/sci/ - Science & Math

Stuck on a problem, quite possibly really simple, but I'd like to know if my approach is correct.

The problem is to find every n in <span class="math"> N_0 [/spoiler] so that

<span class="math"> n + 1 \mid n^2 + 1 [/spoiler]

holds true.
Obviously, the answer is <span class="math"> n = \{0, 1\} [/spoiler], but how to get there?

My idea was to prove it by contradiction. Let n be greater 1, then show that the equation is wrong for every n greater than 1, then prove n = 0 and n = 1, and I'm done. Is this the correct approach or did I miss something?

>>5391153
>>5391153
Hah, my bad. I was in a little hurry and didn't exclude the ones with det(a) = 0, of course you're right.
I'll try out your approach now, then. Thanks!

>>5391159
Thanks. Didn't see that page before for some reason..

Hi, I'm currently trying to prove that G (hereby defined as the set which contains all 2x2 matrices) is a group under matrix multiplication.

Associativity and existence of the identity are done, but I'm stuck on the inverse element. I know it's just the inverse matrix, so that

<span class="math"> A \cdot A^{-1} = id [/spoiler]

(where A, its inverse and id are in G, and id is the unit matrix)

I'm sure it's really simple, but if I want to solve the equation above, how can I ensure that the variables encircled in the image always yield 1 when used in the multiplication? I'm not seeing it.
Thanks for any help!

>>5385972
Alright, thanks!

>>5385961
..because in a bijective function, you can be sure that from every image you can "find back" to a single preimage, meaning there are no open ends?

Hi,

I've got a bit of trouble with understanding the bijectivity in isomorphisms.
Of course I know what a bijective function is, but in this case, I'm not sure if this applies the way I think it does.

Think of two groups <span class="math"> (G_1, \circ_1), (G_2, \circ_2)[/spoiler], where G is the group and <span class="math"> \circ [/spoiler] an operation.

A function f is a homomorphism, if for every element x, y, in <span class="math"> H_1 [/spoiler] it holds true that

<span class="math"> f(x \circ_1 y) = f(x) \circ_2 f(y) [/spoiler]

If f is also bijective, then it's an isomorphism between two groups.
But what does that actually mean? The isomorphism is a function that maps one algebraic structure to another, but does the bijectivity of the function apply to how the elements of the groups are being mapped to each other or to the actual groups?

I'd appreciate it if anyone could shed some light on that!

>>5200780
>>5200784
>>5200790

Wow, that was quick. Thanks!

When you need to seperate two sets of numbers from each other in a proof, how do you express them?

For even and odd numbers, it's pretty easy, since it would just be

<span class="math"> 2k [/spoiler] vs. <span class="math"> 2k + 1[/spoiler],

but what if you only want to express numbers which aren't multiples of 4, for example?

I know it's easy stuff, but I'm only slowly getting into number theory and proofs, so bear with me.

>>5191444
I'd fancy a hint, is there any property I can use to my advantage (if it doesn't give the solution away, since I want to solve it myself).

>>5191437
Forever failing Latex.

<span class="math"> 3 | 2^{2k + 1} -2 [/spoiler]

Is the correct example.

Since Uni has started again, I decided to brush up my proof skills since I've neglected them before.

I was wondering about something simple, such as proving divisibility:

e.g. <span class="math"> 3 | 2^(2k +1) - 2 [/spoiler]

(inb4 homework thread: This is an example problem, I don't care about the solution, but instead about the approach with which you tackle it).

Is it just induction? I tried and didn't find a proper way to use it.

>>4824987
My bad, I'm stupid. It indeed is <span class="math"> y'' + y'<span class="math">, assuming you're the one who asked.[/spoiler][/spoiler]

>>4824973
No, it's <span class="math">\displaystyle y'' - y' [/spoiler]. Not sure if I differentiated wrong. Also, the zero would be on the LHS unless you just turned the equation around (but that wouldn't matter, obviously).

Hi,

I've got a question about the approach to finding a particular solution to a second order linear differential equation. As an example:

<span class="math"> \displaystyle y'' - y' = 8(cos(x) + sin(x)) [/spoiler]

My approach was to say that a particular solution has to be of the form

<span class="math">\displaystyle y_p = 8A (B cos(x) + C sin(x)) [/spoiler]

since the function on the right hand side is a zero-grade polynomial multiplied with trigonometric functions.

When I differentiate <span class="math">\displaystyle y_p [/spoiler] and plug it into the equation, I get the term

<span class="math">\displaystyle 0 = 8 cos(x) + 8 sin(x)[/spoiler]

What went wrong?

Sup.
Does anyone have tips regarding how to "reverse" a definite integral?
Or, in a more understandable way: How do you find the integral that gives you a desired value?
The background of my question is just a practice problem from Uni, but the question itself interests me. In the case of the Uni problem, it was rather simple; The function must not be constant, and the value of the integral must equal 7. Any hints?

>>4821421
Thank you for putting all that effort in it, even though the motivation wasn't all that selfless (hah!).
What you say makes a lot of sense to me, since you can just eliminate the terms you don't need in case the right hand side of the equation doesn't consist of them.

>>4821241
I hope I'd know what exactly my professor meant by it, my notes on that topic are a bit cryptic.

I'll try your idea, though.

Bump.

Hello /sci/,

I'm currently struggling with understanding a solving method for linear 2nd order differential equations, and I hope someone could help.

The problem I'm trying to solve is

<span class="math">\displaystyle y'' - 3y' + 2y = e^{17x}[/spoiler]

The general idea is, if I'm not mistaken, to solve the characteristic equation first, since the solution of the inhomogeneous equation is equal to the sum of the general solution of the homogeneous equation and one particular solution of the inhomogeneous one.

In this case, the characteristic equation would be

<span class="math">\displaystyle \lambda^2 - 3\lambda + 2 = 0 [/spoiler]

which leads to the solutions

<span class="math">\displaystyle \lambda_1 = 1, \lambda_2 = 2 [/spoiler]

and the general solution of the homogeneous equation is

<span class="math"> \displaystyle y = A e^x + B e^{2x} [/spoiler]

Then, my notes say that, if <span class="math">\displaystyle y_p [/spoiler] is a particular solution, then it could be described as

<span class="math"> \displaystyle y_p = e^{cx} * x^r (P_n(x) cos(bx) + Q_n(x) sin(bx)) [/spoiler]

where <span class="math"> \displaystyle P_n(x), Q_n(x) [/spoiler] are polynomials, and <span class="math"> \displaystyle b, c , r [/spoiler] are real numbers.

Also it was said that <span class="math"> \displaystyle c + ib [/spoiler] is an r-fold solution of the characteristic equation.

That last sentence seems to be my main problem of understanding - what exactly do I do with it and why is it that way?

>>4815602
Alright, thanks!

Hi, this might be easy to explain, but I'd be glad for any help.

The differential equation

<span class="math"> y' = \frac {6x -4y +1}{3x -2y} [/spoiler]

is non-linear. Why is that? I've looked at the definition, but didn't really get it.

>>4801314
Duh. Silly question, I admit.

>>4801298

Ah yeah, that. When I checked the way Wolfram calculated it, I saw the

<span class="math"> \int\frac{2\frac{dy(x)}{dx}}{y(x)^2}dx [/spoiler].

What exactly happened there?

>>4801294
Pardon. I shoud've written <span class="math">C_1[/spoiler].