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.

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 …

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) $?

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.

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}) $?

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?

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.