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  1. seanieg89

    Integration by parts

    If we are integrating the arc cosine function our indefinite integral need only be defined on the interval [-1,1], so the original posters substitution has no problems regarding domain and suffers from no flaw other than the fact that a smarter application of IBP yields the answer quicker. cos...
  2. seanieg89

    integration

    substituting u=1-4x yields answer with similar working to drogonski.
  3. seanieg89

    Another challenge question: Irrationality of e

    Nice work with the inequality, and yes contradiction is pretty much the only way to go for the second part. Its not very long, just a little tricky. Hint: Consider the proven inequality for certain special partial sums.
  4. seanieg89

    Another challenge question: Irrationality of e

    $ Euler's constant is commonly defined by the infinite series: $ e:= \sum_{n=0}^{\infty}\frac{1}{n!}$. By comparison to a suitable geometric series prove that $\frac{1}{(n+1)!} < e - s_n < \frac{1}{n \cdot n!}$ where $s_n$ is the $n$-th partial sum. Hence deduce that $e$ is irrational. $
  5. seanieg89

    Integration

    alternatively break it up as: x \sin^n(x) = x \sin^{n-1}(x)\sin(x)=-x \sin^{n-1}(x)\frac{d}{dx}(\cos(x)), then only one application of IBP will be required. more or less the same working though.
  6. seanieg89

    Polynomials

    a further hint: You can make use of the following factorisation (a^2+b^2)(c^2+d^2)=(ac+bd)^2 + (ad-bc)^2
  7. seanieg89

    Polynomials

    scroates is correct, you are all missing the 'if and only if'. The fact that the sum of two squares of polynomials is non-negative is indeed trivial, but the fact that every non-negative polynomial is expressible thus is not. Hint: Try induction in some form. And this is not my uni work, I...
  8. seanieg89

    Polynomials

    Prove that a real polynomial is non-negative everywhere if and only if it is expressible as the sum of two squares of polynomials. eg 2x^2 + 2x + 1 = x^2 + (x+1)^2, and the LHS is always non-negative. Don't see how it can be phrased much clearer than that...
  9. seanieg89

    Polynomials

    Prove that a polynomial with real coefficients P(x) \geq 0 for all real x if and only if P is expressible as the sum of two squares of real polynomials.
  10. seanieg89

    Integration proofs help

    You've created an extra factor of x in the integrand of the RHS of your final line trebla.
  11. seanieg89

    Harder Parabola Question

    yeah thats correct. Its a hard problem definately, but not beyond a 4u student (in theory) if properly guided. See: http://community.boredofstudies.org/14/mathematics-extension-2/244395/polynomials.html . The converse of that result would be the ideal lead-up question. Admittedly I did not...
  12. seanieg89

    Harder Parabola Question

    Only one normal to the parabola passes through the origin, namely the y axis.
  13. seanieg89

    Harder Parabola Question

    huh? what are you suggesting/asking?
  14. seanieg89

    integration question

    I'm not sure which courses don't require it, but it is technically correct. For example f(x) := 1/x is integrable over the interval [-2,-1] but our primitive F(x) = ln (x) is undefined for non-positive x. We need to use the primitive F(x) = ln |x| (this works because f is an odd function).
  15. seanieg89

    Harder Parabola Question

    One of the harder questions from my self-written quizzes for 4u tutees. Sketch the set of all points in the cartesian plane from which three different normals can be drawn to the parabola x^2=4ay
  16. seanieg89

    integration question

    Absolute values inside the logarithm trebla. And hscishard, no you cannot. You cannot in general take non-constant factors of the integrand to the outside of the integral.
  17. seanieg89

    Permutation and Combinations Help

    Each arrangement corresponds to a triple of non-negative integers (r,u,b) the number of red, blue, and black balls respectively in box A. Counting these triples is an easy application of the rule of product, keeping in mind the possible values these integers can take. The second part follows by...
  18. seanieg89

    Mathematics Specialist Tutoring

    Hey, My name is Sean and I have tutored high school mathematics for the last five years, both privately and as an employee of several tutoring agencies. I am currently accepting new students for private tuition. I am currently in my third and final year of a BSc (Advanced Mathematics) at the...
  19. seanieg89

    Conics Query

    Solution without Distance Formula: Let P' be the projection of P onto the y-axis. ie P'= (0,a \cdot sec(\theta) It is fairly simple to show that \triangle OXY is similar to \triangle P'PY \implies \frac{PX}{PY}=\frac{PY-XY}{PY}=1-\frac{XY}{PY}=1-\frac{OX}{P'P}=1-\frac{(a^2-b^2)\cdot...
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