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# δοτζο

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• ### Christians: Assuming you've read the bible...?

Assuming you've read the bible, how did you determine that God was the good one and Satan was the evil one?

I've yet to have the opportunity have ask someone this face to face, so I thought I'd try here... brave proposition, I know. But I welcome all answers.

15 AnswersReligion & Spirituality7 years ago
• ### Proving property of Fibonacci Numbers using Binet's Formula?

Let α and β be the roots of x² - x - 1 = 0 (the Golden ratio and it's complement). Then Binet's formula states that

F_n = (αⁿ - βⁿ) / (α - β)

where F_n is the nth Fibonacci Number.

Now, what I want to know is there any way besides just some crazy algebraic trick to prove that

F_{n+m} = F_{m+1}F_n + F_mF_{n - 1}

using Binet's Formula. I can do it through induction quite easily, but the instructions state that Binet's Formula is to be used, and even that with induction is just messy... unless I'm missing something.

I know

aⁿ - bⁿ = (a - b)∑{i = 1,n} a^(n - i) b^(i - 1)

but that doesn't seem to be of much help.

Any assistance is greatly appreciated.

• ### So what's with the new rating system for answers?

What happened to the simple Thumbs up vs. Thumbs down rating system? And moreover how do we rate bad answers? Are they supposed to be "Amusing" simply because they're wrong?

Please return the Thumbs system. (Listens to the silence)

• ### What does a person, who doesn't really like to read, read next?

I am a person who generally doesn't like to read all that much, but when I begin a book that grabs my interest I can't put it down. I've even been known to stay up 36+ hours reading. The problem is that I don't often find books which can do this; so I ask of you all to recommend some books that may do this for me. The things that I've read recently that I like the most include The Dark Tower series by Stephen King, the Harry Potter books by J.K. Rowling, and stories about King Arthur and all associated characters (Merlin, Lancelot, etc.). Obviously I tend toward fantastic, epic stories.

Thank you good people.

3 AnswersBooks & Authors9 years ago
• ### Oddities of the Möbius strip?

Hi there Yahoo! Answers people. I once again turn to your wisdom with a problem that has completely gotten me stumped and had my mind tied up for nearly a week now.

So, we're studying developable surfaces and we're trying to prove that for a Möbius strip to be constructible out of paper (non-compressing, non-stretching, bounded) if it has width 1 it must have length between π/2 and √3. At the end of the proof we see that for it to be π/2 the paper must past through itself, but that's not my main problem. Our book is horrible at doing proofs in a way people can actually understand them. It first goes over rulings, straight intervals on the surface that pass through a given point, and says that if a surface is developable these rulings have parallel tangent planes at each point.

So here's my question: when concerning a Möbius strip given by the parametrization

M(u,v) = <cos(u), sin(u), 0> + v <cos(u/2) cos(u), cos(u/2) sin(u), sin(u/2)>

it is a ruled surface as it satisfies the form

x(u,v) = b(u) + vδ(u)

but is that ruling, i.e. x(u₀,v) for some fixed u₀, equivalent to the developable ruling above?

Our book shows a flat representation of the Möbius strip which has rulings that are trapezoidal in shape, but the rulings given by the parameterization are square, each line is perpendicular to the 3-d surface, even if it was cut and unrolled it would be perpendicular, wouldn't it? So are the trapezoids projections of the parametric rulings onto the plane? Or are the parameterization and making the Möbius strip out of paper (from what I understand is an embedding of the plane into 3-space) completely different?

If someone could shed some light onto this that'd be fantastic.

http://www.ucl.ac.uk/~ucesgvd/moebius.pd…

That link has pictures of what I'm talking about as far as the trapezoidal ruling. From what I can tell, in the context of this paper, those lines are "straight generators" for the shape, but I have no idea what that means as I can't find a definition of a "straight generator" anywhere.

As far as my math background (so you can make your explanation more understandable to me): I am currently taking Topology and we're studying homeomorphisms now; I've taken up through Vector Calculus (not analysis) and have also taken abstract algebra.

I know it's long already, but 1 more thing. I found that half the edge length (0 to π) in the parameterization is more than the mid-line circumference. Should that happen when you make a Möbius strip out of paper in the real world, and is that where the discrepancy between the rulings come from?

Thanks again.

• ### Oddities of the Möbius strip (high level math warning)?

Hi there Yahoo! Answers people. I once again turn to your wisdom with a problem that has completely gotten me stumped and had my mind tied up for nearly a week now.

So, we're studying developable surfaces and we're trying to prove that for a Möbius strip to be constructible out of paper (non-compressing, non-stretching, bounded) if it has width 1 it must have length between π/2 and √3. At the end of the proof we see that for it to be π/2 the paper must past through itself, but that's not my main problem. Our book is horrible at doing proofs in a way people can actually understand them. It first goes over rulings, straight intervals on the surface that pass through a given point, and says that if a surface is developable these rulings have parallel tangent planes at each point.

So here's my question: when concerning a Möbius strip given by the parametrization

M(u,v) = <cos(u), sin(u), 0> + v <cos(u/2) cos(u), cos(u/2) sin(u), sin(u/2)>

it is a ruled surface as it satisfies the form

x(u,v) = b(u) + vδ(u)

but is that ruling, i.e. x(u₀,v) for some fixed u₀, equivalent to the developable ruling above?

Our book shows a flat representation of the Möbius strip which has rulings that are trapezoidal in shape, but the rulings given by the parameterization are square, each line is perpendicular to the 3-d surface, even if it was cut and unrolled it would be perpendicular, wouldn't it? So are the trapezoids projections of the parametric rulings onto the plane? Or are the parameterization and making the Möbius strip out of paper (from what I understand is an embedding of the plane into 3-space) completely different?

If someone could shed some light onto this that'd be fantastic.

http://www.ucl.ac.uk/~ucesgvd/moebius.pd…

That link has pictures of what I'm talking about as far as the trapezoidal ruling. From what I can tell, in the context of this paper, those lines are "straight generators" for the shape, but I have no idea what that means as I can't find a definition of a "straight generator" anywhere.

As far as my math background (so you can make your explanation more understandable to me): I am currently taking Topology and we're studying homeomorphisms now; I've taken up through Vector Calculus (not analysis) and have also taken abstract algebra.

I know it's long already, but 1 more thing. I found that half the edge length (0 to π) in the parameterization is more than the mid-line circumference. Should that happen when you make a Möbius strip out of paper in the real world, and is that where the discrepancy between the rulings come from?

Thanks again.

• ### Oddities of the Möbius strip (high level math warning)?

Hi there Yahoo! Answers people. I once again turn to your wisdom with a problem that has completely gotten me stumped and had my mind tied up for nearly a week now.

So, we're studying developable surfaces and we're trying to prove that for a Möbius strip to be constructible out of paper (non-compressing, non-stretching, bounded) if it has width 1 it must have length between π/2 and √3. At the end of the proof we see that for it to be π/2 the paper must past through itself, but that's not my main problem. Our book is horrible at doing proofs in a way people can actually understand them. It first goes over rulings, straight intervals on the surface that pass through a given point, and says that if a surface is developable these rulings have parallel tangent planes at each point.

So here's my question: when concerning a Möbius strip given by the parametrization

M(u,v) = <cos(u), sin(u), 0> + v <cos(u/2) cos(u), cos(u/2) sin(u), sin(u/2)>

it is a ruled surface as it satisfies the form

x(u,v) = b(u) + vδ(u)

but is that ruling, i.e. x(u₀,v) for some fixed u₀, equivalent to the developable ruling above?

Our book shows a flat representation of the Möbius strip which has rulings that are trapezoidal in shape, but the rulings given by the parameterization are square, each line is perpendicular to the 3-d surface, even if it was cut and unrolled it would be perpendicular, wouldn't it? So are the trapezoids projections of the parametric rulings onto the plane? Or are the parameterization and making the Möbius strip out of paper (from what I understand is an embedding of the plane into 3-space) completely different?

If someone could shed some light onto this that'd be fantastic.

http://www.ucl.ac.uk/~ucesgvd/moebius.pdf

That link has pictures of what I'm talking about as far as the trapezoidal ruling. From what I can tell, in the context of this paper, those lines are "straight generators" for the shape, but I have no idea what that means as I can't find a definition of a "straight generator" anywhere.

As far as my math background (so you can make your explanation more understandable to me): I am currently taking Topology and we're studying homeomorphisms now; I've taken up through Vector Calculus (not analysis) and have also taken abstract algebra.

• ### Topology and Sequences?

I'm having a real issue with this proof. I'll state the actual problem then show what I've got so far. Any help would be fantastic.

Prove the following: Let A is a subset of ℝⁿ with the standard topology. If x is a limit point of A, then there is a sequence of points in A that converges to x.

Attempted proof:

x being a limit point of A means that every neighborhood of x (open balls) intersects A at some point(s) other than x itself. As ℝⁿ is Hausdorff, every convergent sequence converges to a unique point.

And that's where I get stuck. It's kind of obvious that you could find a sequence, but I just can't think of how to say it or have whatever I say to be convincing.

Thanks for the help.

• ### Abstract Algebra Question involving Quotient Groups?

Question:

Assuming that N ⊲ G, prove that if [G:N] is a prime, then G/N is cyclic. Is the converse true?

Thank you.

• ### Abstract Algebra question(s)?

I've been stuck on this for a while now, and it's really bothering me.

If G is an Abelian group, with a,b ∈ G, with ord(a) = m, ord(b) = n, then (ab)^(mn) = e. Indicate where you use the condition that G is Abelian.

I actually did this problem, but I'm not sure if I put the condition of G being Abelian in the right place.

(ab)^(mn) = ((ab)^m)^n = ((a^m)(b^m))^n = (eb^m)^n = {b^mn = b^nm} = (b^n)^m = e^m = e

The brackets are where I use the condition of G being Abelian.

G is an Abelian group, with a,b ∈ G, with ord(a) = m, ord(b) = n. Prove that ord(ab) divides ord(a)ord(b).

I'm completely stuck on this one. It's obvious if ord(ab) = 0 (mod m) or (mod n), but otherwise I'm completely lost.

• ### Where can I find an answer booklet for Modern Algebra: An Introduction by John R. Durbin?

I don't care if it's in electronic or tangible form.

Thank you.

• ### What does Υαχαι mean?

I was driving to my dad's house and I saw it a few times on buildings and thought it was odd that a company would have it's name or some advertisment in greek.

Thanks!

• ### How to see specific Unicode Characters in Google Chrome for Mac?

I love Chrome to death, it's the best browser I've ever used but there is one detrimental problem that I'm sure has an easy solution that I just can't find. I post on here quite a bit, primarily in the math section, and I rely on unicode characters for simplicity and semantically correct answers. Out of the many characters I use on a daily basis I've notices that ℝ (a double struck R), ℤ (a double struck Z), ℕ (a double struck N), and ℒ (a script L) among other letters in the "Letterlike Symbols" category of unicode don't show up in text boxes when I'm using Chrome. All I see is a ™ in a box. I can highlight the box and right click for a drop down menu in which there is an option to search google for that character, where it shows up correctly. So I know the problem isn't my compute it's just that Chrome for some reason doesn't have this block of unicode.

I'm using the same fonts I did with Safari, where everything worked perfectly. I input the characters using the Character Palette. I'm using Chrome 5.0.342.7 beta for Mac and am running Snow Leopard on an Intel MacBook Pro.

• ### Dropped my MacBook Pro today :(?

Everything seems to be fine performance wise, but there is an obvious problem with the screen. There's a dent in the back of the screen and I believe the LCD has cracked. There are vertical rainbow lines concentrated about the dent and they transfer into horizontal lines at the bottom of the screen. My question is, do I need to replace just the LCD or the whole thing?

Details:

15" MacBook Pro (not unibody)

2.6 GHz Intel Core 2 Duo

Little over a year old

My friend can do the labor so all I need is the part (thank goodness), but I wanted to ask before I spent about 200-500 extra dollars.

I'm currently enrolled as a Mathematics Major in USF. This semester I was not able to take any math classes (just physics II which counts toward pre-reqs) because of late registration (transfer student). Anyway, next semester I need to catch up so I was planning to take 5 classes: Calc III, Linear Algebra, Bridge to Abstract Math, Differential Equations, and Latin II (foreign language requirement). I was just wondering if this would be "do-able."

I get A's in math pretty easily. I'll stay up late at night doing homework that isn't to be due for months because it's so fascinating to me. I know what to expect with Calc III and DE and even taking them at the same times should be fairly breezy. I'm not sure of what Linear Algebra and Bridge to Abstract consist of or how easy they are. I plan to talk to my counselor in the next week or two, but I thought I'd also get the opinion of those here on Yahoo! Answers.

Thank you.

I'm currently enrolled as a Mathematics Major in USF. This semester I was not able to take any math classes (just physics II which counts toward pre-reqs) because of late registration (transfer student). Anyway, next semester I need to catch up so I was planning to take 5 classes: Calc III, Linear Algebra, Bridge to Abstract Math, Differential Equations, and Latin II (foreign language requirement). I was just wondering if this would be "do-able."

I get A's in math pretty easily. I'll stay up late at night doing homework that isn't to be due for months because it's so fascinating to me. I know what to expect with Calc III and DE and even taking them at the same times should be fairly breezy. I'm not sure of what Linear Algebra and Bridge to Abstract consist of or how easy they are. I plan to talk to my counselor in the next week or two, but I thought I'd also get the opinion of those here on Yahoo! Answers.

Thank you.

• ### Solenoid and magnetism help?

Got this question and I think I have the right answer, but the answer isn't in the back so I can't check.

A solenoid 10.0cm in diameter and 75.0cm long is made from copper wire of diameter 0.100 cm, with very thin insulation. The wire is wound onto a cardboard tube in a single layer, with adjacent turns touching each other. What power must be delivered to the solenoid if it is to produce a field of 8.00 mT at its center?

I know that:

B = μ₀nI, where B is the magnetic field, μ₀ is the permeability of free space (4πx10⁻⁷), n is the number of turns per unit length, and I is the current.

So to find N, the number of coils in the length, I did:

75.0 cm / 0.100 cm = 750 coils.

Therefore n, coils per unit length, is:

750 coils / 75cm = 10 coils / cm = 1000 coils / m

So:

I = B / (μ₀n) = 0.008 T / (μ₀(1000)) ≈ 6.367 A

P = I²R and copper has a resistivity of 1.7x10⁻⁸.

1000 coils / m at 10cm for each coil works out to:

C = dπ = .1π m

Length = C*N = .1π(1000) = 100π m ≈ 314.16 m

P = (6.367A)²(1.17x10⁻⁸ ohm*m) / (314.16 m) = 1.5x10⁻⁹ W

Just want to know if my answer is correct and my methods are correct. Not much an application of math kinda guy. I'm definitely more comfortable with abstract concepts.

Thanks!

1.00 g of copper is used to make a wire that has a resistance of 0.500 ohms. Find a) the length of the wire, and b) the diameter of the wire.

I'm not really sure where to start on this one. I think i have to use

R = (rho)*(L / A) where rho is the resistivity, L the length, and A the cross-section diameter. The resistivity of copper is 1.72x10^(-8) ohm-meters.

• ### I need some help with a greek translation?

I found a t-shirt with greek on it and I don't know what it means...

Greek:

O, Αρχιμήδης είναι εξυπνώτερος από σας

What I got for the roman transliteration:

O, Arkimedes einai exypnoteros apo sas.