/SQT/ - Stupid Questions Thread

+/- 1
What reason could you have for rounding up?

what is its wave funktion then?
fucking faggot how are we supposed to help you when you give no context asshole

chill your beans. pic related is all i have to go by. the lecture notes are utterly shit im afraid

Is this an adequate definition?

[math]
\emptyset := \{x \ | \ \text{False} \}
[/math]

you need to know what x and y are. I mean what distribution they have.
Is it just a decreasing exponential?

No

when you use biot-savart law, you almost always get a denominator with a (x^2+something^2)^(3/2)

Why?

Because if the minimum error is 1.2cm, you can't just round down (essentially saying the experiment is more precise than it actually is), instead you should round up so that there is 1 sig fig, but there is a 100% certainty that the value is within that range.

What do ya think?

it almost is. but you're grabbing "all x" from everywhere and that's not allowed. you can only build sets from other sets

How to fix?

[math]
\emptyset := \{ x \in S \ | \ \text{False} \}
[/math]

where S is an unspecified set.

Is it fix?

I got accepted to UCLA and UC Berkeley as a transfer for math.
I'm leaning towards Berkeley because they seem better for algebra, geometry, mathematical physics where UCLA seems better for analysis, number theory, and logic.

What does Veeky Forums think? Anyone at either math program that can give some personal experience?

Berkeley is closers to home and slightly cheaper as well.

>I'm leaning towards Berkeley because they seem better for algebra, geometry, mathematical physics where UCLA seems better for analysis, number theory, and logic.

Nigga it's undergrad; the exact research strengths down to the subfield don't matter. Go to Berkeley since it's cheaper.

I'm looking to do as much (quality) research as I can in the next 2 years and so to me, I felt if I could connect with professors who have similar interests, it'd be much easier.

Both departments are huge, e.g. Berkeley has a shit ton of analysts. You'll find someone working on something you like at either school, and you will probably only work seriously with one prof. Honestly, most incoming grad students aren't even that sure of their specific research interests, and I bet yours will change a lot as you take upper level classes. I'm kind of amazed you're so set on certain subfields this early on. I just think there are better things to consider as an undergrad than research fit.

But yes, Berkeley is very good at algebra and geometry. Good luck, user.

>Suppose we toss a fair coin until we get exactly two heads. Describe the sample space [math]S[/math]. What is the probability that exactly [math]k[/math] tosses are required?

Second part is [math](k - 1)\frac{1}{2}^{k}[/math], right?

If I have an equation f(x), and an equation g(x, y), how do I find the value for g(x, y) that would give me a function that is as close as possible to f(x)?

I mean I have this 3d plot of the function g(x, y) and somewhere I could take a cut that would give me the closest thing to f(x) that exists in g(x, y). Anyone who can help? Ideas are very appreciated too.

What I thought about is getting the integral of |f(x)-g(x, y)| and the global minima should be what I search for, but it doesn't seem to work very well.

More specifically,
f(x)=cos(x)-((12pi-48)/pi^3)*x^2-((72-24pi)/(3pi^2))x-1
g(x, y)=(x^3)y-((3pi)/4)(x^2)y+((pi^2)/8)xy

This is the result of "Oh how hard can it be to approximate cos(x) over an intervall with polynomials, let's try it out".

take the middle of the interval
write taylor series at that point

just use the last line where it says/x^2=y^2=z^2.
THe expected value is linear and r^2=x2+y2+z2 thats all you need

No, go away with your taylor series, it sucks. There's a reason it's not actually used for calculating cos, like ever.

Does a natural number divided by an aleph number mean anything?

Why does raising a number by the 1/2 power square it? I know it does, but what's the mechanics behind it?

Surely 64^1/2 = 32? But it doesn't, it equals 8. Why is that?

>Why does raising a number by the 1/2 power square it

Square root it I should say.

Because you're multiplying it with itself half a time, not multiplying it with a half

You want a function, or a point close to f(x)? A function seems sort of vague, but a point is easy enough. You set up a relation using the square of the distance between the two functions, and you find any critical points.

Yeah but I just don't get it. I understand what you're saying in principle, but I understand why. I can't visualize it.

How do you find the base element in a series?

I've just discovered that hiccups is because of carbon doxide content in the blood. I had the hiccups and I covered my mouth with a blacket and breathed through it and the hiccups went away. How do I get financial backing for a scinece experiemnt?

The reason I'm reasonably certain about my interests is because I self study a lot and have found my interests tend to lean in the direction of algebra topology and geometry.
Thanks for the advice my friend.

>I want the best approximation for a function like cos
>but pls no taylor series, it sucks

how retarded can one person get

? I took the DiffEq for Scientists and Engineers and we hardly talked about applications unless they were some hw/exam questions to give context.

Calculators use en.wikipedia.org/wiki/CORDIC

But Taylor series work and is how the majority of approximations are done in physics (analytically), so stop being a little bitch, acting like you know how things are done.

Can anyone explain how the bounds of integration are computed in the projection of the xz plane?

Is there a drug capable of distorting the brain's perception of time while maintaining consciousness and bodily functions, i.e. the drug makes it feel like a longer amount of time has passed than it actually has but doesn't impair the senses beyond that?

Try low doses of DALT

he wants polynomials
and he wants approximations over an interval.

CORDIC only gives the value at one point

that guy is retarded, he doesn't even deserve any help.

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Hint: the hidden message is Huffman coded.

What number is halfway between 1 and 9? If you take an additive approach, it's 5. If you take a multiplicative approach, it's 3. The square root function is the latter.

So 9/2= (rounded up) 5, but 9^1/2= 1*9*1/3?

I just dont get it. I feel like I understand what you saying, but I can't write it mathematically.

First, you should really write square root as x^0.5 or x^(1/2) or [math]\sqrt{x}[/math].

Anyway, I'll try to be a bit clearer about what I was saying before. There's a slight difference between finding the additive half and the multiplicative half (bullshit terms I just came up with now). For the additive, you should start with zero:
[math]0+c+c=9[/math]
[math]2c=9[/math]
[math]c=\frac{9}{2}[/math]

For the multiplicative, start with one:
[math]1\cdot c \cdot c=9[/math]
[math]c^2=9[/math]
[math]c=3[/math] (ignore the negative root for now)

Now go back and check them:
[math]0+\frac{9}{2}+\frac{9}{2}=9[/math]
[math]1\cdot 3 \cdot 3 =9[/math]

one is called an arithmetic average, the other is a geometric average

Fuck, that was it, thank you.

For a prime p and integer b, if b^2 = -4 mod 4p, does that imply that b^2 = -1 mod p?

What integer are you squaring to get a negative result from modulo with a prime? Do you have a weird definition of prime that includes negative numbers somehow? Are your integers complex?

No, it would imply [math]\frac{b^2}{4}[/math] is congruent to -1 mod p, if [math]\frac{b^2}{4}[/math] is an integer.

4^2=17-1

Well, as an example, if p was 5, then b^2 = -1 mod 5 could have b = 2. They're all real integers.

Wouldn't that mean that b^2 = -4 mod p?

YES.

USE THE DEFINITION

B^2 = -4 + 4P*K FOR A CERTAIN K

DOES IT IMPLY B^2 = -1 + P*M FOR A CERTAIN M?

IN THAT CASE, BOTH B^2+1 AND B^2+4 ARE IN THE SAME CLASS MODULO P

WHICH MEANS 3 DIVIDES P

SO P=3 OR THE IMPLICATION IS INVALID

18) Consider V = IR4 of usual internal product, determine a non-0 vector v Є IR4 that is orthogonal to v1 = (1, 1, 1, −1),v2 = (1, 2, 0, 1) and
v3 = (−4,1,5,2).

Help

...

v= (a,b,c,d)

v.v1=0
v.v2=0
v.v3=0
three equations.

If you want to add one equation, you can fix one of the coordinates of v.
Does it make sense?

Don't be lazy, this is how you fail or succeed.

thank you, quads

i love you

What's a good reference text for learning discrete maths?

How do I do the convolution of two complex exponentials?
[eqn]h_1=1-e^{-j*\omega}[/eqn]
[eqn]h_2=1+e^{-j*\omega}[/eqn]
Calculate [eqn]h_1*h_2[/eqn]

I love you too familia

any context you can give? do h1 and h2 have compact support?

The omage is supposed to be the normalized frequency but I couldn't be bothered putting in the hat.
What does compact support mean anyway?

I got a dumb one.

when squaring an infinite series, i vaguely remember seeing something about making a grid, and summing diagonal cells. i can't for the life of me remember what that's called so i can google it.

I think it's possible to keep a body alive without brain activity, but is it possible to sustain a brain's functions without the original body, i.e. by using a "mechanical" heart/lung machine?

How is the concept of "self" quantified in science; based on how the brain responds to stimuli, or something else?

FOIL

If I wanted to show that a set of vectors have the same angle between them, is it sufficient to just show that their dot products are equal? Or will I need to calculate the angle of all of them?

I believe something like "cauchy summation" might bring it up.

I'd normalize before taking the dot product, and then cite a theorem or definition.

Ha, nobody's actually told you yet.
[math]\displaystyle \pi\langle x^2+y^2\rangle=\pi \langle x^2\rangle + \pi \langle y^2 \rangle[/math]
From distribution of expectation values.
You also have:
[math]\displaystyle\pi\langle r^2\rangle=\pi\langle x^2+y^2+z^2\rangle=\pi\langle x^2\rangle + \pi \langle y^2 \rangle + \pi \langle z^2 \rangle[/math]
Assuming the situation is isotropic, you have:
[math]\displaystyle \langle x^2\rangle=\langle y^2\rangle= \langle z^2\rangle[/math]

This means:
[math]\displaystyle\pi\langle x^2\rangle+\pi\langle y^2 \rangle = 2\pi\langle x^2\rangle=\frac{2}{3}\pi\langle r^2\rangle[/math]

It's easy - you just need to integrate up to the point z=1-x for each point of x you're using.

What happened between

[math](5+2)*7^k-2*2^k[/math]

and

[math]5*7^k+2(7^k-2^k)[/math]

see pic for full solution.

I feel like this is a lot easier then i think it is

[math]j[/math] as complex number. Disgusting engineer.

Just take the fourier transform of each, into time space and then multiply and transform back. Convolutions in frequency space become multiplications in time space, and vice-versa.

both have a common factor of 2

Did someone mention /MIT/?

So, it'll be a basis for a 2D space. Every vector in [math]\mathbf V[/math]can be written in the form:
[math]\displaystyle\mathbf v=\mathbf x - \frac{\mathbf x\cdot\mathbf n}{\sqrt {\mathbf n \cdot \mathbf n}}\mathbf x[/math]
where
[math]\displaystyle\mathbf x \in \mathbf R^3,\ \mathbf n = (2, 3, -1)^T[/math]
Now, just show that [math]\mathbf v_1+\mathbf v_2[/math] and [math]c\mathbf v_3[/math] all can be written as some vector [math]\mathbf v_4\in \mathbf V[/math]

The other axioms of a vector space are already proved, because [math]\mathbf V[/math] inherits from [math]\mathbf R^3[/math]

A basis for [math]\mathbf V[/math] is easy to find. Just pick any 2 vectors orthogonal to [math]\mathbf n[/math], that are also orthogonal to themselves. Normalise them, and you have a basis. The dimension of [math]\mathbf V[/math] is 2 because you've put a constraint on the vectors allowed, which gives you 1 less dimension to play around with.

That feel when you get deferred then ultimately rejected. MIT broke my heart.

I forgot to say that all the vectors have the same magnitude, but I guess I should normalize anyway. But yeah, that's sort of what I was thinking rather than calculating the angle for every one. Though I don't think a definition or anything like that has been taught in my course so far so I'm not really sure.

Cal Tech masterrace here

Is this dictionary actually fruitful to make use of? I've only seen it for the first time in Etingof's "Introduction to Representation Theory" notes and not sure how common it is

O = {}

Are you trying to find a y' such that g (x,y') is closest to f(x)?

If so, I don't think there is a unique solution, it all depend on what norm you want to minimize ||f(x) - g(x,y)|| under.

I suggest minimizing it under the L2 norm, so you integrate (f-g)dx and them minimize this w.r.t. y (i.e. find the critical points etc.)

Oops I meant to say integrate (f-g)^2 dx

I know this is supposed to be an easy question, but why can't I use Bayes' law on the first question?

P(BJ | Ace) = (P(BJ) * P(Ace | BJ)) / (P(Ace))
= (0.0474 * 1) / (32/416)

(P(Ace | BJ)) == 1 right?)

I'm trying to find a g(x, y) that would give me the closest thing to f(x), with which I mean have the smallest area between g(x, y) and f(x).

Here's a shitty mspaint picture, I want to find a value g(x, y), that if plugged into the function would give me a new function g(x) with the smallest possible gray area between it and f(x).

And of course I forgot the picture.

would this ever happen?

Well, trivially, you have g(x)=f(x)

Yes, however, f(x) includes a trigonometric function, while g(x, y) is a polynomial, so that is not the case. So I need to find the function that is closest to f(x).

Perform a taylor expansion of the f(x) function, and use that as g(x, y) to whatever accuracy you require.

No, I want to find a polynomial of third degree that gives me an accurate approximation of cos(x). I found the coefficients of a parabola that closely traces cos(x), with a maximum error of 0.02. If you substract the parabola from cos(x), you get a function that looks very close to 0.02*sin(4x).

Now I want to find the third degree polynomial that traces this function as close as possible. The taylor series is not good because if you take the third degree taylor series for cos(x), you get a maximum error of about 0.2. Which is horrible.

Ok, so you're going to need to fit a function [math]y=ax^3+bx^2+cx+d[/math], with the aim of reducing [math]\int_a^b\{y-f(x)\}dx[/math].

You can fit this on a computer, which is trivial, but maybe time consuming cause of the 4 variables.

Do you want me to explain the process of fitting these variables in a program?

Yes, however, I assumed I need to take the absolute value of f(x)-y, to really get the function closest to f(x). And at that point wolfram alpha tells me I exceeded computational time and I'm at a loss. What should I do?

This is probably easiest to do in python with the SciPy package.

Here's a basic form for the program:

Create a function f(a, b, c, d) that returns:
[math]\int_m^n|ax^3+bx^2+cx+d-cos(x)|dx[/math]

You'll need to do this integration numerically.
Scipy has an optimize function, in which you'll input this function, and some limits for a, b, c, d then it finds the values of them that give the lowest value for the integral.

If you tell me what the range you're integrating [math]f_{a,b,c,d}(x)-cos(x)[/math] over is, I can write a simple example for you?