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52

For a visual intuition.... The inner blue region increments the gray square by one, but to make the next increment you need the extra yellow unit squares to get to the following square.


32

I'm guessing too, but my guess is that it has something to do with the fact that beyond his initial introduction to calculus Turing (in common with many of us) thinks of a function as the entity of which a derivative is taken, either universally or at a particular point. In Leibnitz's notation, $y$ isn't explicitly a function. It's something that has been ...


28

I'll try my hand and give one possible rationale behind Turing's confusion over the notation $ \tfrac{\mathrm{d}y}{\mathrm{d}x} $. The short answer is that he appears to take issue with the notation on the grounds that differentiation is a mapping between two function spaces but $y$ looks like a variable. To answer the first part of your question regarding ...


16

We have the series $$ x_n = n^2 $$ and the differences $$ \Delta x_n = x_{n+1} - x_n = (n+1)^2 - n^2 = n^2 + 2n + 1 - n^2 = 2n + 1 $$ and the difference of differences $$ \Delta x_{n+1} - \Delta x_n = (2(n+1)+1) - (2n+1) = 2n + 3 - 2n - 1 = 2 $$


10

I'm no expert on Alan Turing at all, and the following will not directly answer your question, but it might give some context. Following the link provided, I also found the following page, which might shed some more light on things: It says the following (my apologies for any errors in transcribing, I'm grateful for any corrections): A formal ...


10

Yes. A circle is a special case of an ellipse. The equation of an ellipse is: $$\left(\frac xa \right)^2 + \left(\frac yb \right)^2=1$$ When $a=b=1$, this gives the equation of the unit circle: $x^2+y^2=1$. In general, if $a=b=r$, you get the equation of a circle with radius $r$: $$x^2+y^2=r^2$$ Can you come up the equation of a circle not centered at ...


8

Both the foci of a circle coincide and thus, its eccentricity is zero. So yes, it is an ellipse. It is like that all squares are rectangles but all rectangles are not squares. See this link- [the most intuitive link ever seen!] http://www.mathsisfun.com/geometry/eccentricity.html


7

You shouldn't think of a group as a thing in the same way as a vector space is a thing. Groups are not things, groups act on things. If $V$ is a vector space, then the collection of invertible linear transformations $V\to V$ is a group. If $X$ is a set, then the collection of all permutations of $X$ is a group. If $A$ is any object, in any category, then ...


7

This phrase from the description catches my eye, and prompts me to provide a different perspective from the other answers. This, Turing's wartime notebook on logic, is the first time a manuscript by him has ever come to public market. The concept of a derivative is a rather natural notion, and it is easy for young students of mathematics to intuitively ...


7

To expand upon zoli's reply a bit, the Leibniz notation intuitively makes sense in that the derivative of an explicit formula is algebraically defined as $\frac{\Delta y}{\Delta x}$ for any $(x, y)$ pair around which the function is defined (i.e. 'the relation is laid down'). Of course it's all speculation, but perhaps Turing meant to say that the notation ...


7

We, who learn differentiation based on calculating the limit of quotients of differences, consider the Lebniz notation to be only a notation. We do not try to grab the intuitive meaning of $dx$ and $dy$. We don’t even care. I guess that Turing was trying to understand how Lebniz could reach his results based on this vague notation. I guess that he refers to ...


7

The difference between two adjacent squares, $(a+1)^2$ and $a^2$: $$\begin{align} (a+1)^2 - a^2 &= a^2 + 2a + 1 - a^2 \\ &= 2a + 1 \end{align}$$ So for each increase in $a$ by $1$, the delta between squares gets $2$ bigger.


6

We define an irrational number to be a (real) number that isn't rational, which is equivalent to defining $$\Bbb I := \Bbb R - \Bbb Q,$$ and hence $$\Bbb Q \cup \Bbb I = \Bbb Q \cup (\Bbb R - \Bbb Q) = \Bbb R.$$It is correct, then, to say that $\Bbb Q \cup \Bbb I \subseteq \Bbb R$, but this statement contains less information than our equality. The ...


6

Actually your question is the combination of two questions. About your first question: You certainly know the Euclidean norm, $$\left\|v\right\|_2 = \sqrt{\sum_{k=1}^n \left|v_k\right|^2} = \left(\sum_{k=1}^n \left|v_k\right|^2\right)^{\frac{1}{2}}$$ derived from the Euclidean scalar product. Note that I used absolute value bars here, which are superfluous ...


5

Let's just think about norms in the plane. The ordinary Euclidean norm is $(x^2 + y^2)^{1/2}$. The unit circle for that norm is the ordinary unit circle. There are other useful norms. One is $|x| + |y|$. Its "unit circle" is the square with vertices $(\pm 1, 0)$ and $(0, \pm 1)$. These norms are part of a family of norms defined by $\|(x,y)\|_p = (|x|^p + ...


5

First, a bit of pedantry (sorry, this isn't related to what you ask but somebody had to say it). You write later in the question that the set $\{\Bbb R\times \Bbb R\}$ is mapped to the set $\{\Bbb R\}$. But these are sets of one element, and those elements are the sets $\Bbb R \times\Bbb R$ and $\Bbb R $, respectively. Those two sets (the curly bracket ones) ...


4

Given a real number, if it is rational then it is in $\Bbb Q$, otherwise it is irrational and in $\Bbb I$. So $\Bbb{Q\cup I=R}$. Similarly, if a real number is not algebraic then it is transcendental. The set $\Bbb I$ is essentially defined as $\Bbb {R\setminus Q}$ and $\Bbb T$ is defined as $\Bbb{R\setminus A}$. This is analogous to the following ...


4

For a coordinate transformation $x_i \to x_i'$ the determinant of the matrix of the partial derivatives gives the correction factor for the volumes: $$ dV = dx_1 \cdots dx_n = \mbox{det } \left( \frac{\partial(x_1, \ldots x_n)}{\partial(x_1', \ldots, x_n')} \right) dx_1' \cdots dx_n' = \mbox{det }(J) \, dV' $$ where $J$ is called the Jacobian matrix, with ...


4

Let $\mathcal C$ be a category. A Cartesian square is a commutative square$\require{AMScd}$ \begin{CD} A @>a>> B\\@VVbV @VVcV\\ C @>d>> D \end{CD} that is universal with respect to all commutative squares having the same bottom right hand corner. This means that if we have any commutative square of the form \begin{CD} F @>f>> ...


3

Interestingly the history of Calculus is a bit backwards from the way it is taught in schools. It began with the investigation into finding areas of certain geometric shapes, a field of Mathematics called Quadrature. This led the way to integration theory. One landmark result which is by Fermat, when he demonstrated: $$\int_0^a x^n dx = ...


3

To a question in title No, $2^A \not\subseteq A$ because a power set of any $A$ has a cardinality strictly greater than $A$ itself. To a question in text No, $f(a)\subseteq 2^A$ does not hold in general, as $f(a)$ is by definition an element of $2^A$, and an element of a set is not (in general) a subset of the same set. Althoug in some special cases it ...


3

hint: $\dfrac{3^x}{e^{x-1}} > \dfrac{3^{x-1}}{e^{x-1}} = \left(\dfrac{3}{e}\right)^{x-1}$


3

When you have a set A with $|A|=N$, we have $|\wp(A)|=2^N$, 2 to the power of $N$. That lead to the name power set. The notation $2^N$ came from the fact that this was the case when $A$ is finite. In German, a Potenzfunktion is a fuction in the form of $X^n$. The name Potenzmenge (power set) appears to come form Untersuchungen über die Grundlagen der ...


3

$\sqrt2$ is just an arbitrary number that, when squared, produces $2$. Similarly for $\sqrt 3$. That is, although we define them as the positive roots, there is no algebraic distingtion between the positive and the negative root. Thus for any rational numbers $a,b$ for which $a+b\sqrt 2$ is "a" root of $3$, we should expect that also $a-b\sqrt 2$ (i.e., ...


3

Here's a derivation without trigonometry. There is also a short discussion of the geometric significance of the exponents in the answer after the derivation. Let $L$ be the length of a line touching the inner corner of the two corridors and extending to the outer wall of each corridor. Where this line meets the outer wall of each corridor, draw a line ...


3

It's essentially the very definitions of numbers involved. $\Bbb Q$ is the set of all real numbers that can be expressed as the ratio if two integerss. We define a real number to be irrational if it's not rational, so that $\Bbb I = \Bbb R \setminus \Bbb Q$ by definition, hence $\Bbb R = \Bbb Q \cup \Bbb I$. Similarly, a real number is called algebraic ...


3

As Matt Samuel noted, this is because we use multilinear forms to measure area. Why multilinear forms? Because this takes into account not just area but orientation. Let's take a little jaunt through what an area function should be and why this involves multilinear algebra, and we'll finish with the general formula for a determinant. Properties of an area ...


3

When I learned complex analysis, this integral was called the winding number, and it counts how many times $\gamma(t)$ "goes around" the point $a$ on the way round the curve -- with sign, so going counterclockwise around $a$ counts as $+1$ and going clockwise counts as $-1$. This is most easy to explain in the simple case that $a=0$ and ...


3

Cardinality, Bijections To show that $\mathbb{R}^2$ has the same cardinality as $\mathbb{R}$ you have to show that a bijection $f$ between both sets exists. $$ f : \mathbb{R}^2 \to \mathbb{R} \\ $$ To what particular element from $\mathbb{R}$ a pair from $\mathbb{R}^2$ is assigned does not matter, only that $f$ is bijective. $$ \forall (x,y), (x', y') \in ...



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