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    (Original post by bogstandardname)
    Problem 11* (if you have never seen substitution before, **)

    Find the general solution, by a suitable substitution or otherwise of this differential equation.

    \displaystyle \dfrac{dy}{dx}=\dfrac{sinx(cosx+  y)}{cosx-y}
    Solution 12 (a bit rough-and-ready, I'm afraid, and I feel like there should be a nice "otherwise"):
    Spoiler:
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    Since the denominator involves cos x, I'll use the simplest possible substitution I can think of that will interact nicely with the cos x on the denominator:
    Substitute y = u cos(x), so that \frac{dy}{dx}=\frac{du}{dx} cos(x) - u sin(x).
    Then:
    \frac{du}{dx} cos(x) - u sin(x) = \frac{sin(x) (cos(x)+u cos(x))}{cos(x)-u cos(x)} = \frac{sin(x)(1+u)}{1-u}.
    Hence:
    \frac{du}{dx} cos(x) = u sin(x)+sin(x) \frac{1+u}{1-u}, and so \frac{du}{dx}=tan(x)\frac{-u^2+2u+1}{1-u}.
    This is separable (I'll take out another fraction 1/2, to make the reverse-chain-rule more obvious):
    \frac12 \frac{2-2u}{1+2u-u^2} du = tan(x) dx, and so log(1+2u-u^2) = 2 sec^2(x)+2 C and 1+2u-u^2 = A sec^2(x). Substituting back for y:
    cos^2(x)+2y cos(x) - y^2 = A.
    We can solve this for y, to get:
    \displaystyle y=cos(x)\pm \sqrt{2 cos^2(x)+C}.
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    (Original post by Lord of the Flies)
    Problem 15*/**

    Evaluate \displaystyle \lim_{n\to\infty} \int_1^{\pi} \frac{n}{1+x^n}\,dx
    Hmm not totally sure how to do this (although I have an idea)

    If you let f_n(x) = \dfrac{n}{1+x^n} then \lim_{n \rightarrow \infty} f_n(x) = 0 for all x \in (1, \pi]. I believe the fact it doesn't converge to 0 at one point isn't a problem with this (Lebesgue)

    So by Lebesgue

    \displaystyle \lim_{n\to\infty} \int_1^{\pi} f_n(x)\,dx = \int_1^{\pi} 0\,dx = 0

    Disclaimer: There's a very good chance this is wrong (I haven't covered Lebesgue )
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    Problem 16 **/***

    Let f be a function such that \displaystyle \int_{-\infty}^{\infty} f(x)\ dx < \infty.

    Define f_{\pm}(x) = f(x + \sqrt{x^2 + 1}) \pm f(x - \sqrt{x^2 + 1}) and \displaystyle \mathfrak{I}(f)(x) = f_{+}(x) + \frac{x}{\sqrt{x^2 + 1}} f_{-}(x).

    Show that \displaystyle \int_{-\infty}^{\infty} f(x)\ dx = \int_{-\infty}^{\infty} \mathfrak{I}(f)(x)\ dx.


    Find \displaystyle \int_{-\infty}^{\infty} \frac{\cos(x)\sin(\sqrt{x^2 + 1})}{\sqrt{x^2 + 1}}\ dx.
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    (Original post by Noble.)
    Hmm not totally sure how to do this (although I have an idea)

    If you let f_n(x) = \dfrac{n}{1+x^n} then \lim_{n \rightarrow \infty} f_n(x) = 0 for all x \in (1, \pi]. I believe the fact it doesn't converge to 0 at one point isn't a problem with this (Lebesgue)

    So by Lebesgue

    \displaystyle \lim_{n\to\infty} \int_1^{\pi} f_n(x)\,dx = \int_1^{\pi} 0\,dx = 0

    Disclaimer: There's a very good chance this is wrong (I haven't covered Lebesgue )
    Spoiler:
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    The fact that it doesn't converge to 0 at one is in fact a problem. Example:

    \displaystyle \lim_{n\to\infty} \int_1^{\infty} \frac{n}{x^{2n}}\,dx= \lim_{n\to\infty} \frac{n}{2n-1}=\frac{1}{2}

    Even though \displaystyle\lim_{\infty} f_n =0
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    I'm getting these from somewhere but I think some of these are beautiful problems which hopefully someone with C1-4 can do (except 18, which ironically might be the easiest). Hope you don't mind the influx!

    Problem 17*

    Let {a_n} be an arithmetic sequence and {g_n} be a geometric sequence. The first four terms of {s_n}={g_n + a_n} is 0, 0, 1 and 0.

    Find s_{10}.

    Problem 18**

    Evaluate \lim_{x \rightarrow 0} (x^{x^x}-x^x)

    Problem 19*

    If x+y+z=0, evaluate \frac{xy+yz+zx}{x^2+y^2+z^2}

    Problems 20 and 21 are below. I told you I'd spam this thread

    Problem 22*

    Evaluate \int^{\infty}_1 (\frac{\log x}{x})^n dx

    Problem 23*

    Let S denote the set of triples (i,j,k) such that i+j+k=n.

    Evaluate \sum_{(i,j,k)\in S} ijk

    (I wanted to edit this to restrict the choice of n, but actually it makes little difference. Do people need a hint?)
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    Problem 20 **

    By using a suitable substitution, or otherwise, show that

    \displaystyle \int \sqrt{r^2 - x^2} \ dx = \frac{1}{2}\left(r^2 \arctan \left(\frac{x}{\sqrt{r^2 - x^2}}\right) + x\sqrt{r^2 - x^2}\right) + C

    Show further that the area of a circle, A, satisfies

    A=\pi r^2
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    Problem 21 - */**

    The uniqueness theorem of anti-derivatives states that, if  f’(x) = g’(x) , then  f(x) = g(x)+c .

    By considering the derivatives of  cos^2x and  sin^2x , verify that  cos^2x+sin^2x = 1 .

    By considering suitable derivatives, prove the following identities:

    i)  lnx^n = n lnx

    ii)  ln(f(x)g(x)) = lnf(x) + lng(x)

    iii)  lnx = \dfrac{log_ax}{log_ae} .

    Deduce that the results in parts i) and ii) hold independently of the base of the logarithm.

    Spoiler:
    Show


    This was a question I came up with when there was talk about a user-contributed STEP paper being made by TSR members. However, I think here is a better place for it.

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    Solution 17

    Let a_n=a+(n-1)d and g_n=gr^{n-1}. Then:

    a+g=0

a+d+gr=0

a+2d+gr^2=1

a+3d+gr^3=0

    Solving simultaneously, we obtain a=-\frac{1}{9}, g=\frac{1}{9}, d=\frac{1}{3} and r=-2.

    Hence S_{10}=-\frac{1}{9}+(10-1) \cdot\frac{1}{3}+\frac{1}{9} \cdot (-2)^{10-1}=-54.
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    Solution 19

    x+y+z=0\Rightarrow (x+y+z)^2=x^2+y^2+z^2+2(xy+xz+yz  )=0

\Rightarrow \frac{xy+xz+yz}{x^2+y^2+z^2}=-\frac{1}{2}.
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    (Original post by und)
    Solution 17

    Let a_n=a+(n-1)d and g_n=gr^{n-1}. Then:

    a+g=0

a+d+gr=0

a+2d+gr^2=1

a+3d+gr^3=0

    Solving simultaneously, we obtain a=-\frac{1}{9}, g=\frac{1}{9}, d=\frac{1}{3} and r=-2.

    Hence S_{100}=-\frac{1}{9}+(100-1) \cdot\frac{1}{3}+\frac{1}{9} \cdot (-2)^{100-1}=-70425033346012744527594622488.

    Right, where have I gone wrong??
    I didn't say the answer was pretty! But I meant s_10
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    (Original post by shamika)

    Problem 18**

    Evaluate \lim_{x \rightarrow \infty} (x^{x^x}-x^x)
    Solution 18
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    Since x^x \rightarrow \infty as x becomes large, we have (x^{x^x}-x^x)=x^x (x^{x^x - x}-1) which clearly goes to infinity, being a product of two things which go to infinity. (since x^x-x goes to infinity, so does x to the power of that.)
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    (Original post by Smaug123)
    Solution 18
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    Since x^x \rightarrow \infty as x becomes large, we have (x^{x^x}-x^x)=x^x (x^{x^x - x}-1) which clearly goes to infinity, being a product of two things which go to infinity. (since x^x-x goes to infinity, so does x to the power of that.)
    I really need to stop posting when I'm exhausted - I've corrected the problem, sorry.
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    (Original post by Lord of the Flies)
    Problem 15*/**


    Evaluate \displaystyle \lim_{n\to\infty} \int_1^{n} \frac{n}{1+x^n}\,dx


    What if the upper limit in the integral were fixed, say \pi?

    Hi


    Take the binomial expansion of n/(1+x^2)^-1 (far too tedious to write out all the working!)
    Then simplify and integrate, eliminate the n terms, and you're left with 1 - 1/2 + 1/3 - 1/4... which is of course the taylor expansion of ln(2)


    sorry I don't know how to use LaTex so it would be a lot of effort to type out the working, but it should be straightforward to follow the steps
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    Solution 22

    Let I_m=\int^{\infty}_1 \frac{(logx)^m}{x^n} dx. Then by parts, I_m=\frac{m}{n-1}I_{m-1}. Hence I_n=\frac{n!}{(n-1)^n}\left[ \frac{1}{1-n}x^{1-n} \right]_1^{\infty}=\frac{n!}{(n-1)^{n+1}}
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    (Original post by und)
    Solution 19

    x+y+z=0\Rightarrow (x+y+z)^2=x^2+y^2+z^2+2(xy+xz+yz  )=0

\Rightarrow \frac{xy+xz+yz}{x^2+y^2+z^2}=-\frac{1}{2}.
    yep!
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    (Original post by shamika)
    I really need to stop posting when I'm exhausted - I've corrected the problem, sorry.
    Heh, I did wonder; you did say it might be the easiest :P

    Problem 18**

    Evaluate \lim_{x \rightarrow 0} (x^{x^x}-x^x)

    Solution 18
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    \lim_{x \rightarrow 0} (x^x) = 1 by considering \log(x^x) = x \log(x) and using the continuity of log, and the fact that \lim_{x \rightarrow 0} (x \log(x)) = \lim_{x \rightarrow \infty} (\frac{-\log(x)}x) = 0.
    Therefore, \lim_{x \rightarrow 0} (x^{x^x}-x^x) = \lim_{x \rightarrow 0} (x^1-1) = -1.
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    (Original post by und)
    Solution 22

    Let I_m=\int^{\infty}_1 \frac{(logx)^m}{x^n} dx. Then by parts, I_m=\frac{m}{n-1}I_{m-1}. Hence I_n=\frac{n!}{(n-1)^n}\left[ \frac{1}{1-n}x^{1-n} \right]_1^{\infty}=\frac{n!}{(n-1)^{n+1}}
    Yep!

    (Original post by Smaug123)
    Heh, I did wonder; you did say it might be the easiest :P

    Problem 18**

    Evaluate \lim_{x \rightarrow 0} (x^{x^x}-x^x)

    Solution 18
    Spoiler:
    Show
    \lim_{x \rightarrow 0} (x^x) = 1 by considering \log(x^x) = x \log(x) and using the continuity of log, and the fact that \lim_{x \rightarrow 0} (x \log(x)) = \lim_{x \rightarrow \infty} (\frac{-\log(x)}x) = 0.
    Therefore, \lim_{x \rightarrow 0} (x^{x^x}-x^x) = \lim_{x \rightarrow 0} (x^1-1) = -1.
    Yep!
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    (Original post by und)
    Solution 17

    Let a_n=a+(n-1)d and g_n=gr^{n-1}. Then:

    a+g=0

a+d+gr=0

a+2d+gr^2=1

a+3d+gr^3=0

    Solving simultaneously, we obtain a=-\frac{1}{9}, g=\frac{1}{9}, d=\frac{1}{3} and r=-2.

    Hence S_{10}=-\frac{1}{9}+(10-1) \cdot\frac{1}{3}+\frac{1}{9} \cdot (-2)^{10-1}=-54.
    Yep. This isn't as nice as I'd tried to make it, but oh well, it was meant to entice some lurkers to solve something which is essentially C1/2 standard
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    (Original post by _Izzy)
    ...
    Well done I'll type it out for you (credit to Izzy in the OP):

    \displaystyle\begin{aligned}\int  _1^n  \frac{n}{1+x^n}\,dx &=\int_{1}^n \frac{n}{x^n}\left(\frac{1 }{1+(1/x)^n}\right)dx\\&=\int_1^n\frac{  n}{x^n}\left(1-\frac{1}{x^n}+\frac{1}{x^{2n}}-\cdots\right)dx\\&=n\int_1^n \frac{1}{x^n}-\frac{1}{x^{2n}}+\frac{1}{x^{3n}  }-\cdots\,dx\\&=n\left[\frac{1}{n-1}-\frac{1}{2n-1}+\cdots\right]-\underbrace{n\left[\frac{n^{1-n}}{n-1}-\frac{n^{1-2n}}{2n-1}+\cdots\right]}_{\to\,0}\end{aligned}

    Hence the limit is \ln 2

    Replacing n with \pi obviously makes no difference.
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    Solution 21

    Let f(x)=sin^{2}x and g(x)=cos^{2}x. Then f'(x)+g'(x)=sin2x-sin2x=0\Rightarrow sin^{2}x+cos^{2}x=c. Letting x=0 gives c=1 as required.

    i) Let f(x)=ln(x^n) and g(x)=nln(x). Then f'(x)-g'(x)=\frac{nx^{n-1}}{x^n}-\frac{n}{x}=0\Rightarrow ln(x^n)-nln(x)=c. Letting x=1 gives c=0 as required.

    ii) Let p(x)=ln(f(x)g(x)) and q(x)=ln(f(x))+ln(g(x)). Then p'(x)-q'(x)=\frac{f'(x)g(x)+f(x)g'(x)}  {f(x)g(x)}-(\frac{f'(x)}{f(x)}+\frac{g'(x)}  {g(x)})=0 

\Rightarrow ln(f(x)g(x))-ln(f(x))+ln(g(x))=c. Letting x=1 gives c=0 as required.

    iii) Let f(x)=logx and g(x)=\frac{log_a{x}}{log_{a}e}. Then f'(x)-g'(x)=\frac{1}{x}-\frac{1}{xlog_{e}alog_{a}e}=0 \Rightarrow lnx-\frac{log_a{x}}{log_{a}e}=c. Letting x=1 gives c=0 as required.

    Any logarithm can be mapped onto the natural logarithm using a linear transformation, hence the results still hold.
 
 
 
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