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Variable in upper bound of sum
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I need to find a solution $x$ of the following equation:
$$sum_n=0^left[frac0.9xright] (1-nx) = 45$$
where $[.]$ denotes the nearest integer function.
I am an engineer and I'm currently making an accelerated beam that needs to send a pulse of 100ns after a period of waiting time. That wait period is equal to (1-nx). I want that, after 45 seconds, I want to get a frequency of 10Hz at the end of the 45 seconds (so the wait time has become 0.1). I have 1Hz at the beginning.
I have divided the problem in two: $$sum_n=0^i (1-nx) = 45$$
where $i$ is the integer that comes closest to satisfying
$$1-ix = 0.1,$$
i.e., $i$ is the integer nearest to $0.9over x$.
But now, I'm concerned; is this even possible in discrete form? IS there a way to know if this is a valid equation? I have never encountered variables in sum bounds... and didn't find a way out of it on the internet. I think I might not be using the proper technique, maybe this is something else than a sum.
I love maths, but I might just be bad at it.
Educate me !
discrete-mathematics summation numerical-methods fixed-point-theorems upper-lower-bounds
edited Aug 8 at 2:02
r.e.s.
7,46411852
asked Aug 6 at 10:54
PyThagoras
215
 |Â
show 2 more comments
up vote
4
down vote
favorite
I need to find a solution $x$ of the following equation:
$$sum_n=0^left[frac0.9xright] (1-nx) = 45$$
where $[.]$ denotes the nearest integer function.
I am an engineer and I'm currently making an accelerated beam that needs to send a pulse of 100ns after a period of waiting time. That wait period is equal to (1-nx). I want that, after 45 seconds, I want to get a frequency of 10Hz at the end of the 45 seconds (so the wait time has become 0.1). I have 1Hz at the beginning.
I have divided the problem in two: $$sum_n=0^i (1-nx) = 45$$
where $i$ is the integer that comes closest to satisfying
$$1-ix = 0.1,$$
i.e., $i$ is the integer nearest to $0.9over x$.
But now, I'm concerned; is this even possible in discrete form? IS there a way to know if this is a valid equation? I have never encountered variables in sum bounds... and didn't find a way out of it on the internet. I think I might not be using the proper technique, maybe this is something else than a sum.
I love maths, but I might just be bad at it.
Educate me !
discrete-mathematics summation numerical-methods fixed-point-theorems upper-lower-bounds
edited Aug 8 at 2:02
r.e.s.
7,46411852
asked Aug 6 at 10:54
PyThagoras
215
Sum bounds should be natural numbers
– Davide Morgante
Aug 6 at 11:04
Yes, but I want to find the closest interger. In fact, I just want x. so if we say x equals 0.005, it'll give me a integer, or close to. But x is what i'm searching for here.
– PyThagoras
Aug 6 at 11:13
Oh ok, that makes more sense! Just another clarification: the $x$ variable in the summation bound is the same x as the one in the argument of the sum?
– Davide Morgante
Aug 6 at 11:14
Yes exactly, That's why I showed the two equations I used to make the sum!
– PyThagoras
Aug 6 at 11:15
Well, in that case the equation makes some sense but I think that it'll be very difficult to find the solution. Let me think about it for a moment
– Davide Morgante
Aug 6 at 11:16
 |Â
show 2 more comments
up vote
4
down vote
favorite
up vote
4
down vote
favorite
I need to find a solution $x$ of the following equation:
$$sum_n=0^left[frac0.9xright] (1-nx) = 45$$
where $[.]$ denotes the nearest integer function.
I am an engineer and I'm currently making an accelerated beam that needs to send a pulse of 100ns after a period of waiting time. That wait period is equal to (1-nx). I want that, after 45 seconds, I want to get a frequency of 10Hz at the end of the 45 seconds (so the wait time has become 0.1). I have 1Hz at the beginning.
I have divided the problem in two: $$sum_n=0^i (1-nx) = 45$$
where $i$ is the integer that comes closest to satisfying
$$1-ix = 0.1,$$
i.e., $i$ is the integer nearest to $0.9over x$.
But now, I'm concerned; is this even possible in discrete form? IS there a way to know if this is a valid equation? I have never encountered variables in sum bounds... and didn't find a way out of it on the internet. I think I might not be using the proper technique, maybe this is something else than a sum.
I love maths, but I might just be bad at it.
Educate me !
discrete-mathematics summation numerical-methods fixed-point-theorems upper-lower-bounds
edited Aug 8 at 2:02
r.e.s.
7,46411852
asked Aug 6 at 10:54
PyThagoras
215
I need to find a solution $x$ of the following equation:
$$sum_n=0^left[frac0.9xright] (1-nx) = 45$$
where $[.]$ denotes the nearest integer function.
I am an engineer and I'm currently making an accelerated beam that needs to send a pulse of 100ns after a period of waiting time. That wait period is equal to (1-nx). I want that, after 45 seconds, I want to get a frequency of 10Hz at the end of the 45 seconds (so the wait time has become 0.1). I have 1Hz at the beginning.
I have divided the problem in two: $$sum_n=0^i (1-nx) = 45$$
where $i$ is the integer that comes closest to satisfying
$$1-ix = 0.1,$$
i.e., $i$ is the integer nearest to $0.9over x$.
But now, I'm concerned; is this even possible in discrete form? IS there a way to know if this is a valid equation? I have never encountered variables in sum bounds... and didn't find a way out of it on the internet. I think I might not be using the proper technique, maybe this is something else than a sum.
I love maths, but I might just be bad at it.
Educate me !
discrete-mathematics summation numerical-methods fixed-point-theorems upper-lower-bounds
edited Aug 8 at 2:02
r.e.s.
7,46411852
asked Aug 6 at 10:54
PyThagoras
215
edited Aug 8 at 2:02
r.e.s.
7,46411852
edited Aug 8 at 2:02
r.e.s.
7,46411852
edited Aug 8 at 2:02
r.e.s.
7,46411852
7,46411852
asked Aug 6 at 10:54
PyThagoras
215
asked Aug 6 at 10:54
PyThagoras
215
asked Aug 6 at 10:54
PyThagoras
215
215
Sum bounds should be natural numbers
– Davide Morgante
Aug 6 at 11:04
Yes, but I want to find the closest interger. In fact, I just want x. so if we say x equals 0.005, it'll give me a integer, or close to. But x is what i'm searching for here.
– PyThagoras
Aug 6 at 11:13
Oh ok, that makes more sense! Just another clarification: the $x$ variable in the summation bound is the same x as the one in the argument of the sum?
– Davide Morgante
Aug 6 at 11:14
Yes exactly, That's why I showed the two equations I used to make the sum!
– PyThagoras
Aug 6 at 11:15
Well, in that case the equation makes some sense but I think that it'll be very difficult to find the solution. Let me think about it for a moment
– Davide Morgante
Aug 6 at 11:16
 |Â
show 2 more comments
Sum bounds should be natural numbers
– Davide Morgante
Aug 6 at 11:04
Yes, but I want to find the closest interger. In fact, I just want x. so if we say x equals 0.005, it'll give me a integer, or close to. But x is what i'm searching for here.
– PyThagoras
Aug 6 at 11:13
Oh ok, that makes more sense! Just another clarification: the $x$ variable in the summation bound is the same x as the one in the argument of the sum?
– Davide Morgante
Aug 6 at 11:14
Yes exactly, That's why I showed the two equations I used to make the sum!
– PyThagoras
Aug 6 at 11:15
Well, in that case the equation makes some sense but I think that it'll be very difficult to find the solution. Let me think about it for a moment
– Davide Morgante
Aug 6 at 11:16
Sum bounds should be natural numbers
– Davide Morgante
Aug 6 at 11:04
Sum bounds should be natural numbers
– Davide Morgante
Aug 6 at 11:04
Yes, but I want to find the closest interger. In fact, I just want x. so if we say x equals 0.005, it'll give me a integer, or close to. But x is what i'm searching for here.
– PyThagoras
Aug 6 at 11:13
Yes, but I want to find the closest interger. In fact, I just want x. so if we say x equals 0.005, it'll give me a integer, or close to. But x is what i'm searching for here.
– PyThagoras
Aug 6 at 11:13
Oh ok, that makes more sense! Just another clarification: the $x$ variable in the summation bound is the same x as the one in the argument of the sum?
– Davide Morgante
Aug 6 at 11:14
Oh ok, that makes more sense! Just another clarification: the $x$ variable in the summation bound is the same x as the one in the argument of the sum?
– Davide Morgante
Aug 6 at 11:14
Yes exactly, That's why I showed the two equations I used to make the sum!
– PyThagoras
Aug 6 at 11:15
Yes exactly, That's why I showed the two equations I used to make the sum!
– PyThagoras
Aug 6 at 11:15
Well, in that case the equation makes some sense but I think that it'll be very difficult to find the solution. Let me think about it for a moment
– Davide Morgante
Aug 6 at 11:16
Well, in that case the equation makes some sense but I think that it'll be very difficult to find the solution. Let me think about it for a moment
– Davide Morgante
Aug 6 at 11:16
 |Â
show 2 more comments
3 Answers
3
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oldest
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up vote
4
down vote
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Given your comment that you want the upper bound to be the integer closest to $0.9over x$, the problem can be stated as finding $x$ such that
$$sum_n=0^a(x) (1-nx) = 45,$$
where $a(x)=left[0.9over xright]$ and $[.]$ is the nearest integer function.
Now
$$beginalignsum_n=0^a (1-nx) &= sum_n=0^a 1- sum_n=0^a nx\
&=sum_n=0^a 1- xsum_n=1^a n\
&=(a+1) -x,a(a+1)over 2
endalign$$
where we've used the formula $sum_n=1^a n=a(a+1)over 2$, which Carl Friedrich Gauss supposedly found in his youth (although it was known long before that).
So, we want to solve for $x$ in the following equation:
$$(a(x)+1) -x,a(x)(a(x)+1)over 2 = 45.tag1$$
Approximate solution
An approximate solution can be obtained easily by solving equation (1) with $a(x) = 0.9over x$ (rather than the nearest integer), yielding $xapprox 0.011136ldots.$ To find out how good this approximation is, we now obtain the exact solution.
Exact solution
Rearranging equation (1), we get
$$x = 2a(x)-44over a(x)(a(x)+1)tag2
$$
which provides two observations:
- A solution $x$ (if it exists) must be a rational number, because the RHS of (2) is a ratio of integers.
Fixed-point iteration converges to an exact solution, e.g. if we start with $x_0=0.01$ (say):
$$x_n+1 = 2a(x_n)-44over a(x_n)(a(x_n)+1),quad n=0,1,2,ldots
$$
Thus,
n x_n
-- -------
0 1/100
1 46/4095
2 1/90
3 37/3321
4 37/3321
... ...
giving the exact solution $$x=37over 3321=0.overline011141222523336344474555856669677807889190003$$
where the overline indicates the period of the repeating decimal digits.
Here's a plot from Wolfram Alpha showing the exact LHS of equation (1) in blue and the approximated LHS in orange. The solution in each case is the $x$-coordinate where the curve intersects the horizontal line with ordinate $45$:
NB: You were right to be concerned with the possibility that a solution might not exist, although it happens that the value $45$ is a fortunate choice.
As the above plot shows, there would be no solutions for values that correspond to the infinitely many "gaps" where discontinuities occur (e.g., in the neighborhood of $45.4$ or $44.8$, say).
answered Aug 8 at 1:32
r.e.s.
7,46411852
Extremely rigourous. I assume you study mathematics? Well that was a beautiful problem and the solution you proposed is elegant. At first I was going over it and thought: "this is a kindergarden problem, why can't I resolve it?". Now i'm not so sure ahah. Thank you again.
– PyThagoras
Aug 8 at 3:30
add a comment |Â
up vote
2
down vote
This is mostly a comment on the fine answer by r.e.s.
As r.e.s. shows, if $a$ is the nearest integer to $9over10x$, then
$$x=2a-44over a(a+1)$$
That is, we are looking for (positive) integer solutions to
$$9a(a+1)over20(a-44)=a+r$$
with $|r|le1over2$. (There is a potential ambiguity if there is a solution with $|r|=1over2$, but we'll see that this doesn't occur.) This can be rewritten as
$$-10lea(889-11a)over a-44le10$$
For $agt44$, these inequalities becomes
$$11a^2-899a+440lt0lt11a^2-879a-440$$
while for $alt44$ the inequalities signs are reversed. This gives two intervals in which to look:
$$left(879+sqrt879^2+44cdot440over22,899+sqrt899^2-44cdot440over22 right)approx(80.40656,81.23487)$$
and
$$left(879-sqrt879^2+44cdot440over22,899-sqrt899^2-44cdot440over22 right)approx(-0.49747,0.492399)$$
The first interval contains the relevant integer value $a=81$, with the corresponding value $x=2(81-44)/(81cdot82)=37/3321$. The second interval contains only the irrelevant integer value $a=0$. We thus see that the solution found by r.e.s. is unique.
answered Aug 9 at 11:25
Barry Cipra
56.6k652118
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up vote
-1
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I asked a mathematician friend and that's what he told me:
$$sum_n=0^frac0.9x(1-nx) = sum_n=0^frac0.9x1 -sum_n=0^frac0.9xnx = 45$$
So he splitted it.
$$sum_n=0^frac0.9x1 = frac0.9x+1$$ That one is easy.
$$sum_n=0^frac0.9xnx = 0.45+frac0.405x$$
But that one is tricky. He told me it was some Gauss sum. If anyone can comment or precise what and how the gaussian sum is resolved.
So finally,
$$(frac0.9x+1)-(0.45+frac0.405x)=45$$
$$x = 0.01136...$$
answered Aug 6 at 11:53
PyThagoras
215
You seem to have some digits wrong, as solving that approximating equation gives the approximate solution $0.495over 44.45=0.011136ldots$. As it turns out, this agrees with the exact solution when rounded to five digits after the decimal point! (See my answer for the exact solution.)
– r.e.s.
Aug 8 at 2:47
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
Given your comment that you want the upper bound to be the integer closest to $0.9over x$, the problem can be stated as finding $x$ such that
$$sum_n=0^a(x) (1-nx) = 45,$$
where $a(x)=left[0.9over xright]$ and $[.]$ is the nearest integer function.
Now
$$beginalignsum_n=0^a (1-nx) &= sum_n=0^a 1- sum_n=0^a nx\
&=sum_n=0^a 1- xsum_n=1^a n\
&=(a+1) -x,a(a+1)over 2
endalign$$
where we've used the formula $sum_n=1^a n=a(a+1)over 2$, which Carl Friedrich Gauss supposedly found in his youth (although it was known long before that).
So, we want to solve for $x$ in the following equation:
$$(a(x)+1) -x,a(x)(a(x)+1)over 2 = 45.tag1$$
Approximate solution
An approximate solution can be obtained easily by solving equation (1) with $a(x) = 0.9over x$ (rather than the nearest integer), yielding $xapprox 0.011136ldots.$ To find out how good this approximation is, we now obtain the exact solution.
Exact solution
Rearranging equation (1), we get
$$x = 2a(x)-44over a(x)(a(x)+1)tag2
$$
which provides two observations:
- A solution $x$ (if it exists) must be a rational number, because the RHS of (2) is a ratio of integers.
Fixed-point iteration converges to an exact solution, e.g. if we start with $x_0=0.01$ (say):
$$x_n+1 = 2a(x_n)-44over a(x_n)(a(x_n)+1),quad n=0,1,2,ldots
$$
Thus,
n x_n
-- -------
0 1/100
1 46/4095
2 1/90
3 37/3321
4 37/3321
... ...
giving the exact solution $$x=37over 3321=0.overline011141222523336344474555856669677807889190003$$
where the overline indicates the period of the repeating decimal digits.
Here's a plot from Wolfram Alpha showing the exact LHS of equation (1) in blue and the approximated LHS in orange. The solution in each case is the $x$-coordinate where the curve intersects the horizontal line with ordinate $45$:
NB: You were right to be concerned with the possibility that a solution might not exist, although it happens that the value $45$ is a fortunate choice.
As the above plot shows, there would be no solutions for values that correspond to the infinitely many "gaps" where discontinuities occur (e.g., in the neighborhood of $45.4$ or $44.8$, say).
answered Aug 8 at 1:32
r.e.s.
7,46411852
Extremely rigourous. I assume you study mathematics? Well that was a beautiful problem and the solution you proposed is elegant. At first I was going over it and thought: "this is a kindergarden problem, why can't I resolve it?". Now i'm not so sure ahah. Thank you again.
– PyThagoras
Aug 8 at 3:30
add a comment |Â
up vote
4
down vote
accepted
Given your comment that you want the upper bound to be the integer closest to $0.9over x$, the problem can be stated as finding $x$ such that
$$sum_n=0^a(x) (1-nx) = 45,$$
where $a(x)=left[0.9over xright]$ and $[.]$ is the nearest integer function.
Now
$$beginalignsum_n=0^a (1-nx) &= sum_n=0^a 1- sum_n=0^a nx\
&=sum_n=0^a 1- xsum_n=1^a n\
&=(a+1) -x,a(a+1)over 2
endalign$$
where we've used the formula $sum_n=1^a n=a(a+1)over 2$, which Carl Friedrich Gauss supposedly found in his youth (although it was known long before that).
So, we want to solve for $x$ in the following equation:
$$(a(x)+1) -x,a(x)(a(x)+1)over 2 = 45.tag1$$
Approximate solution
An approximate solution can be obtained easily by solving equation (1) with $a(x) = 0.9over x$ (rather than the nearest integer), yielding $xapprox 0.011136ldots.$ To find out how good this approximation is, we now obtain the exact solution.
Exact solution
Rearranging equation (1), we get
$$x = 2a(x)-44over a(x)(a(x)+1)tag2
$$
which provides two observations:
- A solution $x$ (if it exists) must be a rational number, because the RHS of (2) is a ratio of integers.
Fixed-point iteration converges to an exact solution, e.g. if we start with $x_0=0.01$ (say):
$$x_n+1 = 2a(x_n)-44over a(x_n)(a(x_n)+1),quad n=0,1,2,ldots
$$
Thus,
n x_n
-- -------
0 1/100
1 46/4095
2 1/90
3 37/3321
4 37/3321
... ...
giving the exact solution $$x=37over 3321=0.overline011141222523336344474555856669677807889190003$$
where the overline indicates the period of the repeating decimal digits.
Here's a plot from Wolfram Alpha showing the exact LHS of equation (1) in blue and the approximated LHS in orange. The solution in each case is the $x$-coordinate where the curve intersects the horizontal line with ordinate $45$:
NB: You were right to be concerned with the possibility that a solution might not exist, although it happens that the value $45$ is a fortunate choice.
As the above plot shows, there would be no solutions for values that correspond to the infinitely many "gaps" where discontinuities occur (e.g., in the neighborhood of $45.4$ or $44.8$, say).
answered Aug 8 at 1:32
r.e.s.
7,46411852
Extremely rigourous. I assume you study mathematics? Well that was a beautiful problem and the solution you proposed is elegant. At first I was going over it and thought: "this is a kindergarden problem, why can't I resolve it?". Now i'm not so sure ahah. Thank you again.
– PyThagoras
Aug 8 at 3:30
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
Given your comment that you want the upper bound to be the integer closest to $0.9over x$, the problem can be stated as finding $x$ such that
$$sum_n=0^a(x) (1-nx) = 45,$$
where $a(x)=left[0.9over xright]$ and $[.]$ is the nearest integer function.
Now
$$beginalignsum_n=0^a (1-nx) &= sum_n=0^a 1- sum_n=0^a nx\
&=sum_n=0^a 1- xsum_n=1^a n\
&=(a+1) -x,a(a+1)over 2
endalign$$
where we've used the formula $sum_n=1^a n=a(a+1)over 2$, which Carl Friedrich Gauss supposedly found in his youth (although it was known long before that).
So, we want to solve for $x$ in the following equation:
$$(a(x)+1) -x,a(x)(a(x)+1)over 2 = 45.tag1$$
Approximate solution
An approximate solution can be obtained easily by solving equation (1) with $a(x) = 0.9over x$ (rather than the nearest integer), yielding $xapprox 0.011136ldots.$ To find out how good this approximation is, we now obtain the exact solution.
Exact solution
Rearranging equation (1), we get
$$x = 2a(x)-44over a(x)(a(x)+1)tag2
$$
which provides two observations:
- A solution $x$ (if it exists) must be a rational number, because the RHS of (2) is a ratio of integers.
Fixed-point iteration converges to an exact solution, e.g. if we start with $x_0=0.01$ (say):
$$x_n+1 = 2a(x_n)-44over a(x_n)(a(x_n)+1),quad n=0,1,2,ldots
$$
Thus,
n x_n
-- -------
0 1/100
1 46/4095
2 1/90
3 37/3321
4 37/3321
... ...
giving the exact solution $$x=37over 3321=0.overline011141222523336344474555856669677807889190003$$
where the overline indicates the period of the repeating decimal digits.
Here's a plot from Wolfram Alpha showing the exact LHS of equation (1) in blue and the approximated LHS in orange. The solution in each case is the $x$-coordinate where the curve intersects the horizontal line with ordinate $45$:
NB: You were right to be concerned with the possibility that a solution might not exist, although it happens that the value $45$ is a fortunate choice.
As the above plot shows, there would be no solutions for values that correspond to the infinitely many "gaps" where discontinuities occur (e.g., in the neighborhood of $45.4$ or $44.8$, say).
answered Aug 8 at 1:32
r.e.s.
7,46411852
Given your comment that you want the upper bound to be the integer closest to $0.9over x$, the problem can be stated as finding $x$ such that
$$sum_n=0^a(x) (1-nx) = 45,$$
where $a(x)=left[0.9over xright]$ and $[.]$ is the nearest integer function.
Now
$$beginalignsum_n=0^a (1-nx) &= sum_n=0^a 1- sum_n=0^a nx\
&=sum_n=0^a 1- xsum_n=1^a n\
&=(a+1) -x,a(a+1)over 2
endalign$$
where we've used the formula $sum_n=1^a n=a(a+1)over 2$, which Carl Friedrich Gauss supposedly found in his youth (although it was known long before that).
So, we want to solve for $x$ in the following equation:
$$(a(x)+1) -x,a(x)(a(x)+1)over 2 = 45.tag1$$
Approximate solution
An approximate solution can be obtained easily by solving equation (1) with $a(x) = 0.9over x$ (rather than the nearest integer), yielding $xapprox 0.011136ldots.$ To find out how good this approximation is, we now obtain the exact solution.
Exact solution
Rearranging equation (1), we get
$$x = 2a(x)-44over a(x)(a(x)+1)tag2
$$
which provides two observations:
- A solution $x$ (if it exists) must be a rational number, because the RHS of (2) is a ratio of integers.
Fixed-point iteration converges to an exact solution, e.g. if we start with $x_0=0.01$ (say):
$$x_n+1 = 2a(x_n)-44over a(x_n)(a(x_n)+1),quad n=0,1,2,ldots
$$
Thus,
n x_n
-- -------
0 1/100
1 46/4095
2 1/90
3 37/3321
4 37/3321
... ...
giving the exact solution $$x=37over 3321=0.overline011141222523336344474555856669677807889190003$$
where the overline indicates the period of the repeating decimal digits.
Here's a plot from Wolfram Alpha showing the exact LHS of equation (1) in blue and the approximated LHS in orange. The solution in each case is the $x$-coordinate where the curve intersects the horizontal line with ordinate $45$:
NB: You were right to be concerned with the possibility that a solution might not exist, although it happens that the value $45$ is a fortunate choice.
As the above plot shows, there would be no solutions for values that correspond to the infinitely many "gaps" where discontinuities occur (e.g., in the neighborhood of $45.4$ or $44.8$, say).
answered Aug 8 at 1:32
r.e.s.
7,46411852
edited Aug 8 at 23:49
answered Aug 8 at 1:32
r.e.s.
7,46411852
answered Aug 8 at 1:32
r.e.s.
7,46411852
answered Aug 8 at 1:32
r.e.s.
7,46411852
7,46411852
Extremely rigourous. I assume you study mathematics? Well that was a beautiful problem and the solution you proposed is elegant. At first I was going over it and thought: "this is a kindergarden problem, why can't I resolve it?". Now i'm not so sure ahah. Thank you again.
– PyThagoras
Aug 8 at 3:30
add a comment |Â
Extremely rigourous. I assume you study mathematics? Well that was a beautiful problem and the solution you proposed is elegant. At first I was going over it and thought: "this is a kindergarden problem, why can't I resolve it?". Now i'm not so sure ahah. Thank you again.
– PyThagoras
Aug 8 at 3:30
Extremely rigourous. I assume you study mathematics? Well that was a beautiful problem and the solution you proposed is elegant. At first I was going over it and thought: "this is a kindergarden problem, why can't I resolve it?". Now i'm not so sure ahah. Thank you again.
– PyThagoras
Aug 8 at 3:30
Extremely rigourous. I assume you study mathematics? Well that was a beautiful problem and the solution you proposed is elegant. At first I was going over it and thought: "this is a kindergarden problem, why can't I resolve it?". Now i'm not so sure ahah. Thank you again.
– PyThagoras
Aug 8 at 3:30
add a comment |Â
up vote
2
down vote
This is mostly a comment on the fine answer by r.e.s.
As r.e.s. shows, if $a$ is the nearest integer to $9over10x$, then
$$x=2a-44over a(a+1)$$
That is, we are looking for (positive) integer solutions to
$$9a(a+1)over20(a-44)=a+r$$
with $|r|le1over2$. (There is a potential ambiguity if there is a solution with $|r|=1over2$, but we'll see that this doesn't occur.) This can be rewritten as
$$-10lea(889-11a)over a-44le10$$
For $agt44$, these inequalities becomes
$$11a^2-899a+440lt0lt11a^2-879a-440$$
while for $alt44$ the inequalities signs are reversed. This gives two intervals in which to look:
$$left(879+sqrt879^2+44cdot440over22,899+sqrt899^2-44cdot440over22 right)approx(80.40656,81.23487)$$
and
$$left(879-sqrt879^2+44cdot440over22,899-sqrt899^2-44cdot440over22 right)approx(-0.49747,0.492399)$$
The first interval contains the relevant integer value $a=81$, with the corresponding value $x=2(81-44)/(81cdot82)=37/3321$. The second interval contains only the irrelevant integer value $a=0$. We thus see that the solution found by r.e.s. is unique.
answered Aug 9 at 11:25
Barry Cipra
56.6k652118
add a comment |Â
up vote
2
down vote
This is mostly a comment on the fine answer by r.e.s.
As r.e.s. shows, if $a$ is the nearest integer to $9over10x$, then
$$x=2a-44over a(a+1)$$
That is, we are looking for (positive) integer solutions to
$$9a(a+1)over20(a-44)=a+r$$
with $|r|le1over2$. (There is a potential ambiguity if there is a solution with $|r|=1over2$, but we'll see that this doesn't occur.) This can be rewritten as
$$-10lea(889-11a)over a-44le10$$
For $agt44$, these inequalities becomes
$$11a^2-899a+440lt0lt11a^2-879a-440$$
while for $alt44$ the inequalities signs are reversed. This gives two intervals in which to look:
$$left(879+sqrt879^2+44cdot440over22,899+sqrt899^2-44cdot440over22 right)approx(80.40656,81.23487)$$
and
$$left(879-sqrt879^2+44cdot440over22,899-sqrt899^2-44cdot440over22 right)approx(-0.49747,0.492399)$$
The first interval contains the relevant integer value $a=81$, with the corresponding value $x=2(81-44)/(81cdot82)=37/3321$. The second interval contains only the irrelevant integer value $a=0$. We thus see that the solution found by r.e.s. is unique.
answered Aug 9 at 11:25
Barry Cipra
56.6k652118
add a comment |Â
up vote
2
down vote
up vote
2
down vote
This is mostly a comment on the fine answer by r.e.s.
As r.e.s. shows, if $a$ is the nearest integer to $9over10x$, then
$$x=2a-44over a(a+1)$$
That is, we are looking for (positive) integer solutions to
$$9a(a+1)over20(a-44)=a+r$$
with $|r|le1over2$. (There is a potential ambiguity if there is a solution with $|r|=1over2$, but we'll see that this doesn't occur.) This can be rewritten as
$$-10lea(889-11a)over a-44le10$$
For $agt44$, these inequalities becomes
$$11a^2-899a+440lt0lt11a^2-879a-440$$
while for $alt44$ the inequalities signs are reversed. This gives two intervals in which to look:
$$left(879+sqrt879^2+44cdot440over22,899+sqrt899^2-44cdot440over22 right)approx(80.40656,81.23487)$$
and
$$left(879-sqrt879^2+44cdot440over22,899-sqrt899^2-44cdot440over22 right)approx(-0.49747,0.492399)$$
The first interval contains the relevant integer value $a=81$, with the corresponding value $x=2(81-44)/(81cdot82)=37/3321$. The second interval contains only the irrelevant integer value $a=0$. We thus see that the solution found by r.e.s. is unique.
answered Aug 9 at 11:25
Barry Cipra
56.6k652118
This is mostly a comment on the fine answer by r.e.s.
As r.e.s. shows, if $a$ is the nearest integer to $9over10x$, then
$$x=2a-44over a(a+1)$$
That is, we are looking for (positive) integer solutions to
$$9a(a+1)over20(a-44)=a+r$$
with $|r|le1over2$. (There is a potential ambiguity if there is a solution with $|r|=1over2$, but we'll see that this doesn't occur.) This can be rewritten as
$$-10lea(889-11a)over a-44le10$$
For $agt44$, these inequalities becomes
$$11a^2-899a+440lt0lt11a^2-879a-440$$
while for $alt44$ the inequalities signs are reversed. This gives two intervals in which to look:
$$left(879+sqrt879^2+44cdot440over22,899+sqrt899^2-44cdot440over22 right)approx(80.40656,81.23487)$$
and
$$left(879-sqrt879^2+44cdot440over22,899-sqrt899^2-44cdot440over22 right)approx(-0.49747,0.492399)$$
The first interval contains the relevant integer value $a=81$, with the corresponding value $x=2(81-44)/(81cdot82)=37/3321$. The second interval contains only the irrelevant integer value $a=0$. We thus see that the solution found by r.e.s. is unique.
answered Aug 9 at 11:25
Barry Cipra
56.6k652118
answered Aug 9 at 11:25
Barry Cipra
56.6k652118
answered Aug 9 at 11:25
Barry Cipra
56.6k652118
answered Aug 9 at 11:25
Barry Cipra
56.6k652118
56.6k652118
add a comment |Â
add a comment |Â
up vote
-1
down vote
I asked a mathematician friend and that's what he told me:
$$sum_n=0^frac0.9x(1-nx) = sum_n=0^frac0.9x1 -sum_n=0^frac0.9xnx = 45$$
So he splitted it.
$$sum_n=0^frac0.9x1 = frac0.9x+1$$ That one is easy.
$$sum_n=0^frac0.9xnx = 0.45+frac0.405x$$
But that one is tricky. He told me it was some Gauss sum. If anyone can comment or precise what and how the gaussian sum is resolved.
So finally,
$$(frac0.9x+1)-(0.45+frac0.405x)=45$$
$$x = 0.01136...$$
answered Aug 6 at 11:53
PyThagoras
215
You seem to have some digits wrong, as solving that approximating equation gives the approximate solution $0.495over 44.45=0.011136ldots$. As it turns out, this agrees with the exact solution when rounded to five digits after the decimal point! (See my answer for the exact solution.)
– r.e.s.
Aug 8 at 2:47
add a comment |Â
up vote
-1
down vote
I asked a mathematician friend and that's what he told me:
$$sum_n=0^frac0.9x(1-nx) = sum_n=0^frac0.9x1 -sum_n=0^frac0.9xnx = 45$$
So he splitted it.
$$sum_n=0^frac0.9x1 = frac0.9x+1$$ That one is easy.
$$sum_n=0^frac0.9xnx = 0.45+frac0.405x$$
But that one is tricky. He told me it was some Gauss sum. If anyone can comment or precise what and how the gaussian sum is resolved.
So finally,
$$(frac0.9x+1)-(0.45+frac0.405x)=45$$
$$x = 0.01136...$$
answered Aug 6 at 11:53
PyThagoras
215
You seem to have some digits wrong, as solving that approximating equation gives the approximate solution $0.495over 44.45=0.011136ldots$. As it turns out, this agrees with the exact solution when rounded to five digits after the decimal point! (See my answer for the exact solution.)
– r.e.s.
Aug 8 at 2:47
add a comment |Â
up vote
-1
down vote
up vote
-1
down vote
I asked a mathematician friend and that's what he told me:
$$sum_n=0^frac0.9x(1-nx) = sum_n=0^frac0.9x1 -sum_n=0^frac0.9xnx = 45$$
So he splitted it.
$$sum_n=0^frac0.9x1 = frac0.9x+1$$ That one is easy.
$$sum_n=0^frac0.9xnx = 0.45+frac0.405x$$
But that one is tricky. He told me it was some Gauss sum. If anyone can comment or precise what and how the gaussian sum is resolved.
So finally,
$$(frac0.9x+1)-(0.45+frac0.405x)=45$$
$$x = 0.01136...$$
answered Aug 6 at 11:53
PyThagoras
215
I asked a mathematician friend and that's what he told me:
$$sum_n=0^frac0.9x(1-nx) = sum_n=0^frac0.9x1 -sum_n=0^frac0.9xnx = 45$$
So he splitted it.
$$sum_n=0^frac0.9x1 = frac0.9x+1$$ That one is easy.
$$sum_n=0^frac0.9xnx = 0.45+frac0.405x$$
But that one is tricky. He told me it was some Gauss sum. If anyone can comment or precise what and how the gaussian sum is resolved.
So finally,
$$(frac0.9x+1)-(0.45+frac0.405x)=45$$
$$x = 0.01136...$$
answered Aug 6 at 11:53
PyThagoras
215
edited Aug 8 at 22:25
answered Aug 6 at 11:53
PyThagoras
215
answered Aug 6 at 11:53
PyThagoras
215
answered Aug 6 at 11:53
PyThagoras
215
215
You seem to have some digits wrong, as solving that approximating equation gives the approximate solution $0.495over 44.45=0.011136ldots$. As it turns out, this agrees with the exact solution when rounded to five digits after the decimal point! (See my answer for the exact solution.)
– r.e.s.
Aug 8 at 2:47
add a comment |Â
You seem to have some digits wrong, as solving that approximating equation gives the approximate solution $0.495over 44.45=0.011136ldots$. As it turns out, this agrees with the exact solution when rounded to five digits after the decimal point! (See my answer for the exact solution.)
– r.e.s.
Aug 8 at 2:47
You seem to have some digits wrong, as solving that approximating equation gives the approximate solution $0.495over 44.45=0.011136ldots$. As it turns out, this agrees with the exact solution when rounded to five digits after the decimal point! (See my answer for the exact solution.)
– r.e.s.
Aug 8 at 2:47
You seem to have some digits wrong, as solving that approximating equation gives the approximate solution $0.495over 44.45=0.011136ldots$. As it turns out, this agrees with the exact solution when rounded to five digits after the decimal point! (See my answer for the exact solution.)
– r.e.s.
Aug 8 at 2:47
add a comment |Â
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Sum bounds should be natural numbers
– Davide Morgante
Aug 6 at 11:04
Yes, but I want to find the closest interger. In fact, I just want x. so if we say x equals 0.005, it'll give me a integer, or close to. But x is what i'm searching for here.
– PyThagoras
Aug 6 at 11:13
Oh ok, that makes more sense! Just another clarification: the $x$ variable in the summation bound is the same x as the one in the argument of the sum?
– Davide Morgante
Aug 6 at 11:14
Yes exactly, That's why I showed the two equations I used to make the sum!
– PyThagoras
Aug 6 at 11:15
Well, in that case the equation makes some sense but I think that it'll be very difficult to find the solution. Let me think about it for a moment
– Davide Morgante
Aug 6 at 11:16