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    Problem 205/ **


    At the age of three, Zakee begins to learn how to count, and is faced with a problem in his counting book. He believes it is just a trivial problem, after all, he is only three years of age, and so looks at it and tries to solve it. What was the outcome of his attempts? You decide (see below):

    Sequence of Real numbers:

     x_0, x_1,\ldots, x_n,\ldots

    accede to these mathematical conditions

    1 = x_{0}\leq x_{1}\leq\cdots\leq x_{n}\leq\cdots

    Additionally, let y_{1}, y_{2},\ldots, y_n,\ldots be defined as:

     y_n =\sum_{k=1}^{n}\frac{1-\frac{x_{k-1}}{x_{k}}}{\sqrt{x_k}}

    Prove that:

     0\leq y_n < 2 FOR ALL  n


    Edit: Question amended due to erratum.
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    (Original post by Zakee)
    Back to the topic:

    Problem 205/ **
    Prove that:

     0\leq y_n < 2
    Let x_i=1 \ \forall i,

    \Rightarrow y_n = n

    let n be greater that 2... ?
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    (Original post by Jkn)
    Let x_i=1 \ \forall i,

    \Rightarrow y_n = n

    let n be greater that 2... ?

    I was looking at your solution and thought to myself, either Jkn is a prodigious genius with a bajillion IQ points, or I have made a trivial error in my question. Both are correct.


    Y'may want to redo your solution.
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    (Original post by Zakee)
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    I was looking at your solution and thought to myself, either Jkn is a prodigious genius with a bajillion IQ points, or I have made a trivial error in my question. Both are correct.


    Y'may want to redo your solution.
    You want to make the sequence x_{n} strictly increasing for all n which are greater than a given number N. Otherwise y_{n} tends to \infty as n\to \infty
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    (Original post by Zakee)
    I was looking at your solution and thought to myself, either Jkn is a prodigious genius with a bajillion IQ points, or I have made a trivial error in my question. Both are correct.


    Y'may want to redo your solution.
    I may be being thick again, but doesn't his point still stand? You can still choose n to be greater than 2 if x_i=1....
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    (Original post by Zakee)
    I was looking at your solution and thought to myself, either Jkn is a prodigious genius with a bajillion IQ points, or I have made a trivial error in my question. Both are correct.

    Y'may want to redo your solution.
    Hahhaaha :lol: I wasn't offering a solution, merely pointing out an error :lol:

    The error is either in your y_n (notice how quickly it can be simplified, perhaps you are missing a rogue square terms) or in the definition of the sequence. Perhaps strictly increasing integers? Something that forces x_i to tend to infinity is a necessary (though not sufficient) condition for y_n to converge. i.e. we require that each term in the series tends to zero as the underlying parameter approach infinity.

    Edit: Just checked your latex and say that that was an x_{k-1} rather than an x_k -1 (would be clearer with displaystyle!) Though you would still need a strict increase for some infinite set of x_i (as Mladenov has said)
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    (Original post by Jkn)
    Hahhaaha :lol: I wasn't offering a solution, merely pointing out an error :lol:

    The error is either in your y_n (notice how quickly it can be simplified, perhaps you are missing a rogue square terms) or in the definition of the sequence. Perhaps strictly increasing integers? Something that forces x_i to tend to infinity is a necessary (though not sufficient) condition for y_n to converge. i.e. we require that each term in the series tends to zero as the underlying parameter approach infinity.

    Edit: Just checked your latex and say that that was an x_{k-1} rather than an x_k -1 (would be clearer with displaystyle!) Though you would still need a strict increase for some infinite set of x_i (as Mladenov has said)


    Sorry, my Latex abilities are awful, and I have an exam tomorrow. I rushed with writing up the question. My bad, if you can understand what I was trying to say, then I hope you can do it. If not, I will refine the post tomorrow. Sorry for the problem caused.
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    (Original post by Zakee)
    Sorry, my Latex abilities are awful, and I have an exam tomorrow. I rushed with writing up the question. My bad, if you can understand what I was trying to say, then I hope you can do it. If not, I will refine the post tomorrow. Sorry for the problem caused.
    No worries

    Tbf, I should stop coming on here! Too many exams! :lol:

    Luckily now finished my English and Spanish but now I've got Decision and Physics and **** to worry about Not to mention STEP! Drownnninnnnggggg!
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    (Original post by Jkn)
    No worries

    Tbf, I should stop coming on here! Too many exams! :lol:

    Luckily now finished my English and Spanish but now I've got Decision and Physics and **** to worry about Not to mention STEP! Drownnninnnnggggg!

    I've got Physics tomorrow and half the syllabus to learn. . I'm off man, take care. x_x
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    (Original post by Zakee)
    I've got Physics tomorrow and half the syllabus to learn. . I'm off man, take care. x_x
    ****, I've got my physics next week and haven't finished the syllabus either The answers to most questions can be figured out form the context though :lol: On the past paper I tried though there were all of these questions asking tricky mathematical things where they, in fact, want you to count squares on a graph, fml I always end up whipping out the calculus and taking longer than intended!

    Good luck!
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    (Original post by Jkn)
    ****, I've got my physics next week and haven't finished the syllabus either The answers to most questions can be figured out form the context though :lol: On the past paper I tried though there were all of these questions asking tricky mathematical things where they, in fact, want you to count squares on a graph, fml I always end up whipping out the calculus and taking longer than intended!

    Good luck!

    Precisely man. One of them was to do with Hooke's law, and I worked out the area of the graph by approximation through integrals. Instead they wanted counting squares. Damn Physics exams.
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    (Original post by Zakee)
    I've got Physics tomorrow and half the syllabus to learn. . I'm off man, take care. x_x
    I'll join that club as well
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    (Original post by metaltron)
    I'll join that club as well
    Same o_O
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    I'm not impressed guys :l
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    (Original post by bananarama2)
    I'm not impressed guys :l

    I actually never ended up revising in the end. Damn.
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    (Original post by bananarama2)
    I'm not impressed guys :l
    Yeah, physics is fun as well though! Just a bit late to start revising...
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    Well, I see what kind of problems people prefer.

    Problem 206*


    Evaluate \displaystyle \int_{0}^{\pi} \frac{\ln(1+a\cos x)}{\cos x} \, dx, where |a| <1.

    Problem 207**

    Let \alpha_{i}, i \in \{1,\cdots,k\} be numbers such that for any two sequences (a_{n})_{n \ge1} and (b_{n})_{n\ge 1}, which satisfy the relation \displaystyle b_{n}=a_{n}+\sum_{i=1}^{k} \alpha_{i}a_{n-i}, for all n>k, from the convergence of (b_{n})_{n\ge 1} follows the convergence of (a_{n})_{n \ge1}. Prove that all the roots of the polynomial \displaystyle x^{k}+ \sum_{i=1}^{k} \alpha_{i}x^{k-i} have absolute values which are less than 1.

    Problem 208***

    Let f : [a,b] \to \mathbb{R} be C^{2}, and suppose that \displaystyle 0<h<\frac{b-a}{2}. Then, \displaystyle \int_{a}^{b} |f'(x)|^{p}dx \le 2^{p}h^{p-1}(b-a)\int_{a}^{b}|f''(x)|^{p}dx+ \frac{2^{2p}}{h^{p}}\int_{a}^{b}  |f(x)|^{p}dx (p>1).
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    Solution 206

    \displaystyle I(a)= \int_0^{\pi} \frac{\ln (1+a \cos x )}{\cos x}dx

    \displaystyle \frac{dI}{da} =  \int_0^{\pi} \frac{1}{1+a \cos x} dx

    Weierstrass sub http://en.wikipedia.org/wiki/Weierstrass_substitution

    \displaystyle  =  \int_0^{\infty} \frac{1}{1+a \frac{1-t^2}{1+t^2}} \frac{2 dt}{1+t^2}

    \displaystyle  =  \int_0^{\infty} \frac{1}{(1+t^2)+a (1-t^2) } 2dt

    \displaystyle  =  \int_0^{\infty} \frac{1}{ (1-a)t^2 +(1+a) } 2dt

    \displaystyle  =  \left[ -\frac{2 \arctan\sqrt{\frac{1-a}{a+1}}t}{\sqrt{(1+a)(1-a)}} \right]_0^\infty

     \displaystyle = \frac{\pi}{\sqrt{(1-a)(1+a)}}

     \displaystyle = \frac{\pi}{\sqrt{1-a^2} }

     \displaystyle I = \pi \arcsin (a) +c a=0 I=0 C=0

     \displaystyle I = \pi \arcsin (a) I'm unconvinced.
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    (Original post by bananarama2)
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    Solution 206

    \displaystyle I(a)= \int_0^{\pi} \frac{\ln (1+a \cos x )}{\cos x}dx

    \displaystyle \frac{dI}{a} =  \int_0^{\pi} \frac{1}{1+a \cos x} dx

    Weierstrass sub http://en.wikipedia.org/wiki/Weierstrass_substitution

    \displaystyle  =  \int_0^{\infty} \frac{1}{1+a \frac{1-t^2}{1+t^2}} \frac{2 dt}{1+t^2}

    \displaystyle  =  \int_0^{\infty} \frac{1}{(1+t^2)+a (1-t^2) } 2dt

    \displaystyle  =  \int_0^{\infty} \frac{1}{ (1-a)t^2 +(1+a) } 2dt

    \displaystyle  =  \left[ -\frac{2 \arctan\sqrt{\frac{1-a}{a+1}}t}{\sqrt{(1+a)(1-a)}} \right]_0^\infty

     \displaystyle = \frac{\pi}{\sqrt{(1-a)(1+a)}}

     \displaystyle = \frac{\pi}{\sqrt{1-a^2} }

     \displaystyle I = \pi \arcsin (a) +c a=0 I=0 C=0

     \displaystyle I = \pi \arcsin (a) I'm unconvinced.
    Well done.

    Problem 209*

    Evaluate \displaystyle \int_{0}^{\infty} \frac{A_{1}\cos a_{1}x + \cdots + A_{k}\cos a_{k}x}{x}dx when a_{i}>0 for all i \in \{1,2, \cdots,k \} and A_{1}+\cdots+A_{k}=0.

    Problem 210**

    Find \displaystyle \int_{0}^{1} \frac{(1-x^{\alpha})(1-x^{\beta})}{(1-x)\ln x} \, dx, for \alpha, \beta >-1 and \alpha+\beta >-1.
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    (Original post by Mladenov)
    Problem 210**

    Find \displaystyle \int_{0}^{1} \frac{(1-x^{\alpha})(1-x^{\beta})}{(1-x)\ln x} \, dx, for \alpha, \beta >-1 and \alpha+\beta >-1.
    Solution 210

    Firstly, note that:

     \frac{1-x^{\alpha}}{1-x} = \sum_{i=0}^{\alpha -1} x^i

    When  \alpha = 0 , then we just use the infinite expansion of  (1+x)^{-1} which is allowed since x only varies between 0 to 1 in the integral.

    



\Rightarrow I = \displaystyle\int_0^1 \frac{(1-x^{\alpha})(1-x^{\beta})}{(1-x) \ln x}\ dx = \displaystyle\int_0^1 \frac{(1-x^{\alpha})}{\ln x} \displaystyle\sum_{i=0}^{\beta -1} x^i \ dx 



\beginaligned I =  \displaystyle\sum_{i=0}^{\beta -1} \displaystyle\int_0^1 \frac{(1-x^{\alpha})}{\ln x} x^i \ dx



\beginaligned I = \displaystyle\sum_{i=0}^{\beta -1} \displaystyle\int_0^1 \frac{(x^i-x^{\alpha +i})}{\ln x} \ dx

    Now, just consider the integral, for an arbitrary value of i, and let,  x = e^t

    \displaystyle\int_0^1 \frac{(x^i-x^{\alpha +i})}{\ln x} \ dx = \displaystyle \int_{-\infty}^0 \frac {e^{(i+1)t}- e^{(\alpha + i+1)t}}{t} \ dt

    Here, notice that,

     \displaystyle \int_{\alpha+i+1}^{i+1} e^{ty}\ dy = \frac{e^{(i+1)t}- e^{(\alpha + i+1)t}}{t}

    So,

     \displaystyle \int_{-\infty}^0 \frac {e^{(i+1)t}- e^{(\alpha + i+1)t}}{t} \ dt = \displaystyle\int_{-\infty}^0 \displaystyle\int_{\alpha+i+1}^{  i+1} e^{ty}\ dy\ dt = \displaystyle\int_{\alpha+i+1}^{  i+1} \displaystyle\int_{-\infty}^0 e^{ty}\ dt\ dy



\Rightarrow \displaystyle\int_{\alpha+i+1}^{  i+1} \displaystyle\int_{-\infty}^0 e^{ty}\ dt\ dy = \displaystyle\int_{\alpha+i+1}^{  i+1} \frac{1}{y} \ dy = \ln {(\frac{i+1}{\alpha+i+1})}



\Rightarow \displaystyle\int_0^1 \frac{(x^i-x^{\alpha +i})}{\ln x} \ dx = \ln {(\frac{i+1}{\alpha+i+1})}

    Now, substituting this into the original integral,

    



\beginaligned I = \displaystyle\sum_{i=0}^{\beta -1} \displaystyle\int_0^1 \frac{(x^i-x^{\alpha +i})}{\ln x} \ dx = \displaystyle\sum_{i=0}^{\beta -1} \ln {(\frac{i+1}{\alpha+i+1})}





\beginaligned I = \displaystyle\sum_{i=1}^{\beta} \ln {(\frac{i}{\alpha+i})}





\beginaligned I = \displaystyle \ln {\frac{\beta !}{(\alpha+1)...(\alpha+ \beta)}}





\beginaligned I = - \ln \binom{\alpha+\beta}{\alpha}
 
 
 
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