Mixing
Integral sequence
So we have this sequence, $$I_{n}=int_0^1ln(1+x^n):dx$$ and we have to show that this is monotone and bounded.
My solution is this: $$xin[0,1]implies 0leqln(1+x^n)leqln(2)$$ and if we integrate from $0$ to $1$ we get $$0leq I_{n}leqln(2)$$ which proves that $I_{n}$ is bounded. Is this correct?
To show that the sequence is monotone I was thinking about the difference $I_{n}-I_{n+1}$. Is this correct too?$$ $$
!!!EDIT!!! Now we have to show that $I_{n}leqfrac{1}{n+1},::forall:ninmathbb{N}$ and to find $limlimits_{xtoinfty}I_n$.$$$$So, I guess is obvious that $I_nleqint_{0}^{1}x^n:dx$, right???
integration
asked Nov 20 '18 at 20:44
Numbers
1116
add a comment |
So we have this sequence, $$I_{n}=int_0^1ln(1+x^n):dx$$ and we have to show that this is monotone and bounded.
My solution is this: $$xin[0,1]implies 0leqln(1+x^n)leqln(2)$$ and if we integrate from $0$ to $1$ we get $$0leq I_{n}leqln(2)$$ which proves that $I_{n}$ is bounded. Is this correct?
To show that the sequence is monotone I was thinking about the difference $I_{n}-I_{n+1}$. Is this correct too?$$ $$
!!!EDIT!!! Now we have to show that $I_{n}leqfrac{1}{n+1},::forall:ninmathbb{N}$ and to find $limlimits_{xtoinfty}I_n$.$$$$So, I guess is obvious that $I_nleqint_{0}^{1}x^n:dx$, right???
integration
asked Nov 20 '18 at 20:44
Numbers
1116
For a fixed $xin(0,1)$, $x^{n+1}< x^n$ implies $log(1+x^{n+1})<log(1+x^n)$ and $I_n$ is obviously decreasing.
– Jack D'Aurizio
Nov 20 '18 at 20:48
add a comment |
So we have this sequence, $$I_{n}=int_0^1ln(1+x^n):dx$$ and we have to show that this is monotone and bounded.
My solution is this: $$xin[0,1]implies 0leqln(1+x^n)leqln(2)$$ and if we integrate from $0$ to $1$ we get $$0leq I_{n}leqln(2)$$ which proves that $I_{n}$ is bounded. Is this correct?
To show that the sequence is monotone I was thinking about the difference $I_{n}-I_{n+1}$. Is this correct too?$$ $$
!!!EDIT!!! Now we have to show that $I_{n}leqfrac{1}{n+1},::forall:ninmathbb{N}$ and to find $limlimits_{xtoinfty}I_n$.$$$$So, I guess is obvious that $I_nleqint_{0}^{1}x^n:dx$, right???
integration
asked Nov 20 '18 at 20:44
Numbers
1116
So we have this sequence, $$I_{n}=int_0^1ln(1+x^n):dx$$ and we have to show that this is monotone and bounded.
My solution is this: $$xin[0,1]implies 0leqln(1+x^n)leqln(2)$$ and if we integrate from $0$ to $1$ we get $$0leq I_{n}leqln(2)$$ which proves that $I_{n}$ is bounded. Is this correct?
To show that the sequence is monotone I was thinking about the difference $I_{n}-I_{n+1}$. Is this correct too?$$ $$
!!!EDIT!!! Now we have to show that $I_{n}leqfrac{1}{n+1},::forall:ninmathbb{N}$ and to find $limlimits_{xtoinfty}I_n$.$$$$So, I guess is obvious that $I_nleqint_{0}^{1}x^n:dx$, right???
integration
integration
asked Nov 20 '18 at 20:44
Numbers
1116
asked Nov 20 '18 at 20:44
Numbers
1116
edited Nov 20 '18 at 21:35
asked Nov 20 '18 at 20:44
Numbers
1116
asked Nov 20 '18 at 20:44
Numbers
1116
asked Nov 20 '18 at 20:44
Numbers
1116
1116
For a fixed $xin(0,1)$, $x^{n+1}< x^n$ implies $log(1+x^{n+1})<log(1+x^n)$ and $I_n$ is obviously decreasing.
– Jack D'Aurizio
Nov 20 '18 at 20:48
add a comment |
For a fixed $xin(0,1)$, $x^{n+1}< x^n$ implies $log(1+x^{n+1})<log(1+x^n)$ and $I_n$ is obviously decreasing.
– Jack D'Aurizio
Nov 20 '18 at 20:48
For a fixed $xin(0,1)$, $x^{n+1}< x^n$ implies $log(1+x^{n+1})<log(1+x^n)$ and $I_n$ is obviously decreasing.
– Jack D'Aurizio
Nov 20 '18 at 20:48
For a fixed $xin(0,1)$, $x^{n+1}< x^n$ implies $log(1+x^{n+1})<log(1+x^n)$ and $I_n$ is obviously decreasing.
– Jack D'Aurizio
Nov 20 '18 at 20:48
add a comment |
2 Answers
2
active
oldest
votes
For the last question. you can derivate $ln(1+x^n)-x^n$ prove it's decreasing and then for x=0 it's equal 0 so in [0;1] the difference is negative.
then you need to integrate $x^n$ and for the limite use sandwich theorem( also called gent d'arme)
answered Nov 20 '18 at 21:38
Amin Sassi
162
actually, this is what I did !! thx
– Numbers
Nov 20 '18 at 22:13
add a comment |
What you did is correct. In order to show that $(I_n)_{ninmathbb N}$ is monotonic, you do$$bigl(forall xin[0,1]bigr):1+x^{n+1}leqslant1+x^nimpliesint_0^1ln(1+x^{n+1}),mathrm dxleqslantint_0^1ln(1+x^n),mathrm dx,$$since $ln$ is increasing.
answered Nov 20 '18 at 20:49
José Carlos Santos
151k22123224
Aaah, sure... easy and nice!! thx a lot!
– Numbers
Nov 20 '18 at 21:19
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
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For the last question. you can derivate $ln(1+x^n)-x^n$ prove it's decreasing and then for x=0 it's equal 0 so in [0;1] the difference is negative.
then you need to integrate $x^n$ and for the limite use sandwich theorem( also called gent d'arme)
answered Nov 20 '18 at 21:38
Amin Sassi
162
actually, this is what I did !! thx
– Numbers
Nov 20 '18 at 22:13
add a comment |
For the last question. you can derivate $ln(1+x^n)-x^n$ prove it's decreasing and then for x=0 it's equal 0 so in [0;1] the difference is negative.
then you need to integrate $x^n$ and for the limite use sandwich theorem( also called gent d'arme)
answered Nov 20 '18 at 21:38
Amin Sassi
162
actually, this is what I did !! thx
– Numbers
Nov 20 '18 at 22:13
add a comment |
For the last question. you can derivate $ln(1+x^n)-x^n$ prove it's decreasing and then for x=0 it's equal 0 so in [0;1] the difference is negative.
then you need to integrate $x^n$ and for the limite use sandwich theorem( also called gent d'arme)
answered Nov 20 '18 at 21:38
Amin Sassi
162
For the last question. you can derivate $ln(1+x^n)-x^n$ prove it's decreasing and then for x=0 it's equal 0 so in [0;1] the difference is negative.
then you need to integrate $x^n$ and for the limite use sandwich theorem( also called gent d'arme)
answered Nov 20 '18 at 21:38
Amin Sassi
162
edited Nov 20 '18 at 21:55
answered Nov 20 '18 at 21:38
Amin Sassi
162
answered Nov 20 '18 at 21:38
Amin Sassi
162
answered Nov 20 '18 at 21:38
Amin Sassi
162
162
actually, this is what I did !! thx
– Numbers
Nov 20 '18 at 22:13
add a comment |
actually, this is what I did !! thx
– Numbers
Nov 20 '18 at 22:13
actually, this is what I did !! thx
– Numbers
Nov 20 '18 at 22:13
actually, this is what I did !! thx
– Numbers
Nov 20 '18 at 22:13
add a comment |
What you did is correct. In order to show that $(I_n)_{ninmathbb N}$ is monotonic, you do$$bigl(forall xin[0,1]bigr):1+x^{n+1}leqslant1+x^nimpliesint_0^1ln(1+x^{n+1}),mathrm dxleqslantint_0^1ln(1+x^n),mathrm dx,$$since $ln$ is increasing.
answered Nov 20 '18 at 20:49
José Carlos Santos
151k22123224
Aaah, sure... easy and nice!! thx a lot!
– Numbers
Nov 20 '18 at 21:19
add a comment |
What you did is correct. In order to show that $(I_n)_{ninmathbb N}$ is monotonic, you do$$bigl(forall xin[0,1]bigr):1+x^{n+1}leqslant1+x^nimpliesint_0^1ln(1+x^{n+1}),mathrm dxleqslantint_0^1ln(1+x^n),mathrm dx,$$since $ln$ is increasing.
answered Nov 20 '18 at 20:49
José Carlos Santos
151k22123224
Aaah, sure... easy and nice!! thx a lot!
– Numbers
Nov 20 '18 at 21:19
add a comment |
What you did is correct. In order to show that $(I_n)_{ninmathbb N}$ is monotonic, you do$$bigl(forall xin[0,1]bigr):1+x^{n+1}leqslant1+x^nimpliesint_0^1ln(1+x^{n+1}),mathrm dxleqslantint_0^1ln(1+x^n),mathrm dx,$$since $ln$ is increasing.
answered Nov 20 '18 at 20:49
José Carlos Santos
151k22123224
What you did is correct. In order to show that $(I_n)_{ninmathbb N}$ is monotonic, you do$$bigl(forall xin[0,1]bigr):1+x^{n+1}leqslant1+x^nimpliesint_0^1ln(1+x^{n+1}),mathrm dxleqslantint_0^1ln(1+x^n),mathrm dx,$$since $ln$ is increasing.
answered Nov 20 '18 at 20:49
José Carlos Santos
151k22123224
answered Nov 20 '18 at 20:49
José Carlos Santos
151k22123224
answered Nov 20 '18 at 20:49
José Carlos Santos
151k22123224
answered Nov 20 '18 at 20:49
José Carlos Santos
151k22123224
151k22123224
Aaah, sure... easy and nice!! thx a lot!
– Numbers
Nov 20 '18 at 21:19
add a comment |
Aaah, sure... easy and nice!! thx a lot!
– Numbers
Nov 20 '18 at 21:19
Aaah, sure... easy and nice!! thx a lot!
– Numbers
Nov 20 '18 at 21:19
Aaah, sure... easy and nice!! thx a lot!
– Numbers
Nov 20 '18 at 21:19
add a comment |
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For a fixed $xin(0,1)$, $x^{n+1}< x^n$ implies $log(1+x^{n+1})<log(1+x^n)$ and $I_n$ is obviously decreasing.
– Jack D'Aurizio
Nov 20 '18 at 20:48