Mixing
I have an inequality which is giving me the correct answer, except that the sign is opposite to what it...
$begingroup$
I'm looking at a geometric series that looks like
$A_n = sum_{k=0}^{n}A_0x^k$ where $A_0$ is a constant.
I'm looking to find an $n$ where my error between two terms is less than a certain value $E$ such that
$A_n - A_{n-1} le E$.
Now,
$A_n - A_{n-1} = A_0x^n$
Therefore I need
$A_0x^n le E$,
which I can rewrite as
$ln{A_0} +nln{x} le ln{E}$
$n le frac{ln{E/A_0}}{ln{x}}$
Which gives me what I need (i.e. the correct $n$ for the error I expect), except that it should be that $n$ is greater than this and not less than. Can anyone see what I've missed? Perhaps more caffeine is needed?
Many thanks.
sequences-and-series inequality logarithms geometric-series
asked Jan 29 at 9:19
DreddDredd
62
$endgroup$
|
show 1 more comment
$begingroup$
I'm looking at a geometric series that looks like
$A_n = sum_{k=0}^{n}A_0x^k$ where $A_0$ is a constant.
I'm looking to find an $n$ where my error between two terms is less than a certain value $E$ such that
$A_n - A_{n-1} le E$.
Now,
$A_n - A_{n-1} = A_0x^n$
Therefore I need
$A_0x^n le E$,
which I can rewrite as
$ln{A_0} +nln{x} le ln{E}$
$n le frac{ln{E/A_0}}{ln{x}}$
Which gives me what I need (i.e. the correct $n$ for the error I expect), except that it should be that $n$ is greater than this and not less than. Can anyone see what I've missed? Perhaps more caffeine is needed?
Many thanks.
sequences-and-series inequality logarithms geometric-series
asked Jan 29 at 9:19
DreddDredd
62
$endgroup$
1
$begingroup$
$ln{x}$ is negative, isn’t it?
$endgroup$
– Mindlack
Jan 29 at 9:28
$begingroup$
I think you mistyped the first formula, and it should read $A_n = sum_{k = 0}^n A_0x^k$ is that correct?
$endgroup$
– Vincent
Jan 29 at 9:28
$begingroup$
Thanks, I've just looked at it and $x le 1$, writing the series by hand I can see that greater $n$ is giving me lower error, I just can't get the inequality correct.
$endgroup$
– Dredd
Jan 29 at 9:29
$begingroup$
@Vincent You're correct sorry
$endgroup$
– Dredd
Jan 29 at 9:30
$begingroup$
Ok, good, then please edit the question!
$endgroup$
– Vincent
Jan 29 at 9:32
|
show 1 more comment
$begingroup$
I'm looking at a geometric series that looks like
$A_n = sum_{k=0}^{n}A_0x^k$ where $A_0$ is a constant.
I'm looking to find an $n$ where my error between two terms is less than a certain value $E$ such that
$A_n - A_{n-1} le E$.
Now,
$A_n - A_{n-1} = A_0x^n$
Therefore I need
$A_0x^n le E$,
which I can rewrite as
$ln{A_0} +nln{x} le ln{E}$
$n le frac{ln{E/A_0}}{ln{x}}$
Which gives me what I need (i.e. the correct $n$ for the error I expect), except that it should be that $n$ is greater than this and not less than. Can anyone see what I've missed? Perhaps more caffeine is needed?
Many thanks.
sequences-and-series inequality logarithms geometric-series
asked Jan 29 at 9:19
DreddDredd
62
$endgroup$
I'm looking at a geometric series that looks like
$A_n = sum_{k=0}^{n}A_0x^k$ where $A_0$ is a constant.
I'm looking to find an $n$ where my error between two terms is less than a certain value $E$ such that
$A_n - A_{n-1} le E$.
Now,
$A_n - A_{n-1} = A_0x^n$
Therefore I need
$A_0x^n le E$,
which I can rewrite as
$ln{A_0} +nln{x} le ln{E}$
$n le frac{ln{E/A_0}}{ln{x}}$
Which gives me what I need (i.e. the correct $n$ for the error I expect), except that it should be that $n$ is greater than this and not less than. Can anyone see what I've missed? Perhaps more caffeine is needed?
Many thanks.
sequences-and-series inequality logarithms geometric-series
sequences-and-series inequality logarithms geometric-series
asked Jan 29 at 9:19
DreddDredd
62
asked Jan 29 at 9:19
DreddDredd
62
edited Jan 30 at 10:47
Dredd
asked Jan 29 at 9:19
DreddDredd
62
asked Jan 29 at 9:19
DreddDredd
62
asked Jan 29 at 9:19
DreddDredd
62
62
1
$begingroup$
$ln{x}$ is negative, isn’t it?
$endgroup$
– Mindlack
Jan 29 at 9:28
$begingroup$
I think you mistyped the first formula, and it should read $A_n = sum_{k = 0}^n A_0x^k$ is that correct?
$endgroup$
– Vincent
Jan 29 at 9:28
$begingroup$
Thanks, I've just looked at it and $x le 1$, writing the series by hand I can see that greater $n$ is giving me lower error, I just can't get the inequality correct.
$endgroup$
– Dredd
Jan 29 at 9:29
$begingroup$
@Vincent You're correct sorry
$endgroup$
– Dredd
Jan 29 at 9:30
$begingroup$
Ok, good, then please edit the question!
$endgroup$
– Vincent
Jan 29 at 9:32
|
show 1 more comment
1
$begingroup$
$ln{x}$ is negative, isn’t it?
$endgroup$
– Mindlack
Jan 29 at 9:28
$begingroup$
I think you mistyped the first formula, and it should read $A_n = sum_{k = 0}^n A_0x^k$ is that correct?
$endgroup$
– Vincent
Jan 29 at 9:28
$begingroup$
Thanks, I've just looked at it and $x le 1$, writing the series by hand I can see that greater $n$ is giving me lower error, I just can't get the inequality correct.
$endgroup$
– Dredd
Jan 29 at 9:29
$begingroup$
@Vincent You're correct sorry
$endgroup$
– Dredd
Jan 29 at 9:30
$begingroup$
Ok, good, then please edit the question!
$endgroup$
– Vincent
Jan 29 at 9:32
1
1
$begingroup$
$ln{x}$ is negative, isn’t it?
$endgroup$
– Mindlack
Jan 29 at 9:28
$begingroup$
$ln{x}$ is negative, isn’t it?
$endgroup$
– Mindlack
Jan 29 at 9:28
$begingroup$
I think you mistyped the first formula, and it should read $A_n = sum_{k = 0}^n A_0x^k$ is that correct?
$endgroup$
– Vincent
Jan 29 at 9:28
$begingroup$
I think you mistyped the first formula, and it should read $A_n = sum_{k = 0}^n A_0x^k$ is that correct?
$endgroup$
– Vincent
Jan 29 at 9:28
$begingroup$
Thanks, I've just looked at it and $x le 1$, writing the series by hand I can see that greater $n$ is giving me lower error, I just can't get the inequality correct.
$endgroup$
– Dredd
Jan 29 at 9:29
$begingroup$
Thanks, I've just looked at it and $x le 1$, writing the series by hand I can see that greater $n$ is giving me lower error, I just can't get the inequality correct.
$endgroup$
– Dredd
Jan 29 at 9:29
$begingroup$
@Vincent You're correct sorry
$endgroup$
– Dredd
Jan 29 at 9:30
$begingroup$
@Vincent You're correct sorry
$endgroup$
– Dredd
Jan 29 at 9:30
$begingroup$
Ok, good, then please edit the question!
$endgroup$
– Vincent
Jan 29 at 9:32
$begingroup$
Ok, good, then please edit the question!
$endgroup$
– Vincent
Jan 29 at 9:32
|
show 1 more comment
2 Answers
2
active
oldest
votes
$begingroup$
The sign is correct, with $x > 1$ and $A_0 > 1$, the error term $A_0x^n$ grows with $n$. Therefore if you want the error term to be lower than a constant, you need $n$ to be small
Now if you take $x<1$, I guess your interest, than the error term decreases with $n$, and $ln(x)<0$, therefore when dividing by $ln(x)$ you have to switch sign.
$$ ln A_0 + nln x leq ln E Rightarrow
left{
begin{array}{ll}
n leq frac{ln E/A_0}{ln x} & text{ if } x>1\
n geq frac{ln E/A_0}{ln x} & text{ if } x<1\
end{array}
right.
$$
answered Jan 29 at 9:30
Thomas LesgourguesThomas Lesgourgues
1,285220
$endgroup$
$begingroup$
Perfect! Thank you Thomas
$endgroup$
– Dredd
Jan 29 at 9:37
add a comment |
$begingroup$
$ln(x)$ is negative as $x < 1$. So when you divide by $ln(x)$ you have to flip the sign on the inequality.
answered Jan 29 at 9:30
Erik ParkinsonErik Parkinson
1,17519
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The sign is correct, with $x > 1$ and $A_0 > 1$, the error term $A_0x^n$ grows with $n$. Therefore if you want the error term to be lower than a constant, you need $n$ to be small
Now if you take $x<1$, I guess your interest, than the error term decreases with $n$, and $ln(x)<0$, therefore when dividing by $ln(x)$ you have to switch sign.
$$ ln A_0 + nln x leq ln E Rightarrow
left{
begin{array}{ll}
n leq frac{ln E/A_0}{ln x} & text{ if } x>1\
n geq frac{ln E/A_0}{ln x} & text{ if } x<1\
end{array}
right.
$$
answered Jan 29 at 9:30
Thomas LesgourguesThomas Lesgourgues
1,285220
$endgroup$
$begingroup$
Perfect! Thank you Thomas
$endgroup$
– Dredd
Jan 29 at 9:37
add a comment |
$begingroup$
The sign is correct, with $x > 1$ and $A_0 > 1$, the error term $A_0x^n$ grows with $n$. Therefore if you want the error term to be lower than a constant, you need $n$ to be small
Now if you take $x<1$, I guess your interest, than the error term decreases with $n$, and $ln(x)<0$, therefore when dividing by $ln(x)$ you have to switch sign.
$$ ln A_0 + nln x leq ln E Rightarrow
left{
begin{array}{ll}
n leq frac{ln E/A_0}{ln x} & text{ if } x>1\
n geq frac{ln E/A_0}{ln x} & text{ if } x<1\
end{array}
right.
$$
answered Jan 29 at 9:30
Thomas LesgourguesThomas Lesgourgues
1,285220
$endgroup$
$begingroup$
Perfect! Thank you Thomas
$endgroup$
– Dredd
Jan 29 at 9:37
add a comment |
$begingroup$
The sign is correct, with $x > 1$ and $A_0 > 1$, the error term $A_0x^n$ grows with $n$. Therefore if you want the error term to be lower than a constant, you need $n$ to be small
Now if you take $x<1$, I guess your interest, than the error term decreases with $n$, and $ln(x)<0$, therefore when dividing by $ln(x)$ you have to switch sign.
$$ ln A_0 + nln x leq ln E Rightarrow
left{
begin{array}{ll}
n leq frac{ln E/A_0}{ln x} & text{ if } x>1\
n geq frac{ln E/A_0}{ln x} & text{ if } x<1\
end{array}
right.
$$
answered Jan 29 at 9:30
Thomas LesgourguesThomas Lesgourgues
1,285220
$endgroup$
The sign is correct, with $x > 1$ and $A_0 > 1$, the error term $A_0x^n$ grows with $n$. Therefore if you want the error term to be lower than a constant, you need $n$ to be small
Now if you take $x<1$, I guess your interest, than the error term decreases with $n$, and $ln(x)<0$, therefore when dividing by $ln(x)$ you have to switch sign.
$$ ln A_0 + nln x leq ln E Rightarrow
left{
begin{array}{ll}
n leq frac{ln E/A_0}{ln x} & text{ if } x>1\
n geq frac{ln E/A_0}{ln x} & text{ if } x<1\
end{array}
right.
$$
answered Jan 29 at 9:30
Thomas LesgourguesThomas Lesgourgues
1,285220
edited Jan 29 at 9:39
answered Jan 29 at 9:30
Thomas LesgourguesThomas Lesgourgues
1,285220
answered Jan 29 at 9:30
Thomas LesgourguesThomas Lesgourgues
1,285220
answered Jan 29 at 9:30
Thomas LesgourguesThomas Lesgourgues
1,285220
1,285220
$begingroup$
Perfect! Thank you Thomas
$endgroup$
– Dredd
Jan 29 at 9:37
add a comment |
$begingroup$
Perfect! Thank you Thomas
$endgroup$
– Dredd
Jan 29 at 9:37
$begingroup$
Perfect! Thank you Thomas
$endgroup$
– Dredd
Jan 29 at 9:37
$begingroup$
Perfect! Thank you Thomas
$endgroup$
– Dredd
Jan 29 at 9:37
add a comment |
$begingroup$
$ln(x)$ is negative as $x < 1$. So when you divide by $ln(x)$ you have to flip the sign on the inequality.
answered Jan 29 at 9:30
Erik ParkinsonErik Parkinson
1,17519
$endgroup$
add a comment |
$begingroup$
$ln(x)$ is negative as $x < 1$. So when you divide by $ln(x)$ you have to flip the sign on the inequality.
answered Jan 29 at 9:30
Erik ParkinsonErik Parkinson
1,17519
$endgroup$
add a comment |
$begingroup$
$ln(x)$ is negative as $x < 1$. So when you divide by $ln(x)$ you have to flip the sign on the inequality.
answered Jan 29 at 9:30
Erik ParkinsonErik Parkinson
1,17519
$endgroup$
$ln(x)$ is negative as $x < 1$. So when you divide by $ln(x)$ you have to flip the sign on the inequality.
answered Jan 29 at 9:30
Erik ParkinsonErik Parkinson
1,17519
answered Jan 29 at 9:30
Erik ParkinsonErik Parkinson
1,17519
answered Jan 29 at 9:30
Erik ParkinsonErik Parkinson
1,17519
answered Jan 29 at 9:30
Erik ParkinsonErik Parkinson
1,17519
1,17519
add a comment |
add a comment |
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1
$begingroup$
$ln{x}$ is negative, isn’t it?
$endgroup$
– Mindlack
Jan 29 at 9:28
$begingroup$
I think you mistyped the first formula, and it should read $A_n = sum_{k = 0}^n A_0x^k$ is that correct?
$endgroup$
– Vincent
Jan 29 at 9:28
$begingroup$
Thanks, I've just looked at it and $x le 1$, writing the series by hand I can see that greater $n$ is giving me lower error, I just can't get the inequality correct.
$endgroup$
– Dredd
Jan 29 at 9:29
$begingroup$
@Vincent You're correct sorry
$endgroup$
– Dredd
Jan 29 at 9:30
$begingroup$
Ok, good, then please edit the question!
$endgroup$
– Vincent
Jan 29 at 9:32