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    Problem 519*

    Find the indefinite integral in series form for

    \displaystyle \int \cos(x^2) \, \mathrm{d}x
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    (Original post by Zacken)
    Problem 519*

    Find the indefinite integral in series form for

    \displaystyle \int \cos(x^2) \, \mathrm{d}x
    Solution 519
    \displaystyle \cos(x) = 1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\frac{x^8}{8!}+ \cdots \\

\displaystyle \cos(x^2) = 1-\frac{x^4}{2!}+\frac{x^8}{4!}-\frac{x^{12}}{6!}+\frac{x^{16}}{  8!}+\cdots\\

\displaystyle \int \cos(x^2) \ dx = x-\frac{x^5}{2! \times 5}+\frac{x^9}{4! \times 9}-\frac{x^{13}}{6! \times 13}+\frac{x^{17}}{8! \times 17} + O(x^{21}) + C
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    Problem 520**
    This should be very easy for computer science students, but still qualifies as a math problem:

    A sequence \displaystyle(a_n)_{n\in\mathbb{  N}} is a convergent sequence and is defined recursively as follows:

    \displaystyle a_{n+1} = \frac{1}{2}(a_n+\frac{x}{a_n}) \quad \mathrm{with} \quad a_0 = c

    Such that x and c are constants and the values that they can take are:

    x > c > 1

    Deduce mathematically and/or logically the value of \displaystyle \lim_{n\to\infty}a_n in terms of c and x.
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    (Original post by udvuvvdu)
    Solution 519
    \displaystyle \cos(x) = 1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\frac{x^8}{8!}+ \cdots \\

\displaystyle \cos(x^2) = 1-\frac{x^4}{2!}+\frac{x^8}{4!}-\frac{x^{12}}{6!}+\frac{x^{16}}{  8!}+\cdots\\

\displaystyle \int \cos(x^2) \ dx = x-\frac{x^5}{2! \times 5}+\frac{x^9}{4! \times 9}-\frac{x^{13}}{6! \times 13}+\frac{x^{17}}{8! \times 17} + O(x^{21}) + C

    Well, this is correct, although I must say that when I asked for a series solution, I meant in the much cleaner sense of utilising sigma notation. But this is fine.
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    (Original post by udvuvvdu)
    Problem 520**
    x
    This seems far too easy, have I done something wrong?

    Assume the limit exists and is \ell, then we have (in the limit):

    \displaystyle \ell = \frac{1}{2}\left(\ell+\frac{x}{l  }\right) \iff \ell^2 = x \implies \ell = \sqrt{x}.

    Interestingly enough, this is the ancient method used for calculating square roots, the starting value a_0 = c has no bearing on the limit but does change the computation time for evaluating the square root.
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    (Original post by Zacken)
    This seems far too easy, have I done something wrong?

    Assume the limit exists and is \ell, then we have (in the limit):

    \displaystyle \ell = \frac{1}{2}\left(\ell+\frac{x}{l  }\right) \iff \ell^2 = x \implies \ell = \sqrt{x}.
    Your answer is right... no, you're not wrong. I might have just overestimated the difficulty of the question since I am doing the IB and not A-levels further math

    (Original post by Zacken)
    Well, this is correct, although I must say that when I asked for a series solution, I meant in the much cleaner sense of utilising sigma notation. But this is fine.
    .... sorry about that.
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    (Original post by udvuvvdu)
    Your answer is right... no, you're not wrong. I might have just overestimated the difficulty of the question since I am doing the IB and not A-levels further math



    .... sorry about that.
    It's fine, it's a bit of fun. Do you know much about analysis? Very interesting field, in my opinion.

    By the way, I've done both the IB and A-Levels, so I can empathise with you: how far along your course are you and how are you finding it?

    Edit: this is a bit of a guess in the dark, but you're doing the Calculus option for HL, aren't you?
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    (Original post by Zacken)
    This seems far too easy, have I done something wrong?

    Assume the limit exists and is \ell, then we have (in the limit):

    \displaystyle \ell = \frac{1}{2}\left(\ell+\frac{x}{l  }\right) \iff \ell^2 = x \implies \ell = \sqrt{x}.

    Interestingly enough, this is the ancient method used for calculating square roots, the starting value a_0 = c has no bearing on the limit but does change the computation time for evaluating the square root.
    You have found what the limit will be if it exists, but can you prove the limit exists?
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    (Original post by Zacken)
    It's fine, it's a bit of fun. Do you know much about analysis? Very interesting field, in my opinion.

    By the way, I've done both the IB and A-Levels, so I can empathise with you: how far along your course are you and how are you finding it?

    Edit: this is a bit of a guess in the dark, but you're doing the Calculus option for HL, aren't you?
    I'm on my second year, IB provides a very good course in terms of subjects in my opinion. I do not really know how A-Levels work since I'm not in the UK. I like how flexible the IB is and they have a lot of course options (too bad my school doesn't offer most of them). I find the subjects such as Physics HL and Math HL not hard to understand but each time I practice past papers I can see IB often comes out with very challenging question that takes time to digest or work out somehow. It's better than my local curriculum.

    I don't really enjoy the extra workload for Extended Essay and CAS though, I find TOK quite pointless but sometimes sparks interesting discussions. The only thing I really hate is sometimes in Physics questions we need to follow "IB Definitions" when asnwering questions.

    You guessed right, I'm doing the Calculus options and sadly I barely have knowledge in the topic of analysis, I kind of want to find out more about it though but then I have Internal Assessments that I need to catch up for.
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    (Original post by Gome44)
    You have found what the limit will be if it exists, but can you prove the limit exists?
    The question asks you to assume convergence. However:

    We can prove that it converges by demonstrating that the sequence is monotone-decreasing and we know that monotonic decreasing sequences bounded below converges :

    We have, from simple algebra that \displaystyle a_{k+1} - a_k = \frac{1}{2a_k} \left(x -  a_k^2 \right), now pick some a_0^2 > x then our equation shows us that the sequence a_1 \geq a_2 \geq a_3 \geq \cdots is monotone decreasing and non-negative, hence bounded below by 0, so it converges to some limit \ell.
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    Just for the record, here's the "neat" way

    Alternative solution to 519

    \displaystyle \cos(x) = \sum_{n=0}^{\infty} \dfrac{(-1)^n}{(2n)!} x^{2n} \Rightarrow \cos(x^2)= \sum_{n=0}^{\infty} \dfrac{(-1)^n}{(2n)!} x^{4n}

    Hence


    \displaystyle \int \cos(x^2) \ dx = C + \sum_{n=0}^{\infty} \dfrac{(-1)^n}{(4n+1)(2n)!} x^{4n+1}
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    Problem 521
    Three small, smooth perfectly elastic spheres A, B and C lie at rest, in that order, in a straight line on a smooth horizontal plane. The spheres A and C both have mass 7m and sphere B is of mass m. The sphere B is then projected directly towards sphere C. Given that after all possible collisions have taken place, sphere B is moving with speed u, find the final speeds of spheres A and C.
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    Problem 522.

    A straight tunnel is bored through the Earth (not necessarily through the centre). A particle of mass m is dropped from the surface of the Earth into this tunnel. Show that its motion is simple harmonic and compute its period.

    Posted from TSR Mobile
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    (Original post by A Slice of Pi)
    Problem 521
    Three small, smooth perfectly elastic spheres A, B and C lie at rest, in that order, in a straight line on a smooth horizontal plane. The spheres A and C both have mass 7m and sphere B is of mass m. The sphere B is then projected directly towards sphere C. Given that after all possible collisions have taken place, sphere B is moving with speed u, find the final speeds of spheres A and C.
    I don't suppose that
    Spoiler:
    Show
    \displaystyle v_C = \frac{4u}{9} and \displaystyle v_A = -\frac{u}{3}
    would be correct?
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    (Original post by Zacken)
    I don't suppose that
    Spoiler:
    Show
    \displaystyle v_C = \frac{4u}{9} and \displaystyle v_A = -\frac{u}{3}
    would be correct?
    I've asked other people this question, and everyone seems to have different answers, so I'm unsure. Perhaps you could post your method?
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    Here's another easy one, but it's kind of nice

    Problem 523

    Let the function

    f(t) = e^{-it} (\cos t + i\sin t)

    be defined for real t.

    By first using calculus to show that f(t) is a constant function for all t, show that

    e^{it} = \cos t + i\sin t
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    Solution 523
    Spoiler:
    Show

     f'(t) = -je^{-jt}(\cos t + j\sin t) + e^{-jt}(-\sin t + j\cos t) 

= e^{-jt}(j\cos t  + \sin t - \sin t  + j\cos t) = 0
    Hence f(t) is constant.
    Letting t = 0,  f(0) = 1(\cos 0 + j\sin 0) = 1
    Hence  f(t) = 1 = e^{-jt}(\cos t + j\sin t) \implies e^{jt} = \cos t + j\sin t

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    Problem 524

    Same aim as 523, different method.

    Solve the differential equation

     \frac{dz}{dt} = kz *.

    Show that z = \cos t + j\sin t is a solution to * with  k = j and hence deduce that

     e^{jt} = \cos t + j\sin t **

    State the assumption made in the "proof" of **.
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    Solution 524

    \displaystyle \int \frac{\mathrm{d}z}{z} = kt + c \Rightarrow z = Ae^{kt}

    If z = \cos t + i \sin t then z' = -\sin t + i \cos t = i(\cos t + i \sin t) hence satisfying * with k = i.

    Hence, we have (by the uniqueness of solutions) that \displaystyle e^{it} = \cos t + i\sin t.
 
 
 
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