Proof that function is increasing
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How to prove that the function $$y(x)=x(ln(x+1) - ln(x))$$ is increasing on $[0,1]$?
The derivative test requires to analyze equally challenging function $ln{left(frac{x+1}{x}right)}-frac{1}{x+1}.$ Are there more ways to prove that $y(x)$ is strictly increasing?
monotone-functions
edited Nov 14 at 10:35
Jimmy R.
32.8k42156
asked Nov 14 at 10:29
user314849
525
add a comment |
up vote
2
down vote
favorite
How to prove that the function $$y(x)=x(ln(x+1) - ln(x))$$ is increasing on $[0,1]$?
The derivative test requires to analyze equally challenging function $ln{left(frac{x+1}{x}right)}-frac{1}{x+1}.$ Are there more ways to prove that $y(x)$ is strictly increasing?
monotone-functions
edited Nov 14 at 10:35
Jimmy R.
32.8k42156
asked Nov 14 at 10:29
user314849
525
Use $log(frac{x+1}{x})=-log(1-frac{1}{x+1})=sum_{kgeq 1}frac{1}{k(x+1)^k}$.
– Song
Nov 14 at 11:17
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
How to prove that the function $$y(x)=x(ln(x+1) - ln(x))$$ is increasing on $[0,1]$?
The derivative test requires to analyze equally challenging function $ln{left(frac{x+1}{x}right)}-frac{1}{x+1}.$ Are there more ways to prove that $y(x)$ is strictly increasing?
monotone-functions
edited Nov 14 at 10:35
Jimmy R.
32.8k42156
asked Nov 14 at 10:29
user314849
525
How to prove that the function $$y(x)=x(ln(x+1) - ln(x))$$ is increasing on $[0,1]$?
The derivative test requires to analyze equally challenging function $ln{left(frac{x+1}{x}right)}-frac{1}{x+1}.$ Are there more ways to prove that $y(x)$ is strictly increasing?
monotone-functions
monotone-functions
edited Nov 14 at 10:35
Jimmy R.
32.8k42156
asked Nov 14 at 10:29
user314849
525
edited Nov 14 at 10:35
Jimmy R.
32.8k42156
asked Nov 14 at 10:29
user314849
525
edited Nov 14 at 10:35
Jimmy R.
32.8k42156
edited Nov 14 at 10:35
Jimmy R.
32.8k42156
edited Nov 14 at 10:35
Jimmy R.
32.8k42156
32.8k42156
asked Nov 14 at 10:29
user314849
525
asked Nov 14 at 10:29
user314849
525
asked Nov 14 at 10:29
user314849
525
525
Use $log(frac{x+1}{x})=-log(1-frac{1}{x+1})=sum_{kgeq 1}frac{1}{k(x+1)^k}$.
– Song
Nov 14 at 11:17
add a comment |
Use $log(frac{x+1}{x})=-log(1-frac{1}{x+1})=sum_{kgeq 1}frac{1}{k(x+1)^k}$.
– Song
Nov 14 at 11:17
Use $log(frac{x+1}{x})=-log(1-frac{1}{x+1})=sum_{kgeq 1}frac{1}{k(x+1)^k}$.
– Song
Nov 14 at 11:17
Use $log(frac{x+1}{x})=-log(1-frac{1}{x+1})=sum_{kgeq 1}frac{1}{k(x+1)^k}$.
– Song
Nov 14 at 11:17
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
Write your derivative as $$ln{left(1+frac1xright)}-frac{frac1x}{1+frac1x}$$ and use the inequality $$ln{(1+x)}>frac{x}{1+x}$$ for $0<x<1$.
Edit: By definition $$ln{(1+x)}:=int_{1}^{1+x}frac1tdt>frac1{1+x}int_{1}^{1+x}dt=frac{x}{1+x}$$ where the inequality in the middle comes from the fact that $1/t$ is strictly decreasing.
answered Nov 14 at 10:35
Jimmy R.
32.8k42156
Could you, please, elaborate where this inequality comes from?
– user314849
Nov 14 at 10:40
... which is just a reformulation of $e^xge 1+x$
– Hagen von Eitzen
Nov 14 at 10:40
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Write your derivative as $$ln{left(1+frac1xright)}-frac{frac1x}{1+frac1x}$$ and use the inequality $$ln{(1+x)}>frac{x}{1+x}$$ for $0<x<1$.
Edit: By definition $$ln{(1+x)}:=int_{1}^{1+x}frac1tdt>frac1{1+x}int_{1}^{1+x}dt=frac{x}{1+x}$$ where the inequality in the middle comes from the fact that $1/t$ is strictly decreasing.
answered Nov 14 at 10:35
Jimmy R.
32.8k42156
Could you, please, elaborate where this inequality comes from?
– user314849
Nov 14 at 10:40
... which is just a reformulation of $e^xge 1+x$
– Hagen von Eitzen
Nov 14 at 10:40
add a comment |
up vote
1
down vote
accepted
Write your derivative as $$ln{left(1+frac1xright)}-frac{frac1x}{1+frac1x}$$ and use the inequality $$ln{(1+x)}>frac{x}{1+x}$$ for $0<x<1$.
Edit: By definition $$ln{(1+x)}:=int_{1}^{1+x}frac1tdt>frac1{1+x}int_{1}^{1+x}dt=frac{x}{1+x}$$ where the inequality in the middle comes from the fact that $1/t$ is strictly decreasing.
answered Nov 14 at 10:35
Jimmy R.
32.8k42156
Could you, please, elaborate where this inequality comes from?
– user314849
Nov 14 at 10:40
... which is just a reformulation of $e^xge 1+x$
– Hagen von Eitzen
Nov 14 at 10:40
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Write your derivative as $$ln{left(1+frac1xright)}-frac{frac1x}{1+frac1x}$$ and use the inequality $$ln{(1+x)}>frac{x}{1+x}$$ for $0<x<1$.
Edit: By definition $$ln{(1+x)}:=int_{1}^{1+x}frac1tdt>frac1{1+x}int_{1}^{1+x}dt=frac{x}{1+x}$$ where the inequality in the middle comes from the fact that $1/t$ is strictly decreasing.
answered Nov 14 at 10:35
Jimmy R.
32.8k42156
Write your derivative as $$ln{left(1+frac1xright)}-frac{frac1x}{1+frac1x}$$ and use the inequality $$ln{(1+x)}>frac{x}{1+x}$$ for $0<x<1$.
Edit: By definition $$ln{(1+x)}:=int_{1}^{1+x}frac1tdt>frac1{1+x}int_{1}^{1+x}dt=frac{x}{1+x}$$ where the inequality in the middle comes from the fact that $1/t$ is strictly decreasing.
answered Nov 14 at 10:35
Jimmy R.
32.8k42156
edited Nov 14 at 10:43
answered Nov 14 at 10:35
Jimmy R.
32.8k42156
answered Nov 14 at 10:35
Jimmy R.
32.8k42156
answered Nov 14 at 10:35
Jimmy R.
32.8k42156
32.8k42156
Could you, please, elaborate where this inequality comes from?
– user314849
Nov 14 at 10:40
... which is just a reformulation of $e^xge 1+x$
– Hagen von Eitzen
Nov 14 at 10:40
add a comment |
Could you, please, elaborate where this inequality comes from?
– user314849
Nov 14 at 10:40
... which is just a reformulation of $e^xge 1+x$
– Hagen von Eitzen
Nov 14 at 10:40
Could you, please, elaborate where this inequality comes from?
– user314849
Nov 14 at 10:40
Could you, please, elaborate where this inequality comes from?
– user314849
Nov 14 at 10:40
... which is just a reformulation of $e^xge 1+x$
– Hagen von Eitzen
Nov 14 at 10:40
... which is just a reformulation of $e^xge 1+x$
– Hagen von Eitzen
Nov 14 at 10:40
add a comment |
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Use $log(frac{x+1}{x})=-log(1-frac{1}{x+1})=sum_{kgeq 1}frac{1}{k(x+1)^k}$.
– Song
Nov 14 at 11:17