Mixing
Use the concavity of $log (1-x)+x$ to show that $log (1-x)leq -x$
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I've been reading this post and its accepted answer here. The OP (accepted answer) made a comment that $log(1-x)leq -x$ but I've been having issues proving it.
FULL PROOF (EDIT)
With credits to Kavi Rama Murthy, I provide a full proof.
Let $f: ]0,1[longrightarrow Bbb{R},;x mapsto log (1-x)+x$. Then,
begin{align}f''(x) = -frac{1}{(x-1)^2}leq 0,;forall ;xin ;]0,1[.end{align}
Hence, $f$ is monotone non-increasing and $f(x)leq f(0),forall ;xin ;]0,1[,$ that is
begin{align}log(1-x)leq - x end{align}
real-analysis analysis logarithms
asked Jan 3 at 8:51
Omojola MichealOmojola Micheal
1,802324
$endgroup$
add a comment |
$begingroup$
I've been reading this post and its accepted answer here. The OP (accepted answer) made a comment that $log(1-x)leq -x$ but I've been having issues proving it.
FULL PROOF (EDIT)
With credits to Kavi Rama Murthy, I provide a full proof.
Let $f: ]0,1[longrightarrow Bbb{R},;x mapsto log (1-x)+x$. Then,
begin{align}f''(x) = -frac{1}{(x-1)^2}leq 0,;forall ;xin ;]0,1[.end{align}
Hence, $f$ is monotone non-increasing and $f(x)leq f(0),forall ;xin ;]0,1[,$ that is
begin{align}log(1-x)leq - x end{align}
real-analysis analysis logarithms
asked Jan 3 at 8:51
Omojola MichealOmojola Micheal
1,802324
$endgroup$
3
$begingroup$
$log(1-x)$ is not defined for $xgeq 1$
$endgroup$
– Kavi Rama Murthy
Jan 3 at 8:53
$begingroup$
To show that $f(x)leq f(0),forall ;xin ]0,1[,$ i.e. $log(1-x)+xleq 0$
$endgroup$
– Omojola Micheal
Jan 3 at 9:13
add a comment |
$begingroup$
I've been reading this post and its accepted answer here. The OP (accepted answer) made a comment that $log(1-x)leq -x$ but I've been having issues proving it.
FULL PROOF (EDIT)
With credits to Kavi Rama Murthy, I provide a full proof.
Let $f: ]0,1[longrightarrow Bbb{R},;x mapsto log (1-x)+x$. Then,
begin{align}f''(x) = -frac{1}{(x-1)^2}leq 0,;forall ;xin ;]0,1[.end{align}
Hence, $f$ is monotone non-increasing and $f(x)leq f(0),forall ;xin ;]0,1[,$ that is
begin{align}log(1-x)leq - x end{align}
real-analysis analysis logarithms
asked Jan 3 at 8:51
Omojola MichealOmojola Micheal
1,802324
$endgroup$
I've been reading this post and its accepted answer here. The OP (accepted answer) made a comment that $log(1-x)leq -x$ but I've been having issues proving it.
FULL PROOF (EDIT)
With credits to Kavi Rama Murthy, I provide a full proof.
Let $f: ]0,1[longrightarrow Bbb{R},;x mapsto log (1-x)+x$. Then,
begin{align}f''(x) = -frac{1}{(x-1)^2}leq 0,;forall ;xin ;]0,1[.end{align}
Hence, $f$ is monotone non-increasing and $f(x)leq f(0),forall ;xin ;]0,1[,$ that is
begin{align}log(1-x)leq - x end{align}
real-analysis analysis logarithms
real-analysis analysis logarithms
asked Jan 3 at 8:51
Omojola MichealOmojola Micheal
1,802324
asked Jan 3 at 8:51
Omojola MichealOmojola Micheal
1,802324
edited Jan 3 at 9:25
Omojola Micheal
asked Jan 3 at 8:51
Omojola MichealOmojola Micheal
1,802324
asked Jan 3 at 8:51
Omojola MichealOmojola Micheal
1,802324
asked Jan 3 at 8:51
Omojola MichealOmojola Micheal
1,802324
1,802324
3
$begingroup$
$log(1-x)$ is not defined for $xgeq 1$
$endgroup$
– Kavi Rama Murthy
Jan 3 at 8:53
$begingroup$
To show that $f(x)leq f(0),forall ;xin ]0,1[,$ i.e. $log(1-x)+xleq 0$
$endgroup$
– Omojola Micheal
Jan 3 at 9:13
add a comment |
3
$begingroup$
$log(1-x)$ is not defined for $xgeq 1$
$endgroup$
– Kavi Rama Murthy
Jan 3 at 8:53
$begingroup$
To show that $f(x)leq f(0),forall ;xin ]0,1[,$ i.e. $log(1-x)+xleq 0$
$endgroup$
– Omojola Micheal
Jan 3 at 9:13
3
3
$begingroup$
$log(1-x)$ is not defined for $xgeq 1$
$endgroup$
– Kavi Rama Murthy
Jan 3 at 8:53
$begingroup$
$log(1-x)$ is not defined for $xgeq 1$
$endgroup$
– Kavi Rama Murthy
Jan 3 at 8:53
$begingroup$
To show that $f(x)leq f(0),forall ;xin ]0,1[,$ i.e. $log(1-x)+xleq 0$
$endgroup$
– Omojola Micheal
Jan 3 at 9:13
$begingroup$
To show that $f(x)leq f(0),forall ;xin ]0,1[,$ i.e. $log(1-x)+xleq 0$
$endgroup$
– Omojola Micheal
Jan 3 at 9:13
add a comment |
2 Answers
2
active
oldest
votes
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The function is concave on $(0,1)$ because $f''(x)=-frac 1 {(x-1)^{2}} leq 0$. It is not defined for $x geq 1$.
answered Jan 3 at 8:56
Kavi Rama MurthyKavi Rama Murthy
53.8k32055
$endgroup$
$begingroup$
(+1) But sorry for the stress!
$endgroup$
– Omojola Micheal
Jan 3 at 9:07
add a comment |
$begingroup$
Another way, is to consider the Maclaurin series expansion
begin{align} log(1-x)=-sum^{infty}_{n=1}dfrac{x^n}{n}=-x-dfrac{x^2}{2}-dfrac{x^3}{3}cdots leq -x,;text{for fixed};xin;]0,1[end{align}
answered Jan 3 at 10:29
Omojola MichealOmojola Micheal
1,802324
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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votes
active
oldest
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$begingroup$
The function is concave on $(0,1)$ because $f''(x)=-frac 1 {(x-1)^{2}} leq 0$. It is not defined for $x geq 1$.
answered Jan 3 at 8:56
Kavi Rama MurthyKavi Rama Murthy
53.8k32055
$endgroup$
$begingroup$
(+1) But sorry for the stress!
$endgroup$
– Omojola Micheal
Jan 3 at 9:07
add a comment |
$begingroup$
The function is concave on $(0,1)$ because $f''(x)=-frac 1 {(x-1)^{2}} leq 0$. It is not defined for $x geq 1$.
answered Jan 3 at 8:56
Kavi Rama MurthyKavi Rama Murthy
53.8k32055
$endgroup$
$begingroup$
(+1) But sorry for the stress!
$endgroup$
– Omojola Micheal
Jan 3 at 9:07
add a comment |
$begingroup$
The function is concave on $(0,1)$ because $f''(x)=-frac 1 {(x-1)^{2}} leq 0$. It is not defined for $x geq 1$.
answered Jan 3 at 8:56
Kavi Rama MurthyKavi Rama Murthy
53.8k32055
$endgroup$
The function is concave on $(0,1)$ because $f''(x)=-frac 1 {(x-1)^{2}} leq 0$. It is not defined for $x geq 1$.
answered Jan 3 at 8:56
Kavi Rama MurthyKavi Rama Murthy
53.8k32055
edited Jan 3 at 9:08
answered Jan 3 at 8:56
Kavi Rama MurthyKavi Rama Murthy
53.8k32055
answered Jan 3 at 8:56
Kavi Rama MurthyKavi Rama Murthy
53.8k32055
answered Jan 3 at 8:56
Kavi Rama MurthyKavi Rama Murthy
53.8k32055
53.8k32055
$begingroup$
(+1) But sorry for the stress!
$endgroup$
– Omojola Micheal
Jan 3 at 9:07
add a comment |
$begingroup$
(+1) But sorry for the stress!
$endgroup$
– Omojola Micheal
Jan 3 at 9:07
$begingroup$
(+1) But sorry for the stress!
$endgroup$
– Omojola Micheal
Jan 3 at 9:07
$begingroup$
(+1) But sorry for the stress!
$endgroup$
– Omojola Micheal
Jan 3 at 9:07
add a comment |
$begingroup$
Another way, is to consider the Maclaurin series expansion
begin{align} log(1-x)=-sum^{infty}_{n=1}dfrac{x^n}{n}=-x-dfrac{x^2}{2}-dfrac{x^3}{3}cdots leq -x,;text{for fixed};xin;]0,1[end{align}
answered Jan 3 at 10:29
Omojola MichealOmojola Micheal
1,802324
$endgroup$
add a comment |
$begingroup$
Another way, is to consider the Maclaurin series expansion
begin{align} log(1-x)=-sum^{infty}_{n=1}dfrac{x^n}{n}=-x-dfrac{x^2}{2}-dfrac{x^3}{3}cdots leq -x,;text{for fixed};xin;]0,1[end{align}
answered Jan 3 at 10:29
Omojola MichealOmojola Micheal
1,802324
$endgroup$
add a comment |
$begingroup$
Another way, is to consider the Maclaurin series expansion
begin{align} log(1-x)=-sum^{infty}_{n=1}dfrac{x^n}{n}=-x-dfrac{x^2}{2}-dfrac{x^3}{3}cdots leq -x,;text{for fixed};xin;]0,1[end{align}
answered Jan 3 at 10:29
Omojola MichealOmojola Micheal
1,802324
$endgroup$
Another way, is to consider the Maclaurin series expansion
begin{align} log(1-x)=-sum^{infty}_{n=1}dfrac{x^n}{n}=-x-dfrac{x^2}{2}-dfrac{x^3}{3}cdots leq -x,;text{for fixed};xin;]0,1[end{align}
answered Jan 3 at 10:29
Omojola MichealOmojola Micheal
1,802324
answered Jan 3 at 10:29
Omojola MichealOmojola Micheal
1,802324
answered Jan 3 at 10:29
Omojola MichealOmojola Micheal
1,802324
answered Jan 3 at 10:29
Omojola MichealOmojola Micheal
1,802324
1,802324
add a comment |
add a comment |
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3
$begingroup$
$log(1-x)$ is not defined for $xgeq 1$
$endgroup$
– Kavi Rama Murthy
Jan 3 at 8:53
$begingroup$
To show that $f(x)leq f(0),forall ;xin ]0,1[,$ i.e. $log(1-x)+xleq 0$
$endgroup$
– Omojola Micheal
Jan 3 at 9:13