Mixing
simple inequality need to know how it's obtained
$begingroup$
A First Course in Mathematical Analysis - D. Brannan.
Rules added as they may be needed.
Reciprocal Rule Page 10.
For any positive $a,b$ $a < b Leftrightarrow frac{1}{a} > frac{1}{b}$
Power rule: page 10.
for any non-negative $a,b$ and $p > 0$
$a < b Leftrightarrow {a^p} < {b^p}$.
page 59.
Deduce: $mathop {lim }limits_{n to infty } {a^{{textstyle{1 over n}}}} = 1$
Soln.
If $a > 1$ we can write
$a = 1 + c$ where $c > 0$. Then
$1 le {a^{1/n}} = {(1 + c)^{{textstyle{1 over n}}}} le 1 + frac{c}{n}$ for $n = 1,2,...$ The soln. continues...
Please explain how (from the book) we get LHS inequality ($1 le {a^{{textstyle{1 over n}}}}$).
real-analysis
edited Jan 14 at 21:07
Matt A Pelto
2,602621
asked Jan 14 at 20:34
MegamaticsMegamatics
194
$endgroup$
|
show 4 more comments
$begingroup$
A First Course in Mathematical Analysis - D. Brannan.
Rules added as they may be needed.
Reciprocal Rule Page 10.
For any positive $a,b$ $a < b Leftrightarrow frac{1}{a} > frac{1}{b}$
Power rule: page 10.
for any non-negative $a,b$ and $p > 0$
$a < b Leftrightarrow {a^p} < {b^p}$.
page 59.
Deduce: $mathop {lim }limits_{n to infty } {a^{{textstyle{1 over n}}}} = 1$
Soln.
If $a > 1$ we can write
$a = 1 + c$ where $c > 0$. Then
$1 le {a^{1/n}} = {(1 + c)^{{textstyle{1 over n}}}} le 1 + frac{c}{n}$ for $n = 1,2,...$ The soln. continues...
Please explain how (from the book) we get LHS inequality ($1 le {a^{{textstyle{1 over n}}}}$).
real-analysis
edited Jan 14 at 21:07
Matt A Pelto
2,602621
asked Jan 14 at 20:34
MegamaticsMegamatics
194
$endgroup$
1
$begingroup$
$1 < a implies 1^{1/n} =1 < a^{1/n}$
$endgroup$
– Anthony Ter
Jan 14 at 20:40
$begingroup$
but that is not the inequality there is equality too. Please also add which rule from his book you are using.
$endgroup$
– Megamatics
Jan 14 at 20:47
$begingroup$
are you saying that $1 < a Rightarrow 1 le a$
$endgroup$
– Megamatics
Jan 14 at 20:50
2
$begingroup$
If the statement $P$ is true, then the statement $P lor Q$ is also true. So if $a<b$ ($a$ is less than $b$) is true, then $a leq b$ ($a$ is less than or equal to $b$) is also true. $<$ is simply a stronger condition than $leq$.
$endgroup$
– Matt A Pelto
Jan 14 at 20:55
2
$begingroup$
Nope - just Bernoulli: $(1+c/n)^n ge 1+n(c/n) = 1+c$.
$endgroup$
– marty cohen
Jan 14 at 21:30
|
show 4 more comments
$begingroup$
A First Course in Mathematical Analysis - D. Brannan.
Rules added as they may be needed.
Reciprocal Rule Page 10.
For any positive $a,b$ $a < b Leftrightarrow frac{1}{a} > frac{1}{b}$
Power rule: page 10.
for any non-negative $a,b$ and $p > 0$
$a < b Leftrightarrow {a^p} < {b^p}$.
page 59.
Deduce: $mathop {lim }limits_{n to infty } {a^{{textstyle{1 over n}}}} = 1$
Soln.
If $a > 1$ we can write
$a = 1 + c$ where $c > 0$. Then
$1 le {a^{1/n}} = {(1 + c)^{{textstyle{1 over n}}}} le 1 + frac{c}{n}$ for $n = 1,2,...$ The soln. continues...
Please explain how (from the book) we get LHS inequality ($1 le {a^{{textstyle{1 over n}}}}$).
real-analysis
edited Jan 14 at 21:07
Matt A Pelto
2,602621
asked Jan 14 at 20:34
MegamaticsMegamatics
194
$endgroup$
A First Course in Mathematical Analysis - D. Brannan.
Rules added as they may be needed.
Reciprocal Rule Page 10.
For any positive $a,b$ $a < b Leftrightarrow frac{1}{a} > frac{1}{b}$
Power rule: page 10.
for any non-negative $a,b$ and $p > 0$
$a < b Leftrightarrow {a^p} < {b^p}$.
page 59.
Deduce: $mathop {lim }limits_{n to infty } {a^{{textstyle{1 over n}}}} = 1$
Soln.
If $a > 1$ we can write
$a = 1 + c$ where $c > 0$. Then
$1 le {a^{1/n}} = {(1 + c)^{{textstyle{1 over n}}}} le 1 + frac{c}{n}$ for $n = 1,2,...$ The soln. continues...
Please explain how (from the book) we get LHS inequality ($1 le {a^{{textstyle{1 over n}}}}$).
real-analysis
real-analysis
edited Jan 14 at 21:07
Matt A Pelto
2,602621
asked Jan 14 at 20:34
MegamaticsMegamatics
194
edited Jan 14 at 21:07
Matt A Pelto
2,602621
asked Jan 14 at 20:34
MegamaticsMegamatics
194
edited Jan 14 at 21:07
Matt A Pelto
2,602621
edited Jan 14 at 21:07
Matt A Pelto
2,602621
edited Jan 14 at 21:07
Matt A Pelto
2,602621
2,602621
asked Jan 14 at 20:34
MegamaticsMegamatics
194
asked Jan 14 at 20:34
MegamaticsMegamatics
194
asked Jan 14 at 20:34
MegamaticsMegamatics
194
194
1
$begingroup$
$1 < a implies 1^{1/n} =1 < a^{1/n}$
$endgroup$
– Anthony Ter
Jan 14 at 20:40
$begingroup$
but that is not the inequality there is equality too. Please also add which rule from his book you are using.
$endgroup$
– Megamatics
Jan 14 at 20:47
$begingroup$
are you saying that $1 < a Rightarrow 1 le a$
$endgroup$
– Megamatics
Jan 14 at 20:50
2
$begingroup$
If the statement $P$ is true, then the statement $P lor Q$ is also true. So if $a<b$ ($a$ is less than $b$) is true, then $a leq b$ ($a$ is less than or equal to $b$) is also true. $<$ is simply a stronger condition than $leq$.
$endgroup$
– Matt A Pelto
Jan 14 at 20:55
2
$begingroup$
Nope - just Bernoulli: $(1+c/n)^n ge 1+n(c/n) = 1+c$.
$endgroup$
– marty cohen
Jan 14 at 21:30
|
show 4 more comments
1
$begingroup$
$1 < a implies 1^{1/n} =1 < a^{1/n}$
$endgroup$
– Anthony Ter
Jan 14 at 20:40
$begingroup$
but that is not the inequality there is equality too. Please also add which rule from his book you are using.
$endgroup$
– Megamatics
Jan 14 at 20:47
$begingroup$
are you saying that $1 < a Rightarrow 1 le a$
$endgroup$
– Megamatics
Jan 14 at 20:50
2
$begingroup$
If the statement $P$ is true, then the statement $P lor Q$ is also true. So if $a<b$ ($a$ is less than $b$) is true, then $a leq b$ ($a$ is less than or equal to $b$) is also true. $<$ is simply a stronger condition than $leq$.
$endgroup$
– Matt A Pelto
Jan 14 at 20:55
2
$begingroup$
Nope - just Bernoulli: $(1+c/n)^n ge 1+n(c/n) = 1+c$.
$endgroup$
– marty cohen
Jan 14 at 21:30
1
1
$begingroup$
$1 < a implies 1^{1/n} =1 < a^{1/n}$
$endgroup$
– Anthony Ter
Jan 14 at 20:40
$begingroup$
$1 < a implies 1^{1/n} =1 < a^{1/n}$
$endgroup$
– Anthony Ter
Jan 14 at 20:40
$begingroup$
but that is not the inequality there is equality too. Please also add which rule from his book you are using.
$endgroup$
– Megamatics
Jan 14 at 20:47
$begingroup$
but that is not the inequality there is equality too. Please also add which rule from his book you are using.
$endgroup$
– Megamatics
Jan 14 at 20:47
$begingroup$
are you saying that $1 < a Rightarrow 1 le a$
$endgroup$
– Megamatics
Jan 14 at 20:50
$begingroup$
are you saying that $1 < a Rightarrow 1 le a$
$endgroup$
– Megamatics
Jan 14 at 20:50
2
2
$begingroup$
If the statement $P$ is true, then the statement $P lor Q$ is also true. So if $a<b$ ($a$ is less than $b$) is true, then $a leq b$ ($a$ is less than or equal to $b$) is also true. $<$ is simply a stronger condition than $leq$.
$endgroup$
– Matt A Pelto
Jan 14 at 20:55
$begingroup$
If the statement $P$ is true, then the statement $P lor Q$ is also true. So if $a<b$ ($a$ is less than $b$) is true, then $a leq b$ ($a$ is less than or equal to $b$) is also true. $<$ is simply a stronger condition than $leq$.
$endgroup$
– Matt A Pelto
Jan 14 at 20:55
2
2
$begingroup$
Nope - just Bernoulli: $(1+c/n)^n ge 1+n(c/n) = 1+c$.
$endgroup$
– marty cohen
Jan 14 at 21:30
$begingroup$
Nope - just Bernoulli: $(1+c/n)^n ge 1+n(c/n) = 1+c$.
$endgroup$
– marty cohen
Jan 14 at 21:30
|
show 4 more comments
1 Answer
1
active
oldest
votes
$begingroup$
Take $p = n$.
We want to show that
$a^{1/n} ge 1, tag 0$
right? Well, if not, then
$a^{1/n} < 1, tag 1$
the given power rule implies
$a = (a^{1/n})^n < 1^n = 1, tag 2$
which contradicts the hypothesis $a ge 1$. Therefore,
$a^{1/n} ge 1. tag 3$
We turn to the inequality
$(1 + c)^{1/n} < 1 + dfrac{c}{n}; tag 4$
from the binomial theorem,
$1 + c = 1 + ndfrac{c}{n} < (1 + dfrac{c}{n})^n; tag 5$
now use the power rule with
$p = dfrac{1}{n}; tag 6$
we find
$(1 + c)^{1/n} = (1 + ndfrac{c}{n})^{1/n} < 1 + dfrac{c}{n}, tag 7$
as required.
answered Jan 14 at 23:06
Robert LewisRobert Lewis
46.7k23067
$endgroup$
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
$begingroup$
Take $p = n$.
We want to show that
$a^{1/n} ge 1, tag 0$
right? Well, if not, then
$a^{1/n} < 1, tag 1$
the given power rule implies
$a = (a^{1/n})^n < 1^n = 1, tag 2$
which contradicts the hypothesis $a ge 1$. Therefore,
$a^{1/n} ge 1. tag 3$
We turn to the inequality
$(1 + c)^{1/n} < 1 + dfrac{c}{n}; tag 4$
from the binomial theorem,
$1 + c = 1 + ndfrac{c}{n} < (1 + dfrac{c}{n})^n; tag 5$
now use the power rule with
$p = dfrac{1}{n}; tag 6$
we find
$(1 + c)^{1/n} = (1 + ndfrac{c}{n})^{1/n} < 1 + dfrac{c}{n}, tag 7$
as required.
answered Jan 14 at 23:06
Robert LewisRobert Lewis
46.7k23067
$endgroup$
add a comment |
$begingroup$
Take $p = n$.
We want to show that
$a^{1/n} ge 1, tag 0$
right? Well, if not, then
$a^{1/n} < 1, tag 1$
the given power rule implies
$a = (a^{1/n})^n < 1^n = 1, tag 2$
which contradicts the hypothesis $a ge 1$. Therefore,
$a^{1/n} ge 1. tag 3$
We turn to the inequality
$(1 + c)^{1/n} < 1 + dfrac{c}{n}; tag 4$
from the binomial theorem,
$1 + c = 1 + ndfrac{c}{n} < (1 + dfrac{c}{n})^n; tag 5$
now use the power rule with
$p = dfrac{1}{n}; tag 6$
we find
$(1 + c)^{1/n} = (1 + ndfrac{c}{n})^{1/n} < 1 + dfrac{c}{n}, tag 7$
as required.
answered Jan 14 at 23:06
Robert LewisRobert Lewis
46.7k23067
$endgroup$
add a comment |
$begingroup$
Take $p = n$.
We want to show that
$a^{1/n} ge 1, tag 0$
right? Well, if not, then
$a^{1/n} < 1, tag 1$
the given power rule implies
$a = (a^{1/n})^n < 1^n = 1, tag 2$
which contradicts the hypothesis $a ge 1$. Therefore,
$a^{1/n} ge 1. tag 3$
We turn to the inequality
$(1 + c)^{1/n} < 1 + dfrac{c}{n}; tag 4$
from the binomial theorem,
$1 + c = 1 + ndfrac{c}{n} < (1 + dfrac{c}{n})^n; tag 5$
now use the power rule with
$p = dfrac{1}{n}; tag 6$
we find
$(1 + c)^{1/n} = (1 + ndfrac{c}{n})^{1/n} < 1 + dfrac{c}{n}, tag 7$
as required.
answered Jan 14 at 23:06
Robert LewisRobert Lewis
46.7k23067
$endgroup$
Take $p = n$.
We want to show that
$a^{1/n} ge 1, tag 0$
right? Well, if not, then
$a^{1/n} < 1, tag 1$
the given power rule implies
$a = (a^{1/n})^n < 1^n = 1, tag 2$
which contradicts the hypothesis $a ge 1$. Therefore,
$a^{1/n} ge 1. tag 3$
We turn to the inequality
$(1 + c)^{1/n} < 1 + dfrac{c}{n}; tag 4$
from the binomial theorem,
$1 + c = 1 + ndfrac{c}{n} < (1 + dfrac{c}{n})^n; tag 5$
now use the power rule with
$p = dfrac{1}{n}; tag 6$
we find
$(1 + c)^{1/n} = (1 + ndfrac{c}{n})^{1/n} < 1 + dfrac{c}{n}, tag 7$
as required.
answered Jan 14 at 23:06
Robert LewisRobert Lewis
46.7k23067
answered Jan 14 at 23:06
Robert LewisRobert Lewis
46.7k23067
answered Jan 14 at 23:06
Robert LewisRobert Lewis
46.7k23067
answered Jan 14 at 23:06
Robert LewisRobert Lewis
46.7k23067
46.7k23067
add a comment |
add a comment |
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1
$begingroup$
$1 < a implies 1^{1/n} =1 < a^{1/n}$
$endgroup$
– Anthony Ter
Jan 14 at 20:40
$begingroup$
but that is not the inequality there is equality too. Please also add which rule from his book you are using.
$endgroup$
– Megamatics
Jan 14 at 20:47
$begingroup$
are you saying that $1 < a Rightarrow 1 le a$
$endgroup$
– Megamatics
Jan 14 at 20:50
2
$begingroup$
If the statement $P$ is true, then the statement $P lor Q$ is also true. So if $a<b$ ($a$ is less than $b$) is true, then $a leq b$ ($a$ is less than or equal to $b$) is also true. $<$ is simply a stronger condition than $leq$.
$endgroup$
– Matt A Pelto
Jan 14 at 20:55
2
$begingroup$
Nope - just Bernoulli: $(1+c/n)^n ge 1+n(c/n) = 1+c$.
$endgroup$
– marty cohen
Jan 14 at 21:30