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    Solution 344:

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    Firstly, note that  ln2 = \displaystyle\sum_{r=0}^{\infty}  \dfrac{1}{2n+1} - \dfrac{1}{2n+2} = \displaystyle\sum_{r=0}^{\infty} \dfrac{1}{4n+1}+\dfrac{1}{4n+3} - \dfrac{1}{2n+2}

    And also that  \frac{1}{2} ln2 = \displaystyle\sum_{r=0}^{\infty} \dfrac{1}{4n+2}-\dfrac{1}{4n+4}

    Now, note that  \frac{\pi}{8} = \displaystyle\sum_{r=0}^{\infty} \dfrac{1}{4n+1}-\dfrac{1}{4n+3}

    Now,  \frac{\pi}{4} + ln2 =  \displaystyle\sum_{r=0}^{\infty} \dfrac{2}{4n+1} -\dfrac{1}{2n+2}

    Finally, subtract 1/2 ln2 to get:

     \frac{\pi}{4} + \frac{1}{2} ln2 = 2 \displaystyle\sum_{r=0}^{\infty} \dfrac{1}{4n+1}-\dfrac{1}{4n+2}

    Giving the final answer as  \frac{\pi}{8}+\frac{1}{4}ln2

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    Solution 344

    \displaystyle\begin{aligned}\sum  _{n\geq 0}\frac{1}{4n+1}-\frac{1}{4n+2} =\sum_{n\geq 0}\int_0^1 t^{4n}(1-t)\,dt=\int_0^1 \frac{t-1}{t^4-1}\,dt=\frac{\ln 2}{4}+\frac{\pi}{8}\end{aligned}
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    Just by looking at the diverse ways people have approached Solution 344 validates the beauty of Mathematics.
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    (Original post by Mladenov)
    Let C_{1} and C_{2} be externally tangent circles at M and the radius of C_{2} is greater than the radius of C_{1}. Let A be a point on C_{2} which does not lie on the line joining the centres of C_{1} and C_{2}. Let B and C be points on C_{1} such that AB and AC are tangent to C_{1}. The lines BM and CM intersect C_{2} again at E and F, respectively. Let D be the point of intersection of the tangent to C_{2} at A and the line EF. Prove that as A varies, the locus of D is a line.
    Any sources/books on learning how to tackle questions like this, please? Thanks.
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    (Original post by Lord of the Flies)
    Solution 344

    \displaystyle\begin{aligned}\sum  _{n\geq 0}\frac{1}{4n+1}-\frac{1}{4n+2} =\sum_{n\geq 0}\int_0^1 t^{4n}(1-t)\,dt=\int_0^1 \frac{t-1}{t^4-1}\,dt=\frac{\ln 2}{4}+\frac{\pi}{8}\end{aligned}
    Just out of interest when is switching the order of summations valid? Nice solution too.
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    (Original post by bogstandardname)
    Just out of interest when is switching the order of summations valid? Nice solution too.
    Thanks.

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    If either of \displaystyle \sum_{n\geq 0} \int_a^b |f_n(t)|\,dt,\, \int_a^b \sum_{n\geq 0} |f_n(t)|\,dt converges then \displaystyle \sum_{n\geq 0} \int_a^b f_n(t)\,dt=\int_a^b \sum_{n\geq 0} f_n(t)\,dt
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    (Original post by Mladenov)
    Solution 158. Notice that the function is even.

    Mladenov, which result obtains the second equality of the third line of solution 158? Is it some kind of elliptic integral?
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    (Original post by Mladenov)
    Solution 158. Notice that the function is even.
    In solving 158 (and similar problems), what quality of the integral made it obvious for you that Liebniz differentiation needed to be used, and how did you decide which constants to vary? I'm simply curious what to look out for.
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    Problem 345 **/***

    Evaluate \displaystyle \lim_{x \to 0} \frac{\displaystyle \int_{0}^{\tan x} \tan\sin t\ dt}{\displaystyle \int_{0}^{\sin x} \ln(1 + \sin t)\ dt}.
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    Solution 345

    L'H then \displaystyle\frac{\tan \sin \tan x}{\ln(1+\sin \sin x)}\sim_0 \frac{\tan x}{\ln (1+x)}\sim_0\frac{6x+2x^3}{6x-3x^2}\to 1
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    (Original post by Lord of the Flies)
    Solution 345

    L'H then \displaystyle\frac{\tan \sin \tan x}{\ln(1+\sin \sin x)}\sim_0 \frac{\tan x}{\ln (1+x)}\sim_0\frac{6x+2x^3}{6x-3x^2}\to 1
    I don't like to question the master, but have you differentiated the integral correctly? Don't you have to include the derivative of the limit?
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    (Original post by bananarama2)
    I don't like to question the master, but have you differentiated the integral correctly? Don't you have to include the derivative of the limit?
    Yes, but this does not contribute to the limit, since it offers a factor \sec^3 x which \to 1 as x\to 0
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    (Original post by Lord of the Flies)
    Yes, but this does not contribute to the limit, since it offers a factor \sec^3 x which \to 1 as x\to 0
    Ahh I see. Thanks
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    How would you got about proving that the difference between the square numbers is a series of the odd numbers?
    1 4 9 16 25 36...
    +3 +5 +7 +9 +11...
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    (Original post by Occams Chainsaw)
    How would you got about proving that the difference between the square numbers is a series of the odd numbers?
    1 4 9 16 25 36...
    +3 +5 +7 +9 +11...
    \displaystyle \sum_{n \geq 2} n^2 - (n-1)^2 = \sum_{n \geq 2} 2n - 1
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    (Original post by Felix Felicis)
    \displaystyle \sum_{n \geq 1} n^2 - (n-1)^2 = \sum_{n \geq 1} 2n - 1
    ffs. it's just induction :facepalm:


    I was trying Euler's
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    (Original post by Occams Chainsaw)
    ffs. it's just induction :facepalm:


    I was trying Euler's
    Huh? I don't quite see the induction used...
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    (Original post by Occams Chainsaw)
    ffs. it's just induction :facepalm:


    I was trying Euler's
    Simplicity is often the best. No induction though, just algebra, observe that n^2 - (n-1)^2 = 2n-1.

    Lol - do you mean an Euler-esque method on how he solved the Basel problem?
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    (Original post by bananarama2)
    Huh? I don't quite see the induction used...
    Felix didn't use induction. I saw his answer and realised what my friend was trying to get me to do...
    We just started FP1 so he was trying to play with it!

    (2n - 1) = n^2

n = 1: 1=1 check!



(2k - 1) = k^2 



k+1: 2(k + 1) - 1 = (k + 1)^2

2(k + 1) - 1 = 2(k + 1) - 1

k^2 + (2(k + 1) - 1) = (k + 1)^2 

2(k + 1) - 1 = (k + 1)^2
    Not even sure if that makes sense right now. Brain = fried.
    (Original post by Felix Felicis)
    Simplicity is often the best. No induction though, just manipulation, observe that n^2 - (n-1)^2 = 2n-1.

    Lol - do you mean an Euler-esque method on how he solved the Basel problem?
    Gauss's method, not Euler. My lecturer got it wrong. I just had to google it. Arithmetic series! It's been a long, long day. Had lacrosse training early this morning and then rugby match before lunch!

    EDIT: Felix, I am just not with it tonight. I'll come back tomorrow and try again!
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    (Original post by Lord of the Flies)
    Solution 345
    Nice! You could also use L'hopital's twice. (idk why the rhyme)
 
 
 
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