## Calculation models – Turing: Are the "m configurations" in your original article the same as the "media" in your definition of "computable"?

Are Turing's "m configurations" the same as "media" in their original definition of "computable"?

In the first line of Turing's article "On computable numbers …", define a "computable" number as follows:

A real number "for which the decimal representation can be calculated by finite means".

My question is; Are they "means" to which only the number of configurations m that you define shortly after the first line of your article actually refers?

It seems that the "media" you are mentioning are the real m configurations, and I am trying to understand the difference between these and how the "steps", the "m configurations" and the "calculations" are related.

As I understand it, it works like this;

A real number R is computable <=> There is a "machine" with a finite number of "m configurations" that can be used to print the decimal representation of R, even if the actual number of numbers printed on the "tape" to represent the decimal value is not finite.

– Therefore, the "means" are the number of configurations m, and if R is computable, the machine can still perform an infinite number of actions to calculate the decimal representation of R, provided there is only a finite number of configurations m that produce the (infinite number) of actions.

## linear algebra: calculation of the position of a third-person camera from its right vector

I have a third-person camera, which always looks at the player from a distance. Update its x and z position around the player in relation to the yaw value received from the x-axis mouse input.

I am trying to add collisions to the camera in the third person. The collision algorithm accesses the speed of the objects to update their position, so that for the collision to work in the third-party camera, the position cannot be updated, otherwise the collision will not work.

Therefore, I have a "functional" implementation in which I update the right vector of the camera while it is pointing towards the player and provides the desired orbit effect. But I am scaling this right vector with the x-mouse scroll to recalculate the new camera location, and when I move the mouse quickly, it "cuts corners" in the orbit. Here is a diagram of what I mean, I hope it makes sense. But I would like to get suggestions on how to have a smooth circular orbit and for the orbit to accelerate when the displacement x increases depending on the correct vector.

And here is the code:

``````right = cross(camera.front, camera.up);
right = normalize(right);
right = scale(right, 2.0f * deltaTime * xOffset);
cameraCollider.velocity -= right;
``````

Thanks for your time.

## Calculated column – SharePoint 2010 string calculation

I have the following string stored in a SharePoint column:

11608,613,12,12643,12644,12656

I want to extract the value after the last comma. That is, I just need to extract 12656 and use it for other operations.

How to achieve this using the SP 2010 Designer workflow or the calculated column?

## calculation: is there any geometric explanation for the derivatives of cosecant and cot?

I am trying to understand the calculation intuitively. Therefore, I am looking for a geometric explanation for those derived from trigonometric functions. Through a stack exchange response I understand the geometric explanation of the 4 trigonometric functions, but from that answer I could not understand those derived from sellers and cot functions. Then, give a geometric explanation for the harvesting functions and specifically cot.

## Lunar ray damage calculation for multiple targets

Suppose my character casts a moonbeam and two enemies begin their turn within the radius of the beam. Would each creature's damage be calculated separately (2d10 for one and 2d10 for the other) or would its damage be calculated along with a single roll?

## Entering heavy plating mode does not have a minimum force score required

Warforged Integrated Protection presents states:

[…] You can alter your body to enter different defensive modes; every time you finish a long break, choose a mode to adopt from the Integrated Protection table, as long as it meets the prerequisites of the mode

The prerequisites for each mode are listed in the table provided and the only prerequisites for entering the heavy coating mode are:

Heavy armor competition

This means that this is the only Prerequisite, nothing more. As a result, no force is required to activate heavy plating mode

## calculation and analysis: how to shorten the length of the tangent line

I have all the equations down to manipulate the tangent line in the graph … but the tangent line is too long, what would be the best way to shorten it?

f (x_) = Sin (x);

Plot (f (x), {x, -2 Pi, 2 Pi})

tangent (f_, a_, x_): = f & # 39; (a) (x – a) + f (a)

``````Manipulate(Plot({f(x), tangent(f, p, x)}, {x, -2 Pi, 2 Pi}, Epilog -> {Red, PointSize(.015), Point@{p, Sin(p)}}), {p, -Pi, Pi})
``````

## calculation – Confusion with variables in addition and integration

As for the following sum and integration: I don't understand the difference between t and 𝛕.

When nΔ𝛕 reaches zero, the text says that it becomes 𝛕. But why not t.

I am missing something fundamental here, but I was left without seeing the difference. How can we catch the difference?

## Calculation – How does the rearrangement of the logistic regression derivative work?

I am trying to understand the derivative of the logistic regression loss function described by Dan Jurafsky in his book Speech and Language Processing (Draft for Third Edition Chapter 5.8).

I can follow most of your reasoning. I only have problems with the step in equation 5.41 and how it reaches 5.42:

Rearrange the terms after registration derivation

I guess what I don't understand is how to delete the derived statement.
Why $$frac { partial} { partial w_j} sigma (w cdot x + b)$$ it's not the same as $$frac { partial} { partial w_j} 1- sigma (w cdot x + b)$$.

## Preliminary calculation of algebra: what is n ^ n + 1 cousin for?

This problem is related to the February 2016 algebra problem of HMMT.

https://hmmt-archive.s3.amazonaws.com/tournaments/2016/feb/alg/solutions.pdf

How do you find a test for all the numbers n that satisfy $$n ^ n + 1$$ is he cousin?

First, all odd numbers do not work; it is well known that $$x ^ odd + 1$$ It has a factor of $$x + 1$$.

So, all numbers with some odd factor don't work. For a number $$n$$ which can be factored in $$f1$$ and strange $$f2$$, we can group the expression to become $$(n f1) f2 + 1$$, which has a factor of $$n f1 + 1$$.

Then our number must be a power of $$2$$. But not all the powers of $$2$$ job. If we have $$n = 2 ^ x$$Y $$x$$ can be factored in $$f1$$ and strange $$f2$$, then we can group again, as $$n ^ n + 1$$ is $$(2 ^ x) ^ n + 1$$, which $$2 x * n + 1$$, which $$(2 f1 * n) f2 + 1$$.

Then our number must be of the form $$2 ^ x$$. When testing the first numbers, 1 (technically not of the form), 2 and 4, all give prime numbers. Is there any way to show that all numbers on this form work? Or, if false, find and test a set of solutions that meet the requirement?