Relationship between roots

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Suppose we have the equation $x^2 + 2x + 1, x^2 + 3x + 1, ldots, x^2 + (2 + k) x + 1, k in mathbbN$, consider all the roots less than $-1$. (see Appendix)
Is it possible to find the length of any one of the neighboring segments, knowing one of them? For example, if I know the length of a red segment, can I only use the knowledge of the equation and the length of the segment to find the blue segment?

calculus real-analysis

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asked Jul 18 at 5:48

Vladislav Kharlamov

572216

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Suppose we have the equation $x^2 + 2x + 1, x^2 + 3x + 1, ldots, x^2 + (2 + k) x + 1, k in mathbbN$, consider all the roots less than $-1$. (see Appendix)
Is it possible to find the length of any one of the neighboring segments, knowing one of them? For example, if I know the length of a red segment, can I only use the knowledge of the equation and the length of the segment to find the blue segment?

calculus real-analysis

share|cite|improve this question

asked Jul 18 at 5:48

Vladislav Kharlamov

572216

add a comment | 

up vote
1
down vote

favorite

up vote
1
down vote

favorite

Suppose we have the equation $x^2 + 2x + 1, x^2 + 3x + 1, ldots, x^2 + (2 + k) x + 1, k in mathbbN$, consider all the roots less than $-1$. (see Appendix)
Is it possible to find the length of any one of the neighboring segments, knowing one of them? For example, if I know the length of a red segment, can I only use the knowledge of the equation and the length of the segment to find the blue segment?

calculus real-analysis

share|cite|improve this question

asked Jul 18 at 5:48

Vladislav Kharlamov

572216

Suppose we have the equation $x^2 + 2x + 1, x^2 + 3x + 1, ldots, x^2 + (2 + k) x + 1, k in mathbbN$, consider all the roots less than $-1$. (see Appendix)
Is it possible to find the length of any one of the neighboring segments, knowing one of them? For example, if I know the length of a red segment, can I only use the knowledge of the equation and the length of the segment to find the blue segment?

calculus real-analysis

share|cite|improve this question

asked Jul 18 at 5:48

Vladislav Kharlamov

572216

share|cite|improve this question

share|cite|improve this question

asked Jul 18 at 5:48

Vladislav Kharlamov

572216

asked Jul 18 at 5:48

Vladislav Kharlamov

572216

asked Jul 18 at 5:48

Vladislav Kharlamov

572216

572216

add a comment | 

add a comment | 

1 Answer
1

active

oldest

votes

up vote
1
down vote



accepted

Define
$$f_k(x):=x^2+(2+k)x+1$$
for $x<-1$.

When $f_k(x)=0$,
$$
x=frac-(2+k)pmsqrt(2+k)^2-42
$$

Since $k+2>2$, $$(2+k)^2-4>0$$ so we have two distinct roots. We can show that the case where $x<-1$ is when
$$x=-frac(2+k)+sqrt(2+k)^2-42$$

Define $$r_k:=-frac(2+k)+sqrt(2+k)^2-42$$

Define $$l_k:=r_k-r_k+1$$
as the lengths between the adjacent roots of $y=f_k$ and $y=f_k+1$.

We have
beginalign
l_k
&=frac(3+k)+sqrt(3+k)^2-42-frac(2+k)+sqrt(2+k)^2-42\
&=frac12left(1+sqrt(3+k)^2-4-sqrt(2+k)^2-4right)
endalign
which you can then use to calculate back your unknown $k$ from your known $l_k$ to then calculate $l_k+1$.

share|cite|improve this answer

edited Jul 18 at 6:04

answered Jul 18 at 5:59

Karn Watcharasupat

3,8192426

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you. The problem turned out a bit clumsy.
  – Vladislav Kharlamov
  Jul 18 at 6:06
  

  
  
  
  

add a comment | 

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote



accepted

Define
$$f_k(x):=x^2+(2+k)x+1$$
for $x<-1$.

When $f_k(x)=0$,
$$
x=frac-(2+k)pmsqrt(2+k)^2-42
$$

Since $k+2>2$, $$(2+k)^2-4>0$$ so we have two distinct roots. We can show that the case where $x<-1$ is when
$$x=-frac(2+k)+sqrt(2+k)^2-42$$

Define $$r_k:=-frac(2+k)+sqrt(2+k)^2-42$$

Define $$l_k:=r_k-r_k+1$$
as the lengths between the adjacent roots of $y=f_k$ and $y=f_k+1$.

We have
beginalign
l_k
&=frac(3+k)+sqrt(3+k)^2-42-frac(2+k)+sqrt(2+k)^2-42\
&=frac12left(1+sqrt(3+k)^2-4-sqrt(2+k)^2-4right)
endalign
which you can then use to calculate back your unknown $k$ from your known $l_k$ to then calculate $l_k+1$.

share|cite|improve this answer

edited Jul 18 at 6:04

answered Jul 18 at 5:59

Karn Watcharasupat

3,8192426

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you. The problem turned out a bit clumsy.
  – Vladislav Kharlamov
  Jul 18 at 6:06
  

  
  
  
  

add a comment | 

up vote
1
down vote



accepted

Define
$$f_k(x):=x^2+(2+k)x+1$$
for $x<-1$.

When $f_k(x)=0$,
$$
x=frac-(2+k)pmsqrt(2+k)^2-42
$$

Since $k+2>2$, $$(2+k)^2-4>0$$ so we have two distinct roots. We can show that the case where $x<-1$ is when
$$x=-frac(2+k)+sqrt(2+k)^2-42$$

Define $$r_k:=-frac(2+k)+sqrt(2+k)^2-42$$

Define $$l_k:=r_k-r_k+1$$
as the lengths between the adjacent roots of $y=f_k$ and $y=f_k+1$.

We have
beginalign
l_k
&=frac(3+k)+sqrt(3+k)^2-42-frac(2+k)+sqrt(2+k)^2-42\
&=frac12left(1+sqrt(3+k)^2-4-sqrt(2+k)^2-4right)
endalign
which you can then use to calculate back your unknown $k$ from your known $l_k$ to then calculate $l_k+1$.

share|cite|improve this answer

edited Jul 18 at 6:04

answered Jul 18 at 5:59

Karn Watcharasupat

3,8192426

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you. The problem turned out a bit clumsy.
  – Vladislav Kharlamov
  Jul 18 at 6:06
  

  
  
  
  

add a comment | 

up vote
1
down vote



accepted

up vote
1
down vote



accepted

Define
$$f_k(x):=x^2+(2+k)x+1$$
for $x<-1$.

When $f_k(x)=0$,
$$
x=frac-(2+k)pmsqrt(2+k)^2-42
$$

Since $k+2>2$, $$(2+k)^2-4>0$$ so we have two distinct roots. We can show that the case where $x<-1$ is when
$$x=-frac(2+k)+sqrt(2+k)^2-42$$

Define $$r_k:=-frac(2+k)+sqrt(2+k)^2-42$$

Define $$l_k:=r_k-r_k+1$$
as the lengths between the adjacent roots of $y=f_k$ and $y=f_k+1$.

We have
beginalign
l_k
&=frac(3+k)+sqrt(3+k)^2-42-frac(2+k)+sqrt(2+k)^2-42\
&=frac12left(1+sqrt(3+k)^2-4-sqrt(2+k)^2-4right)
endalign
which you can then use to calculate back your unknown $k$ from your known $l_k$ to then calculate $l_k+1$.

share|cite|improve this answer

edited Jul 18 at 6:04

answered Jul 18 at 5:59

Karn Watcharasupat

3,8192426

Define
$$f_k(x):=x^2+(2+k)x+1$$
for $x<-1$.

When $f_k(x)=0$,
$$
x=frac-(2+k)pmsqrt(2+k)^2-42
$$

Since $k+2>2$, $$(2+k)^2-4>0$$ so we have two distinct roots. We can show that the case where $x<-1$ is when
$$x=-frac(2+k)+sqrt(2+k)^2-42$$

Define $$r_k:=-frac(2+k)+sqrt(2+k)^2-42$$

Define $$l_k:=r_k-r_k+1$$
as the lengths between the adjacent roots of $y=f_k$ and $y=f_k+1$.

We have
beginalign
l_k
&=frac(3+k)+sqrt(3+k)^2-42-frac(2+k)+sqrt(2+k)^2-42\
&=frac12left(1+sqrt(3+k)^2-4-sqrt(2+k)^2-4right)
endalign
which you can then use to calculate back your unknown $k$ from your known $l_k$ to then calculate $l_k+1$.

share|cite|improve this answer

edited Jul 18 at 6:04

answered Jul 18 at 5:59

Karn Watcharasupat

3,8192426

share|cite|improve this answer

share|cite|improve this answer

edited Jul 18 at 6:04

edited Jul 18 at 6:04

edited Jul 18 at 6:04

answered Jul 18 at 5:59

Karn Watcharasupat

3,8192426

answered Jul 18 at 5:59

Karn Watcharasupat

3,8192426

answered Jul 18 at 5:59

Karn Watcharasupat

3,8192426

3,8192426

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you. The problem turned out a bit clumsy.
  – Vladislav Kharlamov
  Jul 18 at 6:06
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you. The problem turned out a bit clumsy.
  – Vladislav Kharlamov
  Jul 18 at 6:06
  

  
  
  
  

Thank you. The problem turned out a bit clumsy.
– Vladislav Kharlamov
Jul 18 at 6:06

Thank you. The problem turned out a bit clumsy.
– Vladislav Kharlamov
Jul 18 at 6:06

add a comment | 

 

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