How can we show that $left|frace^-lambda t-1tright|le|lambda|$ for all $tinmathbb Rsetminusleft0right$ and $lambdainmathbb R$? The Next CEO of Stack OverflowProve that $left|frace^-ht-1hright|le t$ for $h>0$Show that for all $lambda geq 1~$ $fraclambda^ne^lambda < fracClambda^2$Proving that $left(1+lambda/nright)^nto e^lambda$Show that $lim_xto inftyleft( 1-fraclambdax right)^x = e^-lambda$Inequality, $left(frac2x+2right)^n-left(frac2x-2right)^nleq left(frac 4 x right)^n$Show that $fleft(lambda right)=lambda $ .How to prove $left(1+fracxn+1right)^n+1 > left(1+fracxnright)^n$ for all $n in mathbb Z^+$ and $x>0$Show that $bigcap _nepsilon mathbbNleft [ 0, 1+frac1n+1 right ] = left [ 0,1 right ]$$1-|z|^2 le sin^2(lambda+arg(z)) + 1 - left[frac(e^lambda-1)right]^2$solving for $-fracxgamma = lnleft( frac12x + lambdaright)$How can we show that $left|a+b+cright|^p-2left|aright|^ple Cleft(left|bright|^p+left|cright|^pright)$?
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Oldie but Goldie
How can we show that $left|frace^-lambda t-1tright|le|lambda|$ for all $tinmathbb Rsetminusleft0right$ and $lambdainmathbb R$?
The Next CEO of Stack OverflowProve that $left|frace^-ht-1hright|le t$ for $h>0$Show that for all $lambda geq 1~$ $fraclambda^ne^lambda < fracClambda^2$Proving that $left(1+lambda/nright)^nto e^lambda$Show that $lim_xto inftyleft( 1-fraclambdax right)^x = e^-lambda$Inequality, $left(frac2x+2right)^n-left(frac2x-2right)^nleq left(frac 4 x right)^n$Show that $fleft(lambda right)=lambda $ .How to prove $left(1+fracxn+1right)^n+1 > left(1+fracxnright)^n$ for all $n in mathbb Z^+$ and $x>0$Show that $bigcap _nepsilon mathbbNleft [ 0, 1+frac1n+1 right ] = left [ 0,1 right ]$$1-|z|^2 le sin^2(lambda+arg(z)) + 1 - left[fraczzright]^2$solving for $-fracxgamma = lnleft( frac12x + lambdaright)$How can we show that $left|a+b+cright|^p-2left|aright|^ple Cleft(left|bright|^p+left|cright|^pright)$?
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How can we show that $$left|frace^-lambda t-1tright|le|lambda|$$ for all $tinmathbb Rsetminusleft0right$ and $lambdainmathbb R$?
Clearly, $e^-lambda tin[0,1]$ for all $t,lambdage0$, but that doesn't help.
calculus inequality exponential-function
asked Mar 20 at 10:35
0xbadf00d0xbadf00d
1,76141534
$endgroup$
add a comment |
$begingroup$
How can we show that $$left|frace^-lambda t-1tright|le|lambda|$$ for all $tinmathbb Rsetminusleft0right$ and $lambdainmathbb R$?
Clearly, $e^-lambda tin[0,1]$ for all $t,lambdage0$, but that doesn't help.
calculus inequality exponential-function
asked Mar 20 at 10:35
0xbadf00d0xbadf00d
1,76141534
$endgroup$
$begingroup$
@JeanMarie This is not a duplicate of the above post.
$endgroup$
– Kavi Rama Murthy
Mar 20 at 23:09
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You are right. It is more or less a duplicate of math.stackexchange.com/questions/1683328/…
$endgroup$
– Jean Marie
Mar 20 at 23:31
add a comment |
$begingroup$
How can we show that $$left|frace^-lambda t-1tright|le|lambda|$$ for all $tinmathbb Rsetminusleft0right$ and $lambdainmathbb R$?
Clearly, $e^-lambda tin[0,1]$ for all $t,lambdage0$, but that doesn't help.
calculus inequality exponential-function
asked Mar 20 at 10:35
0xbadf00d0xbadf00d
1,76141534
$endgroup$
How can we show that $$left|frace^-lambda t-1tright|le|lambda|$$ for all $tinmathbb Rsetminusleft0right$ and $lambdainmathbb R$?
Clearly, $e^-lambda tin[0,1]$ for all $t,lambdage0$, but that doesn't help.
calculus inequality exponential-function
calculus inequality exponential-function
asked Mar 20 at 10:35
0xbadf00d0xbadf00d
1,76141534
asked Mar 20 at 10:35
0xbadf00d0xbadf00d
1,76141534
asked Mar 20 at 10:35
0xbadf00d0xbadf00d
1,76141534
asked Mar 20 at 10:35
0xbadf00d0xbadf00d
1,76141534
asked Mar 20 at 10:35
0xbadf00d0xbadf00d
1,76141534
1,76141534
$begingroup$
@JeanMarie This is not a duplicate of the above post.
$endgroup$
– Kavi Rama Murthy
Mar 20 at 23:09
$begingroup$
You are right. It is more or less a duplicate of math.stackexchange.com/questions/1683328/…
$endgroup$
– Jean Marie
Mar 20 at 23:31
add a comment |
$begingroup$
@JeanMarie This is not a duplicate of the above post.
$endgroup$
– Kavi Rama Murthy
Mar 20 at 23:09
$begingroup$
You are right. It is more or less a duplicate of math.stackexchange.com/questions/1683328/…
$endgroup$
– Jean Marie
Mar 20 at 23:31
$begingroup$
@JeanMarie This is not a duplicate of the above post.
$endgroup$
– Kavi Rama Murthy
Mar 20 at 23:09
$begingroup$
@JeanMarie This is not a duplicate of the above post.
$endgroup$
– Kavi Rama Murthy
Mar 20 at 23:09
$begingroup$
You are right. It is more or less a duplicate of math.stackexchange.com/questions/1683328/…
$endgroup$
– Jean Marie
Mar 20 at 23:31
$begingroup$
You are right. It is more or less a duplicate of math.stackexchange.com/questions/1683328/…
$endgroup$
– Jean Marie
Mar 20 at 23:31
add a comment |
1 Answer
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The inequality is false. See what happens as $lambda to -infty$ with $t=1$ to see that it is false. [Indeed, when $lambda=-n$ and $t=1$ the inequality says $|e^n-1| leq n$ which is false as you can see from the series expansion of $e^n$]. It is true for $lambda, t geq 0$. For $lambda =0$ this is obvious. For $lambda > 0$ note that $int_0^1 e^-lambda tx dx=frac 1-e^-lambda t lambda t$. Use the fact that the integrand is $leq 1$. The inequality is also true when both parameters are negative.
answered Mar 20 at 10:37
Kavi Rama MurthyKavi Rama Murthy
71.5k53170
$endgroup$
1
$begingroup$
It doesn't matter for the desired conclusion, but shouldn't it be $int_0^1e^-lambda tx:rm dx=frac1-e^-lambda tlambda t$?
$endgroup$
– 0xbadf00d
Mar 20 at 13:24
$begingroup$
@Oxbad00d Yes, that was a silly mistake. Thanks for pointing it out.
$endgroup$
– Kavi Rama Murthy
Mar 20 at 23:08
add a comment |
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1 Answer
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1 Answer
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$begingroup$
The inequality is false. See what happens as $lambda to -infty$ with $t=1$ to see that it is false. [Indeed, when $lambda=-n$ and $t=1$ the inequality says $|e^n-1| leq n$ which is false as you can see from the series expansion of $e^n$]. It is true for $lambda, t geq 0$. For $lambda =0$ this is obvious. For $lambda > 0$ note that $int_0^1 e^-lambda tx dx=frac 1-e^-lambda t lambda t$. Use the fact that the integrand is $leq 1$. The inequality is also true when both parameters are negative.
answered Mar 20 at 10:37
Kavi Rama MurthyKavi Rama Murthy
71.5k53170
$endgroup$
1
$begingroup$
It doesn't matter for the desired conclusion, but shouldn't it be $int_0^1e^-lambda tx:rm dx=frac1-e^-lambda tlambda t$?
$endgroup$
– 0xbadf00d
Mar 20 at 13:24
$begingroup$
@Oxbad00d Yes, that was a silly mistake. Thanks for pointing it out.
$endgroup$
– Kavi Rama Murthy
Mar 20 at 23:08
add a comment |
$begingroup$
The inequality is false. See what happens as $lambda to -infty$ with $t=1$ to see that it is false. [Indeed, when $lambda=-n$ and $t=1$ the inequality says $|e^n-1| leq n$ which is false as you can see from the series expansion of $e^n$]. It is true for $lambda, t geq 0$. For $lambda =0$ this is obvious. For $lambda > 0$ note that $int_0^1 e^-lambda tx dx=frac 1-e^-lambda t lambda t$. Use the fact that the integrand is $leq 1$. The inequality is also true when both parameters are negative.
answered Mar 20 at 10:37
Kavi Rama MurthyKavi Rama Murthy
71.5k53170
$endgroup$
1
$begingroup$
It doesn't matter for the desired conclusion, but shouldn't it be $int_0^1e^-lambda tx:rm dx=frac1-e^-lambda tlambda t$?
$endgroup$
– 0xbadf00d
Mar 20 at 13:24
$begingroup$
@Oxbad00d Yes, that was a silly mistake. Thanks for pointing it out.
$endgroup$
– Kavi Rama Murthy
Mar 20 at 23:08
add a comment |
$begingroup$
The inequality is false. See what happens as $lambda to -infty$ with $t=1$ to see that it is false. [Indeed, when $lambda=-n$ and $t=1$ the inequality says $|e^n-1| leq n$ which is false as you can see from the series expansion of $e^n$]. It is true for $lambda, t geq 0$. For $lambda =0$ this is obvious. For $lambda > 0$ note that $int_0^1 e^-lambda tx dx=frac 1-e^-lambda t lambda t$. Use the fact that the integrand is $leq 1$. The inequality is also true when both parameters are negative.
answered Mar 20 at 10:37
Kavi Rama MurthyKavi Rama Murthy
71.5k53170
$endgroup$
The inequality is false. See what happens as $lambda to -infty$ with $t=1$ to see that it is false. [Indeed, when $lambda=-n$ and $t=1$ the inequality says $|e^n-1| leq n$ which is false as you can see from the series expansion of $e^n$]. It is true for $lambda, t geq 0$. For $lambda =0$ this is obvious. For $lambda > 0$ note that $int_0^1 e^-lambda tx dx=frac 1-e^-lambda t lambda t$. Use the fact that the integrand is $leq 1$. The inequality is also true when both parameters are negative.
answered Mar 20 at 10:37
Kavi Rama MurthyKavi Rama Murthy
71.5k53170
edited Mar 20 at 23:07
answered Mar 20 at 10:37
Kavi Rama MurthyKavi Rama Murthy
71.5k53170
answered Mar 20 at 10:37
Kavi Rama MurthyKavi Rama Murthy
71.5k53170
answered Mar 20 at 10:37
Kavi Rama MurthyKavi Rama Murthy
71.5k53170
71.5k53170
1
$begingroup$
It doesn't matter for the desired conclusion, but shouldn't it be $int_0^1e^-lambda tx:rm dx=frac1-e^-lambda tlambda t$?
$endgroup$
– 0xbadf00d
Mar 20 at 13:24
$begingroup$
@Oxbad00d Yes, that was a silly mistake. Thanks for pointing it out.
$endgroup$
– Kavi Rama Murthy
Mar 20 at 23:08
add a comment |
1
$begingroup$
It doesn't matter for the desired conclusion, but shouldn't it be $int_0^1e^-lambda tx:rm dx=frac1-e^-lambda tlambda t$?
$endgroup$
– 0xbadf00d
Mar 20 at 13:24
$begingroup$
@Oxbad00d Yes, that was a silly mistake. Thanks for pointing it out.
$endgroup$
– Kavi Rama Murthy
Mar 20 at 23:08
1
1
$begingroup$
It doesn't matter for the desired conclusion, but shouldn't it be $int_0^1e^-lambda tx:rm dx=frac1-e^-lambda tlambda t$?
$endgroup$
– 0xbadf00d
Mar 20 at 13:24
$begingroup$
It doesn't matter for the desired conclusion, but shouldn't it be $int_0^1e^-lambda tx:rm dx=frac1-e^-lambda tlambda t$?
$endgroup$
– 0xbadf00d
Mar 20 at 13:24
$begingroup$
@Oxbad00d Yes, that was a silly mistake. Thanks for pointing it out.
$endgroup$
– Kavi Rama Murthy
Mar 20 at 23:08
$begingroup$
@Oxbad00d Yes, that was a silly mistake. Thanks for pointing it out.
$endgroup$
– Kavi Rama Murthy
Mar 20 at 23:08
add a comment |
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$begingroup$
@JeanMarie This is not a duplicate of the above post.
$endgroup$
– Kavi Rama Murthy
Mar 20 at 23:09
$begingroup$
You are right. It is more or less a duplicate of math.stackexchange.com/questions/1683328/…
$endgroup$
– Jean Marie
Mar 20 at 23:31