Oh okay. The fact that the function is unbounded, continuous, and has a global minimum gives you the range by the intermediate value theorem. Onto is the same, the range has to be the whole codomain for the function to be onto so the previous part gives you that answer. For one-to-one, it's pretty obvious from the fact that it's a parabola that it won't be one-to-one, so find two points that map to the same point of the range and that's sufficient to show that it's not one-to-one.
[editline]24th October 2013[/editline]
Homology is difficult. I wish I was better at algebra.
[QUOTE=JohnnyMo1;42633433]Oh okay. The fact that the function is unbounded, continuous, and has a global minimum gives you the range by the intermediate value theorem. Onto is the same, the range has to be the whole codomain for the function to be onto so the previous part gives you that answer. For one-to-one, it's pretty obvious from the fact that it's a parabola that it won't be one-to-one, so find two points that map to the same point of the range and that's sufficient to show that it's not one-to-one.
[editline]24th October 2013[/editline]
Homology is difficult. I wish I was better at algebra.[/QUOTE]
I was expecting an answer involving more detailed set proofing but your answer has helped a lot along with other external resources I was looking at.
Thanks
[QUOTE=B!N4RY;42634828]I was expecting an answer involving more detailed set proofing but your answer has helped a lot along with other external resources I was looking at.
Thanks[/QUOTE]
For any continuous function (especially a polynomial) a graph is the easiest and best way of doing it - all you need are the minima, maxima and whether it tends to +/- infinity in either direction - since then the Intermediate Value Theorem & Rolle's Theorem (or 'it's obvious') let you do exactly what you want to be able to.
In this case, if you want to be absolutely explicit, since it's only a quadratic you can just solve x^2 - 4x + 6 = y
When the discriminant is <0 you have no real solutions
When the discriminant is =0 you have precisely one real solution
When the discriminant is >0 you have two real solutions
You should be able to easily check that these situations correspond to y<2, y=2 and y>2
Thus it's range is [2,infinity) - so it's not surjective, and it doesn't have a unique solution (except at y=2), so isn't injective either.
Yesterday I talked to my topology professor from last semester and I learned about the Mostow rigidity theorem which is kind of amazing. Finite volume hyperbolic manifolds in dimension 3 or higher are determined up to isometry by their fundamental groups. Dayum. The fundamental group is hella powerful in hyperbolic geometry.
Just had my discrete mathematics midterms a few hours ago. Its result statistics is kinda worrying. Last year 40 % failed that exam and the rest got A 3%, B 6%, C 13%, D 16%, E 20%.
All I can think is in sets now.
That is an awful grade distribution.
[img]http://gyazo.com/97d54c7d7b208dc520ffb0f568a44bd8.png[/img]
Footnote in my lecture notes.
How in the world did I pass my math classes in grade school. I feel like in class everyone remembers all the little rules. I think I need to retain this knowledge better... sitting here staring at my Calc homework for a bit just working out different ways I think it can be solved, but I'm definitely missing a rule here.
[QUOTE=Zareox7;42675759]How in the world did I pass my math classes in grade school. I feel like in class everyone remembers all the little rules. I think I need to retain this knowledge better... sitting here staring at my Calc homework for a bit just working out different ways I think it can be solved, but I'm definitely missing a rule here.[/QUOTE]
Post it and maybe we can help.
Well I decided to skip that specific question and come back to it after I reached the end of the assignment, I think it mainly has to do with the way my professor is teaching.
The problem is, [quote]Show that y = (2/3)e^x + e^(-2x) is a solution of the differential equation y' + 2y = 2e^x[/quote]
Now, I go through the steps of placing the variables on their respective sides with their dx and dy.
I get to [quote]-2∫(1/y)dy = 2∫e^(x)dx[/quote]
Which in turn becomes [quote]-2ln|y| = 2e^(x) + c[/quote]
However, that's where I left off. I know that I should know this, seeing as I already have been though calc like this in high school, but I can't seem to remember it. Normally in situations like this, I throw the equation into WolframAlpha to help me through the steps, but I can't find a good wording to have them get me to the final answer that I need to reach.
[editline]28th October 2013[/editline]
And if I made a mistake in how I did things, or if you want to just start over from the start and show step by step, either is fine. I've been doing this homework here for a good few days now. Sometimes I wonder how I'm going to go into my major when sometimes I can breeze through sections, and others I get completely stone-walled.
You don't need to actually solve the differential equation, you already have one solution that you just have to check if it's correct. Differentiate that solution, then plug y and y' into the equation and see if it all checks out.
Holy shit... I sat there staring at that for a long time. Just did what you said, problem done in a minute and a half...
Despite taking AP Stats and Calculus my last two years of high school, I'm in a two semester Pre-Calc class set this year. It's nice and easy and all, but I feel like I'm going to forget all the stuff I learned about differentiation and integration by next year.
Here's hoping my Psychology major won't be too math intensive.
[QUOTE=Mr. Bleak;42679412]Despite taking AP Stats and Calculus my last two years of high school, I'm in a two semester Pre-Calc class set this year. It's nice and easy and all, but I feel like I'm going to forget all the stuff I learned about differentiation and integration by next year.
Here's hoping my Psychology major won't be too math intensive.[/QUOTE]
Practice a little in your spare time!
[QUOTE=Mr. Bleak;42679412]Despite taking AP Stats and Calculus my last two years of high school, I'm in a two semester Pre-Calc class set this year. It's nice and easy and all, but I feel like I'm going to forget all the stuff I learned about differentiation and integration by next year.
Here's hoping my Psychology major won't be too math intensive.[/QUOTE]
Your psych major will not be math intensive.
Yeah, I just looked up University of Oregon's general requirements.
Apparently my single semester of pre-calculus will be more than sufficient. That's quite a surprise.
[QUOTE=Mr. Bleak;42687448]Yeah, I just looked up University of Oregon's general requirements.
Apparently my single semester of pre-calculus will be more than sufficient. That's quite a surprise.[/QUOTE]
Has psych ever been a math intensive field? Most people think the opposite.
I'm not bashing it btw
I didn't assume it to be, but I know some places have general ed requirements of a few semesters. I've always been fairly good at math, took AP Stats my Junior year and Calc my senior year and got near 100s in both, but it frustrates me an insane amount.
I feel like I forget everything pretty soon after the lessons. I'm doing stuff relating to logarithms this semester and, although in theory it's way easier than my second semester calc stuff, some of it is really confusing to me all over again.
[QUOTE=Yahnich;42732314]i know someone who does psychology and it's very intensive on statistics[/QUOTE]
I should really think that's dependent on where you go and on concentration.
I have a strong urge to work TREE(n(G)) into a proof for shits and giggles.
I might be a complete idiot, but how does one, step by step, solve (60/x) - (60/(x+3)) = 1 ?
Multiply 60/x by x+3/x+3 and 60/(x+3) by x/x. Simplify to get a quadratic. Solve.
Okay, so if f(x) is constant, then f(f(f(x))) is constant, right? Is the converse true?
[QUOTE=Krinkels;42835615]Okay, so if f(x) is constant, then f(f(f(x))) is constant, right? Is the converse true?[/QUOTE]
If f is constant then composing it with itself will still be constant. The converse is certainly not true - anything which takes 3 hops to get to the constant (e.g. f is constant zero except f(1)=2, f(2)=3)
Is the converse true if f must be continuous?
[QUOTE=Krinkels;42839727]Is the converse true if f must be continuous?[/QUOTE]
Still no:
f(x) = 0 for x<1
f(x) = 1-(1/x) for x>1
then f(f(x)) = 0 for all x
And a similar (but more complicated) construction would show that it's not even true for smooth functions.
Hey Joey, you're the only person I know of on FP who could potentially answer this.
I was talking to my professor in office hours and he claimed without proof that an isometry from a smooth manifold M to itself can't push a vector V at p forward into to W at p if the geodesic generated by V is closed and the geodesic generated by W is open, or vice versa. Is that true, and more importantly, why? I don't need a proof, just the intuition.
Jesus Johnny you scare the shit out of me for picking physics.
This isn't stuff you're likely to run into unless you want to do some serious GR research. My cosmologist research advisor who I need to know this for doesn't even understand this stuff, I'm basically teaching it to him in my research paper.
Can anyone give me the intuition for why continuous partial derivatives satisfying the Cauchy-Riemann equations is sufficient to establish analyticity for complex functions? I feel like it ought to be really obvious but it's like my brain is on strike :c
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