An inequality for exponentials
$begingroup$
Let $x_1,ldots,x_nin{bf R}$ satisfying $sum_{i=1}^nx_i=0$ and $sum_{i=1}^nx_i^2le1$.
Is it true that for every $rge 1$ one has that $sum_{i=1}^n(r^{x_i}-1)le r-1$?
If so, does anyone have a proof or a reference?
real-analysis inequality
asked Dec 30 '18 at 0:43
Rene SchoofRene Schoof
32115
$endgroup$
add a comment |
$begingroup$
Let $x_1,ldots,x_nin{bf R}$ satisfying $sum_{i=1}^nx_i=0$ and $sum_{i=1}^nx_i^2le1$.
Is it true that for every $rge 1$ one has that $sum_{i=1}^n(r^{x_i}-1)le r-1$?
If so, does anyone have a proof or a reference?
real-analysis inequality
asked Dec 30 '18 at 0:43
Rene SchoofRene Schoof
32115
$endgroup$
add a comment |
$begingroup$
Let $x_1,ldots,x_nin{bf R}$ satisfying $sum_{i=1}^nx_i=0$ and $sum_{i=1}^nx_i^2le1$.
Is it true that for every $rge 1$ one has that $sum_{i=1}^n(r^{x_i}-1)le r-1$?
If so, does anyone have a proof or a reference?
real-analysis inequality
asked Dec 30 '18 at 0:43
Rene SchoofRene Schoof
32115
$endgroup$
Let $x_1,ldots,x_nin{bf R}$ satisfying $sum_{i=1}^nx_i=0$ and $sum_{i=1}^nx_i^2le1$.
Is it true that for every $rge 1$ one has that $sum_{i=1}^n(r^{x_i}-1)le r-1$?
If so, does anyone have a proof or a reference?
real-analysis inequality
real-analysis inequality
asked Dec 30 '18 at 0:43
Rene SchoofRene Schoof
32115
asked Dec 30 '18 at 0:43
Rene SchoofRene Schoof
32115
asked Dec 30 '18 at 0:43
Rene SchoofRene Schoof
32115
asked Dec 30 '18 at 0:43
Rene SchoofRene Schoof
32115
asked Dec 30 '18 at 0:43
Rene SchoofRene Schoof
32115
32115
add a comment |
add a comment |
1 Answer
1
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oldest
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$begingroup$
Yes. Note that for $|x| leq 1$,
$$begin{align}
r^x&=1+xlog r+x^2 sum_{i=2}^{infty}{frac{(log r)^i x^{i-2}}{i!}} \
& leq 1+xlog r+x^2 sum_{i=2}^{infty}{frac{(log r)^i |x|^{i-2}}{i!}} \
& leq 1+xlog r+x^2 sum_{i=2}^{infty}{frac{(log r)^i}{i!}} \
&=1+xlog r+(r-1-log r)x^2 \
end{align}$$
Note clearly each $|x_i| leq 1$. Thus
$$sum_{i=1}^{n}(r^{x_i}-1) leq log r sum_{i=1}^{n}{x_i}+(r-1-log r)sum_{i=1}^{n}{x_i^2} leq r-1-log r leq r-1$$
answered Jan 11 at 8:39
user633720user633720
1
$endgroup$
$begingroup$
Yes! Thank you very much
$endgroup$
– Rene Schoof
Jan 26 at 14:11
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
Yes. Note that for $|x| leq 1$,
$$begin{align}
r^x&=1+xlog r+x^2 sum_{i=2}^{infty}{frac{(log r)^i x^{i-2}}{i!}} \
& leq 1+xlog r+x^2 sum_{i=2}^{infty}{frac{(log r)^i |x|^{i-2}}{i!}} \
& leq 1+xlog r+x^2 sum_{i=2}^{infty}{frac{(log r)^i}{i!}} \
&=1+xlog r+(r-1-log r)x^2 \
end{align}$$
Note clearly each $|x_i| leq 1$. Thus
$$sum_{i=1}^{n}(r^{x_i}-1) leq log r sum_{i=1}^{n}{x_i}+(r-1-log r)sum_{i=1}^{n}{x_i^2} leq r-1-log r leq r-1$$
answered Jan 11 at 8:39
user633720user633720
1
$endgroup$
$begingroup$
Yes! Thank you very much
$endgroup$
– Rene Schoof
Jan 26 at 14:11
add a comment |
$begingroup$
Yes. Note that for $|x| leq 1$,
$$begin{align}
r^x&=1+xlog r+x^2 sum_{i=2}^{infty}{frac{(log r)^i x^{i-2}}{i!}} \
& leq 1+xlog r+x^2 sum_{i=2}^{infty}{frac{(log r)^i |x|^{i-2}}{i!}} \
& leq 1+xlog r+x^2 sum_{i=2}^{infty}{frac{(log r)^i}{i!}} \
&=1+xlog r+(r-1-log r)x^2 \
end{align}$$
Note clearly each $|x_i| leq 1$. Thus
$$sum_{i=1}^{n}(r^{x_i}-1) leq log r sum_{i=1}^{n}{x_i}+(r-1-log r)sum_{i=1}^{n}{x_i^2} leq r-1-log r leq r-1$$
answered Jan 11 at 8:39
user633720user633720
1
$endgroup$
$begingroup$
Yes! Thank you very much
$endgroup$
– Rene Schoof
Jan 26 at 14:11
add a comment |
$begingroup$
Yes. Note that for $|x| leq 1$,
$$begin{align}
r^x&=1+xlog r+x^2 sum_{i=2}^{infty}{frac{(log r)^i x^{i-2}}{i!}} \
& leq 1+xlog r+x^2 sum_{i=2}^{infty}{frac{(log r)^i |x|^{i-2}}{i!}} \
& leq 1+xlog r+x^2 sum_{i=2}^{infty}{frac{(log r)^i}{i!}} \
&=1+xlog r+(r-1-log r)x^2 \
end{align}$$
Note clearly each $|x_i| leq 1$. Thus
$$sum_{i=1}^{n}(r^{x_i}-1) leq log r sum_{i=1}^{n}{x_i}+(r-1-log r)sum_{i=1}^{n}{x_i^2} leq r-1-log r leq r-1$$
answered Jan 11 at 8:39
user633720user633720
1
$endgroup$
Yes. Note that for $|x| leq 1$,
$$begin{align}
r^x&=1+xlog r+x^2 sum_{i=2}^{infty}{frac{(log r)^i x^{i-2}}{i!}} \
& leq 1+xlog r+x^2 sum_{i=2}^{infty}{frac{(log r)^i |x|^{i-2}}{i!}} \
& leq 1+xlog r+x^2 sum_{i=2}^{infty}{frac{(log r)^i}{i!}} \
&=1+xlog r+(r-1-log r)x^2 \
end{align}$$
Note clearly each $|x_i| leq 1$. Thus
$$sum_{i=1}^{n}(r^{x_i}-1) leq log r sum_{i=1}^{n}{x_i}+(r-1-log r)sum_{i=1}^{n}{x_i^2} leq r-1-log r leq r-1$$
answered Jan 11 at 8:39
user633720user633720
1
answered Jan 11 at 8:39
user633720user633720
1
answered Jan 11 at 8:39
user633720user633720
1
answered Jan 11 at 8:39
user633720user633720
1
1
$begingroup$
Yes! Thank you very much
$endgroup$
– Rene Schoof
Jan 26 at 14:11
add a comment |
$begingroup$
Yes! Thank you very much
$endgroup$
– Rene Schoof
Jan 26 at 14:11
$begingroup$
Yes! Thank you very much
$endgroup$
– Rene Schoof
Jan 26 at 14:11
$begingroup$
Yes! Thank you very much
$endgroup$
– Rene Schoof
Jan 26 at 14:11
add a comment |
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