## Show the following: \$ Theta (n log n) cup o (n log n) neq O (n log n) \$

Show that: $$Theta (n log n) cup o (n log n) neq O (n log n)$$

I've tried to do this in many ways, but I do not really know how … intuitively $$Theta cup o = o$$? So that would mean that I just had to show it $$o (n log n) neq O (n log n)$$ Which would be easier, I think. But I do not know how to do that formally.

## Proofing – How to prove the function of a recursive big theta without using repeated substitution, mastering the sentence, or having the closed form?

I have defined a function: $$V (j, k)$$ Where $$j, k in mathbb {N}$$ and $$t> 0 in mathbb {N}$$ and $$1 leq q leq j – 1$$, Note $$mathbb {N}$$ includes $$0$$,

$$V (j, k) = begin {cases} tj & k leq 2 tk & j leq 2 tjk + V (q, k / 2) + T (j – q, k / 2) & j , k> 2 end {cases}$$

I am not allowed to use repeated substitution and I want to prove this by induction. I can not use the main clause because the recursive part is not in that form. Any ideas how I can solve it with given limitations?

When I start induction: I fix $$j, q$$ and introduce $$k$$, Then the base case $$k = 0$$, Then $$V (j, 0) = tj$$, The question indicated that the function may be $$Theta (jk)$$ or maybe $$Theta (j ^ 2k ^ 2)$$ (but it does not necessarily have to be either).

I choose $$Theta (j, k)$$, In the base case, this would mean that I had to prove that $$tj = theta (j, k)$$ when $$j = 0$$, But if I start with the big-oh, I have to show it $$km leq mn = m cdot0 = 0$$ which I currently do not think possible.

I am not sure if I did the basic case wrong or if there is another approach.

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## Calculus and Analysis – How is the derivative of the theta function numerically evaluated?

I come across this derivative, but Mathematica does not evaluate the following:

``````ClearAll[f, g];
a = 0.1;
b = 0.2;
t = Exp[I 2 Pi/3];
f[z_] := SiegelTheta[{{a}, {b}}, {{t}}, z]
g[z_] = D[f[z], z]
g[0.1] // N
``````

I am not sure if this derivation can actually be evaluated to a numeric value …

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## Convergence – finding the probability bound of \$ n ^ {1/4} has { theta} \$

Suppose that $$n ^ {1/3} has { theta} = O_ {p} (1)$$,

I want to find the probability limit of $$n ^ {1/4} has { theta}$$ ,

I have to use the following definition and lemma:

Definition: A sequence of random variables $$left {x_ {n} right }$$ is limited in probability if and only if $$epsilon> 0$$Is there a $$b _ { epsilon} < infty$$ and an integer $$N$$ so that $$P left ( left | x_ {n} right | geq b _ { epsilon} right) < epsilon$$
for all $$N geq N _ { epsilon}$$, We will write $$x_ {N} = O_ {p} (1)$$,

Lemma: If $$x_ {N} overset {p} { rightarrow} a$$ then $$x_ {N} = O_ {p} (1)$$,

I'm not sure how to start, so any help would be welcome.

But we will also have intuitive $$n ^ {1/4} has { theta} = O_ {p} (1)$$?

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## math – Computes elevation, radius, and theta length relative to aspect ratio in Africa

problem
Hi everyone, as the title suggests, I try to render an aframen `a-curvedimage`,
I have the following values ​​from a dynamic data source.

``````    computedX: -365,  //computedX: x + width / 2 - parentArtboardWidth / 2 (centering x origin)
computedY: 737,   //computedY: -(y + height / 2) + parentArtboardHeight (centering y origin for aframe)
height: 350,
name: "white plane",
parentArtboardHeight: 1080,
parentArtboardName: "01",
parentArtboardWidth: 1920,
parentInstanceOf: "Artboard",
width: 686,
x: 252,
y: 168,
zIndex: 3
``````

Now the values ​​work exactly, I want them to work, but with a flat picture. But I need them to bend `height` `radius` and `theta-length` in relation to `image aspect ratio`,
I know how to calculate that `image aspect ratio` but how can I calculate the remaining values?
Here is a Codesanbox. Do not hesitate to try it yourself.

Note that you need to navigate if you want to change the pictures from flat to curved `line 189` and put `showCurved` to `true` or `false`,
Expected issue: I expect to warp the elements a bit, but without disturbing the x and y coordinates. Currently the coordinates are not correct `a-curvedimage` although the `computedX computedY` Values ​​work perfectly in flat images.
Further information

In the codesanbox (https://codesandbox.io/s/zealous-volhard-7y2g1) you can try this problem yourself:
`Line 189` There is a variable `showCurved` you can adjust to `true` `false` switch between `a-curvedimage` and `a-image`,
– The formulas are a mess right now, but they are on `Line 189` to `Line 219`,
`a-image` Entities are generated from `Line 234` and `a-curvedimage` Entities are generated from `Line 249`,

My ultimate goal is to display elements in a curve (the current curve amount is okay), but to put them in the right position, as they are in a plane.

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## Algorithms – Derive a while loop executed in \$ theta ( sqrt {n}) \$

I know this algorithm ONE comes in $$Theta ( sqrt {n})$$But how do you derive this fact?

algorithm ONE

``````i = 0
s = 0
while s <= n:
s += i
i += 1
``````

Here is what I think. We know that ONE is incremented by O (n) $$s$$ by more than 1 in each iteration. We also know that there must be a lower limit $$log n$$how we are incrementing $$s$$ with a little less than $$2 ^ i$$ in every iteration. (Correct me if I'm wrong, that's just my thoughts ...).

Well, what can we say something about? ONE? How can we deduce that its temporal complexity? $$Theta ( sqrt {n})$$?

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## Time Complexity – How Can \$ Theta \$ and \$ O \$ Complexities Be Different?

From the definition of $$Theta$$-Notation,
$$f (x) = theta (g (x)) \ implies c_1g (x) le f (x) le c_2g (x), exists n_0 ; forall c_1, c_2 gt 0 ; forall n nt0_$$

We can see that the inequality is followed for all $$n gt n_0$$and therefore we can say that $$c_2g (x)$$ is the maximum that can achieve this function. Therefore, $$f (x) = O (g (x))$$ also.

If we take Quicksort's example, sources say that $$Theta$$ Complexity is $$n log {n}$$, but $$O$$Complexity is given as $$n ^ 2$$,

If that $$O$$The given complexity is true $$f (x) le c_2n log {n}$$ will not always be true. So in this case, how is $$Theta$$Complexity is different from $$O$$-Complexity?

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## What is the main difference between the theta star algorithm and the phi star algorithm?

I'm trying to work on the pathfinding Phi * algorithm, and I'm confused about it and the theta star. What are the new points of the Phi * algorithm and how could I handle it? Any guidance, please?

## Trigonometry – Eliminate \$ theta \$ from the equation

X $$cos theta$$ – y $$sin theta$$ = $$cos 2 theta$$

X $$sin theta$$ + y $$cos theta$$ = 2 $$sin 2 theta$$

I tried to find with the cross multiplication method $$cos theta$$ and $$sin theta$$ and then the values ​​in $$cos ^ 2 theta$$ + $$sin ^ 2 theta$$ = 1, but could not eliminate $$cos 2 theta$$ or $$sin 2 theta$$, Please help me solve this question.

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