask R-CNN displayed improved AP at higher IoU thresholds (greater than 0.65). Overall, this study demonstrated that DCNNs trained on images independent of breast tissue could accurately segment invasive carcinoma of the breast by implementing transfer learning. One of the main limitations facing DL based researchers and data scientists is the limited availability of training and testing data. As it is impractical to expect that expert pathologists provide the time commitment required to produce large-scale histopathological datasets, various boosting and data synthesis options, such as transfer learning and GANs, should be further explored to augment clinical workflows and ultimately lead to improved patient care.


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# Is Elastic Collision possible if 2 bodies approach each other?

Okay, suppose 2 identical bodies/particles(i.e., of equal masses) moving with same velocities approach each other and have an elastic collision.

Now, as there is elastic collision in between them, both of them will rebound and move in opposite direction with same velocity.(Source: Directs to Quora(Refer to Mohd. Faiz Answer))

Here's the doubt: I don't think so there could be any elastic collision when 2 identical bodies with same speed moves towards each other.

Why there is such doubt in my mind? :-

I know I stated Elastic Collision and thus, so to conserve Kinetic Energy, they should rebound with same velocity.

Consider the same identical bodies/particles, A and B approaching each other with same velocity and have an elastic collision. As there is elastic collision, during time of collision, there will be conservative forces acting on the bodies.

Now equal and opposite conservative forces $F_{AB}$ and $F_{BA}$ acts on bodies A and B respectively. The conservative forces will make the bodies to rebound and thus, the bodies will go away from each other, i.e, the bodies will move in direction of conservative forces.

Hence, both the conservative forces $F_{AB}$ and $F_{BA}$ will do Positive Work on particles A and B respectively.

As per Work-Energy Theorem, Work(Force)=Change in Kinetic Energy of the Body

So, Positive Work By Conservative Forces = Increase in Kinetic Energies of bodies

In other words, $u_A < v_A$ and $u_B < v_B$ such that Initial Total KE < Final Total KE

But this couldn't be possible if the collision is Elastic as, in Elastic Collision Total Kinetic Energy remains conserved.

So, where am i going wrong?

Note: I know in reality there cannot be any Elastic Collision at macroscopic level. But here i have assumed that the bodies are perfectly rigid and no ext. force is acting on them. Also the mechanical energy do not converts into any forms

• You realize that those conservative forces that you are talking about are applied on body for a Rally really short time, that's the reason we call the impact of such large forces with such a short interval of time by the name "Impulse" $\Delta (p)$ – The Dead Legend May 13 '17 at 11:28
• Note: You don't need to assume perfectly rigid object to have ideal elastic collisions. A tennis ball hitting a wall is also an elastic collision, since the deformation that happens in the ball is not permanent but returns to exactly the initial state. – Steeven May 13 '17 at 13:12

Same thing with the elastic collision. The collision happens over a finite time $\Delta t$. During the first half of the collision--from $t = 0$ to $t = \Delta t/2$--the force between the balls does negative work to bring the balls to a stop. At this point, the kinetic energy has been converted into the elastic deformation of the balls (elastic in the spring sense of the word). For the second half of the collision--from $t = \Delta t/2$ to $t = \Delta t$--the forces between the balls do positive work to accelerate the balls away from each other. Since the collision force is elastic, the balls have the same kinetic energy going out that they did going in.


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## Nonisomorphic number fields with the same Galois group and discriminant

It is possible for nonisomorphic fields to have the same discriminant, and even to be arithmetically equivalent, meaning with the same Dedekind zeta function. Previous questions which discussed this are "Number fields with same discriminant and regulator?" and "Are there two non-isomorphic number fields with the same degree, class number and discriminant?". The usual way this comes about involves fields with isomorphic subfields. I would find it useful if the discriminant sufficed to characterize number fields whose splitting field had a simple Galois group, so that there were no subfields. So the question is, can there be two nonisomorphic fields with the same degree, the same discriminant and the same simple Galois group for the splitting field?

-

My answer to one of those previous questions provides an example of two Galois cubic fields of the same discriminant. The cyclic group of order three is simple.

Added: Just to give you an idea, if you look in Cohen's book A course in computational algebraic number theory, Theorem 6.4.6 says that if $e$ is the product of 9 times $t-1$ distinct primes that are congruent to 1 mod 3, then there are exactly $2^{t-1}$ cyclic cubic fields of discriminant $e^2$. There's a similar statement for $9\nmid e$.

And more: So using sage/pari, I just found the following example of two non-isomorphic quintic $A_5$-extensions with the same discriminant. Take $f=x^5 - x^2 - 2x - 3$ and $g=x^5 - x^4 + 5x^3 - 3x^2 + 4x - 3$. The associated fields (and the polynomials themselves) have discriminant 243049 (and 1 real place), which is a square. Hence, the Galois group is $A_5, D_5$, or $C_5$. Their factorization mod 3 are $x(x + 1)(x^3 + 2x^2 + x + 1)$ and $x (x + 2) (x^3 + 2x + 2)$, respectively. Hence, both of their Galois groups contain a 3-cycle and hence must be $A_5$.

In fact, these both have class number 1. However, 5 is inert in the second field, but splits into primes of degree 1,1, and 3 in the first field. Their regulators are 10.7855048337065 and 7.68746944439494, respectively.

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 I'm really just interested in the nonabelian case, if someone has an answer for that, but I didn't see how to use the conductor discriminant formula for the abelian case. That it doesn't work then is not encouraging for nonabelian groups. – Gene Ward Smith Sep 17 at 23:20


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# Question 8f73c

Apr 12, 2017

""_1^3H → ""_2^3He^(1+) + e^− + bar v_e

#### Explanation:

A tritium has 1 electron,1 proton and 2 neutrons

so if you are thinking tritium has 3 neutrons only that's wrong.

Morever if you are thinking of an isotope with 3 neutrons only you are going on the wrong way.

And also if you are looking for a substance composed of 3neutrons per nucleus its name is trineutron which is not detected.

Whatsoever the nucleus of a tritium also called triton has 2 neutrons and 1 proton

Look at the structure of an atom of tritium

And this is the nucleus of tritium atom also called triton

In a beta decay (beta^-)) decay a neutron is converted to a proton and releases an electronantineutrino and an electron.

There are two down quarks and one up quark in a neutron. A free neutron is unstable and one downquark in it is converted in a upp quark releasing a W boson

d = u + W^-)

Where d is the down quark
u is the upper quark
W^-) is the W boson

This is followed by the decay of W boson itself

W^-) rarr e^-) +bar ν_e

Where ${e}^{-}$ is electron

${\overline{v}}_{e}$ is an electron electron antineutrino

One the down quark is converted to upp quark the neutron does not stay neutron any more. It changes into a proton because two upp quarks and 1 down quark make up a proton

"udd" → "uud" + e^− + bar ν_e (Beta decay of neutron, quark notation)

Where u is representing one upp quark
d is representing one down quark
${e}^{-}$ is electron
${\overline{v}}_{e}$

So this actually happens in a beta decay

Only one neutron is converted to a proton and no other neutrons are affected

So for the triton

1 neutron is being converted to a proton and releasing a W boson which decaying into an electron and electron antineutrino

So $\frac{1}{2}$ neutrons is being converted to a proton thus resulting decrease of 1 neutron and increase of 1proton

Count how many neutrons and protons are there in triton

neutrons = 2
protons = 1

After decay

neutrons = 2-1 = 1
protons =1+1 = 2

Recognize the nucleus of the element with 2protons and 1electron.
Yes you figured it right,its a helium nucleus

So now time to represent the decay in an equation

""_1^3H → ""_2^3He^(1+) + e^− + bar v_e#

H is hydrogen
He is helium

${e}^{-}$ is electron
${\overline{v}}_{e}$ is electron antineutrino


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## Distribution with pmf and rand. variables.

I posted this in the wrong section before and meant to put it here, so i apologize if you seen this before.

X=demand for the magazine with pmf

x | 1 2 3 4
p(x)| .1 .2 .4 .3

Shop owner pays $1.00 for each copy of mag. and charges$2.00. If mags. left at end of week are not worth anything, is it better to order two, three, or four copies of the mag.?

I know i need to introduce the random variables:
Y_k = # of mags. sold
R_k= the net profit if k mags are ordered.

I am NOT trying to just get the answer out of someone, I just need advice on how to start this..
Do I need to make another pmf for Y_k and R_k? Or do I need to figure out expected value.
just a hint may help me understand this problem
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 It looks like an open-ended question. First step could be to write down the profit for all 16 combinations {(1 bought, 1 sold), (1 bought, 2 sold), ...} perhaps as a 4x4 table.
 i dont think its open ended, because R_k= -1k+2*Y_k since R_k is the profit could i simply find the expected value is 1, 2, or 3 are sold thats it?

## Distribution with pmf and rand. variables.

That's the open-ended part, it's up to you to choose a selection criteria. Expected value is only one of infinitely many possibilities. It's good that you've got a formula for the profit though it's important to actually look at the values and their relative probabilities (for example, with an appropriate chart) otherwise important details can be hidden.
 okay the profit for k=2 i got 3.8 when i am calculating it for when k=3 is this equation correct? -1(3)+2(.1*1+.2*3.8+.4*3.8+.3*3.8) i may be going off a longshot but i used the profit from k=2 for the values of x in this equation. I just want to make sure im doing k=3 right so i can figure out when k=4...

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