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    (Original post by Smaug123)
    Yay
    Spoiler:
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    Au = \lambda u, Av = \mu v, so v^T (Au) = \lambda v^T u by premultiplying the first equation by v^T; also v^T A u = (A v)^T u by symmetry, and that is \mu v^T u. Hence \lambda v^T u = \mu v^T u; but \lambda, \mu are not the same, so v^T u = 0.
    What's the yay for?

    Also it's not quite what I would have written

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    I would have used the dot product, but maybe that is what you have done?? :P


    But it seems right! :thumbsup:
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    (Original post by rayquaza17)
    What's the yay for?

    Also it's not quite what I would have written

    Spoiler:
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    I would have used the dot product, but maybe that is what you have done?? :P


    But it seems right! :thumbsup:
    Spoiler:
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    v^T u = v.u - that's how we define the dot product

    Yay because there's some linear algebra appearing! I've managed to train myself to like linear algebra, and now I like it
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    (Original post by Smaug123)
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    v^T u = v.u - that's how we define the dot product

    Yay because there's some linear algebra appearing! I've managed to train myself to like linear algebra, and now I like it
    Oh, we defined it using <>'s.

    Linear algebra is nice compared to the nasty integrals that dominate this thread!

    I like finding eigenvalues and eigenvectors of 2x2 matrices.
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    (Original post by Smaug123)
    Spoiler:
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    v^T u = v.u - that's how we define the dot product

    Yay because there's some linear algebra appearing! I've managed to train myself to like linear algebra, and now I like it
    Have some more:

    Problem 463***:

    Let U,V be vector spaces over a field K, \tau: U \times V \to K a bilinear form. For W \subset V, define the \tau-orthogonal complement U^\perp := \{ u \in U : \tau(u,w) = 0 \, \forall w \in W \}, and similarly for W \subset U. Prove that, for any W \subset U, ((W^\perp)^\perp)^\perp = W^\perp.
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    Problem 464 *

    Take the matrix \begin{pmatrix} a & b \\ c & d \end{pmatrix} where a,b,c and d can be any digit from 0-9 inclusive. What is the probability of choosing a singular matrix when values for a,b,c and d are randomly chosen?
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    Problem 465*

    If all the integer numbers from 1 to 1000 inclusive were written out as words, how many letters would be used? Don't count spaces, and make sure to use the word and, e.g. 'seven hundred and forty two', which is 23 letters long.

    For anyone interested, this problem was taken from Project Euler which is a series of challenging mathematical problems that can be solved using clever programming skills. Check it out!

    https://projecteuler.net
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    (Original post by majmuh24)
    Problem 464 *

    Take the matrix \begin{pmatrix} a & b \\ c & d \end{pmatrix} where a,b,c and d can be any digit from 0-9 inclusive. What is the probability of choosing a singular matrix when values for a,b,c and d are randomly chosen?
    Solution 464 *

    Use matlab code

    syms a b c d n s t det


    t=0;
    s=0;
    n=0;

    while n<9999;

    a=floor(n/1000);
    b=floor((n-1000*a)/100);
    c=floor((n-1000*a-100*b)/10);
    d=floor(n-1000*a-100*b-10*c);

    det=a*d-b*c;

    if det==0;
    s=s+1;
    t=t+1;
    else
    t=t+1;
    end
    n=n+1;
    end

    s
    t
    n

    and the answer is 569/9999.
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    (Original post by james22)
    Solution 464 *

    Use matlab code

    syms a b c d n s t det


    t=0;
    s=0;
    n=0;

    while n<9999;

    a=floor(n/1000);
    b=floor((n-1000*a)/100);
    c=floor((n-1000*a-100*b)/10);
    d=floor(n-1000*a-100*b-10*c);

    det=a*d-b*c;

    if det==0;
    s=s+1;
    t=t+1;
    else
    t=t+1;
    end
    n=n+1;
    end

    s
    t
    n

    and the answer is 569/9999.
    Very sneaky, but shouldn't it be out of 10000? Are you missing out a matrix? :confused:

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    (Original post by majmuh24)
    Very sneaky, but shouldn't it be out of 10000? Are you missing out a matrix? :confused:

    Posted from TSR Mobile
    You are right, I should have put <=n not <n so I missed the 9999 case. The actual answer is then the much neater looking 0.057.
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    (Original post by Flauta)
    No of course not But the definite integral from 0 to 1 only needs a substitution to evaluate
    I needed two of 'em :sigh:
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    (Original post by Khallil)
    I needed two of 'em :sigh:
    Maybe I forgot one? I think I used x= \tan \theta and the fact that \int^b_0 f(\theta) d\theta = \int^b_0 f(b-\theta) d\theta
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    (Original post by Flauta)
    ...
    Oh come on. I did the exact same. The second is still a substitution!
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    (Original post by Khallil)
    Oh come on.



    I did the exact same. The second is still a substitution!
    Sorry!
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    (Original post by Flauta)
    Sorry!
    I thought the gif was a bit much.

    After a quick stalk of your profile, MFP3 is your first exam?
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    (Original post by Khallil)
    I thought the gif was a bit much.

    After a quick stalk of your profile, MFP3 is your first exam?
    Nah 'twas fine.

    If you don't include my two music ones next week then yeah Really looking forward to it, differential equations are so easy now I've solved loads and the polar stuff isn't bad if you draw a diagram and remember the power reducing formulae for integration. Which is your first?
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    (Original post by Flauta)
    Nah 'twas fine.

    If you don't include my two music ones next week then yeah Really looking forward to it, differential equations are so easy now I've solved loads and the polar stuff isn't bad if you draw a diagram and remember the power reducing formulae for integration. Which is your first?
    AQA Physics Unit 5C - 19th June

    I haven't started preparing for it at all. Everything's a bit rusty since last year.
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    Problem 465***

    100 rooms each contain countably many boxes labeled with the natural numbers. Inside of each box is a real number. For any natural number n, all 100 boxes labeled n (one in each room) contain the same real number. In other words, the 100 rooms are identical with respect to the boxes and real numbers.

    Knowing the rooms are identical, 100 mathematicians play a game. After a time for discussing strategy, the mathematicians will simultaneously be sent to different rooms, not to communicate with one another again. While in the rooms, each mathematician may open up boxes (perhaps countably many) to see the real numbers contained within. Then each mathematician must guess the real number that is contained in a particular unopened box of his choosing. Notice this requires that each leaves at least one box unopened.

    99 out of 100 mathematicians must correctly guess their real number for them to (collectively) win the game.

    What is a winning strategy?

    This wasn't invented by me, found it and thought it very interesting.
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    Not sure if this has been posted before.

    Proofs that Pi is rational: http://webonastick.com/pi/#940831

    "...submits to us a proof using alternative algebra. Before we show this proof, we must explain alternative algebra to those who do not know of it. Alternative algebra can be summarized as two statements, the Fundamental Postulate and Fundamental Theorem of Alternative Algebra.
    The Fundamental Postulate of Alternative Algebra. One may divide by zero.
    The Fundamental Theorem of Alternative Algebra. Any two numbers are equal."
    Alternative algebra vs symbolic maths?
    :shakehand:
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    Problem 466**

    Determine

     \displaystyle \int_0^{1} \frac{\ln(1 + x^2)}{x} dx
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    (Original post by jjpneed1)
    Problem 466**

    Determine

     \displaystyle \int_0^{1} \frac{\ln(1 + x^2)}{x} dx

    \begin{aligned} \displaystyle \int_0^{1} \frac{\ln(1 + x^2)}{x} \, dx & = \int_{0}^{1} \sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{n+1}\,dx =  \sum_{n=0}^{\infty}\int_{0}^{1} \frac{(-1)^nx^{2n+1}}{n+1}\,dx \\& = \sum_{n=0}^{\infty}\frac{(-1)^n}{(n+1)(2n+2)} = \frac{1}{2} \sum_{n=0}^{\infty}\frac{(-1)^n}{(n+1)^2}  \\& = \frac{1}{2} \left(1-\frac{1}{2}\right)\sum_{n=0}^{ \infty} \frac{1}{(n+1)^2} = \frac{1}{4} \sum_{n=1}^{\infty} \frac{1}{n^2} \\& = \frac{1}{4}\left(\frac{\pi^2}{6} \right) = \frac{\pi^2}{24}. \end{aligned}
 
 
 
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