Mixing
Dividing by greater number
My question is very simple (I think so), but I have a trouble to get the precise understanding of the following question.
Let's take into consideration the following function:
$$f(n,a,s_1,s_2,...s_n)=frac{1}{n(1+s_1)}+frac{1}{n(1+s_2)^2}+...+frac{1}{n(1+s_n)^n}-left(frac{a}{n(1+s_2)^2}+frac{2a}{n(1+s_3)^3}+...frac{(n-1)a}{n(1+s_n)^n}right)-frac{1}{(1+s_n)^n},$$
where $n$ is integer, $ain(0,1), s_iin(0,1), i=1, 2,...,n$ and $s_1<s_2<...<s_n$. The function $f$ attains positive, negative as well as zero values.
If we replace all $s_i$'s $(i=1,...,n-1)$ by $s_n$ is it correct to write the following relationship (note: $s_1<...<s_n$)
$$frac{1}{n(1+s_1)}+frac{1}{n(1+s_2)^2}+...+frac{1}{n(1+s_n)^n}-left(frac{a}{n(1+s_2)^2}+frac{2a}{n(1+s_3)^3}+frac{(n-1)a}{n(1+s_n)^n}right)-frac{1}{(1+s_n)^n}>frac{1}{n(1+s_n)}+frac{1}{n(1+s_n)^2}+...+frac{1}{n(1+s_n)^n}-left(frac{a}{n(1+s_n)^2}+frac{2a}{n(1+s_n)^3}+...frac{(n-1)a}{n(1+s_n)^n}right)-frac{1}{(1+s_n)^n}.$$
calculus inequality
asked Nov 21 '18 at 6:56
David
408
add a comment |
My question is very simple (I think so), but I have a trouble to get the precise understanding of the following question.
Let's take into consideration the following function:
$$f(n,a,s_1,s_2,...s_n)=frac{1}{n(1+s_1)}+frac{1}{n(1+s_2)^2}+...+frac{1}{n(1+s_n)^n}-left(frac{a}{n(1+s_2)^2}+frac{2a}{n(1+s_3)^3}+...frac{(n-1)a}{n(1+s_n)^n}right)-frac{1}{(1+s_n)^n},$$
where $n$ is integer, $ain(0,1), s_iin(0,1), i=1, 2,...,n$ and $s_1<s_2<...<s_n$. The function $f$ attains positive, negative as well as zero values.
If we replace all $s_i$'s $(i=1,...,n-1)$ by $s_n$ is it correct to write the following relationship (note: $s_1<...<s_n$)
$$frac{1}{n(1+s_1)}+frac{1}{n(1+s_2)^2}+...+frac{1}{n(1+s_n)^n}-left(frac{a}{n(1+s_2)^2}+frac{2a}{n(1+s_3)^3}+frac{(n-1)a}{n(1+s_n)^n}right)-frac{1}{(1+s_n)^n}>frac{1}{n(1+s_n)}+frac{1}{n(1+s_n)^2}+...+frac{1}{n(1+s_n)^n}-left(frac{a}{n(1+s_n)^2}+frac{2a}{n(1+s_n)^3}+...frac{(n-1)a}{n(1+s_n)^n}right)-frac{1}{(1+s_n)^n}.$$
calculus inequality
asked Nov 21 '18 at 6:56
David
408
1
replacing with $s_n$ will make all the terms smaller but some are added but some are subtracted so you have no idea how they compare. If you replace all the terms you add with $s_n$ and all the terms you subtract with $s_1$ you will get something definitely smaller. But it might not be the value you want.
– fleablood
Nov 27 '18 at 4:15
add a comment |
My question is very simple (I think so), but I have a trouble to get the precise understanding of the following question.
Let's take into consideration the following function:
$$f(n,a,s_1,s_2,...s_n)=frac{1}{n(1+s_1)}+frac{1}{n(1+s_2)^2}+...+frac{1}{n(1+s_n)^n}-left(frac{a}{n(1+s_2)^2}+frac{2a}{n(1+s_3)^3}+...frac{(n-1)a}{n(1+s_n)^n}right)-frac{1}{(1+s_n)^n},$$
where $n$ is integer, $ain(0,1), s_iin(0,1), i=1, 2,...,n$ and $s_1<s_2<...<s_n$. The function $f$ attains positive, negative as well as zero values.
If we replace all $s_i$'s $(i=1,...,n-1)$ by $s_n$ is it correct to write the following relationship (note: $s_1<...<s_n$)
$$frac{1}{n(1+s_1)}+frac{1}{n(1+s_2)^2}+...+frac{1}{n(1+s_n)^n}-left(frac{a}{n(1+s_2)^2}+frac{2a}{n(1+s_3)^3}+frac{(n-1)a}{n(1+s_n)^n}right)-frac{1}{(1+s_n)^n}>frac{1}{n(1+s_n)}+frac{1}{n(1+s_n)^2}+...+frac{1}{n(1+s_n)^n}-left(frac{a}{n(1+s_n)^2}+frac{2a}{n(1+s_n)^3}+...frac{(n-1)a}{n(1+s_n)^n}right)-frac{1}{(1+s_n)^n}.$$
calculus inequality
asked Nov 21 '18 at 6:56
David
408
My question is very simple (I think so), but I have a trouble to get the precise understanding of the following question.
Let's take into consideration the following function:
$$f(n,a,s_1,s_2,...s_n)=frac{1}{n(1+s_1)}+frac{1}{n(1+s_2)^2}+...+frac{1}{n(1+s_n)^n}-left(frac{a}{n(1+s_2)^2}+frac{2a}{n(1+s_3)^3}+...frac{(n-1)a}{n(1+s_n)^n}right)-frac{1}{(1+s_n)^n},$$
where $n$ is integer, $ain(0,1), s_iin(0,1), i=1, 2,...,n$ and $s_1<s_2<...<s_n$. The function $f$ attains positive, negative as well as zero values.
If we replace all $s_i$'s $(i=1,...,n-1)$ by $s_n$ is it correct to write the following relationship (note: $s_1<...<s_n$)
$$frac{1}{n(1+s_1)}+frac{1}{n(1+s_2)^2}+...+frac{1}{n(1+s_n)^n}-left(frac{a}{n(1+s_2)^2}+frac{2a}{n(1+s_3)^3}+frac{(n-1)a}{n(1+s_n)^n}right)-frac{1}{(1+s_n)^n}>frac{1}{n(1+s_n)}+frac{1}{n(1+s_n)^2}+...+frac{1}{n(1+s_n)^n}-left(frac{a}{n(1+s_n)^2}+frac{2a}{n(1+s_n)^3}+...frac{(n-1)a}{n(1+s_n)^n}right)-frac{1}{(1+s_n)^n}.$$
calculus inequality
calculus inequality
asked Nov 21 '18 at 6:56
David
408
asked Nov 21 '18 at 6:56
David
408
edited Nov 23 '18 at 15:17
asked Nov 21 '18 at 6:56
David
408
asked Nov 21 '18 at 6:56
David
408
asked Nov 21 '18 at 6:56
David
408
408
1
replacing with $s_n$ will make all the terms smaller but some are added but some are subtracted so you have no idea how they compare. If you replace all the terms you add with $s_n$ and all the terms you subtract with $s_1$ you will get something definitely smaller. But it might not be the value you want.
– fleablood
Nov 27 '18 at 4:15
add a comment |
1
replacing with $s_n$ will make all the terms smaller but some are added but some are subtracted so you have no idea how they compare. If you replace all the terms you add with $s_n$ and all the terms you subtract with $s_1$ you will get something definitely smaller. But it might not be the value you want.
– fleablood
Nov 27 '18 at 4:15
1
1
replacing with $s_n$ will make all the terms smaller but some are added but some are subtracted so you have no idea how they compare. If you replace all the terms you add with $s_n$ and all the terms you subtract with $s_1$ you will get something definitely smaller. But it might not be the value you want.
– fleablood
Nov 27 '18 at 4:15
replacing with $s_n$ will make all the terms smaller but some are added but some are subtracted so you have no idea how they compare. If you replace all the terms you add with $s_n$ and all the terms you subtract with $s_1$ you will get something definitely smaller. But it might not be the value you want.
– fleablood
Nov 27 '18 at 4:15
add a comment |
1 Answer
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It is easy to check that the inequality you asking is equivalent to
$$g(n,a,s_1,s_2,dots,s_n)=sum_{i=1}^n (1-(i-1)a)left(frac{1}{(1+s_i)^i}-frac{1}{(1+s_n)^i}right)>0.$$
This may be wrong for $a$ close to 1. For instance, MathCad calculates $g(4,0.99,0.05,0.1,0.15,0.9)simeq –0.07$.
answered Nov 27 '18 at 3:59
Alex Ravsky
39.3k32181
1
Thank you for your answer. Let me continue the work you have done.
– David
Nov 27 '18 at 6:36
If we assume that $1-(n-1)a>0$, therefore this inequality holds for every $i's$. From another side, we have that $frac{1}{(1+s_i)^i}-frac{1}{(1+s_n)^i}>0$ for every $i=1,...,n-1$, therefore the function $g(n,a,s_1,....,s_n)>0$. Do you agree?
– David
Nov 27 '18 at 6:39
1
@David Right. ...
– Alex Ravsky
Nov 27 '18 at 6:41
Please note, that we can write the function $g()$ in the following way: $g(n,a,s_1,s_2,...,s_n)=sum_{i=1}^{n-1} frac{1-(i-1a)}{n}left (frac{1}{(1+s_i)^i}-frac{1}{(1+s_n)^i}right)$
– David
Nov 27 '18 at 6:46
add a comment |
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It is easy to check that the inequality you asking is equivalent to
$$g(n,a,s_1,s_2,dots,s_n)=sum_{i=1}^n (1-(i-1)a)left(frac{1}{(1+s_i)^i}-frac{1}{(1+s_n)^i}right)>0.$$
This may be wrong for $a$ close to 1. For instance, MathCad calculates $g(4,0.99,0.05,0.1,0.15,0.9)simeq –0.07$.
answered Nov 27 '18 at 3:59
Alex Ravsky
39.3k32181
1
Thank you for your answer. Let me continue the work you have done.
– David
Nov 27 '18 at 6:36
If we assume that $1-(n-1)a>0$, therefore this inequality holds for every $i's$. From another side, we have that $frac{1}{(1+s_i)^i}-frac{1}{(1+s_n)^i}>0$ for every $i=1,...,n-1$, therefore the function $g(n,a,s_1,....,s_n)>0$. Do you agree?
– David
Nov 27 '18 at 6:39
1
@David Right. ...
– Alex Ravsky
Nov 27 '18 at 6:41
Please note, that we can write the function $g()$ in the following way: $g(n,a,s_1,s_2,...,s_n)=sum_{i=1}^{n-1} frac{1-(i-1a)}{n}left (frac{1}{(1+s_i)^i}-frac{1}{(1+s_n)^i}right)$
– David
Nov 27 '18 at 6:46
add a comment |
It is easy to check that the inequality you asking is equivalent to
$$g(n,a,s_1,s_2,dots,s_n)=sum_{i=1}^n (1-(i-1)a)left(frac{1}{(1+s_i)^i}-frac{1}{(1+s_n)^i}right)>0.$$
This may be wrong for $a$ close to 1. For instance, MathCad calculates $g(4,0.99,0.05,0.1,0.15,0.9)simeq –0.07$.
answered Nov 27 '18 at 3:59
Alex Ravsky
39.3k32181
1
Thank you for your answer. Let me continue the work you have done.
– David
Nov 27 '18 at 6:36
If we assume that $1-(n-1)a>0$, therefore this inequality holds for every $i's$. From another side, we have that $frac{1}{(1+s_i)^i}-frac{1}{(1+s_n)^i}>0$ for every $i=1,...,n-1$, therefore the function $g(n,a,s_1,....,s_n)>0$. Do you agree?
– David
Nov 27 '18 at 6:39
1
@David Right. ...
– Alex Ravsky
Nov 27 '18 at 6:41
Please note, that we can write the function $g()$ in the following way: $g(n,a,s_1,s_2,...,s_n)=sum_{i=1}^{n-1} frac{1-(i-1a)}{n}left (frac{1}{(1+s_i)^i}-frac{1}{(1+s_n)^i}right)$
– David
Nov 27 '18 at 6:46
add a comment |
It is easy to check that the inequality you asking is equivalent to
$$g(n,a,s_1,s_2,dots,s_n)=sum_{i=1}^n (1-(i-1)a)left(frac{1}{(1+s_i)^i}-frac{1}{(1+s_n)^i}right)>0.$$
This may be wrong for $a$ close to 1. For instance, MathCad calculates $g(4,0.99,0.05,0.1,0.15,0.9)simeq –0.07$.
answered Nov 27 '18 at 3:59
Alex Ravsky
39.3k32181
It is easy to check that the inequality you asking is equivalent to
$$g(n,a,s_1,s_2,dots,s_n)=sum_{i=1}^n (1-(i-1)a)left(frac{1}{(1+s_i)^i}-frac{1}{(1+s_n)^i}right)>0.$$
This may be wrong for $a$ close to 1. For instance, MathCad calculates $g(4,0.99,0.05,0.1,0.15,0.9)simeq –0.07$.
answered Nov 27 '18 at 3:59
Alex Ravsky
39.3k32181
answered Nov 27 '18 at 3:59
Alex Ravsky
39.3k32181
answered Nov 27 '18 at 3:59
Alex Ravsky
39.3k32181
answered Nov 27 '18 at 3:59
Alex Ravsky
39.3k32181
39.3k32181
1
Thank you for your answer. Let me continue the work you have done.
– David
Nov 27 '18 at 6:36
If we assume that $1-(n-1)a>0$, therefore this inequality holds for every $i's$. From another side, we have that $frac{1}{(1+s_i)^i}-frac{1}{(1+s_n)^i}>0$ for every $i=1,...,n-1$, therefore the function $g(n,a,s_1,....,s_n)>0$. Do you agree?
– David
Nov 27 '18 at 6:39
1
@David Right. ...
– Alex Ravsky
Nov 27 '18 at 6:41
Please note, that we can write the function $g()$ in the following way: $g(n,a,s_1,s_2,...,s_n)=sum_{i=1}^{n-1} frac{1-(i-1a)}{n}left (frac{1}{(1+s_i)^i}-frac{1}{(1+s_n)^i}right)$
– David
Nov 27 '18 at 6:46
add a comment |
1
Thank you for your answer. Let me continue the work you have done.
– David
Nov 27 '18 at 6:36
If we assume that $1-(n-1)a>0$, therefore this inequality holds for every $i's$. From another side, we have that $frac{1}{(1+s_i)^i}-frac{1}{(1+s_n)^i}>0$ for every $i=1,...,n-1$, therefore the function $g(n,a,s_1,....,s_n)>0$. Do you agree?
– David
Nov 27 '18 at 6:39
1
@David Right. ...
– Alex Ravsky
Nov 27 '18 at 6:41
Please note, that we can write the function $g()$ in the following way: $g(n,a,s_1,s_2,...,s_n)=sum_{i=1}^{n-1} frac{1-(i-1a)}{n}left (frac{1}{(1+s_i)^i}-frac{1}{(1+s_n)^i}right)$
– David
Nov 27 '18 at 6:46
1
1
Thank you for your answer. Let me continue the work you have done.
– David
Nov 27 '18 at 6:36
Thank you for your answer. Let me continue the work you have done.
– David
Nov 27 '18 at 6:36
If we assume that $1-(n-1)a>0$, therefore this inequality holds for every $i's$. From another side, we have that $frac{1}{(1+s_i)^i}-frac{1}{(1+s_n)^i}>0$ for every $i=1,...,n-1$, therefore the function $g(n,a,s_1,....,s_n)>0$. Do you agree?
– David
Nov 27 '18 at 6:39
If we assume that $1-(n-1)a>0$, therefore this inequality holds for every $i's$. From another side, we have that $frac{1}{(1+s_i)^i}-frac{1}{(1+s_n)^i}>0$ for every $i=1,...,n-1$, therefore the function $g(n,a,s_1,....,s_n)>0$. Do you agree?
– David
Nov 27 '18 at 6:39
1
1
@David Right. ...
– Alex Ravsky
Nov 27 '18 at 6:41
@David Right. ...
– Alex Ravsky
Nov 27 '18 at 6:41
Please note, that we can write the function $g()$ in the following way: $g(n,a,s_1,s_2,...,s_n)=sum_{i=1}^{n-1} frac{1-(i-1a)}{n}left (frac{1}{(1+s_i)^i}-frac{1}{(1+s_n)^i}right)$
– David
Nov 27 '18 at 6:46
Please note, that we can write the function $g()$ in the following way: $g(n,a,s_1,s_2,...,s_n)=sum_{i=1}^{n-1} frac{1-(i-1a)}{n}left (frac{1}{(1+s_i)^i}-frac{1}{(1+s_n)^i}right)$
– David
Nov 27 '18 at 6:46
add a comment |
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replacing with $s_n$ will make all the terms smaller but some are added but some are subtracted so you have no idea how they compare. If you replace all the terms you add with $s_n$ and all the terms you subtract with $s_1$ you will get something definitely smaller. But it might not be the value you want.
– fleablood
Nov 27 '18 at 4:15