I thought so, i dont do any complex analysis as of yet as I am not at Uni yet! But the square at the bottom and the limts made me think of that sub.(Original post by Indeterminate)
That's the solution I was looking for
The complex analysis route is really quite messy in this case.
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Problem 551
A,B are fixed points on a circle, not diametrically opposite each other.P is a variable point of the circle and Q is the point diametrically opposite P. Find the Locus of the point of intersection of AP and BQ.
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 04122015 23:12
(Original post by Gifted)
I am new to this but I had a quick question: Why is it possible to add the two integrals when the first one has the 'dx' operator and the second has the 'du' operator on it ?
as the limits are the same they represent the same number 
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 04122015 23:14
(Original post by Gifted)
I am new to this but I had a quick question: Why is it possible to add the two integrals when the first one has the 'dx' operator and the second has the 'du' operator on it ?
You can see some as transformations aswell ie x>x etc
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 04122015 23:15
(Original post by TeeEm)
∫ f(x) dx = ∫ f(u) du = ∫ f(t) dt = ∫ f(k) dk etc
as the limits are the same they represent the same number 
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 05122015 02:19
(Original post by physicsmaths)
Problem 551
A,B are fixed points on a circle, not diametrically opposite each other.P is a variable point of the circle and Q is the point diametrically opposite P. Find the Locus of the point of intersection of AP and BQ.
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The locus is a circle through A and B centered on the intersection of the tangents to the original circle at A and B.
There are two cases to consider, but I'll write a short argument, the details can be added.
Let the point of intersection of AP and BQ be T.
We know that as Q varies, the angle AQB stays constant.
We also know that the angle QAP, and thus QAT are right angles, as QAP lies in a semicircle.
Thus in the triangle QAT, the remaining angle QTA is constant, thus BTA is constant and the locus is a circle.
Now let the the intersection of the tangents to the original circle through A and B meet at S and let O be the center of the original circle.
ASB = 180  AOB = 180  2AQB = 2(90  AQB) and let T lie outside the original circle (if it doesn't in your sketch, swap the positions of P and Q), so ATB = 90  AQB (considering the triangle AQT), this and symmetry implies that S is the center of the locus circle.
(I haven't done this last part in a particularly beautiful way, but notice that it implies that if we used the locus circle as the generating circle, keeping A and B constant, the locus generated will be the original circle, and notice also that the two circles intersect each other at right angles.)Last edited by EricPiphany; 06122015 at 03:08. 
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 15122015 00:55
Problem 552***
Let and
By considering the Maclaurin series of , or otherwise, show that
for
Show that
Last edited by Indeterminate; 01012016 at 18:34. 
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 19122015 16:55
(Original post by Indeterminate)
Problem 552***
Let and
By considering the Maclaurin series of , or otherwise, show that
for
Show that

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 19122015 18:35
(Original post by EricPiphany)
Solution 551
The locus is a circle through A and B centered on the intersection of the tangents to the original circle at A and B.
There are two cases to consider, but I'll write a short argument, the details can be added.
Let the point of intersection of AP and BQ be T.
We know that as Q varies, the angle AQB stays constant.
We also know that the angle QAP, and thus QAT are right angles, as QAP lies in a semicircle.
Thus in the triangle QAT, the remaining angle QTA is constant, thus BTA is constant and the locus is a circle.
Now let the the intersection of the tangents to the original circle through A and B meet at S and let O be the center of the original circle.
ASB = 180  AOB = 180  2AQB = 2(90  AQB) and let T lie outside the original circle (if it doesn't in your sketch, swap the positions of P and Q), so ATB = 90  AQB (considering the triangle AQT), this and symmetry implies that S is the center of the locus circle.
(I haven't done this last part in a particularly beautiful way, but notice that it implies that if we used the locus circle as the generating circle, keeping A and B constant, the locus generated will be the original circle, and notice also that the two circles intersect each other at right angles.)
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 19122015 19:27
(Original post by qwertzuiop)
What level of mathematics is required for this?
The connoisseurs amongst the Alevel students could do it, though 
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 19122015 19:54
(Original post by Indeterminate)
Problem 552***
Let and
By considering the Maclaurin series of , or otherwise, show that
for
Show that

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 19122015 20:15
(Original post by EricPiphany)
I'm confused about the integral in the second part, because in the first part you prove a formula for x < 1, but the integral ranges from 0 to infinity. Also, when I substitute x = ku, the result seems to imply that the integral is not independent on the value of k.
After substituting , you can split the integral into 2: the first over and the second over
If you can find away around the latter, then the result will be applicable 
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 20122015 02:46
Problem 553 (*)
Here's a nice, easy to start with problem
Let be a sequence of positive integers such that every (positive) integer appears in the sequence only once. Suppose further that divides . Show that such a sequence exist. 
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 20122015 10:16
(Original post by Blazy)
Problem 553 (*)
Here's a nice, easy to start with problem
Let be a sequence of positive integers such that every (positive) integer appears in the sequence only once. Suppose further that divides . Show that such a sequence exist.
We construct a sequence that works. Start with then let be some large number such that we may have , i.e . Now let be some even larger number compared to such that we may have , and such a number exists by CRT. Hence, we keep repeating this, alternating the sequence between large number and the sequence , missing out a part of the sequence if that number has already been used as one of our large numbers, and hence, we have constructed a sequence that works.Last edited by Renzhi10122; 20122015 at 14:06. 
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 20122015 13:06
(Original post by Renzhi10122)
Solution 553
We construct a sequence that works. Start with then let be some large number such that we may have , i.e . Now let be some even large number compared to such that we may have , and such a number exists by CRT. Hence, we keep repeating this, alternating the sequence between large number and the sequence , missing out a part of the sequence if that number has already been used as one of our large numbers, and hence, we have constructed a sequence that works.
I understand that so but then so appears twice in the sequence. 
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 20122015 13:27
(Original post by EricPiphany)
I have never studied the Chinese remainder theorem so this question may be stupid, but I'm trying to understand your construction.
I understand that so but then so appears twice in the sequence.
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 20122015 13:48
(Original post by EricPiphany)
I have never studied the Chinese remainder theorem so this question may be stupid, but I'm trying to understand your construction.
I understand that so but then so appears twice in the sequence. 
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 20122015 14:01
(Original post by Renzhi10122)
I've just used CRT to justify the existence of such a large number with the property I want. We then miss it out in our sequence as physicsmaths said. 
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 20122015 14:06
(Original post by EricPiphany)
OK, I'm still struggling with the construction though. If then which isn't divisible by .
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