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Why do these expression tend to two? [closed]

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-3

$begingroup$

Sorry if I express myself wrongly...

Why in a formula like this

$x^n-x^n-1-x^n-2-x^n-3-x^n-4-...-x^0=0$

with $n to infty$

$x to 2$ ?

calculus geometric-series

share|cite|improve this question

edited Mar 23 at 7:00

Zibri

asked Mar 22 at 14:53

ZibriZibri

953

$endgroup$

closed as unclear what you're asking by José Carlos Santos, Gibbs, Lee Mosher, John Omielan, RRL Mar 23 at 19:48

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  This does not make sense. Can you give more context?
  $endgroup$
  – st.math
  Mar 22 at 14:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  it is unclear, maybe you can give an example straight from textbook? But either way, in the case $n=1$, i dont think it is true since $x=1$
  $endgroup$
  – NazimJ
  Mar 22 at 14:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What you might mean it something like this: Let $x_n$ be the (largest positive?) solution of the equation $x^n-x^n-1-x^n-2-x^n-3-x^n-3-...-1=0$. Then $x_n to 2$ for $n to infty$.
  $endgroup$
  – Martin R
  Mar 22 at 15:01
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Related: math.stackexchange.com/q/648909/42969, math.stackexchange.com/q/976309/42969
  $endgroup$
  – Martin R
  Mar 22 at 15:27
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  sorry.. a typo.... I corrected the formula.
  $endgroup$
  – Zibri
  Mar 23 at 7:01
  

  
  
  
  

 | 
show 1 more comment

-3

$begingroup$

Sorry if I express myself wrongly...

Why in a formula like this

$x^n-x^n-1-x^n-2-x^n-3-x^n-4-...-x^0=0$

with $n to infty$

$x to 2$ ?

calculus geometric-series

share|cite|improve this question

edited Mar 23 at 7:00

Zibri

asked Mar 22 at 14:53

ZibriZibri

953

$endgroup$

closed as unclear what you're asking by José Carlos Santos, Gibbs, Lee Mosher, John Omielan, RRL Mar 23 at 19:48

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  This does not make sense. Can you give more context?
  $endgroup$
  – st.math
  Mar 22 at 14:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  it is unclear, maybe you can give an example straight from textbook? But either way, in the case $n=1$, i dont think it is true since $x=1$
  $endgroup$
  – NazimJ
  Mar 22 at 14:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What you might mean it something like this: Let $x_n$ be the (largest positive?) solution of the equation $x^n-x^n-1-x^n-2-x^n-3-x^n-3-...-1=0$. Then $x_n to 2$ for $n to infty$.
  $endgroup$
  – Martin R
  Mar 22 at 15:01
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Related: math.stackexchange.com/q/648909/42969, math.stackexchange.com/q/976309/42969
  $endgroup$
  – Martin R
  Mar 22 at 15:27
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  sorry.. a typo.... I corrected the formula.
  $endgroup$
  – Zibri
  Mar 23 at 7:01
  

  
  
  
  

 | 
show 1 more comment

$begingroup$

Sorry if I express myself wrongly...

Why in a formula like this

$x^n-x^n-1-x^n-2-x^n-3-x^n-4-...-x^0=0$

with $n to infty$

$x to 2$ ?

calculus geometric-series

share|cite|improve this question

edited Mar 23 at 7:00

Zibri

asked Mar 22 at 14:53

ZibriZibri

953

$endgroup$

share|cite|improve this question

edited Mar 23 at 7:00

Zibri

asked Mar 22 at 14:53

ZibriZibri

953

share|cite|improve this question

edited Mar 23 at 7:00

Zibri

asked Mar 22 at 14:53

ZibriZibri

953

asked Mar 22 at 14:53

ZibriZibri

953

asked Mar 22 at 14:53

ZibriZibri

953

2 Answers
2

active

oldest

votes

1

$begingroup$

Observe that

$$2-1=1,$$
$$2^2-2-1=1,$$
$$2^3-2^2-2-1=1,$$

$$2^4-2^3-2^2-2-1=1,$$
$$cdots$$
$$2^n-2^n-1-2^n-2cdots-2-1=1.$$

In every case, there is a solution $x$ such that the RHS is $0$, which must be close to $2$. As $(2-epsilon)^n$ decreases faster with increasing $n$, $epsilon$ must decrease with $n$.

The plot below shows you the expression for increasing values of $n$.

share|cite|improve this answer

answered Mar 22 at 15:11

Yves DaoustYves Daoust

132k676230

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  note: with n=2 x=φ
  $endgroup$
  – Zibri
  Mar 23 at 7:03
  
  

  
  
  
  

add a comment | 

0

$begingroup$

For $x=2-varepsilon$, the term gets
$$(2-varepsilon)^n-sum_k=0^n-1(2-varepsilon)^k=(2-varepsilon)^n-frac1-(2-varepsilon)^n1-(2-varepsilon)=(2- varepsilon)^n-frac1-(2-varepsilon)^nvarepsilon-1=frac(2-varepsilon)^ncdot(varepsilon-1)-1+(2-varepsilon)^nvarepsilon-1=fracvarepsilon(2-varepsilon)^n-1varepsilon-1.$$
The last term becomes $0$ if $varepsilon(2-varepsilon)^nrightarrow1$ or $$n>log_2-varepsilonfrac1varepsilonrightarrowinfty.$$

So, for $nrightarrowinfty$, the term converges to $0$ if $varepsilon$ is sufficiently small.

share|cite|improve this answer

answered Mar 22 at 15:48

st.mathst.math

1,113115

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I corrected the question... there was a typo in the formula.
  $endgroup$
  – Zibri
  Mar 23 at 7:04
  

  
  
  
  

add a comment | 

1

$begingroup$

Observe that

$$2-1=1,$$
$$2^2-2-1=1,$$
$$2^3-2^2-2-1=1,$$

$$2^4-2^3-2^2-2-1=1,$$
$$cdots$$
$$2^n-2^n-1-2^n-2cdots-2-1=1.$$

In every case, there is a solution $x$ such that the RHS is $0$, which must be close to $2$. As $(2-epsilon)^n$ decreases faster with increasing $n$, $epsilon$ must decrease with $n$.

The plot below shows you the expression for increasing values of $n$.

share|cite|improve this answer

answered Mar 22 at 15:11

Yves DaoustYves Daoust

132k676230

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  note: with n=2 x=φ
  $endgroup$
  – Zibri
  Mar 23 at 7:03
  
  

  
  
  
  

add a comment | 

1

$begingroup$

Observe that

$$2-1=1,$$
$$2^2-2-1=1,$$
$$2^3-2^2-2-1=1,$$

$$2^4-2^3-2^2-2-1=1,$$
$$cdots$$
$$2^n-2^n-1-2^n-2cdots-2-1=1.$$

In every case, there is a solution $x$ such that the RHS is $0$, which must be close to $2$. As $(2-epsilon)^n$ decreases faster with increasing $n$, $epsilon$ must decrease with $n$.

The plot below shows you the expression for increasing values of $n$.

share|cite|improve this answer

answered Mar 22 at 15:11

Yves DaoustYves Daoust

132k676230

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  note: with n=2 x=φ
  $endgroup$
  – Zibri
  Mar 23 at 7:03
  
  

  
  
  
  

add a comment | 

$begingroup$

Observe that

$$2-1=1,$$
$$2^2-2-1=1,$$
$$2^3-2^2-2-1=1,$$

$$2^4-2^3-2^2-2-1=1,$$
$$cdots$$
$$2^n-2^n-1-2^n-2cdots-2-1=1.$$

In every case, there is a solution $x$ such that the RHS is $0$, which must be close to $2$. As $(2-epsilon)^n$ decreases faster with increasing $n$, $epsilon$ must decrease with $n$.

The plot below shows you the expression for increasing values of $n$.

share|cite|improve this answer

answered Mar 22 at 15:11

Yves DaoustYves Daoust

132k676230

$endgroup$

share|cite|improve this answer

answered Mar 22 at 15:11

Yves DaoustYves Daoust

132k676230

answered Mar 22 at 15:11

Yves DaoustYves Daoust

132k676230

answered Mar 22 at 15:11

Yves DaoustYves Daoust

132k676230

0

$begingroup$

For $x=2-varepsilon$, the term gets
$$(2-varepsilon)^n-sum_k=0^n-1(2-varepsilon)^k=(2-varepsilon)^n-frac1-(2-varepsilon)^n1-(2-varepsilon)=(2- varepsilon)^n-frac1-(2-varepsilon)^nvarepsilon-1=frac(2-varepsilon)^ncdot(varepsilon-1)-1+(2-varepsilon)^nvarepsilon-1=fracvarepsilon(2-varepsilon)^n-1varepsilon-1.$$
The last term becomes $0$ if $varepsilon(2-varepsilon)^nrightarrow1$ or $$n>log_2-varepsilonfrac1varepsilonrightarrowinfty.$$

So, for $nrightarrowinfty$, the term converges to $0$ if $varepsilon$ is sufficiently small.

share|cite|improve this answer

answered Mar 22 at 15:48

st.mathst.math

1,113115

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I corrected the question... there was a typo in the formula.
  $endgroup$
  – Zibri
  Mar 23 at 7:04
  

  
  
  
  

add a comment | 

0

$begingroup$

For $x=2-varepsilon$, the term gets
$$(2-varepsilon)^n-sum_k=0^n-1(2-varepsilon)^k=(2-varepsilon)^n-frac1-(2-varepsilon)^n1-(2-varepsilon)=(2- varepsilon)^n-frac1-(2-varepsilon)^nvarepsilon-1=frac(2-varepsilon)^ncdot(varepsilon-1)-1+(2-varepsilon)^nvarepsilon-1=fracvarepsilon(2-varepsilon)^n-1varepsilon-1.$$
The last term becomes $0$ if $varepsilon(2-varepsilon)^nrightarrow1$ or $$n>log_2-varepsilonfrac1varepsilonrightarrowinfty.$$

So, for $nrightarrowinfty$, the term converges to $0$ if $varepsilon$ is sufficiently small.

share|cite|improve this answer

answered Mar 22 at 15:48

st.mathst.math

1,113115

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I corrected the question... there was a typo in the formula.
  $endgroup$
  – Zibri
  Mar 23 at 7:04
  

  
  
  
  

add a comment | 

$begingroup$

For $x=2-varepsilon$, the term gets
$$(2-varepsilon)^n-sum_k=0^n-1(2-varepsilon)^k=(2-varepsilon)^n-frac1-(2-varepsilon)^n1-(2-varepsilon)=(2- varepsilon)^n-frac1-(2-varepsilon)^nvarepsilon-1=frac(2-varepsilon)^ncdot(varepsilon-1)-1+(2-varepsilon)^nvarepsilon-1=fracvarepsilon(2-varepsilon)^n-1varepsilon-1.$$
The last term becomes $0$ if $varepsilon(2-varepsilon)^nrightarrow1$ or $$n>log_2-varepsilonfrac1varepsilonrightarrowinfty.$$

So, for $nrightarrowinfty$, the term converges to $0$ if $varepsilon$ is sufficiently small.

share|cite|improve this answer

answered Mar 22 at 15:48

st.mathst.math

1,113115

$endgroup$

share|cite|improve this answer

answered Mar 22 at 15:48

st.mathst.math

1,113115

answered Mar 22 at 15:48

st.mathst.math

1,113115

answered Mar 22 at 15:48

st.mathst.math

1,113115