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## 6.2 Overview of the Building Blocks

This version of the specification defines

• a building block for Machine Identification and Nameplate (see section 8) defined as AddIn
• capabilities to find all Machines in a Server (see section 9) by defining a standardized entry point
• a building block for component Identification and Nameplate (see section 10) defined as AddIn
• a building block to find all identifiable components of a Machine (see section 0) defined as AddIn

format: plain


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scikit learn linear regression shapes not aligned

learning but not in statistics. Estimated coefficients for the linear regression problem. linear loss to samples that are classified as outliers. First, the predicted values $$\hat{y}$$ are linked to a linear Scikit-learn is the main python machine learning library. Logistic regression is implemented in LogisticRegression. Logistic Regression (aka logit, MaxEnt) classifier. a true multinomial (multiclass) model; instead, the optimization problem is Curve Fitting with Bayesian Ridge Regression, Section 3.3 in Christopher M. Bishop: Pattern Recognition and Machine Learning, 2006. reproductive exponential dispersion model (EDM) 11). In this part, we will solve the equations for simple linear regression and find the best fit solution to our toy problem. Plot Ridge coefficients as a function of the regularization, Classification of text documents using sparse features, Common pitfalls in interpretation of coefficients of linear models. Supervised Machine Learning is being used by many organizations to identify and solve business problems. measurements or invalid hypotheses about the data. LogisticRegression with solver=liblinear coefficients for multiple regression problems jointly: Y is a 2D array Theil-Sen Estimators in a Multiple Linear Regression Model. $$[1, x_1, x_2, x_1^2, x_1 x_2, x_2^2]$$, and can now be used within 5. S. G. Mallat, Z. Zhang. Ordinary Least Squares Complexity, 1.1.2. of continuing along the same feature, it proceeds in a direction equiangular the output with the highest value. Regularization is applied by default, which is common in machine The theory of exponential dispersion models its coef_ member: The Ridge regressor has a classifier variant: In this example, you’ll apply what you’ve learned so far to solve a small regression problem. It is faster https://en.wikipedia.org/wiki/Theil%E2%80%93Sen_estimator. This example uses the only the first feature of the diabetes dataset, in order to illustrate a two-dimensional plot of this regression technique. regularization or no regularization, and are found to converge faster for some The algorithm thus behaves as intuition would expect, and “Regularization Path For Generalized linear Models by Coordinate Descent”, Great, so we did a simple linear regression on the car data. RANSAC (RANdom SAmple Consensus) fits a model from random subsets of For an important sanity check, we compare the $\beta$ values from statsmodels and sklearn to the $\beta$ values that we found from above with our own implementation. because the default scorer TweedieRegressor.score is a function of linear models we considered above (i.e. conjugate prior for the precision of the Gaussian. considering only a random subset of all possible combinations. until one of the special stop criteria are met (see stop_n_inliers and Let's see the structure of scikit-learn needed to make these fits. high-dimensional data. $$\ell_2$$ regularization (it corresponds to the l1_ratio parameter). It is installed by ‘pip install scikit-learn‘. The Ridge regressor has a classifier variant: RidgeClassifier.This classifier first converts binary targets to {-1, 1} and then treats the problem as a regression task, optimizing the same objective as above. variance. The classes SGDClassifier and SGDRegressor provide distributions using the appropriate power parameter. Singer - JMLR 7 (2006). In scikit-learn, an estimator is a Python object that implements the methods fit(X, y) and predict(T) Under certain conditions, it can recover the exact set of non-zero Here we will be using Python to execute Linear Regression. the weights are non-zero like Lasso, while still maintaining of squares: The complexity parameter $$\alpha \geq 0$$ controls the amount


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## 17Calculus - Parametric Equations

Parametrics is a way to plot a graph by specifying the x, y (and z for three dimensions) values separately. We do so using a parameter, another variable used to link equations together. It is simple but deceptively powerful.
If you want a full-length lecture on this topic, we recommend this video.

### Prof Leonard - Introduction to Parametric Equations [1hr-38min-25secs]

video by Prof Leonard

For starters we will use the parameter t. Here is an example.

 $$x = \cos(t)$$ $$y = \sin(t)$$

Notice that, although we have two separate equations, the x and y are linked by the value of t. In this example, we take a value for the parameter t and plug that value into both equations, to get a corresponding point $$(x,y)$$. When $$t=0$$, we get $$x=\cos(0)=1$$ and $$y=\sin(0)=0$$. Writing this as an ordered pair, we have $$(x,y)=(1,0)$$. So, the point $$(1,0)$$ corresponds to the value $$t=0$$.
Before we go on, let's watch a good introductory video.

### Michel vanBiezen - What is a Parametric Equation?

video by Michel vanBiezen

One of the advantages of using parametric equations is that we can describe many more graphs than we could when we had only x and y. Also, we are not going to require the graphs to be functions now, i.e. they may not pass the vertical line test.

Another piece of information we get from parametric equation is direction. The use of the variable t as a parameter is not random. Often, we assign a meaning to the parameter and sometimes that meaning is time. When you graph a set of parametric equations, the graph is swept out in a certain direction. This is an inherent feature of the parametric equations. We will often start at $$t=0$$ and increase t, giving the idea that time is passing. By adjusting the parametric equations, we can reverse the direction that the graph is swept.

Graphing Parametric Equations

As you learned when graphing functions, you need some experience with the equations and what the graphs look like in order to be able to know what a graph looks like just from the equation. As you are learning, I suggest using winplot to graph parametric equations. It is easy to use, has a very short learning curve and, best of all, it's free. You can find out more information on the Tools page.
Here is a video showing how to plot parametric equations.

### MIP4U - Introduction to Parametric Equations

video by MIP4U

Eliminating The Parameter

Sometimes we are given the set of parametric equations and we are asked to write the equation without the parameter by eliminating the parameter. This not always possible but with some equations there are ways to do it.

The easiest technique to try is to solve one of the equations for the parameter and then substitute the result in the other equation. Here is an example.
$$x = 2t$$ and $$y = t^2$$
Solve the first equation for t giving $$t=x/2$$ and substitute into the second equation.
$$y = (x/2)^2 = x^2/4$$

A second technique involves the use of trig identities. For example, given the parametric equations $$x=\cos(t)$$ and $$y=\sin(t)$$, we know that $$\cos^2(x) + \sin^2(x) = 1$$. So we can write $$x^2 + y^2 = 1$$ which eliminates the parameter.

A third way is by inspection. Sometimes it is obvious what a substitution might be. For example, if we have $$x=e^t$$ and $$y=e^{3t}+1$$ then we can rewrite $$y=(e^t)^3+1$$. Then we can replace $$e^t$$ with $$x$$ in the last equation (because our first parametric equation was $$x=e^t$$) to get $$y=x^3+1$$. Of course, the first technique would have worked too by solving $$x=e^t$$ for $$t$$ and then substituting and simplifying but by standing back and looking at the equations more carefully, the solution was much easier.
Okay, time for some practice problems.

Practice

Basic

Eliminate the parameter $$t$$ from $$x=3t$$ and $$y=t^2$$.

Problem Statement

Eliminate the parameter $$t$$ from $$x=3t$$ and $$y=t^2$$.

$$y = x^2/9$$

Problem Statement

Eliminate the parameter $$t$$ from $$x=3t$$ and $$y=t^2$$.

Solution

Solve the first equation for t giving $$t=x/3$$ and substitute into the second equation.
$$y = (x/3)^2 = x^2/9$$

$$y = x^2/9$$

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Eliminate the parameter $$t$$ from $$x=3\cos(t)$$ and $$y=2\sin(t)$$.

Problem Statement

Eliminate the parameter $$t$$ from $$x=3\cos(t)$$ and $$y=2\sin(t)$$.

$$\displaystyle{ \frac{x^2}{9} + \frac{y^2}{4} = 1 }$$

Problem Statement

Eliminate the parameter $$t$$ from $$x=3\cos(t)$$ and $$y=2\sin(t)$$.

Solution

We will use the trig identity $$\cos^2(t) + \sin^2(t) = 1$$.

$$x = 3\cos(t) \to x/3 = \cos(t)$$

$$y = 2\sin(t) \to y/2 = \sin(t)$$

$$(x/3)^2 + (y/2)^2 = 1$$

$$\displaystyle{ \frac{x^2}{9} + \frac{y^2}{4} = 1 }$$

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Eliminate the parameter t from $$x=5\cos(t)$$ and $$y=3\sin(t)$$.

Problem Statement

Eliminate the parameter t from $$x=5\cos(t)$$ and $$y=3\sin(t)$$.

$$9x^2 + 25y^2 = 225$$

Problem Statement

Eliminate the parameter t from $$x=5\cos(t)$$ and $$y=3\sin(t)$$.

Solution

In the video, she shows a really difficult way to solve this problem. Her solution is correct but it is overly tedious and unnecessarily complicated. Here is an easier and more common solution.
In this problem, you are given that $$x=5\cos(t)$$ and $$y=3\sin(t)$$. We solve each of these for the sine and cosine terms since we use $$\cos^2(t)+\sin^2(t)=1$$. So we have $$x/5=\cos(t) \to (x/5)^2=\cos^2(t)$$ and $$y/3=\sin(t) \to (y/3)^2=\sin^2(t)$$.
$$\begin{array}{rcl} 1 & = & \cos^2(t)+\sin^2(t) \\ 1 & = & (x/5)^2+(y/3)^2 \\ 1 & = & x^2/25+y^2/9 \\ 225 & = & 9x^2 + 25y^2 \end{array}$$

### 719 video

video by Krista King Math

$$9x^2 + 25y^2 = 225$$

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Eliminate the parameter from the parametric equations $$x=t-1$$, $$y=4t+1$$.

Problem Statement

Eliminate the parameter from the parametric equations $$x=t-1$$, $$y=4t+1$$.

Solution

### 48 video

video by PatrickJMT

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Eliminate the parameter from the parametric equations $$x=\sqrt{t+1}$$, $$y=3t+2$$.

Problem Statement

Eliminate the parameter from the parametric equations $$x=\sqrt{t+1}$$, $$y=3t+2$$.

Solution

### 49 video

video by PatrickJMT

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Graph the parametric equations $$x=1+\sqrt{t}$$, $$y=t^2-4t$$ on $$0\leq t\leq4$$.

Problem Statement

Graph the parametric equations $$x=1+\sqrt{t}$$, $$y=t^2-4t$$ on $$0\leq t\leq4$$.

Solution

### 51 video

video by PatrickJMT

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Graph the parametric equations $$x=\sqrt{t}$$, $$y=1-t$$ and eliminate the parameter.

Problem Statement

Graph the parametric equations $$x=\sqrt{t}$$, $$y=1-t$$ and eliminate the parameter.

Solution

### 52 video

video by PatrickJMT

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Eliminate the parameter t from $$x=\ln t$$ and $$y=\sqrt{t}$$.

Problem Statement

Eliminate the parameter t from $$x=\ln t$$ and $$y=\sqrt{t}$$.

Solution

### 2466 video

video by Michel vanBiezen

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Intermediate

Eliminate the parameter $$t$$ from $$x=e^t-1$$ and $$y=e^{2t}$$.

Problem Statement

Eliminate the parameter $$t$$ from $$x=e^t-1$$ and $$y=e^{2t}$$.

$$y=x^2+2x+1$$

Problem Statement

Eliminate the parameter $$t$$ from $$x=e^t-1$$ and $$y=e^{2t}$$.

Solution

### 2464 video

video by Michel vanBiezen

$$y=x^2+2x+1$$

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For the parametric equations $$x=1+\sin(t)$$, $$y=-2+\cos(t)$$, find the point that corresponds to $$t=\pi/2$$, graph the equations and eliminate the parameter.

Problem Statement

For the parametric equations $$x=1+\sin(t)$$, $$y=-2+\cos(t)$$, find the point that corresponds to $$t=\pi/2$$, graph the equations and eliminate the parameter.

Solution

### 47 video

video by PatrickJMT

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Eliminate the parameter from the parametric equations $$x=4e^{t/4}$$, $$y=3e^t$$.

Problem Statement

Eliminate the parameter from the parametric equations $$x=4e^{t/4}$$, $$y=3e^t$$.

Solution

### 50 video

video by PatrickJMT

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Find all points of intersection of the parametric curves $$C_1: x=t+1; y=t^2$$ and $$C_2: x=3t+1; y=t^2+1$$.

Problem Statement

Find all points of intersection of the parametric curves $$C_1: x=t+1; y=t^2$$ and $$C_2: x=3t+1; y=t^2+1$$.

Solution

### 53 video

video by PatrickJMT

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You CAN Ace Calculus

Wikipedia - Parametric Equations

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text:
## Institute of Mathematical Statistics Lecture Notes - Monograph Series

### Empirical graph Laplacian approximation of Laplace–Beltrami operators: Large sample results

#### Abstract

Let ${M}$ be a compact Riemannian submanifold of ${{\bf R}^m}$ of dimension $\scriptstyle{d}$ and let ${X_1,\dots, X_n}$ be a sample of i.i.d. points in ${M}$ with uniform distribution. We study the random operators

center$${\Delta_{h_n,n}f(p):={1\over {nh_n^{d+2}}}\sum_{i=1}^n K({p-X_i\over h_n})(f(X_i)-f(p)),\ p\in M}$$

where ${K(u):={1\over (4\pi)^{d/2}}e^{-\|u\|^2/4}}$ is the Gaussian kernel and ${h_n\to 0}$ as ${n\to\infty.}$ Such operators can be viewed as graph laplacians (for a weighted graph with vertices at data points) and they have been used in the machine learning literature to approximate the Laplace-Beltrami operator of ${M,}$ ${\Delta_M f}$ (divided by the Riemannian volume of the manifold). We prove several results on a.s. and distributional convergence of the deviations ${\Delta_{h_n,n}f(p)-{1\over |\mu|}\Delta_Mf(p)}$ for smooth functions ${f}$ both pointwise and uniformly in ${f}$ and ${p}$ (here ${|\mu|=\mu(M)}$ and ${\mu}$ is the Riemannian volume measure). In particular, we show that for any class ${{\cal F}}$ of three times differentiable functions on ${M}$ with uniformly bounded derivatives

center$${ \sup_{p\in M}\sup_{f\in {\cal F}}\Big|\Delta_{h_n,p}f(p)-{1\over|\mu|}\Delta_Mf(p)\Big|=O\Big(\sqrt{\log(1/h_n)\over nh_n^{d+2}}\Big)\ \ {\rm a.s.}}$$

as soon as

center$${nh_n^{d+2}/\log h_n^{-1}\to \infty\ \ {\rm and}\ \nh^{d+4}_n/\log h_n^{-1}\to 0,}$$

and also prove asymptotic normality of ${\Delta_{h_n,p}f(p)-{1\over |\mu|}\Delta_Mf(p)}$ (functional CLT) for a fixed ${p\in M}$ and uniformly in ${f}$.

#### Chapter information

Source
Evarist Giné, Vladimir Koltchinskii, Wenbo Li, Joel Zinn, eds., High Dimensional Probability: Proceedings of the Fourth International Conference (Beachwood, Ohio, USA: Institute of Mathematical Statistics, 2006), 238-259

Dates
First available in Project Euclid: 28 November 2007

https://projecteuclid.org/euclid.lnms/1196284116

Digital Object Identifier
doi:10.1214/074921706000000888

Mathematical Reviews number (MathSciNet)
MR2387773

Zentralblatt MATH identifier
1124.60030

Rights


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Broad Topics > Measures and Mensuration > Area

Shear Magic

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What are the areas of these triangles? What do you notice? Can you generalise to other "families" of triangles?

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Do you know how to find the area of a triangle? You can count the squares. What happens if we turn the triangle on end? Press the button and see. Try counting the number of units in the triangle now. . . .

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What can you say about these shapes? This problem challenges you to create shapes with different areas and perimeters.

Overlap

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A red square and a blue square overlap so that the corner of the red square rests on the centre of the blue square. Show that, whatever the orientation of the red square, it covers a quarter of the. . . .

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How can you change the area of a shape but keep its perimeter the same? How can you change the perimeter but keep the area the same?

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This rectangle is cut into five pieces which fit exactly into a triangular outline and also into a square outline where the triangle, the rectangle and the square have equal areas.

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A hallway floor is tiled and each tile is one foot square. Given that the number of tiles around the perimeter is EXACTLY half the total number of tiles, find the possible dimensions of the hallway.

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Can you help the children find the two triangles which have the lengths of two sides numerically equal to their areas?

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Use the interactivity to find all the different right-angled triangles you can make by just moving one corner of the starting triangle.

A Square in a Circle

Stage: 2 Challenge Level:

What shape has Harry drawn on this clock face? Can you find its area? What is the largest number of square tiles that could cover this area?

Tiling

Stage: 2 Challenge Level:

An investigation that gives you the opportunity to make and justify predictions.

Numerically Equal

Stage: 2 Challenge Level:

Can you draw a square in which the perimeter is numerically equal to the area?

Fit These Shapes

Stage: 1 and 2 Challenge Level:

What is the largest number of circles we can fit into the frame without them overlapping? How do you know? What will happen if you try the other shapes?

Two Squared

Stage: 2 Challenge Level:

What happens to the area of a square if you double the length of the sides? Try the same thing with rectangles, diamonds and other shapes. How do the four smaller ones fit into the larger one?

Tiles on a Patio

Stage: 2 Challenge Level:

How many ways can you find of tiling the square patio, using square tiles of different sizes?

Wrapping Presents

Stage: 2 Challenge Level:

Choose a box and work out the smallest rectangle of paper needed to wrap it so that it is completely covered.

Kissing Triangles

Stage: 3 Challenge Level:

Determine the total shaded area of the 'kissing triangles'.

Torn Shapes

Stage: 2 Challenge Level:

These rectangles have been torn. How many squares did each one have inside it before it was ripped?

Warmsnug Double Glazing

Stage: 3 Challenge Level:

How have "Warmsnug" arrived at the prices shown on their windows? Which window has been given an incorrect price?

Square Areas

Stage: 3 Challenge Level:

Can you work out the area of the inner square and give an explanation of how you did it?

Making Boxes

Stage: 2 Challenge Level:

Cut differently-sized square corners from a square piece of paper to make boxes without lids. Do they all have the same volume?

My New Patio

Stage: 2 Challenge Level:

What is the smallest number of tiles needed to tile this patio? Can you investigate patios of different sizes?

Geoboards

Stage: 2 Challenge Level:

This practical challenge invites you to investigate the different squares you can make on a square geoboard or pegboard.

Making Squares

Stage: 2 Challenge Level:

Investigate all the different squares you can make on this 5 by 5 grid by making your starting side go from the bottom left hand point. Can you find out the areas of all these squares?

Can They Be Equal?

Stage: 3 Challenge Level:

Can you find rectangles where the value of the area is the same as the value of the perimeter?

Perimeter Possibilities

Stage: 3 Challenge Level:

I'm thinking of a rectangle with an area of 24. What could its perimeter be?

Muggles Magic

Stage: 3 Challenge Level:

You can move the 4 pieces of the jigsaw and fit them into both outlines. Explain what has happened to the missing one unit of area.

Take Ten

Stage: 3 Challenge Level:

Is it possible to remove ten unit cubes from a 3 by 3 by 3 cube made from 27 unit cubes so that the surface area of the remaining solid is the same as the surface area of the original 3 by 3 by 3. . . .

Tiling Into Slanted Rectangles

Stage: 2 and 3 Challenge Level:

A follow-up activity to Tiles in the Garden.

Inside Seven Squares

Stage: 2 Challenge Level:

What is the total area of the four outside triangles which are outlined in red in this arrangement of squares inside each other?

Overlapping Squares

Stage: 2 Challenge Level:

Have a good look at these images. Can you describe what is happening? There are plenty more images like this on NRICH's Exploring Squares CD.

Rati-o

Stage: 3 Challenge Level:

Points P, Q, R and S each divide the sides AB, BC, CD and DA respectively in the ratio of 2 : 1. Join the points. What is the area of the parallelogram PQRS in relation to the original rectangle?

Kite

Stage: 3 Challenge Level:

Derive a formula for finding the area of any kite.

Cylinder Cutting

Stage: 2 and 3 Challenge Level:

An activity for high-attaining learners which involves making a new cylinder from a cardboard tube.

Isosceles

Stage: 3 Challenge Level:

Prove that a triangle with sides of length 5, 5 and 6 has the same area as a triangle with sides of length 5, 5 and 8. Find other pairs of non-congruent isosceles triangles which have equal areas.

The Pi Are Square

Stage: 3 Challenge Level:

A circle with the radius of 2.2 centimetres is drawn touching the sides of a square. What area of the square is NOT covered by the circle?

Triangle Relations

Stage: 2 Challenge Level:

What do these two triangles have in common? How are they related?


text:
# Is the “wing area” of a biplane the projected area or the total area?

When you specify the wing area of a biplane, do you specify the planform area (the projected area) or do you specify the effective area of both the wings?

• I could have sworn we had this exact question, or a very similar one about interference and effective wing area, but I can't find it. – AEhere supports Monica Oct 25 '19 at 7:59


text:
# Fraction Brackets size

I wanted to write the following equation but somehow my first bracket before the fraction is shorter, last one is fine. but I have no clue what happened to the first one,

also is it possible to write the stuffs under sup as two separate lines, rather than in one line?

\begin{aligned} \Lambda \left( \phi \right) \coloneqq \adjustlimits\sup_{x,y \in E, x \neq y} \left({\dfrac{ d \left( \phi \left(x\right) , \phi \left(y\right) \right) } {d \left(x, y \right)} } \right) \end{aligned}

• Welcome to TeX.SX! You shouldn't use \adjustlimits in this case; no blank lines inside aligned either. – egreg Apr 26 '15 at 17:48
• but then how shall I write under supremum??, actually that is why I used adjust limits....sorry to ask again – Shaikh Tanvir Hossain Apr 26 '15 at 17:49
• Try without it! For the two lines, write \sup_{\substack{x,y\in E\\ x\neq y}} – egreg Apr 26 '15 at 17:50

The \adjustlimits macro of mathtools is used for getting vertically aligned limits when two operators taking limits follow one another. In this case it's exactly the reason why \left( doesn't do what it's supposed to.

When in a display, \sup behaves exactly like \lim, so there's no special thing to do.

\documentclass{article}
\usepackage{mathtools}

\begin{document}

$$\Lambda(\phi) \coloneqq \sup_{\substack{x,y \in E\\ x \neq y}} \left( \dfrac{d(\phi(x), \phi(y))}{d(x, y)}\right)$$

\end{document}

Note that \left and \right are necessary only around the fraction (but in this case I'd omit those parentheses altogether).

Also aligned is not necessary.

• Thanks sorry I made a stupid mistake again, but finally worked – Shaikh Tanvir Hossain Apr 26 '15 at 18:00
• @ShaikhTanvirHossain Put \DeclareMathOperator*{\argmax}{arg\,max} in the document preamble and use \argmax in the document. – egreg Apr 27 '15 at 9:40
• Ah! ok...Thanks man, many thanks...I was deleting my question, thought you might not check again......really appreciate this – Shaikh Tanvir Hossain Apr 27 '15 at 9:41
• so my question was ...(sorry I deleted it) !!Sorry @egreg I got another similar problem, couldn't solve, rather than \sup now I have \arg\max, problem is now stuff below is coming under max, can I make it in center?? – – Shaikh Tanvir Hossain Apr 27 '15 at 9:42


text:
Modify bibliography style of @Online

I'm using biblatex with the following parameters:

• style=footnote-dw
• bibstyle=numeric
• citestyle=numeric
• url=false
• sorting=none
• backend=biber

Then I've online sources, for example:

@Online{online:def_commerce,
author  = {{Oxford Dictionaries}},
title   = {{Definition of commerce}},
url     = {https://en.oxforddictionaries.com/definition/commerce},
urldate = {2017-01-24},
}

In the bibliography, the source is printed as follows:

I'd like to modify the output. I'd prefer it like this:

Oxford Dictionaries. Definition of commerce. https://en.oxforddictionaries.com/definition/commerce. Accessed: 2017-01-24.

To sum it up:

• I'd like to remove the "URL:"
• After the URL should be a period.
• Instead of "(visited on xxx )" should be "Accessed: xxx."
• The date should be of format "yyyy-mm-dd"
• Welcome to TeX.SX! Please help us help you and add a minimal working example (MWE) that illustrates your problem. Reproducing the problem and finding out what the issue is will be much easier when we see compilable code, starting with \documentclass{...} and ending with \end{document}. – user36296 May 6 '17 at 20:31
• Hi samcarter, thanks for the warm welcome. Problem with a minimal working example is that I'm using a non-standard documentclass (provided by my university). However, I've found a solution for my question, and will edit the question to add the answer. – theb May 6 '17 at 21:26
• You can use a standard class to make your MWE unless you can't reproduce the problem that way. (In that case, we'd need a link for the class, as it is essential to the issue.) Please can you post your solution as an answer? The site likes to keep questions separate from answers so we can see which questions don't yet have answers. It is fine to answer your own question. – cfr May 6 '17 at 23:15

Use

\documentclass[british]{article}
\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}
\usepackage{babel}
\usepackage{csquotes}
\usepackage[backend=biber, style=numeric, sorting=none, url=false, urldate=edtf]{biblatex}

\usepackage{filecontents}
\begin{filecontents}{\jobname.bib}
@Online{online:def_commerce,
author  = {{Oxford Dictionaries}},
title   = {{Definition of commerce}},
url     = {https://en.oxforddictionaries.com/definition/commerce},
urldate = {2017-01-24},
}
\end{filecontents}

\DeclareFieldFormat{url}{\url{#1}}

\renewbibmacro*{url+urldate}{%
\usebibmacro{url}%
\iffieldundef{urlyear}
{}
\usebibmacro{urldate}}}

\DefineBibliographyStrings{english}{urlseen = {accessed}}

\begin{document}
\cite{online:def_commerce}
\printbibliography
\end{document}

Note in particular that style=footnote-dw, bibstyle=numeric, citestyle=numeric can be collapsed into style=numeric.

• Thank you, exactly what I was looking for. And thanks for pointing out style=numeric. – theb May 7 '17 at 8:25


text:
• 2 Vote(s) - 4 Average
• 1
• 2
• 3
• 4
• 5
 Searching for an asymptotic to exp[0.5] tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 07/15/2014, 08:22 PM (This post was last modified: 07/15/2014, 08:59 PM by tommy1729.) Conjecture T0 : Let $f(z) = a_0 + a_1 z + a_2 z^2 + ...$ be a real entire function where $a_i > 0$. If for all $n$ we have $a_{n+1} < a_n/n$ then $f(z)$ is stable. This has probably been proved already. (edit) More specificly : Conjecture T1 : Let $f_1(z) = A_0 + A_1 z + A_2 z^2 + ...$ be a real entire function where $0. Let $f_2(z) = B_0 + B_1 z + B_2 z^2 + ...$ be a real entire function where $B_i >= 0$ such that at least $B_i$ is not equal to $0$ for 2 values of $i$. Let $f_3(z) = C_0 + C_1 z + C_2 z^2 + ...$ be a real entire function where $C_i = [f_2({A_i}^{-1})]^{-1}$. Then if $f_1(z)$ is stable , then so is $f_3(z)$. regards tommy1729 tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 07/15/2014, 09:43 PM NOTE : For those who want to conjecture the inverse of conjecture T1 ; If f_3(z) is stable , then so is f_1(z). THIS IS FALSE ! For instance if f_2(z) = eps + z^2. Notice 1 + z + z^2/sqrt(2) + z^3/sqrt(6) + z^4/sqrt(24) = 0 has the zero : 0.2026 + 1.5304 i and 1 + z + z^2/sqrt(2) + z^3/sqrt(6) + z^4/sqrt(24) + z^5/sqrt(120) = 0 has the zero : 0.545368 + 1.61261 i Hence disproving the inverse of conjecture T1. Btw taking more terms to avoid polynomials : g(z) = 1 + z + z^2/sqrt(2) + z^3/sqrt(6) + z^4/sqrt(24) + z^5/sqrt(120) + ... only makes the case more Obvious. g(z) has zero's with an unbounded large real part. ( or so it appears , anyway the trends seems growing , unbounded is a " mini conjecture / exercise " for those who are intrested ). regards tommy1729 tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 07/15/2014, 09:48 PM (This post was last modified: 07/15/2014, 09:51 PM by tommy1729.) Also to warn others , I want to mention gottfriend " dream of a sequence " again. It " pseudofactors " the exp(z). Why does that matter ? Well because theorems about stable polynomials assume factorizations to hold everywhere. Hence the analogue theorems for (stable) Taylor series might fail strongly ! To refresh your memory : http://www.go.helms-net.de/math/musings/...quence.pdf regards tommy1729 sheldonison Long Time Fellow Posts: 640 Threads: 22 Joined: Oct 2008 07/17/2014, 05:46 AM (This post was last modified: 08/02/2014, 02:47 PM by sheldonison.) (07/14/2014, 12:20 AM)tommy1729 Wrote: ..... This leads to 3 ideas and a remark. remark : you say " oscillating between positive and negative ". Now I know that if a closed jordan curve path has all arguments n times then we have n zero's within that closed path. So that suggests that all real roots have multiplicity 1 and you are in the possesion of a proof ? Also you have claimed that there are no zero's off the real line ? Does that have a proof ? I think it should be provable for Re(z) > 0.I'm getting close to a proof that there are no zeros off the real line, all zeros multiplicity 1. It seems to me that the details come to me very slowly, since I keep thinking about it, and keep making steady progress. Basically, it comes down to how Kneser's half iterate behaves in the complex plane, along with the definition used for the entire asymptotic function. Interestingly, to understand the negative real axis, you mostly have to understand how Kneser's half iterate behaves for positive $z+\pi i$, due to the equation $\exp^{0.5}(z)=\exp(\exp^{0.5}(\log(z)))$. And the definition I'm using for negative axis comes down to $\text{asymptotic}(-z) \approx \exp(\exp^{0.5}(\log(z)+\pi i))+\exp(\exp^{0.5}(\log(z)- \pi i))$ plus stuff, where stuff can be shown too small to create zeros anywhere else, since halfk(z) can be shown to have a large absolute value at the negative axis, so stuff winds up acting like a small perturbation of the zeros of the approximation. More later. - Sheldon tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 07/24/2014, 12:10 PM Im getting close to a proof of zero-free regions. I assume the ideas will be different from sheldon. It is funny , when I talked about tetration on math forums I got alot of "resistance". It was called " nonsense " or " useless " and people who were into it were insulted. And a few of those strong " opponents " where professors in ... complex analysis. And now here we are using complex analysis in the majority of posts of the tetration forum. And it seems that the use of complex analysis here is just beginning , its use is accelerating. Im just a few lemma's away from a proof , so that is what " close " means. How easy those lemma's are is not yet clear to me though. Im Always fighting against the clock. Its a shame that society pulls people away from doing research without financial benefit. regards tommy1729 tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 07/24/2014, 10:47 PM The conclusion is that if | f(f(z)) - exp(z) | << 1 where sheldon's picture claims it to be , then there are no nonreal zero's for f(z). Also the converse : If there are no nonreal zero's for f(z) then | f(f(z)) - exp(z) | << 1 where sheldon's picture claims it to be. I assume this was already known by sheldon but he did not post the proof. I assumed him to be correct but double checked it. Now its certain. Its intresting , f(z) is now unique by the region where | f(f(z)) - exp(z) | << 1. Hmm well probably. Seems like another mission. It is unique (by that criterion above) if f(z) is unique by the condition : no nonreal zero's and the zero's at the real line are the only possible. Im already considering pro and contra arguments for that mission. The majority of pro arguments are based on the periodicity of exp(z). The contra arguments are based on small perturbations of the zero's. All proofs are based on arg(f(z)). And NO use is made of ln(f(z)) is close to f^[-1](z). Generalization ideas and conjectures are underway. However the initial problem is reduced to an equivalence statement : | f(f(z)) - exp(z) | << 1 where sheldon's picture claims it to be <=> there are no nonreal zero's for f(z). But a full solution is still not here. Although Im happy with this new insight. Not sure how kneser is suppose to be related as sheldon talks about. We seem to be getting a strong foundation for " fake function theory " regards tommy1729 tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 07/24/2014, 10:54 PM At the moment I dream about the zero's of fake exp^[1/4](z) and the zero's of its integral. This could be hard and might then be considered as the analogue of the Riemann Hypothesis for tetration. Or maybe not regards tommy1729 sheldonison Long Time Fellow Posts: 640 Threads: 22 Joined: Oct 2008 07/25/2014, 02:46 PM (This post was last modified: 08/02/2014, 02:48 PM by sheldonison.) (07/24/2014, 10:47 PM)tommy1729 Wrote: The conclusion is that if | f(f(z)) - exp(z) | << 1 where sheldon's picture claims it to be , then there are no nonreal zero's for f(z). Also the converse : If there are no nonreal zero's for f(z) then | f(f(z)) - exp(z) | << 1 where sheldon's picture claims it to be. I assume this was already known by sheldon but he did not post the proof. I assumed him to be correct but double checked it. Now its certain. I like the approach. It might be an elegent simplification of the last step of my proof. The boundary of convergence of |f(f(z))/exp(z)| would be where $f^{-1}(-z)$. Here, the amplitude of $|f(f(z))-\exp(z)|\approx\exp(z)$, but it is smaller everywhere else. Here is a sample calculation, where f(z) is halfway between the 2nd and 3rd zeros. Here, $f(f(z))\approx 2\exp(z)$. If f(z) is one of the zeros, then f(f(z))=0 by definition, and $f(f(z))-\exp(z)=-\exp(z)$. $z \approx -0.183411676 + 6.28413080i$ $f(z) \approx -8.38653002$ $f(f(z)) \approx 1.66174551$ $\exp(z) \approx 0.832425023 + 0.000787053836i$ For my proof, I am focused on the error terms for the assymptotic approximation, and a proof that all of the Taylor series terms (and Laurent series terms) converge. I have a mostly complete version of of the proofs, but its a bit involved; requires some pictures and quite a few equations. Also, there are two different dueling error terms, both of which get arbitrarily small as |z| gets larger. One of the error terms is the 1/x part of the Laurent series for the entire half iterate, which is approximately 0.003898/z-0.00172/z^2 and the other is the number of terms required in the approximation. Then I got side tracked into investigating this Laurent series, which I calculated accurate to 32 decimal digits for terms>=-20. The terms>=0 for the Laurent series are exactly the same as the entire half iterate approximation. All of the terms of the Laurent series converge. But the infinite Laurent series itself probably does not converge anywhere. Anyway, except for all of the pictures required for the proofs, that is where I'm at. Would you be interested in the partial details so far? The last step would be to show that the approximation error term are too small to allow zeros anywhere else other than the real axis for, $\text{asymptotic}(-z) \approx \exp(\exp^{0.5}(\log(z)+\pi i))+\exp(\exp^{0.5}(\log(z)- \pi i))$ - Sheldon tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 07/26/2014, 12:21 PM (This post was last modified: 07/26/2014, 12:24 PM by tommy1729.) Im considering that f(f(z)) - exp(z) is small in a region without holes. The idea of holes fascinates me. Many conjectures come to my mind though the quality of them is not yet certain. For instance CONJECTURE HOLE A If g(z) is a fake (entire) half-iterate of exp(z) on the positive real line and all the derivatives are positive then g(g(z)) - exp(z) is small in a region without holes. Problem is also that things are only defined intuitive. For instance the presence of a hole or not depends on your " definition of small ". Only depending on your definition of small is if there is 1 hole or 2 holes close together. ( perhaps name that a pseudohole or fake hole ?) Im not sure if the concept of holes is relevant to the location of zero's of f(z) , some (conjectured) uniqueness conditions , equivalence principles and other stuff discussed sofar , or if it is just a new subject. Maybe I did to much topology lately. regards tommy1729 tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 07/27/2014, 08:37 AM Another thing. Maybe trivial sorry. But maybe not. In sheldon's picture of f(z) he has colored negative blue and positive red. It seems that between the zero's on the negative real line , we get for every red/blue line on the real line an intersection with another red/blue line coming from the upper resp lower complex plane. This intersection seems to Always be perpendicular !!? Is that an optical illusion or is it true ? Why is that true ? Is it a property of polynomials that carries on because of the hadamard product ? Or is this some complex analysis 101 that I forgot about ? regards tommy1729 « Next Oldest | Next Newest »

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# meromorphic

## English

### Etymology

meromorphic (not comparable)

1. (complex analysis, of a function) That is the ratio of two holomorphic functions (and so possibly infinite at a discrete set of points).
• 1993, Joel L. Schiff, Normal Families, Springer, page 71,
Normal families of meromorphic functions are most naturally studied using the spherical metric (§1.2), an approach initiated by Ostrowski [1926]. Some results for meromorphic functions, such as the FNT, are immediate extensions from the analytic case, whereas others, such as Landau's or Julia's theorem are set in a much broader context than their analytic counterparts. Normality criteria pertinent to families of meromorphic functions, such as Marty's theorem, have not yet been encountered.
• 2000, Werner Balser, Formal Power Series and Linear Systems of Meromorphic Ordinary Differential Equations, Springer, page 39,
Note that such a transformation is holomorphic at the origin, but is essentially singular at infinity. However, since ${\displaystyle T(z)}$ commutes with ${\displaystyle A(z)}$, the transformed system has coefficient matrix ${\displaystyle A(z)-zq'(z)I}$ and hence is again meromorphic at infinity.
• 2012, Marius van der Put, Michael F. Singer, Galois Theory of Linear Differential Equations, Springer, page 147,
A point ${\displaystyle p\in P^{1}}$ is singular for ${\displaystyle \textstyle {\frac {d}{dz}}-A}$ if the equation cannot be made regular at ${\displaystyle p}$ with a local meromorphic transformation.


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# Integral $\int_{0}^{1}\frac{\log^{2}(x^{2}-x+1)}{x}dx$

Here is an integral I derived while evaluating another. It appears to be rather tough, but some here may not be so challenged :)

Show that: $$\int_{0}^{1}\frac{\log^{2}(x^{2}-x+1)}{x}dx=\frac{11}{9}\zeta(3)-\frac{\pi}{72\sqrt{3}}\left(5\psi_{1}\left(\frac13\right)+4\psi_{1}\left(\frac23\right)-3\psi_{1}\left(\frac56\right)\right)$$ $$=\frac{11}{9}\zeta(3)+\frac{4\pi^{3}}{27\sqrt{3}}-\frac{2\pi}{9\sqrt{3}}\psi_{1}\left(\frac13\right)=\frac{11}{9}\zeta(3)-\frac{4\pi}{9}\operatorname{Cl}_{2}\left(\frac{\pi}{3}\right)$$ $$=\operatorname{Cl}_{2}\left(\frac{\pi}{3}\right)-2\operatorname{Cl}_{2}\left(\frac{2\pi}{3}\right)-\frac{4\pi}{9}\operatorname{Cl}_{2}\left(\frac{\pi}{3}\right)$$

I attempted all kinds of 'starts' to no satisfactory end, but things look promising. There are some mighty sharp folks here that may be better at deriving the solution.

I thought perhaps the identity:

$$\frac{\log^{2}(1-(x-x^{2}))}{x}=2\sum_{n=1}^{\infty}\frac{H_{n}}{n+1}x^{n}(1-x)^{n+1}$$

or the Beta function could be used if given enough ingenuity.

This led me to the no-less-imposing Euler/reciprocal of central binomial coefficients sum below. It would be great to just show the middle sum is equivalent to the right sum:

$$1/4\sum_{n=1}^{\infty}\frac{H_{n}n\Gamma^{2}(n)}{(n+1)(2n+1)\Gamma(2n)}=1/2\sum_{n=1}^{\infty}\frac{H_{n}}{(n+1)(2n+1)\binom{2n}{n}}=1/3\zeta(3)-2/3\sum_{n=1}^{\infty}\frac{1}{n^{3}\binom{2n}{n}}$$

Is there a general form for $$\sum_{n=1}^{\infty}\frac{H_{n}}{\binom{2n}{n}}x^{n}?$$

I tried starting with the identity: $$\sum_{n=1}^{\infty}\frac{\Gamma^{2}(n)}{\Gamma(2n)}x^{n-1}=\frac{4\sin^{-1}\left(\frac{\sqrt{x}}{2}\right)}{\sqrt{x(4-x)}}$$ and using various manipulations to hammer into the needed form. This, too, turned monstrous.

There appears to be a relation to Clausen functions (as with other log integrals such as $$\int_{0}^{1}\frac{\log(x)}{x^{2}-x+1}dx$$), to wit:

I use Cl for sin and CL for cos Clausen functions

$$\operatorname{Cl}_{2}\left(\frac{\pi}{3}\right)=\sum_{k=1}^{\infty}\frac{\sin(\frac{\pi k}{3})}{k^{2}}=\frac{\sqrt{3}}{72}\left(\psi_{1}(1/6)+\psi_{1}(1/3)-\psi_{1}(2/3)-\psi_{1}(5/6)\right)$$

$$=\frac{\sqrt{3}}{6}\psi_{1}(1/3)-\frac{\pi^{2}\sqrt{3}}{9}$$

and

$$\operatorname{Cl}_{3}\left(\frac{\pi}{3}\right)-\operatorname{Cl}_{3}\left(\frac{2\pi}{3}\right)=\sum_{k=1}^{\infty}\frac{\cos(\frac{\pi k}{3})}{k^{3}}-2\sum_{k=1}^{\infty}\frac{\cos(\frac{2\pi k}{3})}{k^{3}}=\frac{11}{9}\zeta(3)$$

Another approach. I also broke the integral up as such:

$$\int_{0}^{1}\frac{\log^{2}(x^{2}-x+1)}{x}dx=\int_{0}^{1}\frac{\log^{2}(1-xe^{\frac{\pi i}{3}})}{x}dx+2\int_{0}^{1}\frac{\log(1-xe^{\pi i/3})\log(1-xe^{-\pi i/3})}{x}dx+\int_{0}^{1}\frac{\log^{2}(1-xe^{-\pi i/3})}{x}dx$$

The middle integral right of the equal sign is the one that has given me the fit.

I think this is a fun and head-scratchin' integral that has led me to other discoveries. Maybe a generalization could be obtained with other powers of log such as n = 3, 4, etc.

I wonder if they can also be evaluated in terms of Clausens and then into closed forms involving $$\zeta(n+1)$$ and derivatives of digamma, $$\psi_{n-1}(z)?$$.

Another easier one is $$\int_{0}^{1}\frac{\log(x^{2}-x+1)}{x}dx=\frac{-\pi^{2}}{18}=\frac{-1}{3}\zeta(2)?$$

• I don't see how Cauchy's theorem can help here. I have been able to find that the integral is equal to another integral: $$\int_0^1 dx \frac{\log{\left (1-\frac{x}{4} \right )}}{x \sqrt{1-x}}$$ But again, we can derive a sum from this and given what the result is, it is going to be a matter of manipulating the terms in the sum to match the trigammas. – Ron Gordon Sep 30 '15 at 10:58
• Since you have forgotten more about contour integration than I know, I thought perhaps you knew something I was not aware of . Anyway, may I ask how you arrived at that integrand?. I get a different numerical result. Maybe I messed up. If anyone can make headway, it'd be you RG :) I was going to try and manipulate the Clausen sums and see if I hit upon anything. I have to leave now, but I will probably play with it this evening. – Cody Sep 30 '15 at 12:52
• Hey Ron, I think that integral is equivalent to $\int_{0}^{1}\frac{\log(x^{2}-x+1)}{x}dx=\frac{-\pi^{2}}{18}$. I think you have a typo and need a square on your log as well. Then it is the same. That's a cool derivation. If you get time, please show how you arrived at that . :) – Cody Sep 30 '15 at 13:04

Asset at our disposal: $$\sum\limits_{n=0}^{\infty} \frac{x^{2n+2}}{(n+1)(2n+1)\binom{2n}{n}} = 4(\arcsin (x/2))^2$$

Differentiation followed by the substitution $x \to \sqrt{x}$ gives:

$\displaystyle \sum\limits_{n=0}^{\infty} \frac{x^{n}}{(2n+1)\binom{2n}{n}} = \frac{2\arcsin (\sqrt{x}/2)}{\sqrt{x}\sqrt{1-(\sqrt{x}/2)^2}}$

Thus, we split the series as: $$\sum\limits_{n=0}^{\infty} \frac{H_n}{(n+1)(2n+1)\binom{2n}{n}} \\= \sum\limits_{n=0}^{\infty} \frac{H_{n+1}}{(n+1)(2n+1)\binom{2n}{n}} - \sum\limits_{n=0}^{\infty} \frac{1}{(n+1)^2(2n+1)\binom{2n}{n}}$$

The first series can be dealt with using, $\displaystyle\frac{H_{n+1}}{n+1} = -\int_0^1 x^n\log(1-x)\,dx$

\begin{align*}\sum\limits_{n=0}^{\infty} \frac{H_{n+1}}{(n+1)(2n+1)\binom{2n}{n}}&= -\sum\limits_{n=0}^{\infty} \int_0^1 \frac{x^n\log(1-x)}{(2n+1)\binom{2n}{n}}\,dx\\ &= -2\int_0^1 \frac{\arcsin (\sqrt{x}/2)\log (1-x)}{\sqrt{x}\sqrt{1-(\sqrt{x}/2)^2}}\,dx\\ &= -8\int_0^{1/2} \frac{\arcsin x \cdot \log (1-4x^2)}{\sqrt{1-x^2}}\,dx\\ &= -8\int_0^{\pi/6} \theta \log (1-4\sin^2 \theta)\,d\theta\\ &= -8\int_0^{\pi/6} \theta \log \left(4\sin\left(\theta + \frac{\pi}{6}\right)\sin\left(\frac{\pi}{6}-\theta\right)\right) \end{align*}

Using the Fourier Series, $\displaystyle \log (2\sin \theta) = -\sum\limits_{n=1}^{\infty} \frac{\cos 2n\theta}{n}$ we get:

\begin{align*}&\int_0^{\pi/6} \theta\log \left(2\sin\left(\frac{\pi}{6}+\theta\right)\right)\,d\theta \\&= -\sum\limits_{n=1}^{\infty} \int_0^{\pi/6} \frac{\theta\cos \left(\dfrac{n\pi}{3}+2n\theta\right)}{n}\,d\theta\\&= -\frac{\pi}{12}\sum\limits_{n=1}^{\infty} \frac{\sin (2n\pi/3)}{n^2}-\frac{1}{4}\sum\limits_{n=1}^{\infty} \frac{\cos (2n\pi/3)}{n^3} +\frac{1}{4}\sum\limits_{n=1}^{\infty} \frac{\cos (n\pi/3)}{n^3}\end{align*}

and, \begin{align*}&\int_0^{\pi/6} \theta\log \left(2\sin\left(\frac{\pi}{6}-\theta\right)\right)\,d\theta \\&= -\sum\limits_{n=1}^{\infty} \int_0^{\pi/6} \frac{(\pi/6 - \theta)\cos \left(2n\theta\right)}{n}\,d\theta\\&= -\frac{1}{4}\zeta(3)+\frac{1}{4}\sum\limits_{n=1}^{\infty}\frac{\cos (n\pi/3)}{n^3}\end{align*}

Hence, $$\sum\limits_{n=0}^{\infty}\frac{H_{n+1}}{(n+1)(2n+1)\binom{2n}{n}} = -\frac{2}{9}\zeta(3) + \frac{2\pi}{3}\sum\limits_{n=1}^{\infty}\frac{\sin (2n\pi/3)}{n^2}$$

Similarly we may deal with the second series:

\begin{align*}\sum\limits_{n=0}^{\infty} \frac{1}{(n+1)^2(2n+1)\binom{2n}{n}} &= 8\int_0^{1/2} \frac{\arcsin^2 (x)}{x}\,dx \\&= -4\zeta(3)+4\sum\limits_{n=1}^{\infty}\frac{\cos (n\pi/3)}{n^3}+\frac{4\pi}{3}\sum\limits_{n=1}^{\infty} \frac{\sin (2n\pi/3)}{n^2}\end{align*}

Combining the results we get:

\begin{align*}\sum\limits_{n=1}^{\infty} \frac{H_n}{(n+1)(2n+1)\binom{2n}{n}} &= \frac{22}{9}\zeta(3) - \frac{2\pi}{3}\sum\limits_{n=1}^{\infty} \frac{\sin (2n\pi/3)}{n^2} \\&= \frac{22}{9}\zeta(3) - \frac{\pi}{9\sqrt{3}}\left(\psi'\left(\frac{1}{3}\right) - \psi'\left(\frac{2}{3}\right)\right)\end{align*}

• Very nice, r9m. Turning it into a log trig integral is the ticket. :) – Cody Oct 24 '15 at 7:08
• @Cody Thanks! couldn't have done it without your help! :-) – r9m Oct 24 '15 at 7:25
• Good job!:D (+1) – user 1357113 Oct 24 '15 at 8:08

The easier one is in fact very much easier: just write \begin{align} \int_0^1\frac{\ln\left(x^2-x+1\right)dx}{x}&=\int_0^1\frac{\ln\left(1+x^3\right)dx}{x}-\int_0^1\frac{\ln\left(1+x\right)dx}{x}=\\ &=\int_0^1\frac{\ln\left(1+x^3\right)d\left(x^3\right)}{3x^3}-\int_0^1\frac{\ln\left(1+x\right)dx}{x}=\\ &=-\frac23\int_0^1\frac{\ln\left(1+x\right)dx}{x}=-\frac23\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k}\int_0^1 x^{k-1}dx=\\ &=-\frac23\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k^2}=-\frac{\zeta(2)}{3}=-\frac{\pi^2}{18}. \end{align}

• Nice swp. I derived it a different way by using $$Li_{2}(e^{\pm \pi i/3})$$, but your method is nice and straightforward. – Cody Oct 17 '15 at 8:03

Using the series $$2\sum_{k=1}^\infty\frac{H_kx^k}{k+1}=\frac{\log(1-x)^2}x$$ and the integral $$\int_0^{1/2}\left(\frac14-x^2\right)^k\,\mathrm{d}x=\frac{k!}{2^{k+1}(2k+1)!!}$$ we get \begin{align} \int_0^1\frac{\log\left(x^2-x+1\right)^2}x\,\mathrm{d}x &=\int_{-1/2}^{1/2}\frac{\log\left(x^2+\frac34\right)^2}{x+\frac12}\,\mathrm{d}x\\ &=\frac12\int_{-1/2}^{1/2}\frac{\log\left(x^2+\frac34\right)^2}{\frac14-x^2}\,\mathrm{d}x\\ &=\int_0^{1/2}\frac{\log\left(x^2+\frac34\right)^2}{\frac14-x^2}\,\mathrm{d}x\\ &=\int_0^{1/2}2\sum_{k=1}^\infty\frac{H_k\left(\frac14-x^2\right)^k}{k+1}\,\mathrm{d}x\\ &=\sum_{k=1}^\infty\frac{H_k}{k+1}\frac{k!}{2^k(2k+1)!!}\\ &=\sum_{k=1}^\infty\frac{2H_k}{(k+1)^2\binom{2k+2}{k+1}} \end{align} which converges at over $0.6$ digits per term to $$0.1041096792619493789449118629712286069593$$

• Thanks robjohn. Very insightful. I would not have thought of using that integral second from the top. Very clever, as usual. May I ask, is there some way we can now relate this to $$2/3\zeta(3)-4/3\sum_{n=1}^{\infty}\frac{1}{n^{3}\binom{2n}{n}}$$?. – Cody Oct 17 '15 at 4:35


text:
# A and B together can finish a work in 6 days. A alone can finish the work in 18 days. In how many days will B alone finish the work?

1. 6 days
2. 9 days
3. 12 days
4. 4 days

## Answer (Detailed Solution Below)

Option 2 : 9 days

## Detailed Solution

Given:

A and B complete a work in = 6 days

A complete the work in = 18 days

Calculation:

One day's work of (A + B) = 1/6

One day's work of B = 1/18

According to the question:

B's one day's work = 1/6 – 1/18 = (3-1)/18 = 1/9

∴ B can complete the work 9 days.

Shortcut Trick

If A and B together can do a piece of work in x day and A alone can do it in y days, then B alone
can do the work in xy/(y – x) days.
Given:

A and B together finish a piece of work = x = 6 days

A alone finish a piece of  work = y = 18 days

B alone can do the whole work in

⇒ 6 × 18/(18 – 6) = (6 × 18)/12 = 9 days

∴ B can complete the work 9 days.


text:
# zbMATH — the first resource for mathematics

##### Examples
 Geometry Search for the term Geometry in any field. Queries are case-independent. Funct* Wildcard queries are specified by * (e.g. functions, functorial, etc.). Otherwise the search is exact. "Topological group" Phrases (multi-words) should be set in "straight quotation marks". au: Bourbaki & ti: Algebra Search for author and title. The and-operator & is default and can be omitted. Chebyshev | Tschebyscheff The or-operator | allows to search for Chebyshev or Tschebyscheff. "Quasi* map*" py: 1989 The resulting documents have publication year 1989. so: Eur* J* Mat* Soc* cc: 14 Search for publications in a particular source with a Mathematics Subject Classification code (cc) in 14. "Partial diff* eq*" ! elliptic The not-operator ! eliminates all results containing the word elliptic. dt: b & au: Hilbert The document type is set to books; alternatively: j for journal articles, a for book articles. py: 2000-2015 cc: (94A | 11T) Number ranges are accepted. Terms can be grouped within (parentheses). la: chinese Find documents in a given language. ISO 639-1 language codes can also be used.

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Elliptic functions. (English) Zbl 0575.33001
Grundlehren der Mathematischen Wissenschaften, 281. Berlin etc.: Springer-Verlag. XI, 189 p. DM 138.00 (1985).

Nowadays there seems to be a renewed interest in the theory of elliptic functions. At the end of the 19-th and the beginning of the 20-th century there were classical books on the subject. We recall only H. Burkhardt and G. Faber, Elliptische Funktionen (1920); R. Fricke, Die Elliptischen Funktionen und ihre Anwendungen. I,II (1916- 1922); J. Tannery and J. Molk, Eléments de la théorie des fonctions elliptiques. I,II,III,IV (1893-1902).

Lateron we studied elliptic functions as a special subject in classical books on analysis, more special complex function theory; such as E. T. Whittaker and G. N. Watson’s, A course of modern analysis (1952; Zbl 0108.269), K. Knopp’s Funktionentheory II (1981; Zbl 0452.30001) and the many cours d’Analyse of the French mathematicians. And nearly always elliptic functions were also used as an introduction to modular functions, the last being an essential tool for number theory. In the classical book there was the application to the spherical pendulum.

In recent times interest was asked for the history of elliptic functions by the reedition of a chapter out of C. Jordan’s, Fonctions elliptiques. Reprint of ”Cours d’Analyse de l’école Polytechnique” (1981; Zbl 0472.33002) and the historical study of A. Weil, Elliptic functions according to Eisenstein and Kronecker (1976; Zbl 0318.33004).

Apart from these several modern books concerning elliptic functions were published. The volume under review can be compared with S. Lang, Elliptic functions (1973; Zbl 0316.14001) and B. Schoeneberg, Elliptic modular functions (1974; Zbl 0285.10016). Lang’s book is quite modern, some people will call it high-low, it deals also with recent research, we mention his chapters on Tate-parametrization, on Ihara’s theory, on modern aspects of complex multiplication and of Kronecker’s limit formula. Schoeneberg’s book is written on the more classical topics in the tradition set by Erich Hecke.

The author’s book is largely inspired by many ideas and contributions to the field made by C. L. Siegel. The book is published in the series ”Grundlehren”, its title could have been (following the French tradition) ”Elements of the theory of elliptic functions”. It gives a nice elementary introduction in the subject, presuppositing only the fundamentals of complex analysis. It deals with the elliptic functions as well in the Weierstraß setting, starting with the $\wp$-function and the thereof derived functions $\zeta$ and $\sigma$, as with the $\vartheta$-functions, as with the Jacobian functions sn, cn and dn and all the relations between these elliptic functions.

In the direction of modular functions the author deals with the modular invariant J($\tau \right)$ and with Dedekind’s $\eta$-function. And, of course, having this material at hand, the author gives a number of applications in number theory, e.g. the law of quadratic reciprocity using Gaußian sums, the representation of integers by a positive-definite quadratic form, such as the sum of four squares. For the beginning student and for the specialist in the field very extensive notes at the end of every chapter are of great interest. There is a lot of explicit mentioning of the work of many classical authors and of new proofs of classical theorems given in our generation. As was said above many of the new, shorter and more elegant proofs are due to C. L. Siegel. The last sentence of the author’s book reads:

It is these papers of Siegel, in which the methods of Minkowski and of Hardy are combined and adapted, about which H. Weyl (in his Ges. Abhandlungen, No.136 (1944/45), Bd. IV, p. 232) wrote: ”In a series of publications which continue the great tradition of Gauß, Dirichlet, Hermite, Minkowski and which I am sure will be admired by generations of mathematicians to come Siegel has vastly added to our knowledge in the theory of definite and indefinite quadratic forms and of arithmetical reduction”.

One could add that the work of A. Weil continuing in the great tradition of Siegel has once more vastly added to our knowledge in the theory of definite and indefinite quadratic forms and to our insight of the rôle of this theory as a special (and simple) case of a more general, group theoretical, adelic and algebraic-geometrical approach of number theory.

Reviewer: F.van der Blij
##### MSC:
 33E05 Elliptic functions and integrals 11F03 Modular and automorphic functions 11E16 General binary quadratic forms 11-02 Research monographs (number theory) 11P05 Waring’s problem and variants


text:
Article Text

Henry Currey FRIBA (1820–1900): leading Victorian hospital architect, and early exponent of the “pavilion principle”
Free
1. G C Cook
1. Correspondence to:
 Dr G C Cook, Wellcome Trust Centre for the History of Medicine at UCL, 183 Euston Road, London, NW1 2BE, UK;
 g.cook{at}ucl.ac.uk

## Abstract

The “pavilion plan” for hospital design originated in France in the 18th century and was popularised in England by John Roberton and George Godwin in the mid-19th century; the underlying rationale was that with improved ventilation the mortality rate (at that time exceedingly high) was significantly reduced. Among the enthusiasts for this new style, was Florence Nightingale (herself a miasmatist)—who had experienced astronomically high death rates in the hospital at Scutari during the Crimean War (1854–6). One of the leading exponents of this style of hospital architecture was Henry Currey (1820–1900) whose greatest achievement was undoubtedly the design for the new St Thomas's Hospital on the Lambeth Palace Road.

• Henry Currey
• hospital architect
• St Thomas's Hospital
• “pavilion principle” in hospital design

## Statistics from Altmetric.com

The Architect and Contract Reporter for 30 November 1900, recorded1:

“The profession of architecture in London has been deprived of a prominent representative by the death of Mr HENRY CURREY (fig 1). His St. Thomas's Hospital [built in 1868–71] is an enduring evidence of his skill, for, as it was one of the first to exemplify the advantages of the pavilion principle, it continues to be, after a quarter of a century, referred to as the [my italics] type of a modern hospital. Numerous as were the buildings Mr CURREY designed, they represented only a part of his activity. He was recognised as a sound adviser on all questions relating to property in building, and as an arbitrator his experience was of much value. Mr CURREY was indisposed to seek after prominence outside the professional limits he had fixed for himself, but as an architect he won the respect of all who consulted him or were associated with him in business”.

Figure 1

Henry Currey FRIBA, 1820–1900 (The Building News 1890;58:169).

The “pavilion principle” (which was introduced into England in the mid-19th century—before the “germ theory” of disease was accepted—and continued, without serious challenge well into the 20th) had been advocated by, among others, Florence Nightingale (1820–1910) (who was essentially a “miasmatist”) following her nursing experiences in the Crimean War (1854–6).2 This form of construction significantly improved the patients' chance(s) of survival (which had hitherto been depressingly poor in hospitals throughout Britain). The “pavilion plan” was both sanitary for the patient, and convenient for the nurse.2 It was dependent on the fact that there was a greater degree of separation and segregation, than was provided by earlier designs, together with greatly improved “ventilation”.

## BIOGRAPHICAL NOTES

Henry Currey was born in October 18203–8; he was the third son of Benjamin Currey, a solicitor of Old Palace Yard, Westminster, who had for “many years [been] one of the Clerks of the Table” in the House of Lords.7 Educated at Dr Pinckney's School at East Sheen, and later at Eton, he held the distinction of having rowed in the school eight against Westminster. For five years, he was articled to Decimus Burton (1800–81)9 who had himself designed the new Charing Cross Hospital, built in 1831–4; following this he “went into the office of . . . William Cubitt [1791–1863] & Co, Gray's Inn Road”, for nine months.6,7 He then travelled in Germany and Italy, and following his return began practice as an architect (from his residence in Brook Street, Grosvenor Square) in 1843. He “obtained the first premium” in competitions for “the erection of houses and terraces in Toxteth Park, Liverpool”, and also “the enlargement of the Surrey County Lunatic Asylum”.6,7

In 1847, Currey was appointed architect and surveyor to the Governors of St Thomas's Hospital (a post which he held until his death) and moved to offices at 4 Lancaster Place, Strand; when that building was acquired by the Metropolitan Board of Works (for an approach to the Embankment), he transferred to 37 Norfolk Street, Strand.5–7 Currey's architectural practice was “very considerable and varied”,7 his major work being the new St Thomas's Hospital following its removal from the Borough (as a result of the enlargement of London Bridge Station). This new terminus had been designed by Henry Roberts in 1843–4, and the expansion “occasioned the destruction of nearly all the St. Thomas's Hospital buildings . . . to which [Sir Christopher] Wren, in 1692–1711, [the hospital's surveyor] Samuel Robinson, [and his partner] James Field in 1814–42, Sir Robert Smirke [1781–1867] . . ., and Sir Charles Barry [1795–1860] . . ., had in succession contributed various portions”.5 He also became architect and surveyor of the Foundling Hospital—where he “carried out extensive works in the schools, chapel, and new infirmary”.6,7 Another appointment was as architect and surveyor to the Magdalen Hospital, Streatham.

Among other works (not connected with hospitals) were: “sundry country houses”—including one at Leigh, near Reigate (for James Freshfield), and one at Buxton (for John Shaw, 1803–70); he was also responsible for hotels at Buxton, Eastbourne, and London Bridge (1861–2),6,10 a large bathing establishment and pump room at Buxton (1875), and the Peninsular and Oriental Company's offices in Leadenhall Street, together with “many other commercial buildings in the City and Southwark”.6,7 Currey also erected “churches at Burbage [Derbyshire], Buxton, Chiswick, Notting Hill, and St. Peter's, Eastbourne”. He laid out the Duke of Devonshire's building estate at Eastbourne, and “executed large works on sea walls, terraces, &c”.6,7 He built “sundry houses at Eastbourne, the Pavilion and Theatre at Devonshire Park [also at Eastbourne], the College Building Chapel, &c”. Among other works, Currey designed “some warehouses in St. Thomas-street, Southwark . . .; St. Paul's Church [near Chiswick railway station]; and a lych-gate at St. Bartholomew's Church [Sydenham]”.5 Together with his son (see below), “he carried out, in 1898, repairs and superintended the restoration of the church of St. Clement Danes [Strand], and prepared the plans and designs for the . . . reconstruction and extensive enlargement of the Queen Anne's Bounty Offices in [Westminster], established in 1704”.5 This father-son partnership “acted also as honorary architects in respect of the planning and designing of the St. Clement Danes and Clare-market Parish Hall, 1897–8”.5

Currey also seems to have been in great demand as an “assessor [to the Metropolitan Asylums Board (MAB) which had come into existence in 1867]5 in important competitions, in valuations for compensation, and as an arbitrator”.6,7

His association with the (Royal) Institute of British Architects (RIBA) began in 1848; he was elected Associate on 20 November of that year, and on 14 April 1856 the fellowship [he was proposed by Decimus Burton,9 John Shaw and Charles Barry Jr] followed; he was a member of Council for many years, and served as Vice President on two occasions: 1874–7 and 1889–93.5 He was also a Fellow of the Surveyors' Institution and an Associate of the Institution of Civil Engineers (from 4 April 1848).5,6 Currey was later a trustee of the RIBA, and “took an active interest in the work of the Architects' Benevolent Society”.

In 1845, Currey married the youngest daughter of Sir Charles Price Bt (1800–72).6,7 Percivall Currey (1851–1918),8 one of his sons, also became an architect. Currey died at his home, The Chestnuts, Lawrie Park, Sydenham on 23 November 1900.4–6

Although it is widely assumed that architects and members of the medical profession have always worked closely together, satisfactory examples of close cooperation are few and far between2; Currey was an excellent example!

## SANITATION, VENTILATION, AND THE “PAVILION PRINCIPLE”

Nightingale was anxious to incorporate sanitary reforms (especially improved ventilation) into hospitals (earlier advocated by Southwood Smith11), probably largely as a result of the very high mortality rate (due principally to poor hygiene—as first suggested by the statistician William Farr FRS [1807–83]12) in her hospital at Scutari. She had initially attributed the fact that the mortality rate at that hospital was up to three times higher than in primitive front line hospitals in the Crimea, to overcrowding and/or the advanced state of the cases, rather than to poor sanitation.13

Nightingale was by no means the originator of the “pavilion principle” however; in Britain this must go to the surgeon (appointed to the Manchester Lying-in Hospital in 1827) John Roberton MRCSE (1797–1876)14,15 together with the editor of The Builder, George Godwin FRS (1815–88)—who publicised the “plan”; both blamed inadequate ventilation and poor planning for the prevailing high mortality in British hospitals. Roberton began his first paper14:

“the air we breathe is called in the Scriptures the breath of life'; when we cease to respire it, we die. If we should inhale some other kind of air, or atmospheric air much contaminated, this would prove to us, not the breath of life', but that of sickness or of death. How significant the words! The atmosphere is God's common gift, as free to the peasant as to the prince; and yet, strange to say, in a large proportion of human dwellings, and of vessels as they navigate the sea, the admission and circulation of the air are unduly limited and impeded; and not seldom, the supply, such as it is, is allowed to become so impure as actually to generate disease”.

Plentiful fresh air and separation would (in Roberton's view) “rid the hospital environment of the fatal miasmas responsible for disease”.2 The inspiration behind this plan, in England, had been provided by the Royal Naval Hospital, Plymouth (1758–62)—the design of which was published by John Howard (1726–90) in 1784. This “principle” had developed in France; when fire destroyed much of the Hôtel Dieu in Paris (in 1772) the designers for its replacement were Jacques Tenon (a medical practitioner) and Bernard Poyet (an architect).2 Rebuilding was delayed by several decades however, by the French Revolution. In the period 1820–50, that is, before the Hôtel Dieu could be completed, several French hospitals—including the Lariboisière (the best known example)—had been built on these lines (fig 2). The essential prerequisites were: sufficient space between the pavilions to provide good light, coupled with a free circulation of air.

Figure 2

Ground plan of the Lariboisière Hospital.18

Nightingale's Notes on Hospitals (1859) (fig 3),16 much of which can be attributed to her disastrous experience(s) at Scutari, included as an “appendix”, three articles reproduced from The Builder—which advocated this revolutionary principle. It was, she wrote, then acknowledged by careful observers that “the origin and spread of fever in a hospital, or the appearance and spread of hospital gangrene, erysipelas and pyaemia generally, are much better tests of the defective state of a hospital than its mortality returns [my italics]”. The four basic defects in hospital design were, in her opinion: (i) the “agglomeration of a large number of sick under the same roof”, (ii) a deficiency of space (which correlated closely with deficiency of ventilation), (iii) deficiency of ventilation, and (iv) inadequate light. In part II of this book, Nightingale departed from general principles, and focused on many details of hospital design—most of which were taken into account by Currey in his design(s) for the new St Thomas's Hospital (see below).

Figure 3

Title page of Florence Nightingale's Notes on Hospitals published in 1859.16

In 1858, the first two pavilion plan hospitals in England were begun: Blackburn Infirmary (architect: James Turnbull)15 and the Royal Marine Barracks Hospital, Woolwich. Very rapidly, the new concept was adapted to accommodate various categories of patient. The key was separation; cross ventilation became the watchword of hospital architects2! The two most important examples of early pavilion plan hospitals (both of which were of substantial size) were: the (Royal) Herbert Hospital, Woolwich (begun in 1861), and the new St Thomas's Hospital (begun in 1868) (see below); at the outer ends of the pavilions were twin towers—which housed the sanitary facilities (and in the latter years of the 19th century were considered to provide the key to public health). Very rapidly, use of the “pavilion plan” spread throughout the country, and was also incorporated into numerous hospitals throughout Britain's colonies; it could be either grand (Italianate or Gothic) or “cheaply functional” in style; the latter option apparently appealed to parsimonious Poor Law guardians.2 Initially, beds were placed in pairs between the windows (as recommended by Tenon and Poyet); however, later only one bed occupied the window pier, and the sanitary facilities (bath, sinks, and lavatories) became increasingly separated from the ward itself. Introduction of the principle failed, however, to abort the relentless flow of articles, pamphlets, and books on: ventilation, sanitation, and heating. By judicious use of fires, it was considered possible to produce air drafts which would remove “noxious vapours” or miasmas from the ward(s). Cleanliness and fire prevention were also high on the agenda of hospital design.

The “pavilion plan” therefore became (in the latter half of the 19th century) the universally accepted design for hospitals throughout the country, and “inevitably produced a similarity in contemporary [hospital] appearance”.2 An enormous national campaign—involving medical (including Roberton) and architectural authorities—was mounted in 1856 in opposition to the plans for the new army hospital at Netley (on Southampton Water) which was based on the prevailing (early 19th century) hospital architectural design, and consisted of numerous small wards which opened off corridors (the “corridor plan”).17 This led to the appointment of a Sanitary Commission to “inquire into the Regulations affecting the Sanitary Condition of the Army, the Organisation of Military Hospitals, and the Treatment of the Sick and Wounded”; it reported in 185818—and severely criticised the plans for the new Netley Hospital; it also gave official sanction to the “pavilion principle”.2

The advent of the “pavilion principle”, which coincided with a sharp decline in hospital mortality rates, led to a major influx of inpatients (many of whom had of necessity to be excluded from admission); this gave rise to the establishment of specialist hospitals—which proliferated in the late 19th century. “General” hospitals, were at this time, also forced to open their own specialist departments.

Although John Bristowe FRCP, FRS (1827–95) (Physician to St Thomas's Hospital) and Timothy Holmes FRCS (1825–1907) (Surgeon to St George's Hospital) had already written a report for the Privy Council (1864) on the hospitals of Britain, a far more thorough text, Hospitals and Asylums of the World, was published in four volumes by (Sir) Henry Burdett (1847–1920)19 in 1891–3; this work was accompanied by a portfolio of plans compiled by the architects Keith Young FRIBA (1848–1929) and Henry Hall FRIBA (1826–1909).

## THE NEW ST THOMAS'S PROJECT

Currey's magnum opus was undoubtedly the new St Thomas's Hospital (built in the Italianate style) established on the Albert Embankment “of which the long range of pavilions [to this day], constitutes a prominent feature of the riverside view as seen from Westminster Bridge, and forms a striking contrast to the opposite Houses of Parliament”.5 The long narrow site (much of it reclaimed land) of course ruled out a courtyard arrangement (as exemplified by the Lariboisière hospital, fig 2), and instead Currey was to set six ward pavilions side-by-side at right angles to the river—each with “paired sanitary towers facing the seat of government”.2 The axial corridor, services, theatres, outpatient department, and staff accommodation were arranged on the eastern side of the site—on either side of a huge entrance hall, above which was the chapel. The main administrative offices (governors' hall and committee room included) were to be in a separate block to the north. The medical school (with all of its facilities, including the museum) would be built on the southern wedge of the site.

In June 1865, Currey presented an overview of his intended design (accompanied by a great deal of detail) in a memorandum addressed to the “Grand Committee” of St Thomas's Hospital20; “The prominent defect of the Lariboisière Hospital—the too close proximity of the blocks with reference to their height—is however avoided”, wrote Currey in his opening remarks. The Governors (who had already decided on the “detached pavilion” style) had already bought the site (8½12 acres), half of it having been “reclaimed from the foreshore at Stangate . . . for 90,000 l [£]” from the “late Metropolitan Board of Works”.5 The total frontage of the hospital was to be 900 ft; the blocks were to stand 125 ft apart; and the central space 200 ft in length. It was “designed for 600 beds in wards measuring 28 ft. by 120 ft., and 15 ft. high, for twenty-eight beds apiece, allowing each patient a space of 1,800 cubic feet”.5,19 “At the south end [were to be the medical] schools and a museum [the dimensions of which were to be] 85. ft by 30 ft. and 34 ft. high, with two galleries”. At the footbridge there would be “offices and separate residences for the treasurer and four resident officers”.5

Ventilation was thus preoccupying hospital architects at this time, although by the early years of the 20th century the beneficial effects of sunshine and fresh air also entered the equation; open-air treatment of pulmonary tuberculosis was but one example.21 It was slowly to become clear that “no amount of cross ventilation could kill germs”2; with the gradual acceptance of the “germ theory” this was to play a lesser part in 20th century hospital design.

Figure 4

Currey's plan for the new St Thomas's Hospital (The Builder 1865;23:556).

Figure 5

Laying the Foundation Stone of the new St Thomas's Hospital by Queen Victoria, 13 May 1868 (Illustrated London News 1868;52:508–9).

“It was determined to depend as much as possible on natural [my italics] ventilation, avoiding all costly arrangements and fanciful theories, at the same time providing the means of changing the air during cold and boisterous weather and at night.22 . . . The boilers for warming purposes would not [he stressed] be available in the summer, but the furnace for the supply of hot water and baths would be continuous in its operation. In the upper part of [the main extraction] shaft [was] also placed the hot water cistern, and if found necessary, hot water coils [would] be added to assist the rarefaction. Shafts [were] carried from the ends of all the wards, both at the ceiling and floor level ... To replace the air thus extracted, fresh air [would be] introduced by means of zinc tubes laid between the Dennett' [concrete] arching23 and the floor boards communicating with the stoves and hot water coils, thus passing over a cool surface in summer, and tempered in winter by contact with the heated surfaces before entering the wards, the whole admitting of regulation by valves”.

“Each pavilion [would have] its independent means of warming and ventilating, avoiding as much as possible all complication in the arrangement, but the pipes [would be so] arranged that in the event of a break-down in any one block, its neighbour [would] come to its assistance during [the] temporary failure. . . . The wards generally are warmed [he continued] by three open fireplaces, aided in cold weather by an auxiliary system of hot water. The corridors and staircases are also warmed by hot water. The open fireplaces [which stood in the middle of the wards] might have been arranged against the outer walls, but bed space would thus have been sacrificed. The hot metal [of the shafts would] not . . . come in contact with the atmosphere of the wards, but the space between the two tubes [would be] an efficient ventilating shaft, . . . connected as before with the main trunk in [the] roof. . . . A cast iron socket [was] built into each floor, supported on two small bearers running from girder to girder, and the “Dennett” arching23 . . . made good to the same all round, thus avoiding any communication from floor to floor”.

“There [was also, Currey continued] one food lift to each block [arranged] upon the rack and piston principle. . . . and these lifts [would be] worked from the same tank as the passenger lifts. [Currey concluded this section of his address:] the whole of the work [is] of the most substantial character, and nothing has been neglected to ensure thorough efficiency, combined with perfect safety”.

The overall cost of the project (excluding the site) amounted to about £400 000 in total (or £650 per bed).

Figure 6 shows the opening ceremony, carried out by Queen Victoria on 21 June 1871, and fig 7 the “pavilion style” St Thomas's Hospital shortly after completion.

Figure 6

Opening ceremony of the new St Thomas's Hospital by Queen Victoria, 21 June 1871 (Illustrated London News 1871;58:640).

Figure 7

The completed “pavilion planned” St Thomas's Hospital (Illustrated London News 1871;58:616–7).

Despite all Currey's reassurances, a report on the sanitary state of the hospital (in 1878) “revealed that the new building was far from hygienic”.24 In this “most expensive-to-run hospital, windows were difficult to open, chutes for soiled linen were not used, buckets were without lids, lifts were not used because they took so long and chamber-pots were left under beds hidden by the pinned-down quilts”. Figure 8 shows a letter from Currey to Henry Bonham-Carter (1827–1921), outlining a sanitation problem which had occurred 14 years after the opening of the new St Thomas's Hospital!

Figure 8

Letter from Currey to Henry Bonham-Carter dated 1885 which refers to on-going sanitation problems at the new St Thomas's Hospital—some 14 years after its official opening (London Metropolitan archive document: A/NFC 83/19, reproduced with permission).

## INVOLVEMENT WITH THE SEAMEN'S HOSPITAL SOCIETY

In 1867, the Seamen's Hospital Society (SHS) had decided to remove its clinical facilities from the third (and last) of the Hospital-ships to dry land25; however, their committee had not yet determined the site for a definitive location. A plot of ground had been purchased (to the east of Trinity Hospital [now Norfolk College], Greenwich, and bounded by Woolwich Old Road and ground belonging to Morden College; fig 9); the following is recorded in an SHS minute dated 8 November 186726:

Figure 9

Plan of the Seamen's Hospital Society's site (c 1867) on which it was intended to build a new land based hospital (Seamen's Hospital Society archive, reproduced with permission). This strategy failed to materialise, and the plot of land was sold in 1880.

“The Secretary reported that he had only this morng. obtained the plan of the Hospital Site . . . It was . . . arranged that Mr. McGrouther would bring with him to the Committee at their next Meeting, the Treasurer or Architect of Saint Thomas's Hospital . . . to give the Committee some advice as to the Course they should persue”.

At the following meeting27:

“The Committee saw Mr [Henry] Currey, the Architect of Saint Thomas's Hospital now building—a plan of the site belonging to this Society was given him and he promised to go over the ground and give the Committee an outline sketch of what he might Consider the best mode of arranging a Hospital thereon and an estimate of the probable Expense of its Erection . . .”.

Currey again attended at one of the three December meetings28 when he:

“laid before them plans for a hospital on the Society's site and he read a statement explanatory whereof. Ordered that the same [that is, the plans, which seem unfortunately to have been lost] do lie on the table for the Consideration of the Members of the Committee for the next few weeks”.

The fact that this (like St Thomas's) was a difficult site on which to build a hospital, coupled with the desires of the SHS Committee for a “pavilion planned” hospital, undoubtedly lay behind the decision to invite Currey (and/or the Treasurer of St Thomas's Hospital) for advice; being of a generous disposition, he seems to have been pleased to be of assistance to this great charitable foundation—apparently at no charge!

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