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  1. J

    How is this graph Hamiltonian and Eulerian?

    Thanks. Guess the website has it wrong then.
  2. J

    How is this graph Hamiltonian and Eulerian?

    Is the graph Hamilton and Eulerian? The website says the graph is Hamilton and Eulerian but I think it's wrong. Ref: https://scanftree.com/Graph-Theory/Eulerian-and-Hamiltonian-Graphs There is no path that covers all paths only once. Any help? I think the graph is drawn wrongly.
  3. J

    Continuous or differentiable

    But these were the kind of functions in the examples to give us examples of discontinuous functions.
  4. J

    Continuous or differentiable

    One is 10 and the other is 12. This is function: f(x) = 2x for x = 0 to 5 so for x = 5, f(x) = 10 and f(x) = 0 for x < 0 and f(x) = 12 for x more than equal to 5, so f(x) = 12 for x = 5. So we have 2 values for x = 5. So function is discontinuous.
  5. J

    Continuous or differentiable

    But this function has multiple values at x = 5.
  6. J

    Continuous or differentiable

    But if a function has multiple values then it's not continous. Like there must not be any ambiguity. Example: f(x) = 2x for x = 0 to 5 and f(x) = 0 for x < 0 and f(x) = 12 for x more than equal to 5 is discontinuous at x = 5
  7. J

    Continuous or differentiable

    Ok. Got it. So a function is continuous if there is no place where the function is not defined.
  8. J

    Continuous or differentiable

    At x = 0, I get (-1)/(-1) = 1 I didn't understand this 'cancellation is not application only if factors aren't 0' Could you please explain?
  9. J

    Continuous or differentiable

    Homework Statement Homework Equations Solve using limits. Function is continuous if it's graph is continous throughout. here the (x-1) term gets cancelled in numerator and denominator. So we have a continuous graph of (x+1). The Attempt at a Solution The (x-1) term gets cancelled from...
  10. J

    Probability that more girls than boys go on the field trip

    This is horrible. I had this question in exam and in the answer sheet they gave the above solution. I wasted ten minutes in the exam trying to solve this question and ended up losing precious time. Now I got 43 out of 100 and won't get admission. Will have to apply again and study again for next...
  11. J

    Probability that more girls than boys go on the field trip

    Homework Statement Homework Equations First use equations to find out number of boys n number of girls. Then use probability concepts. The Attempt at a Solution Number of boys + number of girls = 12 Boys = Girls + 2 So we get Boys = 7, girls = 5. Select more girls than boys in a group of 3...
  12. J

    Tan (a + ib) = x + iy, so how is tan (a-ib) = x - iy

    Yeah that's actually pretty easy. I went ahead with that method and got the answer. Thanks.
  13. J

    Tan (a + ib) = x + iy, so how is tan (a-ib) = x - iy

    I didn't understand how Tan(X + iy) can be a + ib N then Tan(x-iy) be a-ib I didn't know tan is linear function.
  14. J

    Tan (a + ib) = x + iy, so how is tan (a-ib) = x - iy

    Thanks. It's a solved example. Have you also used the method in this link http://mathumatiks.org/subpage-556-Real%20and%20Imaginary%20Parts%20of%20Circular%20.htm Mathumatiks Website.
  15. J

    Number of ways 6 boys and 6 girls. No 2 B or 2 G together

    Oh yeah. I can't have _G_G_G_G_G_G_G_ I can't have boys such that between any blank is empty. Like no: B G B G G B G B G B G B So there aren't 7 places for girls to go. There are only 6 places with 2 possibilities 1) Leftmost seat is occupied by boy 2) Rightmost seat is by boy.
  16. J

    Number of ways 6 boys and 6 girls. No 2 B or 2 G together

    Homework Statement Number of ways in which 6 boys and 6 girls can sit such that no 2 boys and 2 girls are together in a row. Homework Equations If there are n different objects in a row, then to place the we have n! ways If n objects have n1 object of 1 kind then we have n!/(n1)! The Attempt...
  17. J

    Residue at poles of complex function

    Yeah you're right. In solved example they've broken 0.5z - 1.5i to z - 3j and multiplied 2 into numerator. Thanks.
  18. J

    Residue at poles of complex function

    The (3z+2)3 will get cancelled when I find residue at z = -2/3, just like the (3z-2)3 part will get cancelled when I find residue at z = 2/3.
  19. J

    Singularity of e^(-1/z^2)

    Wow thanks. I hope this question comes in exam now. Direct 10 marks, ten such questions. :D
  20. J

    Singularity of e^(-1/z^2)

    Should answer be C or D? I think it's C. Non-isolated singularity means like there are many other singularities neighboring 0 but that's not the case here. At z = 1, the f(z) becomes 0.368.
  21. J

    Singularity of e^(-1/z^2)

    z = it as t tend to 0 Means e^(+1/t^2) at t tend to 0 means e(1/0) means e^(inf) means inf. So there is a singularity at t or z = 0 But book answer is B So can i assume that book answer is wrong?
  22. J

    Singularity of e^(-1/z^2)

    This is getting more and more complicated. From what I know, singularity means function fails to be analytic at the point. Analytic has 2 conditions: Ux = Vy and Vx = -Uy Ux is derivative w.r.t x Uv is derivate w.r.t y Z = U + iV Short cut to find out analytic is see if function is going to...
  23. J

    Singularity of e^(-1/z^2)

    I'm sorry I didn't understand this.
  24. J

    Singularity of e^(-1/z^2)

    Also when z = 0, then e^(-1/0) = e^(-inf) = 1/[e^inf] = 1/inf = 0. This should be infinite if this is a singularity, right?
  25. J

    Singularity of e^(-1/z^2)

    But book answer is B.
  26. J

    Singularity of e^(-1/z^2)

    That's e^(100) Yeah you're right. So how to solve this question?
  27. J

    Singularity of e^(-1/z^2)

    I guess there is no singularity. For singularity e^(-1/z^2) has to go to infinite. That happens when So there is no singularity. Book answer is right.
  28. J

    Tan (a + ib) = x + iy, so how is tan (a-ib) = x - iy

    I guess it's for general function then. Like if f(x + iy) = real + imaginary times i then f(x - iy) = real - imaginary times i right?
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