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PAT Maths question but I think I can do better? Watch

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    The "markscheme" answer is from at least 2 different sources so :\

    So yeah am I right or not?
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    (Original post by DrSebWilkes)
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    The "markscheme" answer is from at least 2 different sources so :\

    So yeah am I right or not?
    "the other rules"? Note that for 0<x\le 1, \sum \frac 1 {x^n} diverges, as you had previously said.
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    The RHS does not exist for 0 < x < 1.
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    (Original post by DFranklin)
    The RHS does not exist for 0 < x < 1.
    Why not? It might be undefined because it diverges... but surely to a mathematician that's good enough right?

    (Original post by _gcx)
    "the other rules"? Note that for 0&lt;x\le 1, \sum \frac 1 {x^n} diverges, as you had previously said.
    I basically meant the other stuff I said. If something diverges that means it gets pretty huwowage so surely that means it'd be bigger than the measly x+2 ?
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    (Original post by DrSebWilkes)
    Why not? It might be undefined because it converges ... but surely to a mathematician that's good enough right?
    I'm not sure what you're trying to say here.

    I basically meant the other stuff I said. If something diverges that means it gets pretty huwowage so surely that means it'd be bigger than the measly x+2 ?
    I guess the point is this:

    1. You have summed the series to find that S=\frac{x}{x-1}. However that sum only converges to that expression for x &lt;-1, x&gt;1. The sum does not spit out the same value as the expression for values of x in the range -1 &lt;x&lt;1 so they are not the same thing in that range e.g. x=0.1 \Rightarrow S=\frac{0.1}{0.1-1}=-\frac{1}{9} but the series does not give that value

    2. You have done some algebraic inequality solving with the expression for S to find that -\sqrt{2}&lt;x&lt;\sqrt{2}. However, we are working firstly only with +ve x so 0 &lt; x, and that expression is only a valid representation of the sum if the +ve value of x is in the range x&gt;1, and hence your inequality solution is only valid for values of x in that range.

    So putting those points together we get the markscheme result.
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    (Original post by atsruser)
    I'm not sure what you're trying to say here.
    Sorry I meant diverges; this means that the value gets really big so surely that's enough to justify that the function on the RHS > LHS

    (Original post by atsruser)
    I guess the point is this:

    1. You have summed the series to find that S=\frac{x}{x-1}. However that sum only converges to that expression for x &lt;-1, x&gt;1. The sum does not spit out the same value as the expression for values of x in the range -1 &lt;x&lt;1 so they are not the same thing in that range e.g. x=0.1 \Rightarrow S=\frac{0.1}{0.1-1}=-\frac{1}{9} but the series does not give that value

    2. You have done some algebraic inequality solving with the expression for S to find that -\sqrt{2}&lt;x&lt;\sqrt{2}. However, we are working firstly only with +ve x so 0 &lt; x, and that expression is only a valid representation of the sum if the +ve value of x is in the range x&gt;1, and hence your inequality solution is only valid for values of x in that range.

    So putting those points together we get the markscheme result.
    But there are 3 possible ranges of values which require different functions in order to solve. When x=1, the RHS function diverges so we can say it equals infinity. Last time I checked, infinity > 3. And if 0<x<1 it still goes to infinity which means, again, RHS > LHS. The first is just the just 1+1+1.... and the second is ... a(1-r^n)/1-r where n gets huwowge. Valid expressions, but different nevertheless to the one you mentioned. When we get to x>1, it becomes clear that there will be a point where the RHS gets smaller than the LHS which is + root 2.

    So, using 3 different, valid functions, the RHS is larger than the LHS between 0<x<root 2
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    (Original post by DrSebWilkes)
    Sorry I meant diverges; this means that the value gets really big so surely that's enough to justify that the function on the RHS > LHS

    But there are 3 possible ranges of values which require different functions in order to solve. When x=1, the RHS function diverges so we can say it equals infinity. Last time I checked, infinity > 3. And if 0<x<1 it still goes to infinity which means, again, RHS > LHS.
    The expression x &lt;y can only be assigned a true or false value if both x,y are real numbers. \infty is not a real number so the expression x &lt; \infty is not a valid expression - it's not well-formed. In order to decide for which values of x the following expression is true:

    E_1=x+2&lt;1+\frac{1}{x}+\frac{1}{x  ^2}+\cdots=E_2

    we must restrict ourselves to values of +ve x where both E_1,E_2 \in \mathbb{R} else the expression is not well-formed.

    If 0&lt;x&lt;1 the expression E_2 is not equal to any real number, so we can't work with values in that range at all. For values of x&gt;1 then E_2 is correctly represented by x/(x-1), so we replace it with that expression which allows us to solve the inequality easily with no need to play around with an infinite series.
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    (Original post by atsruser)
    The expression x &lt;y can only be assigned a true or false value if both x,y are real numbers. \infty is not a real number so the expression x &lt; \infty is not a valid expression - it's not well-formed. In order to decide for which values of x the following expression is true:

    E_1=x+2&lt;1+\frac{1}{x}+\frac{1}{x  ^2}+\cdots=E_2

    we must restrict ourselves to values of +ve x where both E_1,E_2 \in \mathbb{R} else the expression is not well-formed.

    If 0&lt;x&lt;1 the expression E_2 is not equal to any real number, so we can't work with values in that range at all. For values of x&gt;1 then E_2 is correctly represented by x/(x-1), so we replace it with that expression which allows us to solve the inequality easily with no need to play around with an infinite series.
    Hahaha never knew that! Thank you very much
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    Take x=2.

    RHS = 1 + 1/2 + 1/4 + 1/8 + ... = 2

    Your formula claims it converges to x/(x+1) = 2/3
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    (Original post by SerBronn)
    Take x=2.

    RHS = 1 + 1/2 + 1/4 + 1/8 + ... = 2

    Your formula claims it converges to x/(x+1) = 2/3
    I thought the OP wrote \displaystyle \sum \frac 1 {x^n} \left(= \frac 1 {1-\frac 1 x}\right) = \frac x {x-1}, but they wrote \displaystyle \sum \frac 1 {x^n} = \frac x {x+1}, you're right. Their interval, however, is of course correct, so they might have just copied it wrong.
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    (Original post by DrSebWilkes)
    Why not? It might be undefined because it diverges... but surely to a mathematician that's good enough right?
    Did you mean to say "to a physicist that's good enough right"?

    For a Physics exam, I wouldn't actually want to go out on a limb and say they [i]weren't[/b] expecting the 0<x<=1 range of values to be included.

    But undoubtedly the safest answer is to say the LHS < RHS for 1 < x < sqrt(2), and the RHS is undefined for 0 < x < 1.
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    (Original post by DFranklin)
    Did you mean to say "to a physicist that's good enough right"?

    For a Physics exam, I wouldn't actually want to go out on a limb and say they [i]weren't[/b] expecting the 0<x<=1 range of values to be included.

    But undoubtedly the safest answer is to say the LHS < RHS for 1 < x < sqrt(2), and the RHS is undefined for 0 < x < 1.
    Hahah yes indeed!

    Well thanks then, I'm glad I asked because it's brought up an important difference and understanding the actual maths behind things is what I enjoy anyway!
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    (Original post by DFranklin)
    Did you mean to say "to a physicist that's good enough right"?

    For a Physics exam, I wouldn't actually want to go out on a limb and say they [i]weren't[/b] expecting the 0<x<=1 range of values to be included.
    I would agree. I wouldn't be entirely sure what they were expecting myself if the mark scheme wasn't there.

    But undoubtedly the safest answer is to say the LHS < RHS for 1 < x < sqrt(2), and the RHS is undefined for 0 < x < 1.
    Yes, however, I think that it's not entirely easy to explain why this is the case. The OPs argument about infinity being bigger than any real number has some obvious attractions, and though I tried to justify the markscheme result above, I'm not entirely sure that I convinced myself, let alone anyone else.

    I guess part of the problem is that at A level, students get used to working very informally with (some version of) the extended reals, but without any good definitions, so all kinds of confusions can arise when the question of dealing with divergent values arises.
 
 
 
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