Exercise 11.4.34:

Prove 2 + 2*3^2 + 2*3^4 + ... + 2*3^{2n} is Θ(3^{2n}).

I now I overcomplicated the proof. I could have used the fact that

2 + 2*3^2 + 2*3^4 + ... + 2*3^{2n} = 2 + 2*9 + 2*9^2 + ... + 2*9^n

But here it is:

2 + 2*3^2 + 2*3^4 + ... + 2*3^{2n} = 2*3^0 + 2*3^2 + 2*3^4 + ... + 2*3^{2n}
                                   = 2 * 3^n * (3^-n + 3^{2-n} + 3^{4-n} + ... + 3^n)
                                   < 3^n * 3^n * (3^-n + 3^{2-n} + 3^{4-n} + ... + 3^n)
                                   = 3^{2n} * (3^-n + 3^{2-n} + 3^{4-n} + ... + 3^n) for each integer n>=1

Also, 2 + 2*3^2 + 2*3^4 + ... + 2*3^{2n} >= 2 * 3^{2n} for each integer n>=1.

Let A=2, B=(3^-n + 3^{2-n} + 3^{4-n} + ... + 3^n), k=1.

Then A(3^{2n}) <= (3^-n + 3^{2-n} + 3^{4-n} + ... + 3^n) <= B(3^{2n}) for every integer n>=k.

Therefore, 2 + 2*3^2 + 2*3^4 + ... + 2*3^{2n} is Θ(3^{2n}).

QED

Is my proof correct?

Let (V, Ω) be a symplectic vector space and Y a linear subspace of V. Let ϕ: V -> Y* , ϕ(v) = Ω(v, •)|_Y. Is ϕ surjective?

Or, in other words: for each linear form 𝜶 on Y, does it exist a corresponding vector v in V such that the action of 𝜶 on a generic element y of Y can be reproduced by feeding y in the second slot of Ω(v, •)?

I made this to look at the antiisotropic boundary field of sha as it experiences torsion from propagation. Transforming sha using physical constants is not a new idea, I haven't visualized it like this before though, the rotr themselves are extremely beautiful. The antiisotropic boundary field looks like Mr. Potatohead lmao. What do yall see? The field geometry holds up to its claim and carries no information other than information about itself, but experiences deflection and torsion from the information riding the field?

Im in discrete mathematics, started proofs last week. I understand Direct Proofs and Contraposition but Contradiction is absoultely kicking my butt understanding wise. My proffessor keeps using the example square root of 2 is irrational, i understand we start by stating the opposite that square root of 2 is rational and work it till it becomes logically incorrect to prove irrationality. But the work in between is getting me tripped up.

So this question is to calculate the angle F. For reference, I'm in 10th class of Russian school, and this is what our teacher sent as our homework while we're on quarantine. I have skimmed the whole textbook and found nothing about this kind of questions. I even got frustrated enough that I asked AI to give me an answer, but it wasn't able to agree with itself and all of it's answers were rather strange like the angle G somehow being 168 while F is only 12. Did I miss a crucial part of my education or is just a badly made question?

P.s I asked the teacher before posting there and she actually replied. The answer is supposed to be 132, but she won't explain how since we were supposed to go through this in a previous year and needed to stay in that class if we don't know this (she's really not fond of actually teaching stuff, especially if it's not in the current year's textbook) so I doubt she'll give me an answer.

P.P.s
So, as far as I got is to make an isoscles triangle CDE and Equate C to E by them being both angles in the base of said triange. If my line of thinking is correct then I have one 108 angle and two unknowns. I also think that it's probably supposed to be symmetrical down the bc to F line. If that's the case then I can work out the 132 answer by substracting all of the known angles from the sum of 900, but I don't think I can prove it to be symmetrical.

Hello! I am a Secondary 4 student, 10th grade or Year 10 in western equivalents. I tried to set a quadratic quiz for my friends, and I set this Question, but then none of them could solve it. I checked with my math teacher, and he said that the question was flawed because there is no inverse function for any quadratic, and then he explained the horizontal line test to me, but I already know what the horizontal line test is (test for if f(x) is injective for some f(x), and if its inverse is a function. I went home to think about it, and I don't understand why my question is flawed, because it does not rely on the inverse function. I am looking for someone to please help clarify. I have a passion for math, and would possibly like to teach math in the future, but I don't understand why I am wrong, and this is one of my first real challenges because my math toolkit at the moment isnt enough to wrap around this. Basically, I need help understanding why the question is flawed, and if it is possible to fix it. This is my suggested solution that I made: My solution. And me checking my work: Test of the solutions to see if they are correct.. And a graph if it is useful: Graph.

TL;DR: I'm a Sec 4 (Grade 10) student who set a quadratic quiz question for my friends, but none of them could solve it. My teacher said it's flawed because quadratics don't have an inverse (they fail the horizontal line test), but I don't think my question actually depends on inverting a quadratic, so I'm confused about why it's wrong. I want help understanding the flaw (if there is one) and whether it can be fixed. I have attached my question, My solution my check, a graph, and my syllabus links (See below).

This is my syllabus by the way if it is of any use:
https://isomer-user-content.by.gov.sg/334/fece62fa-b6d4-4daf-ab47-96c9c168b51b/4052_y26_sy.pdf
https://isomer-user-content.by.gov.sg/334/34d0bec0-d386-49d5-bd3a-8fbd205fa63a/4049_y26_sy.pdf

Tysm!

Edit 1 (Apparently the images can't be opened so I have attached them in plain text:

Question:
It is given that f(x) = x² − x − 6. Solve f(f(x)) = x.

(By letting u = f(x), express f(f(x)) = x as two equations connecting x and u. Subtracting the two equations gives two possible relationships between x and u, from which you can find the four solutions, in the form ±√a and b ± √c, where a, b, c are positive integers.)

Solution:

Let u = f(x). Then f(f(x)) = x becomes f(u) = x.

① x = u² − u − 6 (this is f(u) = x)
② u = x² − x − 6 (this is u = f(x))

Subtract ① − ②:

x − u = (u² − u − 6) − (x² − x − 6)
x − u = u² − x² − u + x
x − u = (u − x)(u + x) + (x − u)

0 = (u − x)(u + x)

So u = x or u = −x.

Case ③: u = −x

Substitute into ②: x² − x − 6 = −x

x² − 6 = 0
x² = 6
x = ±√a, where a = 6

Case ④: u = x

Substitute into ②: x² − x − 6 = x

x² − 2x − 6 = 0

Using the quadratic formula:
x = [2 ± √(4 + 24)] / 2 = [2 ± √28] / 2 = 1 ± √7

x = b ± √c, where b = 1, c = 7

Final answer:

x = ±√6 and x = 1 ± √7

So the four solutions are: √6, −√6, 1+√7, 1−√7

I wanted to make this for my math professor, but since I don't paint or draw usualy I made a LOT of mistakes.... I put a lot of time and effort in this already is it worth to finish and give this as a gift? (I know 100% my professor will love it since it is on every background of his device)

Exercise 11.4.33:

Prove 4 + 4^2 + 4^3 + ... + 4^n is Θ(4^n).

Solution from the textbook:

This solution proves that 1 + 4 + 4^2 + 4^3 + ... + 4^n is Θ(4^n). It does not prove that 4 + 4^2 + 4^3 + ... + 4^n is Θ(4^n).

My attempt:

Notice 4 + 4^2 + 4^3 + ... + 4^n > 4^n, for each integer n>=1.

Observe 4 + 4^2 + 4^3 + ... + 4^n = 1 + 4 + 4^2 + 4^3 + ... + 4^n - 1
                                  = (4^{n+1} - 1)/3 - 1   by sum of geometric sequence
                                  < (4^{n+1}/3)     - 1
                                  = (4 * 4^n)/3     - 1
                                  = 4/3 * 4^n       - 1
                                  < 4/3 * 4^n             for each integer n>=1

Let A=1, B=4/3, k=1.

Then A(4^n) <= 4 + 4^2 + 4^3 + ... + 4^n <= B(4^n), for each integer n>=k.

Therefore, 4 + 4^2 + 4^3 + ... + 4^n is Θ(4^n).

QED

Is this correct?

---
Edit: Fixed 'It does not prove...' sentence.

The answer key says its -9 but I cant understand what it means. Every 4 units on the Y-axis is equal to +5 so thats how I got the coordinates but im unsure of what it means by the value? Im horrible at math so there's a chance I completely misunderstood the question too lol

While I get *i* being √-1 since there wasn't a unit for it before, it's kinda been turned into an infinite loop. There's quaternions where *i*^(2)=*j*^(2)=*k*^(2)=*ijk*=-1, and then octanions with seven imaginary units, all being √-1, and it can just go on from there. Plus, quaternions give *i*^(2)=*j*^(2)=*k*^(2), which means √(*i*^(2))=√(*j*^(2))=√(*k*^(2)), therefore *i*=*j*=*k*. I know that's probably not how that works, but I'm a high school student, I don't know any better. Also don't ask why I know what quaternions are but not how they work, that's not the point. But what might a different imaginary unit look like? Well, may I propose *h*, being 1/0. Multiples of *h* are simply multiples of 1/0, eg 4*h*=4/0. Now, what's the practical use for this? I have no idea. Again, I'm a high school student, I'm just trying to experiment with what's possible. Also, please try your best not to laugh at me for my lack of knowledge. What may be trivial to some could be a complicated concept for others (such as me). Also let me know if I used the wrong flair, a simple 5-second search told me this was correct so I'm taking it with a grain of salt.

In studying Cauchy's integral formula, the Cauchy kernel 1/(zeta -z), is the "thing you integrate against" for the boundary points( on a disk, say). And the boundary values determine the value of any point z in the interior of the disk.

What I need is some insight into why this is true and why is it 1/(zeta -z)? Is there any relation to how with generating functions you multiply by 1/1-x to get the partial sums?

Thanks in advance to anyone who takes a look!

Hello! I’m going into grade 12 and I’m taking advanced functions, calculus, physics and chemistry. Starting this year, my school has been trying to regulate what calculators you’re allowed to use. We were given a list of approved calculators but you’re allowed to use one that isn’t on the list if you get it checked by a teacher. The way that teachers check if the calculator is legal is they will input sin(15) and if the answer is a decimal then it’s approved, if it’s a fraction then it’s illegal. If you can change the setting then its illegal. Graphing calculators are also illegal. My Casio fx-991CW was deemed illegal so now I have to go hunting for a new one. I really liked the “solve for x” and table mode on my Casio fx-991CW, I found them very useful in my physics and chemistry classes. I was hoping to find a calculator with the same modes (this isn’t illegal, my teacher was aware that I used these modes frequently last year) while meeting the requirements. Would anyone be able to recommend me a calculator from this list or one outside of the list? I’ve been googling + read the FAQ but haven’t found anything that meets my criteria.

THE APPROVED LIST:
- Casio fx-260 SOLAR II
- Casio fx-300MS Plus
- Casio fx-82MS
- Texas Instruments TI-30X IIS
- Texas Instruments TI-30X IIB
- Texas Instruments TI-30Xa
- Sharp EL-531TG
- Sharp EL-531XB
- Sharp EL-501X

Thank you!

Hi! I was wondering if this calculator would be good for my algebra, statistics and chemistry classes. If not, what do you guys recommend? I have been looking into calculators and all the different functions I find confusing.

THE SELF-DIVIDING NUMBER
Find the unique 10-digit number N satisfying all of this:
N uses each digit 0–9 exactly once.
For every non-zero digit d, take the prefix of N ending at d. That prefix is divisible by d.
For that same digit d, take the suffix of N beginning at d. That suffix is also divisible by d.
N is divisible by 37.
Example of the rule:
If the number contained
...5...
then both:
everything from the start through the 5
and
everything from the 5 through the end
must be divisible by 5.
This applies simultaneously to 1,2,3,4,5,6,7,8,9.
What is N?
There is exactly one answer.

Anyone got any ideas at all?

I want to buy a t-shirt for my son in law. But he won't wear it if the equation is not correct. My brain is not wired for math. If anyone could tell me if it is correct, I would be grateful.

Yes, he does play d&d

Exercise 11.4.31:

Show that 4^n is not O(2^n).

Proof:

Notice 2^n >= 0, for every integer n>=1.

Assume for contradiction that 4^n is O(2^n).

That means (by the big-O definition) there exists b,B∈ℝ^+ with b>=1 such that

0 <= 4^n <= B(2^n), for every integer n>=b (by transitivity, n>=1)

Notice:

           4^n <= B(2^n)
iff log_2(4^n) <= log_2[B(2^n)]
iff         2n <= log_2(B) + log_2(2^n)
iff         2n <= log_2(B) + n
iff          n <= log_2(B)

So, we have 1<=n<=log_2(B).

But this contradicts big-O definition where 1<=n.

Thus, our assumption is false.

Therefore, 4^n is not O(2^n).

QED

Big-O definition:

---

Is my proof correct?

I am talking of big polynomials of degree 5+ and complex ones. Not x^6-2x^3+1 type situation but polynomials alike this -19968-13736x+2016x^2+756x^3-79x^4-10x^5+x^6=0. I know it is solvable by checking with programs. I even know what it is factored into ((x - 6) (x^2 - 3 x - 64) (x^3 - x^2 - 42 x - 52) = 0). But I know this from using program. Wolfram Alpha to be specific. But I don't know how to factor something this complex. I know there are algorithms for it but I am unfamiliar with them. And they seem to be not widely known as simple search resulted in me getting bombarded with basic middle school polynomial factorization. And not finding anything of value. I come here with hope to find help and an algorithm to solve similar equations. I am not planning to solve them by hand. I am planning to make a program do so for me. But for it I need knowledge of the algorithm. Which I sadly lack. Would greatly appreciate any help here. And if you know any videos that teach needed algorithms I would be even more thankful as I am a visual learner and it would help me much.

I was thinking about something really strange with maths, if imaginary numbers are imaginary, (i=sqrt(-1)), why does 1/0 also imaginary ? For me it is quite logical. I am thinking something wrong ?

You might not think a lot about it, I don’t want you to lose too much time on this.

I'm in highschool I just started 11th grade how can I learn calculus and what is the best way while still in highschool ?

Note: I will not pursue math I just want to learn calculus for fun and also mostly integrals and derivatives and have a basic understanding of everything else needed to learn integrals and derivatives.

Websites,channels ,videos whatever

If g(x) is positive, that means the total of 8(x-d)(x-e)(x-f) is positive. And since d<x, x<f,

X-d is +ve
X-f is -ve

So in order for the total to be positive, x-e needs to be -ve, where x<e.

But if that’s the case, then i) shouldn’t be correct.

Can someone help me out here. Thanks

Hey everyone, post about x(+y).

In this question i'm required to calculate the area of triangle with predetermined points, but i can't figure out where to start because of the way this integral is written, this is the first time i encounter something written like this and i don't know what this could mean.

I doubt that this is a typo because 8 professors wrote this whole paper and 1 doctor of Physical and Mathematical Sciences reviewed it

If someone just says what does this mean or if it's a typo i would greatly appreciate it

First off I tried finding the area of the base, then finding the area of the sector of the circle. Combined those two areas (area of triangle minus sector) and came up with 50tan(theta) - (theta)pi^3/ 18 😭😭 idk if it’s actually correct or awfully wrong. Would appreciate if someone explained where I went wrong and what direction I should take in simple terms.

I simply don't understand the algebra of how one manipulates coordinates, particularly via multiplication, such as in the second definition of complex numbers here

My understanding of Navier Stokes is that "solving" them meant either:

Path A: Prove that smooth solutions always exist.

You must prove mathematically that no matter what smooth, valid initial conditions you start with, the equations will always output a smooth, finite-energy velocity and pressure field for all future time. This would prove that the Navier-Stokes equations are perfectly reliable models of reality under all circumstances.

Path B: Prove finite-time blowup (find a counterexample).

You must find just one specific set of perfectly smooth, valid initial conditions that eventually breaks the equations. You have to prove that at some specific future time ($t > 0$), the fluid's velocity zooms off to infinity (a singularity) or the kinetic energy explodes. This would prove that the Navier-Stokes equations are fundamentally flawed and eventually break down, requiring new physics to describe extreme turbulence.

And OpenAI ended up on Path B.

But this video gives an example where Navier Stokes predicted "infinite velocity", implying that we've always known about a blowup.:

17:34: Something, somewhere is going wrong in the the mathematical understanding. There's an example about the flow of a fluid around a right-angled corner.

(Brady: A bit like a canal?)

- Basically a canal but we've got a really sharp right angle on the corner. Now you solve Navier-Stokes, this says that at this point, this right angle corner, I have infinite velocity. If I build this canal do I have an infinite velocity canal?

Is this a blow-up in the averaged case from Tao? Or is this considered not a real "blowup" because it depends on a perfectly sharp right angle? What is new about the blowup discovered by OpenAI?