Mixing
Proof of Inequality by Binomial Theorem
Clash Royale CLAN TAG#URR8PPP
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I have been stuck in this proof for a while. This should not be a long shot.
Claim If $x_n>0$ for all $nin N$, then $1+nx_nleq (1+x_n)^n$.
My strategy is to use Induction Theorem and Binomial Theorem, which does not work quite well.
My proof If $n=1$, $1+x_1leq (1+x_1)^1$ (True).
Assume this claim is true for $n$. Consider $n+1$:
$(1+x_n+1)^n+1=x_n+1^n+1+(n+1)x_n+1^n+dots+2x_n+1+1$. (**)
Need to show $(**)geq 1+(n+1)x_n+1$, given $(1+nx_n)leq (1+x_n)^n$, i.e. $(1+nx_n)leq x_n^n+nx_n^n-1+dots + 2x_n+1$.
Could someone give me a hint on how to deal with the $n+1$ case? In particular, how to link $x_n$ to $x_n+1$, if induction is the right way to go? Thanks in advance!
real-analysis binomial-theorem
asked Aug 1 at 3:16
James Wang
876
add a comment |Â
up vote
1
down vote
favorite
I have been stuck in this proof for a while. This should not be a long shot.
Claim If $x_n>0$ for all $nin N$, then $1+nx_nleq (1+x_n)^n$.
My strategy is to use Induction Theorem and Binomial Theorem, which does not work quite well.
My proof If $n=1$, $1+x_1leq (1+x_1)^1$ (True).
Assume this claim is true for $n$. Consider $n+1$:
$(1+x_n+1)^n+1=x_n+1^n+1+(n+1)x_n+1^n+dots+2x_n+1+1$. (**)
Need to show $(**)geq 1+(n+1)x_n+1$, given $(1+nx_n)leq (1+x_n)^n$, i.e. $(1+nx_n)leq x_n^n+nx_n^n-1+dots + 2x_n+1$.
Could someone give me a hint on how to deal with the $n+1$ case? In particular, how to link $x_n$ to $x_n+1$, if induction is the right way to go? Thanks in advance!
real-analysis binomial-theorem
asked Aug 1 at 3:16
James Wang
876
4
$(1+x)^n+1=x^n+1+cdots+(n+1)x+1$.
– Lord Shark the Unknown
Aug 1 at 3:19
1
Try multiplying both sides of $(1+x)^nge1+nx$ by $1+x$.
– Simply Beautiful Art
Aug 1 at 3:20
1
Just prove that $forall x>0;forall nin Bbb N;(1+nxleq (1+x)^n))$ by induction on $n$.
– DanielWainfleet
Aug 1 at 4:18
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I have been stuck in this proof for a while. This should not be a long shot.
Claim If $x_n>0$ for all $nin N$, then $1+nx_nleq (1+x_n)^n$.
My strategy is to use Induction Theorem and Binomial Theorem, which does not work quite well.
My proof If $n=1$, $1+x_1leq (1+x_1)^1$ (True).
Assume this claim is true for $n$. Consider $n+1$:
$(1+x_n+1)^n+1=x_n+1^n+1+(n+1)x_n+1^n+dots+2x_n+1+1$. (**)
Need to show $(**)geq 1+(n+1)x_n+1$, given $(1+nx_n)leq (1+x_n)^n$, i.e. $(1+nx_n)leq x_n^n+nx_n^n-1+dots + 2x_n+1$.
Could someone give me a hint on how to deal with the $n+1$ case? In particular, how to link $x_n$ to $x_n+1$, if induction is the right way to go? Thanks in advance!
real-analysis binomial-theorem
asked Aug 1 at 3:16
James Wang
876
I have been stuck in this proof for a while. This should not be a long shot.
Claim If $x_n>0$ for all $nin N$, then $1+nx_nleq (1+x_n)^n$.
My strategy is to use Induction Theorem and Binomial Theorem, which does not work quite well.
My proof If $n=1$, $1+x_1leq (1+x_1)^1$ (True).
Assume this claim is true for $n$. Consider $n+1$:
$(1+x_n+1)^n+1=x_n+1^n+1+(n+1)x_n+1^n+dots+2x_n+1+1$. (**)
Need to show $(**)geq 1+(n+1)x_n+1$, given $(1+nx_n)leq (1+x_n)^n$, i.e. $(1+nx_n)leq x_n^n+nx_n^n-1+dots + 2x_n+1$.
Could someone give me a hint on how to deal with the $n+1$ case? In particular, how to link $x_n$ to $x_n+1$, if induction is the right way to go? Thanks in advance!
real-analysis binomial-theorem
asked Aug 1 at 3:16
James Wang
876
asked Aug 1 at 3:16
James Wang
876
asked Aug 1 at 3:16
James Wang
876
asked Aug 1 at 3:16
James Wang
876
876
4
$(1+x)^n+1=x^n+1+cdots+(n+1)x+1$.
– Lord Shark the Unknown
Aug 1 at 3:19
1
Try multiplying both sides of $(1+x)^nge1+nx$ by $1+x$.
– Simply Beautiful Art
Aug 1 at 3:20
1
Just prove that $forall x>0;forall nin Bbb N;(1+nxleq (1+x)^n))$ by induction on $n$.
– DanielWainfleet
Aug 1 at 4:18
add a comment |Â
4
$(1+x)^n+1=x^n+1+cdots+(n+1)x+1$.
– Lord Shark the Unknown
Aug 1 at 3:19
1
Try multiplying both sides of $(1+x)^nge1+nx$ by $1+x$.
– Simply Beautiful Art
Aug 1 at 3:20
1
Just prove that $forall x>0;forall nin Bbb N;(1+nxleq (1+x)^n))$ by induction on $n$.
– DanielWainfleet
Aug 1 at 4:18
4
4
$(1+x)^n+1=x^n+1+cdots+(n+1)x+1$.
– Lord Shark the Unknown
Aug 1 at 3:19
$(1+x)^n+1=x^n+1+cdots+(n+1)x+1$.
– Lord Shark the Unknown
Aug 1 at 3:19
1
1
Try multiplying both sides of $(1+x)^nge1+nx$ by $1+x$.
– Simply Beautiful Art
Aug 1 at 3:20
Try multiplying both sides of $(1+x)^nge1+nx$ by $1+x$.
– Simply Beautiful Art
Aug 1 at 3:20
1
1
Just prove that $forall x>0;forall nin Bbb N;(1+nxleq (1+x)^n))$ by induction on $n$.
– DanielWainfleet
Aug 1 at 4:18
Just prove that $forall x>0;forall nin Bbb N;(1+nxleq (1+x)^n))$ by induction on $n$.
– DanielWainfleet
Aug 1 at 4:18
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
The way I look at this one is to temporarily forget about the index $n$ on the $x_n$ and simply note that, for any real $y > 0$ and $1 le n in Bbb N$,
$1 + y le 1 + ny le sum_0^n dfracn!i!(n - i)! 1^n - i y^i = (1 + y)^n, tag 1$
which follows from the binomial theorem applied to $(1 + y)^n$; since (1) holds for every $n$, we may successively set $y = x_n$ for any positive real sequence $x_n$ and obtain
$1 + x_n le (1 + x_n)^n tag 2$
for each $n$.
The fact is that the problem specifies no relationship 'twixt $x_n$ and $x_n + 1$, etc., so attempting to use one as a means of solution is ill-founded; the $x_n$ are arbitrary and independent.
So it's really all about (1). The essential relaitionship
$1 + y le (1 + y)^n tag 3$
can in fact be proved by induction without resorting to the binomial theorem: if
$1 + y le (1 + y)^k, tag 4$
then since
$1 < 1 + y, tag 5$
we have
$1 + y < (1 + y)(1 + y) le (1 + y)^k (1 + y) = (1 + y)^k + 1. tag 6$
"About binomial theorem I am teeming with a lot of news,
With many cheerful facts about the square of the hypoteneuse!"
--Gilbert & Sullivan, "Modern Major General".
answered Aug 1 at 3:56
Robert Lewis
36.7k22155
add a comment |Â
up vote
1
down vote
Note that the induction doesn't work here because you have not a relation between $x_n$ and $x_n+1$. You need to prove the inequality for every fixed $n$.
answered Aug 1 at 3:28
user 108128
18.9k41544
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
The way I look at this one is to temporarily forget about the index $n$ on the $x_n$ and simply note that, for any real $y > 0$ and $1 le n in Bbb N$,
$1 + y le 1 + ny le sum_0^n dfracn!i!(n - i)! 1^n - i y^i = (1 + y)^n, tag 1$
which follows from the binomial theorem applied to $(1 + y)^n$; since (1) holds for every $n$, we may successively set $y = x_n$ for any positive real sequence $x_n$ and obtain
$1 + x_n le (1 + x_n)^n tag 2$
for each $n$.
The fact is that the problem specifies no relationship 'twixt $x_n$ and $x_n + 1$, etc., so attempting to use one as a means of solution is ill-founded; the $x_n$ are arbitrary and independent.
So it's really all about (1). The essential relaitionship
$1 + y le (1 + y)^n tag 3$
can in fact be proved by induction without resorting to the binomial theorem: if
$1 + y le (1 + y)^k, tag 4$
then since
$1 < 1 + y, tag 5$
we have
$1 + y < (1 + y)(1 + y) le (1 + y)^k (1 + y) = (1 + y)^k + 1. tag 6$
"About binomial theorem I am teeming with a lot of news,
With many cheerful facts about the square of the hypoteneuse!"
--Gilbert & Sullivan, "Modern Major General".
answered Aug 1 at 3:56
Robert Lewis
36.7k22155
add a comment |Â
up vote
1
down vote
accepted
The way I look at this one is to temporarily forget about the index $n$ on the $x_n$ and simply note that, for any real $y > 0$ and $1 le n in Bbb N$,
$1 + y le 1 + ny le sum_0^n dfracn!i!(n - i)! 1^n - i y^i = (1 + y)^n, tag 1$
which follows from the binomial theorem applied to $(1 + y)^n$; since (1) holds for every $n$, we may successively set $y = x_n$ for any positive real sequence $x_n$ and obtain
$1 + x_n le (1 + x_n)^n tag 2$
for each $n$.
The fact is that the problem specifies no relationship 'twixt $x_n$ and $x_n + 1$, etc., so attempting to use one as a means of solution is ill-founded; the $x_n$ are arbitrary and independent.
So it's really all about (1). The essential relaitionship
$1 + y le (1 + y)^n tag 3$
can in fact be proved by induction without resorting to the binomial theorem: if
$1 + y le (1 + y)^k, tag 4$
then since
$1 < 1 + y, tag 5$
we have
$1 + y < (1 + y)(1 + y) le (1 + y)^k (1 + y) = (1 + y)^k + 1. tag 6$
"About binomial theorem I am teeming with a lot of news,
With many cheerful facts about the square of the hypoteneuse!"
--Gilbert & Sullivan, "Modern Major General".
answered Aug 1 at 3:56
Robert Lewis
36.7k22155
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
The way I look at this one is to temporarily forget about the index $n$ on the $x_n$ and simply note that, for any real $y > 0$ and $1 le n in Bbb N$,
$1 + y le 1 + ny le sum_0^n dfracn!i!(n - i)! 1^n - i y^i = (1 + y)^n, tag 1$
which follows from the binomial theorem applied to $(1 + y)^n$; since (1) holds for every $n$, we may successively set $y = x_n$ for any positive real sequence $x_n$ and obtain
$1 + x_n le (1 + x_n)^n tag 2$
for each $n$.
The fact is that the problem specifies no relationship 'twixt $x_n$ and $x_n + 1$, etc., so attempting to use one as a means of solution is ill-founded; the $x_n$ are arbitrary and independent.
So it's really all about (1). The essential relaitionship
$1 + y le (1 + y)^n tag 3$
can in fact be proved by induction without resorting to the binomial theorem: if
$1 + y le (1 + y)^k, tag 4$
then since
$1 < 1 + y, tag 5$
we have
$1 + y < (1 + y)(1 + y) le (1 + y)^k (1 + y) = (1 + y)^k + 1. tag 6$
"About binomial theorem I am teeming with a lot of news,
With many cheerful facts about the square of the hypoteneuse!"
--Gilbert & Sullivan, "Modern Major General".
answered Aug 1 at 3:56
Robert Lewis
36.7k22155
The way I look at this one is to temporarily forget about the index $n$ on the $x_n$ and simply note that, for any real $y > 0$ and $1 le n in Bbb N$,
$1 + y le 1 + ny le sum_0^n dfracn!i!(n - i)! 1^n - i y^i = (1 + y)^n, tag 1$
which follows from the binomial theorem applied to $(1 + y)^n$; since (1) holds for every $n$, we may successively set $y = x_n$ for any positive real sequence $x_n$ and obtain
$1 + x_n le (1 + x_n)^n tag 2$
for each $n$.
The fact is that the problem specifies no relationship 'twixt $x_n$ and $x_n + 1$, etc., so attempting to use one as a means of solution is ill-founded; the $x_n$ are arbitrary and independent.
So it's really all about (1). The essential relaitionship
$1 + y le (1 + y)^n tag 3$
can in fact be proved by induction without resorting to the binomial theorem: if
$1 + y le (1 + y)^k, tag 4$
then since
$1 < 1 + y, tag 5$
we have
$1 + y < (1 + y)(1 + y) le (1 + y)^k (1 + y) = (1 + y)^k + 1. tag 6$
"About binomial theorem I am teeming with a lot of news,
With many cheerful facts about the square of the hypoteneuse!"
--Gilbert & Sullivan, "Modern Major General".
answered Aug 1 at 3:56
Robert Lewis
36.7k22155
answered Aug 1 at 3:56
Robert Lewis
36.7k22155
answered Aug 1 at 3:56
Robert Lewis
36.7k22155
answered Aug 1 at 3:56
Robert Lewis
36.7k22155
36.7k22155
add a comment |Â
add a comment |Â
up vote
1
down vote
Note that the induction doesn't work here because you have not a relation between $x_n$ and $x_n+1$. You need to prove the inequality for every fixed $n$.
answered Aug 1 at 3:28
user 108128
18.9k41544
add a comment |Â
up vote
1
down vote
Note that the induction doesn't work here because you have not a relation between $x_n$ and $x_n+1$. You need to prove the inequality for every fixed $n$.
answered Aug 1 at 3:28
user 108128
18.9k41544
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Note that the induction doesn't work here because you have not a relation between $x_n$ and $x_n+1$. You need to prove the inequality for every fixed $n$.
answered Aug 1 at 3:28
user 108128
18.9k41544
Note that the induction doesn't work here because you have not a relation between $x_n$ and $x_n+1$. You need to prove the inequality for every fixed $n$.
answered Aug 1 at 3:28
user 108128
18.9k41544
answered Aug 1 at 3:28
user 108128
18.9k41544
answered Aug 1 at 3:28
user 108128
18.9k41544
answered Aug 1 at 3:28
user 108128
18.9k41544
18.9k41544
add a comment |Â
add a comment |Â
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4
$(1+x)^n+1=x^n+1+cdots+(n+1)x+1$.
– Lord Shark the Unknown
Aug 1 at 3:19
1
Try multiplying both sides of $(1+x)^nge1+nx$ by $1+x$.
– Simply Beautiful Art
Aug 1 at 3:20
1
Just prove that $forall x>0;forall nin Bbb N;(1+nxleq (1+x)^n))$ by induction on $n$.
– DanielWainfleet
Aug 1 at 4:18