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  • Know of a good free lightweight python IDE?

    I want a good IDE for developing python when i'm out of the office. We use proprietary software there and i'm not interested in buying a program if i don't have to. A good eclipse plugin would be nice, but I have a net-book who is not a big fan on the super bulky eclipse program, so something more lightweight would be better. I work with a bunch of java developers so none of us know a ton about python/jython....

    4 AnswersProgramming & Design10 years ago
  • |x - a|^(n+1)/(n+1)! Prove that this converges to zero.?

    a is an arbitrary constant.

    This is from the Taylor polynomial,

    it would be, f^n+1(c) * |x - a|^(n+1)/(n+1)!

    but i can show the the f^(n+1)(c) is bdd, so now I just need to show convergence to prove the approximation. Instinctively i can tell that this is true.

    (I know the o(n^n) contains o(n!) but not vice versa so this must be true, but i'm quite sure asymptotic notation is a terrible way to structure this argument haha.)

    I just need help structuring this argument i suppose. my background is computer science not math, so i have trouble with the proofs.

    1 AnswerMathematics1 decade ago
  • How does one create a bijection between the rational numbers and the rational numbers.?

    Saw a video online that said that rational numbers were isomorphic to the natural numbers, but i also know that rational numbers are dense, where natural numbers are not. I understand that this should work since we cant use the diagonalization argument to disprove this like on would with the irrationals. I'm just wondering what the argument for the bijection / isomorphism would look like? Is it far to simply map

    1 -> 1/1

    2 -> 1/2

    3 -> 1/3

    4 -> 2/3

    5 -> 1/4

    6 -> 2/4

    etc... ?

    ** I kinda think I answered my own question in the process of asking it, but gonna post it anyway to see what people think.

    2 AnswersMathematics1 decade ago
  • explain this paradox?

    Let: T = 1 + 2 + 4 + 8 + 16 + 32 + . . .

    Thus: 2T = 2 + 4 + 16 + 32 . . .

    Thus: 2T - T = -1 + 0 + 0 + 0 + 0 . . .

    Thus: T = -1

    Therfore : -1 = 1 + 2 + 4 + 8 + 16 + 32 + . . .

    This is clearly impossible yet, i see no contradictions in the proof?

    does anyone understand this?

    2 AnswersMathematics1 decade ago
  • What compiler/editor should i use for c++?

    been doing java for about three years now, time to move on. What do you recommend?

    6 AnswersProgramming & Design1 decade ago
  • Does any one else ever dream about math equations or is grad school driving me crazy...?

    I have dreams about solving equations in my dreams, And when I'm stressed I usually wake up thinking about some equation that just doesn't work/ is unsolvable and I can't fall back asleep until i get myself to forget about it. The funny things is how bad my unconscious mind is at math, because when i do remember whatever it was i was thinking about, I always do retarded things that don't make any sense.

    Any analysis / useful insight

    1 AnswerPsychology1 decade ago
  • Momentum and Kinertic Energy?

    Okay so this is a problem i think i have an answer to but I cant tell... can anyone confirm or refute this?

    A weight (17.0kg) hangs from a wire (4m) with negligible mass. The wire is free to rotate 360 degrees. A dart (5.5kg) is shot horizontally at the weight and embeds itself into the mass. What is the minimum speed of the dart required to propel the weight in a full loop.

    a.)

    Assumptions:

    --The initial energy = Ki = (1/2)M*Vi^2

    --If the weight make it so the top of the circle it finishes the loop.

    --As the velocity approaches the minimum the energy at the top of the loop Kf approaches 0.

    --The work done is equal to work of lifting a weight twice the radius of the circle --> W = M*g*10

    --Final energy equals initial energy plus work Kf = Ki + W

    Thus 0 = (1/2)M*Vi^2 + M*g*10

    M = 17+5.5 = 22.5; g = -9.8.

    Vi = SQRT((2)(9.8)(8))

    b.)

    |Vi| is the speed of the weight.

    momentum of the weight = mass of the weight * velocity of the weight

    momentum of the dart = mass of the dart * velocity of the dart

    --> Mw*Vi=Md*Vd so

    --> Vd = Md*Vi/Mw

    --> Vd = Md*(SQRT(2*g*2R))/Mw

    --> Vd = 5.5*(SQRT(2*9.8*8))/22.5

    -->3.2 M/s

    This seems too low, but i cant see what would be wrong...

    4 AnswersPhysics1 decade ago