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Finding value

  • #1
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Homework Statement



Let ## S_k , k = 1,2,3,…,100 ## denote the sum of the infinite geometric series whose first term is ## \frac{k-1}{k!} ## and the common ratio is ##\frac {1}{k}##. Then value of ##\frac {100^2}{100!} + \sum\limits_{k=1}^{100} | (k^2 - 3k + 1)S_k | ## is

Homework Equations


Sum of infinite geometric series is a/(1-r) where a is first term and r the common ratio.


The Attempt at a Solution


Got ## S_k= \frac{1}{(k-1)!}##
 

Answers and Replies

  • #2
BvU
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You sure about this problem statement ? 100 ! is an extremely big number !

And: S1 = 1, even if all terms are 0 ?
 
  • #3
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You sure about this problem statement ? 100 ! is an extremely big number !

And: S1 = 1, even if all terms are 0 ?
Yeah, the problem statement is correct. 100! Is a great number indeed but it may reduce I think so by solving.
Didn't understand And: S1 = 1, even if all terms are 0 ?
 
  • #4
BvU
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S1: first term (1-1)/1! all other terms: (1-1) / (1!*1n) n = 2,3,4...

whereas 1/(1-1)! = 1
 
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  • #5
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S1: first term (1-1)/1! all other terms: (1-1) / (1!*1n) n = 2,3,4...

whereas 1/(1-1)! = 1
You are right,
So, ## S_k=\frac{1}{(k-1)!} ## is relevant for k >= 2 for k=1 the sum is 0.
 
  • #6
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S1: first term (1-1)/1! all other terms: (1-1) / (1!*1n) n = 2,3,4...

whereas 1/(1-1)! = 1
That's true but that doesn't mean my question is wrong.What about next?
We have got Sk as 0 for k=1
Well the solution to the problem is 3, if it makes one easy to solve.
But I don't know how to get it.
 
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  • #7
Dick
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That's true but that doesn't mean my question is wrong.What about next?
We have got Sk as 0 for k=1
Well the solution to the problem is 3, if it makes one easy to solve.
But I don't know how to get it.
The sum is 3. I did it in maxima. It must telescope in some clever way. But I still don't see how.
 
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  • #8
BvU
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Makes two [edit:] three of us. I can cheat with a spreadsheet and see that even after 11 terms the summation is already at 3 to within 6 significant digits (using Sk = 0 for k = 1), but that's probably not the idea for this exercise.

Especially the presence of this 1002/100! is intriguing: if the correct, exact answer is a Taylor series, that would mean that this expression is the ##\displaystyle \sum_{k=101}^\infty## (with a value of about 1.07 * 10-154 ) . But a Taylor series of what function ? I can't find out.
 
  • #9
Dick
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Makes two [edit:] three of us. I can cheat with a spreadsheet and see that even after 11 terms the summation is already at 3 to within 6 significant digits (using Sk = 0 for k = 1), but that's probably not the idea for this exercise.

Especially the presence of this 1002/100! is intriguing: if the correct, exact answer is a Taylor series, that would mean that this expression is the ##\displaystyle \sum_{k=101}^\infty## (with a value of about 1.07 * 10-154 ) . But a Taylor series of what function ? I can't find out.
I don't think it's a Taylor series. It's some clever way of expressing f(k)=(k^2-3k+1)/(k-1)! as f(k)=g(k)-g(k-1) so it telescopes. What's g(k)?
 
  • #10
haruspex
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Seems fairly straightforward. First, sum the infinite series to get Sk into closed form.
You can get rid of the modulus signs by recognising which few terms would be negative otherwise.
Then use the closed form of Sk to simplify the expression in the finite sum.
As Dick suspected, you get cancellation of all but the first few and last few terms.
 
  • #11
haruspex
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I don't think it's a Taylor series. It's some clever way of expressing f(k)=(k^2-3k+1)/(k-1)! as f(k)=g(k)-g(k-1) so it telescopes. What's g(k)?
My post crossed with yours. Don't assume it'll be g(k-1).
 
  • #12
Dick
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My post crossed with yours. Don't assume it'll be g(k-1).
Well, it's 5AM here. g(k-2)?
 
  • #13
haruspex
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Well, it's 5AM here. g(k-2)?
Yes (10pm).
 
  • #14
Dick
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  • #15
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Makes two [edit:] three of us. I can cheat with a spreadsheet and see that even after 11 terms the summation is already at 3 to within 6 significant digits (using Sk = 0 for k = 1), but that's probably not the idea for this exercise.

Especially the presence of this 1002/100! is intriguing: if the correct, exact answer is a Taylor series, that would mean that this expression is the ##\displaystyle \sum_{k=101}^\infty## (with a value of about 1.07 * 10-154 ) . But a Taylor series of what function ? I can't find out.
I am a high school student. So Taylor series questions are not usually given to that advanced level.
So something other might be applied here.
I don't think it's a Taylor series. It's some clever way of expressing f(k)=(k^2-3k+1)/(k-1)! as f(k)=g(k)-g(k-1) so it telescopes. What's g(k)?
Seems fairly straightforward. First, sum the infinite series to get Sk into closed form.
You can get rid of the modulus signs by recognising which few terms would be negative otherwise.
Then use the closed form of Sk to simplify the expression in the finite sum.
As Dick suspected, you get cancellation of all but the first few and last few terms.
Sorry to both of you as I can't figure out how the cancellation is taking place.
Well if one is finding it difficult to explain by latex ( as much time is consumed in that ). He/She can write in a paper and upload that as this question is very important for me.
Please if one is not able to figure out the solution can he/she ask his/her friend/neighbor/ or anybody.
Well this might be also my last question for PF and any further help would be appreciated.
To some people it might be seeming why I am giving so reasons for the sake of this single question.
But it is really important for me.
Hope one might understand
 
  • #16
BvU
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Well, as Dick and Haru indicated, I was thinking in the wrong direction. So forget Taylor and think what a summation ∑ f(k) will do if term f(k) can be written as a difference g(k) - g(k-2) ... a lot of stuff cancels (the 'telescope' description)

Just so you know: to D&H it seems to be obvious, but I still don't have a clue what g(k) could be ! o_O
 
  • #17
Dick
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Well, as Dick and Haru indicated, I was thinking in the wrong direction. So forget Taylor and think what a summation ∑ f(k) will do if term f(k) can be written as a difference g(k) - g(k-2) ... a lot of stuff cancels (the 'telescope' description)

Just so you know: to D&H it seems to be obvious, but I still don't have a clue what g(k) could be ! o_O
Here's kind of a large hint. $$k^2-3k+1=(k-2)(k-1)-1$$.
 
  • #18
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Here's kind of a large hint. $$k^2-3k+1=(k-2)(k-1)-1$$.
Well this was the masterpiece for solving the question. How did you think of it? I got the answer. The rest work was of manipulation. I am not posting my solving method as it would require a lot of latex. Thanks Dick. I would like to thanks others also for replying (BvU, haruspex). This also proved to me for the time being at high school that PF could solve any type of question. It's a matter of understanding. Thanks all.
 
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  • #19
Dick
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Well this was the masterpiece for solving the question. How did you think of it? I got the answer. The rest work was of manipulation. I am not posting my solving method as it would require a lot of latex. Thanks Dick. I would like to thanks others also for replying (BvU, haruspex). This also proved to me for the time being at high school that PF could solve any type of question. It's a matter of understanding. Thanks all.
I thought of it because if I see a finite series that isn't arithmetic or geometric that has a nice sum then it must telescope somehow. I.e. I can rewrite the summand in such a way that there will be a lot of cancellation. I was being a little thick about how to do it until haruspex gave me a nudge.
 
  • #20
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I thought of it because if I see a finite series that isn't arithmetic or geometric that has a nice sum then it must telescope somehow. I.e. I can rewrite the summand in such a way that there will be a lot of cancellation. I was being a little thick about how to do it until haruspex gave me a nudge.
:)
Seems fairly straightforward. First, sum the infinite series to get Sk into closed form.
You can get rid of the modulus signs by recognising which few terms would be negative otherwise.
Then use the closed form of Sk to simplify the expression in the finite sum.
As Dick suspected, you get cancellation of all but the first few and last few terms.
This might have given you the nudge.
 
  • #21
Dick
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My post crossed with yours. Don't assume it'll be g(k-1).
This was the 'nudge'.
 
  • #22
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Okay.
It is amazing sometimes that without posting big latex, one figures out solution. Isn't it?
 
  • #23
haruspex
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Further thought:
Show that if we replace the quadratic in the question by some other polynomial P(k), degree n, then such a collapse can be found if and only if ##\Sigma _{i=0}^n \frac{P(i)}{i!} \Sigma _{j=0}^{n-i} \frac{(-1)^j}{j!} =0##.
Btw, seems like Dick and I arrived at the solution by slightly different routes. Dick explicitly looked for cancelling pairs (right?). I looked at the quadratic/factorial expression and treated it like a partial fractions problem. First extract the remainder modulo k-1 to get P(k)(k-1)+constant, giving constant/(k-1)! + P(k)/(k-2)!, then reduce the second term in the same way wrt k-2. That way you can find reductions involving any number of 'g()' terms. For collapse to occur, the sum of the constants must turn out to be zero.
 

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