Mixing
How to find the solutions of the following equation?
$begingroup$
Let we have the following equation with the unknown $x$
$$lfloorln(x+1)rfloor - lfloorln(x)rfloor = 1$$
Where $lfloor xrfloor$ means the integer part of $x$
real-analysis number-theory discrete-mathematics
edited Jan 1 at 10:26
Shubham Johri
4,666717
asked Jan 1 at 10:04
Neil hawkingNeil hawking
48619
$endgroup$
add a comment |
$begingroup$
Let we have the following equation with the unknown $x$
$$lfloorln(x+1)rfloor - lfloorln(x)rfloor = 1$$
Where $lfloor xrfloor$ means the integer part of $x$
real-analysis number-theory discrete-mathematics
edited Jan 1 at 10:26
Shubham Johri
4,666717
asked Jan 1 at 10:04
Neil hawkingNeil hawking
48619
$endgroup$
$begingroup$
Use formula for difference of logs.
$endgroup$
– coffeemath
Jan 1 at 10:09
1
$begingroup$
Why did you enclose in square parentheses [ ] each one of those expressions? Why not merely write $;ln(x+1) - ln x;$ ? Does this mean anything special? And after this, what properties of logarithms you know?
$endgroup$
– DonAntonio
Jan 1 at 10:09
$begingroup$
It is [x] integer part of x, I think
$endgroup$
– greedoid
Jan 1 at 10:11
2
$begingroup$
Nice. The OP hasn't even addressed the comment, but someone already decided that he can read minds and edited the question...
$endgroup$
– DonAntonio
Jan 1 at 11:22
$begingroup$
but there is two answers already.
$endgroup$
– onepound
Jan 1 at 13:55
add a comment |
$begingroup$
Let we have the following equation with the unknown $x$
$$lfloorln(x+1)rfloor - lfloorln(x)rfloor = 1$$
Where $lfloor xrfloor$ means the integer part of $x$
real-analysis number-theory discrete-mathematics
edited Jan 1 at 10:26
Shubham Johri
4,666717
asked Jan 1 at 10:04
Neil hawkingNeil hawking
48619
$endgroup$
Let we have the following equation with the unknown $x$
$$lfloorln(x+1)rfloor - lfloorln(x)rfloor = 1$$
Where $lfloor xrfloor$ means the integer part of $x$
real-analysis number-theory discrete-mathematics
real-analysis number-theory discrete-mathematics
edited Jan 1 at 10:26
Shubham Johri
4,666717
asked Jan 1 at 10:04
Neil hawkingNeil hawking
48619
edited Jan 1 at 10:26
Shubham Johri
4,666717
asked Jan 1 at 10:04
Neil hawkingNeil hawking
48619
edited Jan 1 at 10:26
Shubham Johri
4,666717
edited Jan 1 at 10:26
Shubham Johri
4,666717
edited Jan 1 at 10:26
Shubham Johri
4,666717
4,666717
asked Jan 1 at 10:04
Neil hawkingNeil hawking
48619
asked Jan 1 at 10:04
Neil hawkingNeil hawking
48619
asked Jan 1 at 10:04
Neil hawkingNeil hawking
48619
48619
$begingroup$
Use formula for difference of logs.
$endgroup$
– coffeemath
Jan 1 at 10:09
1
$begingroup$
Why did you enclose in square parentheses [ ] each one of those expressions? Why not merely write $;ln(x+1) - ln x;$ ? Does this mean anything special? And after this, what properties of logarithms you know?
$endgroup$
– DonAntonio
Jan 1 at 10:09
$begingroup$
It is [x] integer part of x, I think
$endgroup$
– greedoid
Jan 1 at 10:11
2
$begingroup$
Nice. The OP hasn't even addressed the comment, but someone already decided that he can read minds and edited the question...
$endgroup$
– DonAntonio
Jan 1 at 11:22
$begingroup$
but there is two answers already.
$endgroup$
– onepound
Jan 1 at 13:55
add a comment |
$begingroup$
Use formula for difference of logs.
$endgroup$
– coffeemath
Jan 1 at 10:09
1
$begingroup$
Why did you enclose in square parentheses [ ] each one of those expressions? Why not merely write $;ln(x+1) - ln x;$ ? Does this mean anything special? And after this, what properties of logarithms you know?
$endgroup$
– DonAntonio
Jan 1 at 10:09
$begingroup$
It is [x] integer part of x, I think
$endgroup$
– greedoid
Jan 1 at 10:11
2
$begingroup$
Nice. The OP hasn't even addressed the comment, but someone already decided that he can read minds and edited the question...
$endgroup$
– DonAntonio
Jan 1 at 11:22
$begingroup$
but there is two answers already.
$endgroup$
– onepound
Jan 1 at 13:55
$begingroup$
Use formula for difference of logs.
$endgroup$
– coffeemath
Jan 1 at 10:09
$begingroup$
Use formula for difference of logs.
$endgroup$
– coffeemath
Jan 1 at 10:09
1
1
$begingroup$
Why did you enclose in square parentheses [ ] each one of those expressions? Why not merely write $;ln(x+1) - ln x;$ ? Does this mean anything special? And after this, what properties of logarithms you know?
$endgroup$
– DonAntonio
Jan 1 at 10:09
$begingroup$
Why did you enclose in square parentheses [ ] each one of those expressions? Why not merely write $;ln(x+1) - ln x;$ ? Does this mean anything special? And after this, what properties of logarithms you know?
$endgroup$
– DonAntonio
Jan 1 at 10:09
$begingroup$
It is [x] integer part of x, I think
$endgroup$
– greedoid
Jan 1 at 10:11
$begingroup$
It is [x] integer part of x, I think
$endgroup$
– greedoid
Jan 1 at 10:11
2
2
$begingroup$
Nice. The OP hasn't even addressed the comment, but someone already decided that he can read minds and edited the question...
$endgroup$
– DonAntonio
Jan 1 at 11:22
$begingroup$
Nice. The OP hasn't even addressed the comment, but someone already decided that he can read minds and edited the question...
$endgroup$
– DonAntonio
Jan 1 at 11:22
$begingroup$
but there is two answers already.
$endgroup$
– onepound
Jan 1 at 13:55
$begingroup$
but there is two answers already.
$endgroup$
– onepound
Jan 1 at 13:55
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The solution set is a sequence of intervals. For each positive integer $N$, we can construct an interval containing $e^N$. Let
$$epsilon = lnleft(frac{e^N}{e^N+1}right).$$
Then the interval from $e^{N-epsilon}$ to $e^N$ is a set of solutions.
You need
$$ln(e^{N-epsilon}+1)geq N$$
and solving this inequality gives the value for $epsilon.$
answered Jan 1 at 11:12
B. GoddardB. Goddard
18.4k21340
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The solution set is a sequence of intervals. For each positive integer $N$, we can construct an interval containing $e^N$. Let
$$epsilon = lnleft(frac{e^N}{e^N+1}right).$$
Then the interval from $e^{N-epsilon}$ to $e^N$ is a set of solutions.
You need
$$ln(e^{N-epsilon}+1)geq N$$
and solving this inequality gives the value for $epsilon.$
answered Jan 1 at 11:12
B. GoddardB. Goddard
18.4k21340
$endgroup$
add a comment |
$begingroup$
The solution set is a sequence of intervals. For each positive integer $N$, we can construct an interval containing $e^N$. Let
$$epsilon = lnleft(frac{e^N}{e^N+1}right).$$
Then the interval from $e^{N-epsilon}$ to $e^N$ is a set of solutions.
You need
$$ln(e^{N-epsilon}+1)geq N$$
and solving this inequality gives the value for $epsilon.$
answered Jan 1 at 11:12
B. GoddardB. Goddard
18.4k21340
$endgroup$
add a comment |
$begingroup$
The solution set is a sequence of intervals. For each positive integer $N$, we can construct an interval containing $e^N$. Let
$$epsilon = lnleft(frac{e^N}{e^N+1}right).$$
Then the interval from $e^{N-epsilon}$ to $e^N$ is a set of solutions.
You need
$$ln(e^{N-epsilon}+1)geq N$$
and solving this inequality gives the value for $epsilon.$
answered Jan 1 at 11:12
B. GoddardB. Goddard
18.4k21340
$endgroup$
The solution set is a sequence of intervals. For each positive integer $N$, we can construct an interval containing $e^N$. Let
$$epsilon = lnleft(frac{e^N}{e^N+1}right).$$
Then the interval from $e^{N-epsilon}$ to $e^N$ is a set of solutions.
You need
$$ln(e^{N-epsilon}+1)geq N$$
and solving this inequality gives the value for $epsilon.$
answered Jan 1 at 11:12
B. GoddardB. Goddard
18.4k21340
answered Jan 1 at 11:12
B. GoddardB. Goddard
18.4k21340
answered Jan 1 at 11:12
B. GoddardB. Goddard
18.4k21340
answered Jan 1 at 11:12
B. GoddardB. Goddard
18.4k21340
18.4k21340
add a comment |
add a comment |
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$begingroup$
Use formula for difference of logs.
$endgroup$
– coffeemath
Jan 1 at 10:09
1
$begingroup$
Why did you enclose in square parentheses [ ] each one of those expressions? Why not merely write $;ln(x+1) - ln x;$ ? Does this mean anything special? And after this, what properties of logarithms you know?
$endgroup$
– DonAntonio
Jan 1 at 10:09
$begingroup$
It is [x] integer part of x, I think
$endgroup$
– greedoid
Jan 1 at 10:11
2
$begingroup$
Nice. The OP hasn't even addressed the comment, but someone already decided that he can read minds and edited the question...
$endgroup$
– DonAntonio
Jan 1 at 11:22
$begingroup$
but there is two answers already.
$endgroup$
– onepound
Jan 1 at 13:55