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    (Original post by Arieisit)
    Looks like my system of differential equations was a bit beyond you guys I guess. I'll post the solution in the very near future.

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    You've gotta be kidding...

    With trivial substitutions we arrive at y_{1}'''-6y_{1}''+12y_{1}'-8=0; hence, P(z)=(z-2)^{3} is the characteristic polynomial with a triple root z=2, so the general solution to this equation is y_{1}=c_{1}e^{2x}+c_{2}xe^{2x}+c  _{3}x^{2}e^{2x}. Now it is trivial to find y_{2} and y_{3}.

    I really have no time at the moment to do problems on this thread.
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    (Original post by Mladenov)
    You've gotta be kidding...

    With trivial substitutions we arrive at y_{1}'''-6y_{1}''+12y_{1}'-8=0; hence, P(z)=(z-2)^{3} is the characteristic polynomial with a triple root z=2, so the general solution to this equation is y_{1}=c_{1}e^{2x}+c_{2}xe^{2x}+c  _{3}x^{2}e^{2x}. Now it is trivial to find y_{2} and y_{3}.

    I really have no time at the moment to do problems on this thread.
    I'm sorry I doubted you
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    As many of us will have pre-interview tests looming (which may include some physics),

    Problem 407**
    A satellite is projected towards a massive body (mass  m) from infinity at an angle  \theta to the line joining the centre of masses of the body and satellite at a speed v. What shape will the motion be? What is the minimum distance that the satellite gets to the body?

    Problem 408**
    A hole is made in a table and a string, length  l , with two equal masses at either end is passed through this hole. The mass on the table is pulled out as far as possible so that the mass under the table is at the hole. Gravity exists and the table is smooth. When does the mass on the table reach the hole if:
    1. The mass on the table is initially at rest?
    2. The mass on the table is influentially moving with tangential velocity  v.

    Find the (polar) equation of motion for the mass on the table in both cases, and solve it.
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    (Original post by Lord of the Flies)
    Problem 3*

    Show that for positive integers m,n:

    \displaystyle\int_0^1 \sqrt[m]{1-x^n}-\sqrt[n]{1-x^m}\,dx=0
    Solution 3*

    Consider the curves y^m+x^n=1, y^n+x^n=1. The second is a reflection in the line  y=x of the first, so the area over the interval  [0,1] is the same.
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    Problem 409**
    Using Cauchy's formula for repeated integration (check it at http://en.wikipedia.org/wiki/Cauchy_...ed_integration) :

     \int_d^2 \int_0^c \int_b^2 \int_0^a (2-x) \mathrm{d}x\, \mathrm{d}a\, \mathrm{d}b\, \mathrm{d}c}\,

    Then generalize the same, using the formula or otherwise, for:

     \int_z^2 \int_0^y \int_x^2 \int_0^v \cdots\int_d^2 \int_0^c \int_b^2 \int_0^a (2-x) \mathrm{d}x\, \mathrm{d}a\, \mathrm{d}b\, \mathrm{d}c}\,\cdots \, \mathrm{d}y

    It's very difficult thats what my teacher said! I didn't get it myself though.

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    Problem 410**

    Let F_n(x) = \dfrac{1}{n!} \displaystyle\int_0^x t^n e^-^t dt.

    Show that if M is an integer greater than 1, then

    e^xF_M(x) = -(x + \dfrac{x^2}{2!} + \dfrac{x^3}{3!} + ... + \dfrac{x^M}{M!}) + (e^x - 1).
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    (Original post by Arieisit)
    Problem 410**

    Let F_n(x) = \dfrac{1}{n!} \displaystyle\int_0^x t^n e^-^t dt.

    Show that if M is an integer greater than 1, then

    e^xF_M(x) = -(x + \dfrac{x^2}{2!} + \dfrac{x^3}{3!} + ... + \dfrac{x^M}{M!}) + (e^x - 1).
    Solution 410**

    Spoiler:
    Show
    More integration by parts :sleep:
    Subtitutions
    \begin{array}{ccc} u = t^n & \implies & \dfrac{du}{dt} = nt^{n-1} \\ \dfrac{dv}{dt} = e^{-t} & \implies & v = -e^{-t} \end{array}

    \begin{aligned} F_{n} (x) & \ \overset{\text{I.B.P}}= \ \dfrac{1}{n!} \left( \Big[ -t^{n}e^{-t} \Big]_{0}^{x} + \displaystyle n \int_{0}^{x}  t^{n-1} e^{-t} dt \right) \\ & \  \  = \ \dfrac{-x^{n}e^{-x}}{n!} + \dfrac{1}{(n-1)!} \ \int_{0}^{x}  t^{n-1} e^{-t} dt \\ & \  \  = \dfrac{-x^{n}e^{-x}}{n!} + F_{n-1} (x) \end{aligned}

    \begin{aligned} e^x F_{n} (x) & = \dfrac{-x^n}{n!} + e^x F_{n-1} (x) \\ & = \dfrac{-x^n}{n!} + \dfrac{-x^{n-1}}{(n-1)!} + e^x F_{n-2} (x) \\ &  ... \\ & = - \left( \dfrac{x^n}{n!} + \dfrac{x^{n-1}}{(n-1)!} + ... + \dfrac{x^{2}}{2!} + x  \right) + e^x F_{0} (x) \\ & = - \left( \dfrac{x^n}{n!} + \dfrac{x^{n-1}}{(n-1)!} + ... + \dfrac{x^{2}}{2!} + x  \right) + e^x \left( 1 - e^{-x} \right) \\ & = - \left( \dfrac{x^n}{n!} + \dfrac{x^{n-1}}{(n-1)!} + ... + \dfrac{x^{2}}{2!} + x  \right) + \left( e^x - 1 \right) \end{aligned}

    \text{Let } n = M, \  M \text{ is an integer and } >1

    \therefore \ e^{x} F_{M} (x) & = - \left( \dfrac{x^M}{M!} + \dfrac{x^{M-1}}{(M-1)!} + ... + \dfrac{x^{2}}{2!} + x  \right) + \left( e^x - 1 \right)
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    Problem 411**

    Prove that

    \forall n \in N,



(a + b)^n = \displaystyle \sum_{r=0} ^n C(n, r)a^{n-r}b^r
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    (Original post by Arieisit)
    Problem 411**

    Prove that

    \forall n \in N,



(a + b)^n = \displaystyle \sum_{r=0} ^n C(n, r)a^{n-r}b^r
    Solution 411**

    Spoiler:
    Show
    Base case
    When n=1
    LHS = (a+b)^1 = a+b, \text{  RHS} = \displaystyle \sum_{r=0}^{r=n=1} \binom{1}{r} a^{1-r} b^r = \binom{1}{0} a^1 b^0 + \binom{1}{1} a^0 b^1 = a+b

    \therefore \text{ LHS} = \text{RHS} so the binomial expansion is true for n=1
    Assumption step
    Assume the binomial expansion is valid when n=k, \ n \in \mathbb{N}
    (a+b)^k = \displaystyle \sum_{r=0}^{r=n=k} \binom{k}{r} a^{k-r} b^{r}
    Inductive step
    We've now gotta show that the binomial expansion holds true for the next natural number, n=k+1:

    \displaystyle \begin{aligned} (a+b)^{k+1} & = (a+b)(a+b)^k \\ & = (a+b) \left[ \sum_{r=0}^{r=k} \binom{k}{r} a^{k-r} b^{r} \right] \\ & = \sum_{r=0}^{r=k} \binom{k}{r} a^{k+1-r} b^{r} + \sum_{r=0}^{r=k} \binom{k}{r} a^{k-r} b^{r+1} \\ \text{ Let } s=r+1 \\ & = \sum_{r=0}^{r=k} \binom{k}{r} a^{k+1-r} b^{r} + \sum_{s=1}^{s=k+1} \binom{k}{s-1} a^{k+1-s} b^{s} \\ & = \sum_{r=0}^{r=k} \binom{k}{r} a^{k+1-r} b^{r} + \sum_{r=1}^{r=k+1} \binom{k}{r-1} a^{k+1-r} b^{r} \\ & = \left( \binom{k}{0} a^{k+1} + \sum_{r=1}^{r=k} \binom{k}{r} a^{k+1-r} b^{r} \right) + \left( \binom{k}{(k+1)-1} b^{k+1} + \sum_{r=1}^{r=k} \binom{k}{r-1} a^{k+1-r} b^{r} \right) \\ & = \binom{k}{0} a^{k+1} + \sum_{r=1}^{r=k} \left[ \binom{k}{r} + \binom{k}{r-1} \right] a^{k+1-r} b^{r} + \binom{k}{k} b^{k+1} \end{aligned}

    The binomial coefficient amalgamation!
    \begin{aligned} \binom{k}{r} + \binom{k}{r-1} & = \dfrac{k!}{r! (k-r)!} + \dfrac{k!}{(r-1)! (k-(r-1))!} \\ & = \dfrac{k!}{r! (k-r)!} + \dfrac{k!}{(r-1)! ((k-r)+1))!} \\ & = \dfrac{k! \times (k-r+1)}{r! (k-r)!} + \dfrac{k! \times r}{r! ((k-r)+1))!} \\ & = \dfrac{k! \times (k-r+1 + r)}{r! (k+1-r)!} \\ & = \binom{k+1}{r} \end{aligned}

    \begin{aligned} \displaystyle (a+b)^{k+1} & = \binom{k}{0} a^{k+1} + \sum_{r=1}^{r=k}  \binom{k+1}{r} a^{k+1-r} b^{r} + \binom{k+1}{k+1} b^{k+1} \\ & = \sum_{r=0}^{k+1} \binom{k+1}{r} a^{k+1-r} b^r \end{aligned}
    Conclusion
    We've shown the binomial expansion is correct for n=1. Afterwards, we went as far as assuming it's true for a natural number k, then proving its correctness for the next natural number k+1. By this logic it's true for n=1, 2, 3, ... i.e. \forall n \in \mathbb{N}
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    (Original post by Khallil)
    Solution 411**

    Spoiler:
    Show
    Base case
    When n=1
    LHS = (a+b)^1 = a+b, \text{  RHS} = \displaystyle \sum_{r=0}^{r=n=1} \binom{1}{r} a^{1-r} b^r = \binom{1}{0} a^1 b^0 + \binom{1}{1} a^0 b^1 = a+b

    \therefore \text{ LHS} = \text{RHS} so the binomial expansion is true for n=1
    Assumption step
    Assume the binomial expansion is valid when n=k, \ n \in \mathbb{N}
    (a+b)^k = \displaystyle \sum_{r=0}^{r=n=k} \binom{k}{r} a^{k-r} b^{r}
    Inductive step
    We've now gotta show that the binomial expansion holds true for the next natural number, n=k+1:

    \displaystyle \begin{aligned} (a+b)^{k+1} & = (a+b)(a+b)^k \\ & = (a+b) \left[ \sum_{r=0}^{r=k} \binom{k}{r} a^{k-r} b^{r} \right] \\ & = \sum_{r=0}^{r=k} \binom{k}{r} a^{k+1-r} b^{r} + \sum_{r=0}^{r=k} \binom{k}{r} a^{k-r} b^{r+1} \\ \text{ Let } s=r+1 \\ & = \sum_{r=0}^{r=k} \binom{k}{r} a^{k+1-r} b^{r} + \sum_{s=1}^{s=k+1} \binom{k}{s-1} a^{k+1-s} b^{s} \\ & = \sum_{r=0}^{r=k} \binom{k}{r} a^{k+1-r} b^{r} + \sum_{r=1}^{r=k+1} \binom{k}{r-1} a^{k+1-r} b^{r} \\ & = \left( \binom{k}{0} a^{k+1} + \sum_{r=1}^{r=k} \binom{k}{r} a^{k+1-r} b^{r} \right) + \left( \binom{k}{(k+1)-1} b^{k+1} + \sum_{r=1}^{r=k} \binom{k}{r-1} a^{k+1-r} b^{r} \right) \\ & = \binom{k}{0} a^{k+1} + \sum_{r=1}^{r=k} \left[ \binom{k}{r} + \binom{k}{r-1} \right] a^{k+1-r} b^{r} + \binom{k}{k} b^{k+1} \end{aligned}

    The binomial coefficient amalgamation!
    \begin{aligned} \binom{k}{r} + \binom{k}{r-1} & = \dfrac{k!}{r! (k-r)!} + \dfrac{k!}{(r-1)! (k-(r-1))!} \\ & = \dfrac{k!}{r! (k-r)!} + \dfrac{k!}{(r-1)! ((k-r)+1))!} \\ & = \dfrac{k! \times (k-r+1)}{r! (k-r)!} + \dfrac{k! \times r}{r! ((k-r)+1))!} \\ & = \dfrac{k! \times (k-r+1 + r)}{r! (k+1-r)!} \\ & = \binom{k+1}{r} \end{aligned}

    \begin{aligned} \displaystyle (a+b)^{k+1} & = \binom{k}{0} a^{k+1} + \sum_{r=1}^{r=k}  \binom{k+1}{r} a^{k+1-r} b^{r} + \binom{k+1}{k+1} b^{k+1} \\ & = \sum_{r=0}^{k+1} \binom{k+1}{r} a^{k+1-r} b^r \end{aligned}
    Conclusion
    We've shown the binomial expansion is correct for n=1. Afterwards, we went as far as assuming it's true for a natural number k, then proving its correctness for the next natural number k+1. By this logic it's true for n=1, 2, 3, ... i.e. [tex]\forall n \in \mathbb{N}
    How did you organise your solution in such an elegant way? Whenever I do proof by induction it looks awful
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    (Original post by Jkn)
    Problem 181*

    Evaluate \displaystyle \int \frac{1}{\sqrt{x}-1} dx
    Solution 181*

    \text{Let } u = \sqrt{x} \implies dx = 2u\ du

    \displaystyle \begin{aligned} \therefore \int \frac{1}{\sqrt{x}-1} dx & = 2 \int \frac{u}{u-1}\ du \\ & = 2\int du \ + \ 2\int \dfrac{1}{u-1} du \\ & = 2u + 2\ln (u-1) + C \\ & = 2\sqrt{x} + 2\ln (\sqrt{x} -1) + C \end{aligned}

    (Original post by Flauta)
    How did you organise your solution in such an elegant way? Whenever I do proof by induction it looks awful
    I guess the sigma notation and expand tags make everything look a bit neater!
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    (Original post by Khallil)
    I guess the sigma notation and expand tags make everything look a bit neater!
    And the aligns :sexface:
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    Problem 412**

    Prove that

    \displaystyle \sum_{n>m \ge 1}\frac{1}{n^2m^2}= \frac{\pi^4}{120}

    (that is, sum \frac{1}{n^2m^2} over all distinct integer pairs (n,m) (distinct here means (n,m)=(m,n), i.e. don't double count)).

    Problem 412b**

    Prove similar identities for

    \displaystyle \sum_{a>b>c \ge 1}\frac{1}{a^2b^2c^2}

    and

    \displaystyle \sum_{a_1>\cdots >a_n \ge 1}\frac{1}{a_0^2 \cdots a_n^2}.

    Problem 412c**


    Solve 412 and 412b with the additional requirement that n, m \equiv 1 \mod 2.
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    (Original post by Indeterminate)
    Problem 402*

    Find, for \psi, \eta \neq 0

    I = \displaystyle \int e^{\eta x} \sin(\psi x) dx

    and give a condition on x such that I>0
    It turns out there was a less tedious and more interesting way to find the integral! Consider the imaginary part of:

    \displaystyle \begin{aligned} \int e^{(\eta + i\psi)x} dx & = \dfrac{e^{(\eta + i\psi)x}}{\eta + i\psi} = \dfrac{e^{(\eta + i\psi)x}(\eta - i\psi)}{\eta^2 + \psi^2} = \dfrac{e^{\eta x} \left( \eta \cos \psi x + \eta i \sin \psi x - \psi i \cos \psi x + \psi \sin \psi x \right)}{\eta^2 + \psi^2}

     \displaystyle \begin{aligned} I & = \int e^{\eta x} \sin(\psi x) dx \ = \ \operatorname{Im} \left[ \dfrac{e^{\eta x} \left( \eta \cos \psi x + \eta i \sin \psi x - \psi i \cos \psi x + \psi \sin \psi x \right)}{\eta^2 + \psi^2} \right] = \dfrac{e^{\eta x} \Big( \eta \sin(\psi x) - \psi \cos(\psi x) \Big)}{\psi^2 + \eta^2}
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    (Original post by Arieisit)
    Problem 411**

    Prove that

    \forall n \in N,



(a + b)^n = \displaystyle \sum_{r=0} ^n C(n, r)a^{n-r}b^r

    (Original post by Khallil)
    ...
    Alt. Solution 411**

    Spoiler:
    Show

    WLOG let |a| \geq |b| and set x = \dfrac{b}{a}. Hence \left(a+b\right)^n = a^n \left(1+x\right)^n.

    But the Maclaurin series (Taylor series centred at zero) of \left(1+x\right)^n w.r.t. x is:

    \displaystyle \left(1+x\right)^n = \sum^\infty_{k=0} \left[\frac{d^k \left(1+x\right)^n}{dx^k}\right]_{x=0} \frac{x^k}{k!}

    which converges for all natural n since by definition |x| \leq 1. In particular, by repeated differentiation each \dfrac{d^k \left(1+x\right)^n}{dx^k} \bigg|_{x=0} term is clearly:

    \displaystyle \frac{d^k \left(1+x\right)^n}{dx^k} \bigg|_{x=0} = \frac{n!\left(1+x\right)^{n-k}}{\left(n-k\right)!} \bigg|_{x=0} = \frac{n!}{\left(n-k\right)!}

    for all 0 \leq k \leq n and zero when k > n. Hence:

    \displaystyle \left(1+x\right)^n = \sum^n_{k=0} \frac{n! x^k}{k!\left(n-k\right)!} = \sum^n_{k=0} \binom{n}{k} x^k

    Therefore:

    \displaystyle \left(a+b\right)^n = \sum^n_{k=0} \binom{n}{k}a^n x^k = \sum^n_{k=0} \binom{n}{k}a^{n-k} b^k
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    Here we go, have an analysis question that doesn't start with the word "evaluate":

    Problem 413***:

    Let T > 0 and define
    \displaystyle G: C[0,T] \to C[0,T] by G(f)(t) := 1 + \int_0^t 2cos(s(f(s))^2)ds \forall t \in [0,T].

    Find T > 0 such that G is a contraction on C[0,T].



    If you want one that's a little less horrible, instead try this version (vastly easier):

    Problem 413***:

    Let T > 0 and define
    \displaystyle H: C[0,T] \to C[0,T] by H(f)(t) := 1 + \int_0^t 2cos(sf(s))ds \forall t \in [0,T].

    Find T > 0 such that H is a contraction on (C[0,T],\|\cdot\|_\infty).




    Some definitional/notational things for those that aren't certain of them/haven't met them:

    Spoiler:
    Show
    If (A, \|.\|) is a normed vector space, then f: A \to A is a contraction iff \|f(x)-f(y)\| \leq \|x-y\|.

    C[0,T] denotes the set of continuous functions on [0,T].
    \|\cdot \|_\infty is defined (on C[0,T] - the same notation is used for different norms on different spaces) by \displaystyle\|f\|_\infty = \sup_{x\in[0,T]}|f(x)|.





    The former of these questions was an assessed question for 2nd year maths students, but the deadline is now past.
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    409 anyone? B) :ahee::colone:

    Posted from TSR Mobile
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    (Original post by henpen)
    Problem 412**

    Prove that

    \displaystyle \sum_{n>m \ge 1}\frac{1}{n^2m^2}= \frac{\pi^4}{120}

    (that is, sum \frac{1}{n^2m^2} over all distinct integer pairs (n,m) (distinct here means (n,m)=(m,n), i.e. don't double count)).

    Problem 412b**

    Prove similar identities for

    \displaystyle \sum_{a>b>c \ge 1}\frac{1}{a^2b^2c^2}

    and

    \displaystyle \sum_{a_1>\cdots >a_n \ge 1}\frac{1}{a_0^2 \cdots a_n^2}.

    Problem 412c**


    Solve 412 and 412b with the additional requirement that n, m \equiv 1 \mod 2.
    EDITED:
    I guess \displaystyle \sum_{a_1>\cdots >a_n \ge 1}\frac{1}{a_0^2 \cdots a_n^2}.=\frac{ {\pi}^{2n+2} - 6^{n+1} \zeta(2^{n+1}) }{6^{n+1} {(n+1)}!}. where  \zeta(n) is the Riemann Zeta function.
    Is that right?
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    (Original post by journeyinwards)
    EDITED:
    I guess \displaystyle \sum_{a_1>\cdots >a_n \ge 1}\frac{1}{a_0^2 \cdots a_n^2}.=\frac{ {\pi}^{2n+2} - 6^{n+1} \zeta(2^{n+1}) }{6^{n+1} {(n+1)}!}. where  \zeta(n) is the Riemann Zeta function.
    Is that right?
    Provide your proof. There's certainly a nicer way to write the result (a constant multiplied by a power of pi).
    • Political Ambassador
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    Political Ambassador
    (Original post by Khallil)
    It turns out there was a less tedious and more interesting way to find the integral! Consider the imaginary part of:

    \displaystyle \begin{aligned} \int e^{(\eta + i\psi)x} dx & = \dfrac{e^{(\eta + i\psi)x}}{\eta + i\psi} = \dfrac{e^{(\eta + i\psi)x}(\eta - i\psi)}{\eta^2 + \psi^2} = \dfrac{e^{\eta x} \left( \eta \cos \psi x + \eta i \sin \psi x - \psi i \cos \psi x + \psi \sin \psi x \right)}{\eta^2 + \psi^2}

     \displaystyle \begin{aligned} I & = \int e^{\eta x} \sin(\psi x) dx \ = \ \operatorname{Im} \left[ \dfrac{e^{\eta x} \left( \eta \cos \psi x + \eta i \sin \psi x - \psi i \cos \psi x + \psi \sin \psi x \right)}{\eta^2 + \psi^2} \right] = \dfrac{e^{\eta x} \Big( \eta \sin(\psi x) - \psi \cos(\psi x) \Big)}{\psi^2 + \eta^2}
    That's the way I'd do it. I think you grow out of using techniques such as IBP as you progress, or certainly try to avoid them and use something you've more recently discovered
 
 
 
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