Mixing
Proving with induction $(1-x)^n<frac 1 {1+nx}$
$begingroup$
Prove using induction that $forall ninmathbb N, forall xin mathbb R: 0<x<1: (1-x)^n<frac 1 {1+nx}$
My attempt:
Base: for $n=1: 1-x<frac 1 {1+x}iff 1-x^2<1$, true since $0<x<1$.
Suppose the statement is true for $n$, prove for $n+1$:
$(1-x)^{n+1}=(1-x)(1-x)^{n}overset{i.h}<frac{(1-x)}{1+nx}$
Now I got stuck, maybe another induction to show that $1+nx+x<1+nx$? Is there another way?
Moreover, I was told it's wrong to begin with $(1-x)^{n+1}$ and reach to $frac 1 {1+(n+1)x}$ but why? Is it assuming what I need to prove?
induction
asked Jan 28 '15 at 20:51
shinzoushinzou
2,0561635
$endgroup$
|
show 1 more comment
$begingroup$
Prove using induction that $forall ninmathbb N, forall xin mathbb R: 0<x<1: (1-x)^n<frac 1 {1+nx}$
My attempt:
Base: for $n=1: 1-x<frac 1 {1+x}iff 1-x^2<1$, true since $0<x<1$.
Suppose the statement is true for $n$, prove for $n+1$:
$(1-x)^{n+1}=(1-x)(1-x)^{n}overset{i.h}<frac{(1-x)}{1+nx}$
Now I got stuck, maybe another induction to show that $1+nx+x<1+nx$? Is there another way?
Moreover, I was told it's wrong to begin with $(1-x)^{n+1}$ and reach to $frac 1 {1+(n+1)x}$ but why? Is it assuming what I need to prove?
induction
asked Jan 28 '15 at 20:51
shinzoushinzou
2,0561635
$endgroup$
$begingroup$
The first $to$ should rather be an $iff$.
$endgroup$
– Pedro Tamaroff♦
Jan 28 '15 at 20:52
$begingroup$
@PedroTamaroff why is it wrong to write just $rightarrow$? isn't it enough for this case?
$endgroup$
– shinzou
Jan 28 '15 at 20:53
1
$begingroup$
You want to show that if $0<x<1$, $1-x<frac{1}{1+x}$. You cannot assume what you want to show is true, show it implies a true fact and conclude that what you assumed is true is... true. Your logic is "Let's show $P$. Assume $P$. Since $P$ implies $Q$ and $Q$ is true, $P$ is true."
$endgroup$
– Pedro Tamaroff♦
Jan 28 '15 at 20:54
$begingroup$
Oh yes I see that $frac{1-x}{1+nx}<frac{1}{1+nx}$ the problem is with the denominator.
$endgroup$
– shinzou
Jan 28 '15 at 20:57
$begingroup$
Note that $$frac{1-x}{1+nx}<frac{1}{1+(n+1)x}$$ is true if and only if $-(n+1)x^2<0$ (just cross multiply and play around a while). Since $x^2>0$ and $n+1>0$, you're done.
$endgroup$
– Pedro Tamaroff♦
Jan 28 '15 at 20:57
|
show 1 more comment
$begingroup$
Prove using induction that $forall ninmathbb N, forall xin mathbb R: 0<x<1: (1-x)^n<frac 1 {1+nx}$
My attempt:
Base: for $n=1: 1-x<frac 1 {1+x}iff 1-x^2<1$, true since $0<x<1$.
Suppose the statement is true for $n$, prove for $n+1$:
$(1-x)^{n+1}=(1-x)(1-x)^{n}overset{i.h}<frac{(1-x)}{1+nx}$
Now I got stuck, maybe another induction to show that $1+nx+x<1+nx$? Is there another way?
Moreover, I was told it's wrong to begin with $(1-x)^{n+1}$ and reach to $frac 1 {1+(n+1)x}$ but why? Is it assuming what I need to prove?
induction
asked Jan 28 '15 at 20:51
shinzoushinzou
2,0561635
$endgroup$
Prove using induction that $forall ninmathbb N, forall xin mathbb R: 0<x<1: (1-x)^n<frac 1 {1+nx}$
My attempt:
Base: for $n=1: 1-x<frac 1 {1+x}iff 1-x^2<1$, true since $0<x<1$.
Suppose the statement is true for $n$, prove for $n+1$:
$(1-x)^{n+1}=(1-x)(1-x)^{n}overset{i.h}<frac{(1-x)}{1+nx}$
Now I got stuck, maybe another induction to show that $1+nx+x<1+nx$? Is there another way?
Moreover, I was told it's wrong to begin with $(1-x)^{n+1}$ and reach to $frac 1 {1+(n+1)x}$ but why? Is it assuming what I need to prove?
induction
induction
asked Jan 28 '15 at 20:51
shinzoushinzou
2,0561635
asked Jan 28 '15 at 20:51
shinzoushinzou
2,0561635
edited Jan 28 '15 at 21:26
shinzou
asked Jan 28 '15 at 20:51
shinzoushinzou
2,0561635
asked Jan 28 '15 at 20:51
shinzoushinzou
2,0561635
asked Jan 28 '15 at 20:51
shinzoushinzou
2,0561635
2,0561635
$begingroup$
The first $to$ should rather be an $iff$.
$endgroup$
– Pedro Tamaroff♦
Jan 28 '15 at 20:52
$begingroup$
@PedroTamaroff why is it wrong to write just $rightarrow$? isn't it enough for this case?
$endgroup$
– shinzou
Jan 28 '15 at 20:53
1
$begingroup$
You want to show that if $0<x<1$, $1-x<frac{1}{1+x}$. You cannot assume what you want to show is true, show it implies a true fact and conclude that what you assumed is true is... true. Your logic is "Let's show $P$. Assume $P$. Since $P$ implies $Q$ and $Q$ is true, $P$ is true."
$endgroup$
– Pedro Tamaroff♦
Jan 28 '15 at 20:54
$begingroup$
Oh yes I see that $frac{1-x}{1+nx}<frac{1}{1+nx}$ the problem is with the denominator.
$endgroup$
– shinzou
Jan 28 '15 at 20:57
$begingroup$
Note that $$frac{1-x}{1+nx}<frac{1}{1+(n+1)x}$$ is true if and only if $-(n+1)x^2<0$ (just cross multiply and play around a while). Since $x^2>0$ and $n+1>0$, you're done.
$endgroup$
– Pedro Tamaroff♦
Jan 28 '15 at 20:57
|
show 1 more comment
$begingroup$
The first $to$ should rather be an $iff$.
$endgroup$
– Pedro Tamaroff♦
Jan 28 '15 at 20:52
$begingroup$
@PedroTamaroff why is it wrong to write just $rightarrow$? isn't it enough for this case?
$endgroup$
– shinzou
Jan 28 '15 at 20:53
1
$begingroup$
You want to show that if $0<x<1$, $1-x<frac{1}{1+x}$. You cannot assume what you want to show is true, show it implies a true fact and conclude that what you assumed is true is... true. Your logic is "Let's show $P$. Assume $P$. Since $P$ implies $Q$ and $Q$ is true, $P$ is true."
$endgroup$
– Pedro Tamaroff♦
Jan 28 '15 at 20:54
$begingroup$
Oh yes I see that $frac{1-x}{1+nx}<frac{1}{1+nx}$ the problem is with the denominator.
$endgroup$
– shinzou
Jan 28 '15 at 20:57
$begingroup$
Note that $$frac{1-x}{1+nx}<frac{1}{1+(n+1)x}$$ is true if and only if $-(n+1)x^2<0$ (just cross multiply and play around a while). Since $x^2>0$ and $n+1>0$, you're done.
$endgroup$
– Pedro Tamaroff♦
Jan 28 '15 at 20:57
$begingroup$
The first $to$ should rather be an $iff$.
$endgroup$
– Pedro Tamaroff♦
Jan 28 '15 at 20:52
$begingroup$
The first $to$ should rather be an $iff$.
$endgroup$
– Pedro Tamaroff♦
Jan 28 '15 at 20:52
$begingroup$
@PedroTamaroff why is it wrong to write just $rightarrow$? isn't it enough for this case?
$endgroup$
– shinzou
Jan 28 '15 at 20:53
$begingroup$
@PedroTamaroff why is it wrong to write just $rightarrow$? isn't it enough for this case?
$endgroup$
– shinzou
Jan 28 '15 at 20:53
1
1
$begingroup$
You want to show that if $0<x<1$, $1-x<frac{1}{1+x}$. You cannot assume what you want to show is true, show it implies a true fact and conclude that what you assumed is true is... true. Your logic is "Let's show $P$. Assume $P$. Since $P$ implies $Q$ and $Q$ is true, $P$ is true."
$endgroup$
– Pedro Tamaroff♦
Jan 28 '15 at 20:54
$begingroup$
You want to show that if $0<x<1$, $1-x<frac{1}{1+x}$. You cannot assume what you want to show is true, show it implies a true fact and conclude that what you assumed is true is... true. Your logic is "Let's show $P$. Assume $P$. Since $P$ implies $Q$ and $Q$ is true, $P$ is true."
$endgroup$
– Pedro Tamaroff♦
Jan 28 '15 at 20:54
$begingroup$
Oh yes I see that $frac{1-x}{1+nx}<frac{1}{1+nx}$ the problem is with the denominator.
$endgroup$
– shinzou
Jan 28 '15 at 20:57
$begingroup$
Oh yes I see that $frac{1-x}{1+nx}<frac{1}{1+nx}$ the problem is with the denominator.
$endgroup$
– shinzou
Jan 28 '15 at 20:57
$begingroup$
Note that $$frac{1-x}{1+nx}<frac{1}{1+(n+1)x}$$ is true if and only if $-(n+1)x^2<0$ (just cross multiply and play around a while). Since $x^2>0$ and $n+1>0$, you're done.
$endgroup$
– Pedro Tamaroff♦
Jan 28 '15 at 20:57
$begingroup$
Note that $$frac{1-x}{1+nx}<frac{1}{1+(n+1)x}$$ is true if and only if $-(n+1)x^2<0$ (just cross multiply and play around a while). Since $x^2>0$ and $n+1>0$, you're done.
$endgroup$
– Pedro Tamaroff♦
Jan 28 '15 at 20:57
|
show 1 more comment
3 Answers
3
active
oldest
votes
$begingroup$
Apply again the base case: $1-x<displaystylefrac1{1+x}$ and that $x^2>0$ to get
$$frac{1-x}{1+nx} < frac1{(1+nx)(1+x)}=frac1{1+(n+1)x+nx^2}<frac1{1+(n+1)x},.$$
answered Jan 28 '15 at 20:57
BerciBerci
60.6k23672
$endgroup$
$begingroup$
Is there nothing wrong with starting with $(1-x)^{n+1}$?
$endgroup$
– shinzou
Jan 28 '15 at 21:00
add a comment |
$begingroup$
Notice that there is no need of induction. By the AM-GM inequality,
$$ (1+nx)(1-x)^n < left(frac{(1+nx)+n(1-x)}{n+1}right)^{n+1} =1.$$
answered Jan 28 '15 at 22:15
Jack D'AurizioJack D'Aurizio
289k33281661
$endgroup$
1
$begingroup$
Ingenious. Similar to the proofs that $(1+1/n)^n$ is increasing and $(1+1/n)^{n+1}$ is decreasing (and so both approach a common limit, which I propose to call "$e$").
$endgroup$
– marty cohen
Jan 10 at 21:59
$begingroup$
The proofs are in N.S Mendelsohn, An application of a famous inequality, Amer. Math. Monthly 58 (1951), 563.
$endgroup$
– marty cohen
Jan 10 at 22:03
add a comment |
$begingroup$
Write it in the form
$(1+nx)(1-x)^n
lt 1$.
If true for $n$, then
$begin{array}\
(1+(n+1)x)(1-x)^{n+1}
&=(1+(n+1)x)(1-x)^{n+1}\
&=dfrac{(1+(n+1)x)}{1+nx}(1+nx)(1-x)^{n+1}\
&=dfrac{(1+(n+1)x)(1-x)}{1+nx}(1+nx)(1-x)^{n}\
&<dfrac{(1+(n+1)x)(1-x)}{1+nx}\
&=dfrac{1+nx-(n+1)x^2}{1+nx}\
< 1\
end{array}
$
answered Jan 10 at 22:08
marty cohenmarty cohen
73.5k549128
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Apply again the base case: $1-x<displaystylefrac1{1+x}$ and that $x^2>0$ to get
$$frac{1-x}{1+nx} < frac1{(1+nx)(1+x)}=frac1{1+(n+1)x+nx^2}<frac1{1+(n+1)x},.$$
answered Jan 28 '15 at 20:57
BerciBerci
60.6k23672
$endgroup$
$begingroup$
Is there nothing wrong with starting with $(1-x)^{n+1}$?
$endgroup$
– shinzou
Jan 28 '15 at 21:00
add a comment |
$begingroup$
Apply again the base case: $1-x<displaystylefrac1{1+x}$ and that $x^2>0$ to get
$$frac{1-x}{1+nx} < frac1{(1+nx)(1+x)}=frac1{1+(n+1)x+nx^2}<frac1{1+(n+1)x},.$$
answered Jan 28 '15 at 20:57
BerciBerci
60.6k23672
$endgroup$
$begingroup$
Is there nothing wrong with starting with $(1-x)^{n+1}$?
$endgroup$
– shinzou
Jan 28 '15 at 21:00
add a comment |
$begingroup$
Apply again the base case: $1-x<displaystylefrac1{1+x}$ and that $x^2>0$ to get
$$frac{1-x}{1+nx} < frac1{(1+nx)(1+x)}=frac1{1+(n+1)x+nx^2}<frac1{1+(n+1)x},.$$
answered Jan 28 '15 at 20:57
BerciBerci
60.6k23672
$endgroup$
Apply again the base case: $1-x<displaystylefrac1{1+x}$ and that $x^2>0$ to get
$$frac{1-x}{1+nx} < frac1{(1+nx)(1+x)}=frac1{1+(n+1)x+nx^2}<frac1{1+(n+1)x},.$$
answered Jan 28 '15 at 20:57
BerciBerci
60.6k23672
answered Jan 28 '15 at 20:57
BerciBerci
60.6k23672
answered Jan 28 '15 at 20:57
BerciBerci
60.6k23672
answered Jan 28 '15 at 20:57
BerciBerci
60.6k23672
60.6k23672
$begingroup$
Is there nothing wrong with starting with $(1-x)^{n+1}$?
$endgroup$
– shinzou
Jan 28 '15 at 21:00
add a comment |
$begingroup$
Is there nothing wrong with starting with $(1-x)^{n+1}$?
$endgroup$
– shinzou
Jan 28 '15 at 21:00
$begingroup$
Is there nothing wrong with starting with $(1-x)^{n+1}$?
$endgroup$
– shinzou
Jan 28 '15 at 21:00
$begingroup$
Is there nothing wrong with starting with $(1-x)^{n+1}$?
$endgroup$
– shinzou
Jan 28 '15 at 21:00
add a comment |
$begingroup$
Notice that there is no need of induction. By the AM-GM inequality,
$$ (1+nx)(1-x)^n < left(frac{(1+nx)+n(1-x)}{n+1}right)^{n+1} =1.$$
answered Jan 28 '15 at 22:15
Jack D'AurizioJack D'Aurizio
289k33281661
$endgroup$
1
$begingroup$
Ingenious. Similar to the proofs that $(1+1/n)^n$ is increasing and $(1+1/n)^{n+1}$ is decreasing (and so both approach a common limit, which I propose to call "$e$").
$endgroup$
– marty cohen
Jan 10 at 21:59
$begingroup$
The proofs are in N.S Mendelsohn, An application of a famous inequality, Amer. Math. Monthly 58 (1951), 563.
$endgroup$
– marty cohen
Jan 10 at 22:03
add a comment |
$begingroup$
Notice that there is no need of induction. By the AM-GM inequality,
$$ (1+nx)(1-x)^n < left(frac{(1+nx)+n(1-x)}{n+1}right)^{n+1} =1.$$
answered Jan 28 '15 at 22:15
Jack D'AurizioJack D'Aurizio
289k33281661
$endgroup$
1
$begingroup$
Ingenious. Similar to the proofs that $(1+1/n)^n$ is increasing and $(1+1/n)^{n+1}$ is decreasing (and so both approach a common limit, which I propose to call "$e$").
$endgroup$
– marty cohen
Jan 10 at 21:59
$begingroup$
The proofs are in N.S Mendelsohn, An application of a famous inequality, Amer. Math. Monthly 58 (1951), 563.
$endgroup$
– marty cohen
Jan 10 at 22:03
add a comment |
$begingroup$
Notice that there is no need of induction. By the AM-GM inequality,
$$ (1+nx)(1-x)^n < left(frac{(1+nx)+n(1-x)}{n+1}right)^{n+1} =1.$$
answered Jan 28 '15 at 22:15
Jack D'AurizioJack D'Aurizio
289k33281661
$endgroup$
Notice that there is no need of induction. By the AM-GM inequality,
$$ (1+nx)(1-x)^n < left(frac{(1+nx)+n(1-x)}{n+1}right)^{n+1} =1.$$
answered Jan 28 '15 at 22:15
Jack D'AurizioJack D'Aurizio
289k33281661
answered Jan 28 '15 at 22:15
Jack D'AurizioJack D'Aurizio
289k33281661
answered Jan 28 '15 at 22:15
Jack D'AurizioJack D'Aurizio
289k33281661
answered Jan 28 '15 at 22:15
Jack D'AurizioJack D'Aurizio
289k33281661
289k33281661
1
$begingroup$
Ingenious. Similar to the proofs that $(1+1/n)^n$ is increasing and $(1+1/n)^{n+1}$ is decreasing (and so both approach a common limit, which I propose to call "$e$").
$endgroup$
– marty cohen
Jan 10 at 21:59
$begingroup$
The proofs are in N.S Mendelsohn, An application of a famous inequality, Amer. Math. Monthly 58 (1951), 563.
$endgroup$
– marty cohen
Jan 10 at 22:03
add a comment |
1
$begingroup$
Ingenious. Similar to the proofs that $(1+1/n)^n$ is increasing and $(1+1/n)^{n+1}$ is decreasing (and so both approach a common limit, which I propose to call "$e$").
$endgroup$
– marty cohen
Jan 10 at 21:59
$begingroup$
The proofs are in N.S Mendelsohn, An application of a famous inequality, Amer. Math. Monthly 58 (1951), 563.
$endgroup$
– marty cohen
Jan 10 at 22:03
1
1
$begingroup$
Ingenious. Similar to the proofs that $(1+1/n)^n$ is increasing and $(1+1/n)^{n+1}$ is decreasing (and so both approach a common limit, which I propose to call "$e$").
$endgroup$
– marty cohen
Jan 10 at 21:59
$begingroup$
Ingenious. Similar to the proofs that $(1+1/n)^n$ is increasing and $(1+1/n)^{n+1}$ is decreasing (and so both approach a common limit, which I propose to call "$e$").
$endgroup$
– marty cohen
Jan 10 at 21:59
$begingroup$
The proofs are in N.S Mendelsohn, An application of a famous inequality, Amer. Math. Monthly 58 (1951), 563.
$endgroup$
– marty cohen
Jan 10 at 22:03
$begingroup$
The proofs are in N.S Mendelsohn, An application of a famous inequality, Amer. Math. Monthly 58 (1951), 563.
$endgroup$
– marty cohen
Jan 10 at 22:03
add a comment |
$begingroup$
Write it in the form
$(1+nx)(1-x)^n
lt 1$.
If true for $n$, then
$begin{array}\
(1+(n+1)x)(1-x)^{n+1}
&=(1+(n+1)x)(1-x)^{n+1}\
&=dfrac{(1+(n+1)x)}{1+nx}(1+nx)(1-x)^{n+1}\
&=dfrac{(1+(n+1)x)(1-x)}{1+nx}(1+nx)(1-x)^{n}\
&<dfrac{(1+(n+1)x)(1-x)}{1+nx}\
&=dfrac{1+nx-(n+1)x^2}{1+nx}\
< 1\
end{array}
$
answered Jan 10 at 22:08
marty cohenmarty cohen
73.5k549128
$endgroup$
add a comment |
$begingroup$
Write it in the form
$(1+nx)(1-x)^n
lt 1$.
If true for $n$, then
$begin{array}\
(1+(n+1)x)(1-x)^{n+1}
&=(1+(n+1)x)(1-x)^{n+1}\
&=dfrac{(1+(n+1)x)}{1+nx}(1+nx)(1-x)^{n+1}\
&=dfrac{(1+(n+1)x)(1-x)}{1+nx}(1+nx)(1-x)^{n}\
&<dfrac{(1+(n+1)x)(1-x)}{1+nx}\
&=dfrac{1+nx-(n+1)x^2}{1+nx}\
< 1\
end{array}
$
answered Jan 10 at 22:08
marty cohenmarty cohen
73.5k549128
$endgroup$
add a comment |
$begingroup$
Write it in the form
$(1+nx)(1-x)^n
lt 1$.
If true for $n$, then
$begin{array}\
(1+(n+1)x)(1-x)^{n+1}
&=(1+(n+1)x)(1-x)^{n+1}\
&=dfrac{(1+(n+1)x)}{1+nx}(1+nx)(1-x)^{n+1}\
&=dfrac{(1+(n+1)x)(1-x)}{1+nx}(1+nx)(1-x)^{n}\
&<dfrac{(1+(n+1)x)(1-x)}{1+nx}\
&=dfrac{1+nx-(n+1)x^2}{1+nx}\
< 1\
end{array}
$
answered Jan 10 at 22:08
marty cohenmarty cohen
73.5k549128
$endgroup$
Write it in the form
$(1+nx)(1-x)^n
lt 1$.
If true for $n$, then
$begin{array}\
(1+(n+1)x)(1-x)^{n+1}
&=(1+(n+1)x)(1-x)^{n+1}\
&=dfrac{(1+(n+1)x)}{1+nx}(1+nx)(1-x)^{n+1}\
&=dfrac{(1+(n+1)x)(1-x)}{1+nx}(1+nx)(1-x)^{n}\
&<dfrac{(1+(n+1)x)(1-x)}{1+nx}\
&=dfrac{1+nx-(n+1)x^2}{1+nx}\
< 1\
end{array}
$
answered Jan 10 at 22:08
marty cohenmarty cohen
73.5k549128
answered Jan 10 at 22:08
marty cohenmarty cohen
73.5k549128
answered Jan 10 at 22:08
marty cohenmarty cohen
73.5k549128
answered Jan 10 at 22:08
marty cohenmarty cohen
73.5k549128
73.5k549128
add a comment |
add a comment |
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$begingroup$
The first $to$ should rather be an $iff$.
$endgroup$
– Pedro Tamaroff♦
Jan 28 '15 at 20:52
$begingroup$
@PedroTamaroff why is it wrong to write just $rightarrow$? isn't it enough for this case?
$endgroup$
– shinzou
Jan 28 '15 at 20:53
1
$begingroup$
You want to show that if $0<x<1$, $1-x<frac{1}{1+x}$. You cannot assume what you want to show is true, show it implies a true fact and conclude that what you assumed is true is... true. Your logic is "Let's show $P$. Assume $P$. Since $P$ implies $Q$ and $Q$ is true, $P$ is true."
$endgroup$
– Pedro Tamaroff♦
Jan 28 '15 at 20:54
$begingroup$
Oh yes I see that $frac{1-x}{1+nx}<frac{1}{1+nx}$ the problem is with the denominator.
$endgroup$
– shinzou
Jan 28 '15 at 20:57
$begingroup$
Note that $$frac{1-x}{1+nx}<frac{1}{1+(n+1)x}$$ is true if and only if $-(n+1)x^2<0$ (just cross multiply and play around a while). Since $x^2>0$ and $n+1>0$, you're done.
$endgroup$
– Pedro Tamaroff♦
Jan 28 '15 at 20:57