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Solving $1-x = (1-ax)^n$ for $x$ in general for some values of $a$ and $n$.

How to determine if a polynomial is of a particular order: 3rd degree (cubic), 4th degree (quartic) etc.Solving Yoshida equationsSolving $y^2 - yx - y + x = 0$ for $y$?Let $f(x)=7x^{32}+5x^{22}+3x^{12}+x^2$. Find the remainder when $x^2+1$ divides $f(x)$ and $xf(x)$.Find the values of $p$ where two lines intersectSolving a pair of quadratic equations for two unknowns efficientlySolving $x^4-15x^2-10x+ 24 = 0$ using Ferrari’s methodIf $(x+1)(x+3)(x+5)(x+7)= 5760$, what are the possible values of $x$?Solving polynomial equations by decompositionthe roots of the equation $ax^2 +bx+c=0$ are $alpha$ and $beta$ find an equation with roots $alpha + beta$ and $alpha beta$.

0

$begingroup$

I have an equation of the form $1-x = (1-ax)^n$ for $0<a<1$ and some value $n in mathbb{N}$. When substituting in the values I have for $a$ and $n$ (which are $a=0.002$ and $n=2999$ I cannot seem to solve the equation. So I tried writing it in a more general form to see if there was anything I could do with that, but I cannot think of anything. The only way I can seem to get an answer is by graphing both sides and looking where they intersect, but this is not accurate (it gives $x=1$ as a solution). Is there a trick when solving equations like this? Thank you for all your help.

polynomials

share|cite|improve this question

edited Mar 14 at 0:44

Rócherz

2,9863821

asked Mar 14 at 0:08

Andrew MarkovicAndrew Markovic

32

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  For x=1, the left side =0, while the right side is extremely small (.000006), so graph isn't too far off.
  $endgroup$
  – herb steinberg
  Mar 14 at 0:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for your quick response. I accidently mis-wrote the equation as per update. I can solve this using Wolfram and I get $x=0.9975$. The value for $x$ is a ratio so I need to find an exact value for it.
  $endgroup$
  – Andrew Markovic
  Mar 14 at 0:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Obviously, $x=0$ is a solution. And it seems that the equation has no algebraic solution by Abel–Ruffini theorem (en.wikipedia.org/wiki/Abel%E2%80%93Ruffini_theorem ). You can only get the numerical solutions by iterative methods (en.wikipedia.org/wiki/Numerical_analysis ) most of the time which is implemented by mathematical software, such as Wolfram and matlab.
  $endgroup$
  – zongxiang yi
  Mar 14 at 6:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for mentioning this Theorem I didn't realise this applied to the problem.
  $endgroup$
  – Andrew Markovic
  Mar 14 at 12:26
  

  
  
  
  

add a comment | 

0

$begingroup$

I have an equation of the form $1-x = (1-ax)^n$ for $0<a<1$ and some value $n in mathbb{N}$. When substituting in the values I have for $a$ and $n$ (which are $a=0.002$ and $n=2999$ I cannot seem to solve the equation. So I tried writing it in a more general form to see if there was anything I could do with that, but I cannot think of anything. The only way I can seem to get an answer is by graphing both sides and looking where they intersect, but this is not accurate (it gives $x=1$ as a solution). Is there a trick when solving equations like this? Thank you for all your help.

polynomials

share|cite|improve this question

edited Mar 14 at 0:44

Rócherz

2,9863821

asked Mar 14 at 0:08

Andrew MarkovicAndrew Markovic

32

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  For x=1, the left side =0, while the right side is extremely small (.000006), so graph isn't too far off.
  $endgroup$
  – herb steinberg
  Mar 14 at 0:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for your quick response. I accidently mis-wrote the equation as per update. I can solve this using Wolfram and I get $x=0.9975$. The value for $x$ is a ratio so I need to find an exact value for it.
  $endgroup$
  – Andrew Markovic
  Mar 14 at 0:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Obviously, $x=0$ is a solution. And it seems that the equation has no algebraic solution by Abel–Ruffini theorem (en.wikipedia.org/wiki/Abel%E2%80%93Ruffini_theorem ). You can only get the numerical solutions by iterative methods (en.wikipedia.org/wiki/Numerical_analysis ) most of the time which is implemented by mathematical software, such as Wolfram and matlab.
  $endgroup$
  – zongxiang yi
  Mar 14 at 6:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for mentioning this Theorem I didn't realise this applied to the problem.
  $endgroup$
  – Andrew Markovic
  Mar 14 at 12:26
  

  
  
  
  

add a comment | 

$begingroup$

I have an equation of the form $1-x = (1-ax)^n$ for $0<a<1$ and some value $n in mathbb{N}$. When substituting in the values I have for $a$ and $n$ (which are $a=0.002$ and $n=2999$ I cannot seem to solve the equation. So I tried writing it in a more general form to see if there was anything I could do with that, but I cannot think of anything. The only way I can seem to get an answer is by graphing both sides and looking where they intersect, but this is not accurate (it gives $x=1$ as a solution). Is there a trick when solving equations like this? Thank you for all your help.

polynomials

share|cite|improve this question

edited Mar 14 at 0:44

Rócherz

2,9863821

asked Mar 14 at 0:08

Andrew MarkovicAndrew Markovic

32

$endgroup$

share|cite|improve this question

edited Mar 14 at 0:44

Rócherz

2,9863821

asked Mar 14 at 0:08

Andrew MarkovicAndrew Markovic

32

share|cite|improve this question

edited Mar 14 at 0:44

Rócherz

2,9863821

asked Mar 14 at 0:08

Andrew MarkovicAndrew Markovic

32

edited Mar 14 at 0:44

Rócherz

2,9863821

edited Mar 14 at 0:44

Rócherz

2,9863821

asked Mar 14 at 0:08

Andrew MarkovicAndrew Markovic

32

asked Mar 14 at 0:08

Andrew MarkovicAndrew Markovic

32

1 Answer
1

active

oldest

votes

0

$begingroup$

Consider that you look for the zero of function $$f(x)=1-x - (1-ax)^n$$ Assuming that $n$ is a positive integer $n > 1$, this represents a polynomial of degree $n$ which cannot be solved if $n>4$. So, you need either numerical methods (such as Newton) or approximations.

If you consider the derivatives
$$f'(x)=a n (1-a x)^{n-1}-1$$
$$f''(x)=-a^2 (n-1) n (1-a x)^{n-2}$$ the first derivative cancels at
$$x_*=frac{1-left(frac{1}{a n}right)^{frac{1}{n-1}}}{a}$$ and notice that $f(0)=0$. So, assuming that this is a minimum and that $f(x_*)<0$, ther is a root which is $ > x_*$.

To get a first approximation, you could build a Taylor series around $x=x_*$ to get
$$f(x)=f(x_*)+frac 12 f''(x_*)(x-x_*)^2+O((x-x_*)^3)$$ and ignoring the higher order terms, get an estimate
$$x_0=x_*+sqrt{-2frac{f(x_*)}{f''(x_*)}}$$

Appliead to the case $a=0.002$, $n=2999$, this would give $x_0 approx 0.721$ and, using this Newton iterates would be
$$left(
begin{array}{cc}
n & x_n \
0 & 0.720871 \
1 & 1.009707 \
2 & 0.997500 \
3 & 0.997494
end{array}
right)$$ which is the solution for six significant figures.

If you know that the solution is close to $x=1$, you could expand $f(x)$ as a Taylor series at $x=1$ and get
$$f(x)=-(1-a)^n+(x-1) left(a n (1-a)^{n-1}-1right)+Oleft((x-1)^2right)$$ Using the first term only would give
$$x=1+frac{(1-a)^{n+1}}{a n (1-a)^n+a-1}$$ which, for the worked example, would immediately give $x=0.997494$.

share|cite|improve this answer

answered Mar 14 at 7:04

Claude LeiboviciClaude Leibovici

124k1158135

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  That makes perfect sense, thank you your response.
  $endgroup$
  – Andrew Markovic
  Mar 14 at 12:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @AndrewMarkovic. I am sorry ! I did not notice that you are a very new user : so, Welcome to the site ! I thank you for the problem; it is interesting and I had fun wtih it. I think that we could make better : could you tell the practical ranges of interest for $a$ and $n$ ? Cheers :-)
  $endgroup$
  – Claude Leibovici
  Mar 14 at 14:27
  
  
  

  
  
  
  

add a comment | 

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0

$begingroup$

Consider that you look for the zero of function $$f(x)=1-x - (1-ax)^n$$ Assuming that $n$ is a positive integer $n > 1$, this represents a polynomial of degree $n$ which cannot be solved if $n>4$. So, you need either numerical methods (such as Newton) or approximations.

If you consider the derivatives
$$f'(x)=a n (1-a x)^{n-1}-1$$
$$f''(x)=-a^2 (n-1) n (1-a x)^{n-2}$$ the first derivative cancels at
$$x_*=frac{1-left(frac{1}{a n}right)^{frac{1}{n-1}}}{a}$$ and notice that $f(0)=0$. So, assuming that this is a minimum and that $f(x_*)<0$, ther is a root which is $ > x_*$.

To get a first approximation, you could build a Taylor series around $x=x_*$ to get
$$f(x)=f(x_*)+frac 12 f''(x_*)(x-x_*)^2+O((x-x_*)^3)$$ and ignoring the higher order terms, get an estimate
$$x_0=x_*+sqrt{-2frac{f(x_*)}{f''(x_*)}}$$

Appliead to the case $a=0.002$, $n=2999$, this would give $x_0 approx 0.721$ and, using this Newton iterates would be
$$left(
begin{array}{cc}
n & x_n \
0 & 0.720871 \
1 & 1.009707 \
2 & 0.997500 \
3 & 0.997494
end{array}
right)$$ which is the solution for six significant figures.

If you know that the solution is close to $x=1$, you could expand $f(x)$ as a Taylor series at $x=1$ and get
$$f(x)=-(1-a)^n+(x-1) left(a n (1-a)^{n-1}-1right)+Oleft((x-1)^2right)$$ Using the first term only would give
$$x=1+frac{(1-a)^{n+1}}{a n (1-a)^n+a-1}$$ which, for the worked example, would immediately give $x=0.997494$.

share|cite|improve this answer

answered Mar 14 at 7:04

Claude LeiboviciClaude Leibovici

124k1158135

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  That makes perfect sense, thank you your response.
  $endgroup$
  – Andrew Markovic
  Mar 14 at 12:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @AndrewMarkovic. I am sorry ! I did not notice that you are a very new user : so, Welcome to the site ! I thank you for the problem; it is interesting and I had fun wtih it. I think that we could make better : could you tell the practical ranges of interest for $a$ and $n$ ? Cheers :-)
  $endgroup$
  – Claude Leibovici
  Mar 14 at 14:27
  
  
  

  
  
  
  

add a comment | 

0

$begingroup$

Consider that you look for the zero of function $$f(x)=1-x - (1-ax)^n$$ Assuming that $n$ is a positive integer $n > 1$, this represents a polynomial of degree $n$ which cannot be solved if $n>4$. So, you need either numerical methods (such as Newton) or approximations.

If you consider the derivatives
$$f'(x)=a n (1-a x)^{n-1}-1$$
$$f''(x)=-a^2 (n-1) n (1-a x)^{n-2}$$ the first derivative cancels at
$$x_*=frac{1-left(frac{1}{a n}right)^{frac{1}{n-1}}}{a}$$ and notice that $f(0)=0$. So, assuming that this is a minimum and that $f(x_*)<0$, ther is a root which is $ > x_*$.

To get a first approximation, you could build a Taylor series around $x=x_*$ to get
$$f(x)=f(x_*)+frac 12 f''(x_*)(x-x_*)^2+O((x-x_*)^3)$$ and ignoring the higher order terms, get an estimate
$$x_0=x_*+sqrt{-2frac{f(x_*)}{f''(x_*)}}$$

Appliead to the case $a=0.002$, $n=2999$, this would give $x_0 approx 0.721$ and, using this Newton iterates would be
$$left(
begin{array}{cc}
n & x_n \
0 & 0.720871 \
1 & 1.009707 \
2 & 0.997500 \
3 & 0.997494
end{array}
right)$$ which is the solution for six significant figures.

If you know that the solution is close to $x=1$, you could expand $f(x)$ as a Taylor series at $x=1$ and get
$$f(x)=-(1-a)^n+(x-1) left(a n (1-a)^{n-1}-1right)+Oleft((x-1)^2right)$$ Using the first term only would give
$$x=1+frac{(1-a)^{n+1}}{a n (1-a)^n+a-1}$$ which, for the worked example, would immediately give $x=0.997494$.

share|cite|improve this answer

answered Mar 14 at 7:04

Claude LeiboviciClaude Leibovici

124k1158135

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  That makes perfect sense, thank you your response.
  $endgroup$
  – Andrew Markovic
  Mar 14 at 12:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @AndrewMarkovic. I am sorry ! I did not notice that you are a very new user : so, Welcome to the site ! I thank you for the problem; it is interesting and I had fun wtih it. I think that we could make better : could you tell the practical ranges of interest for $a$ and $n$ ? Cheers :-)
  $endgroup$
  – Claude Leibovici
  Mar 14 at 14:27
  
  
  

  
  
  
  

add a comment | 

$begingroup$

Consider that you look for the zero of function $$f(x)=1-x - (1-ax)^n$$ Assuming that $n$ is a positive integer $n > 1$, this represents a polynomial of degree $n$ which cannot be solved if $n>4$. So, you need either numerical methods (such as Newton) or approximations.

If you consider the derivatives
$$f'(x)=a n (1-a x)^{n-1}-1$$
$$f''(x)=-a^2 (n-1) n (1-a x)^{n-2}$$ the first derivative cancels at
$$x_*=frac{1-left(frac{1}{a n}right)^{frac{1}{n-1}}}{a}$$ and notice that $f(0)=0$. So, assuming that this is a minimum and that $f(x_*)<0$, ther is a root which is $ > x_*$.

To get a first approximation, you could build a Taylor series around $x=x_*$ to get
$$f(x)=f(x_*)+frac 12 f''(x_*)(x-x_*)^2+O((x-x_*)^3)$$ and ignoring the higher order terms, get an estimate
$$x_0=x_*+sqrt{-2frac{f(x_*)}{f''(x_*)}}$$

Appliead to the case $a=0.002$, $n=2999$, this would give $x_0 approx 0.721$ and, using this Newton iterates would be
$$left(
begin{array}{cc}
n & x_n \
0 & 0.720871 \
1 & 1.009707 \
2 & 0.997500 \
3 & 0.997494
end{array}
right)$$ which is the solution for six significant figures.

If you know that the solution is close to $x=1$, you could expand $f(x)$ as a Taylor series at $x=1$ and get
$$f(x)=-(1-a)^n+(x-1) left(a n (1-a)^{n-1}-1right)+Oleft((x-1)^2right)$$ Using the first term only would give
$$x=1+frac{(1-a)^{n+1}}{a n (1-a)^n+a-1}$$ which, for the worked example, would immediately give $x=0.997494$.

share|cite|improve this answer

answered Mar 14 at 7:04

Claude LeiboviciClaude Leibovici

124k1158135

$endgroup$

share|cite|improve this answer

answered Mar 14 at 7:04

Claude LeiboviciClaude Leibovici

124k1158135

answered Mar 14 at 7:04

Claude LeiboviciClaude Leibovici

124k1158135

answered Mar 14 at 7:04

Claude LeiboviciClaude Leibovici

124k1158135