Mixing
Linear Approximation of x/ (1-x)
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
I am trying to linearize the following function, but, having difficulties.
Let,
$x = fraclm,$
where $l,m in R^+$ and $l<m$
Assume $l$ is a variable, while, $m$ is a constant (parameter), which makes $x$ a variable. I want to find a linear approximation for the following:
$f(x) = fracx2m(1-x)$
In other words, as $2$ and $m$ are constant, I am interested in
$g(x)=fracx1-x$
I plotted the graph, but, it did not really help to derive something useful. Any help is appreciated.
linearization
asked Jul 15 at 2:45
user8028576
277
 |Â
show 2 more comments
up vote
1
down vote
favorite
I am trying to linearize the following function, but, having difficulties.
Let,
$x = fraclm,$
where $l,m in R^+$ and $l<m$
Assume $l$ is a variable, while, $m$ is a constant (parameter), which makes $x$ a variable. I want to find a linear approximation for the following:
$f(x) = fracx2m(1-x)$
In other words, as $2$ and $m$ are constant, I am interested in
$g(x)=fracx1-x$
I plotted the graph, but, it did not really help to derive something useful. Any help is appreciated.
linearization
asked Jul 15 at 2:45
user8028576
277
Linear Approximation? You want to approximate the function using a straight line?
– W. mu
Jul 15 at 2:50
Yes. Does it sound impossible?
– user8028576
Jul 15 at 2:51
Of course, it's possible, but the error may be large. en.wikipedia.org/wiki/Approximation_theory
– W. mu
Jul 15 at 3:04
If you are looking for the usual calculus approximation, then it will be $h(x)=x$. This can be found by the formula $f(x)≈f'(x_0)(x-x_0)+f(x_0)$ when $x≈x_0$.
– user496634
Jul 15 at 9:53
@user496634 thanks for your comment. In my search for solutions, I confronted with the calculus approximation. Yet, I could not quite understand how it can make the function linear because the first derivative of the function still remains to be non-linear with squared $x$ terms. Additionally, I am lost with the term $x_0$. I am using this function in a MILP problem and I do solve for $x$ and other thousands of variables while there are several inequality constraints including $x$. If you may put your solution approach down, I will be thrilled to read it.
– user8028576
Jul 16 at 1:53
 |Â
show 2 more comments
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I am trying to linearize the following function, but, having difficulties.
Let,
$x = fraclm,$
where $l,m in R^+$ and $l<m$
Assume $l$ is a variable, while, $m$ is a constant (parameter), which makes $x$ a variable. I want to find a linear approximation for the following:
$f(x) = fracx2m(1-x)$
In other words, as $2$ and $m$ are constant, I am interested in
$g(x)=fracx1-x$
I plotted the graph, but, it did not really help to derive something useful. Any help is appreciated.
linearization
asked Jul 15 at 2:45
user8028576
277
I am trying to linearize the following function, but, having difficulties.
Let,
$x = fraclm,$
where $l,m in R^+$ and $l<m$
Assume $l$ is a variable, while, $m$ is a constant (parameter), which makes $x$ a variable. I want to find a linear approximation for the following:
$f(x) = fracx2m(1-x)$
In other words, as $2$ and $m$ are constant, I am interested in
$g(x)=fracx1-x$
I plotted the graph, but, it did not really help to derive something useful. Any help is appreciated.
linearization
asked Jul 15 at 2:45
user8028576
277
asked Jul 15 at 2:45
user8028576
277
asked Jul 15 at 2:45
user8028576
277
asked Jul 15 at 2:45
user8028576
277
277
Linear Approximation? You want to approximate the function using a straight line?
– W. mu
Jul 15 at 2:50
Yes. Does it sound impossible?
– user8028576
Jul 15 at 2:51
Of course, it's possible, but the error may be large. en.wikipedia.org/wiki/Approximation_theory
– W. mu
Jul 15 at 3:04
If you are looking for the usual calculus approximation, then it will be $h(x)=x$. This can be found by the formula $f(x)≈f'(x_0)(x-x_0)+f(x_0)$ when $x≈x_0$.
– user496634
Jul 15 at 9:53
@user496634 thanks for your comment. In my search for solutions, I confronted with the calculus approximation. Yet, I could not quite understand how it can make the function linear because the first derivative of the function still remains to be non-linear with squared $x$ terms. Additionally, I am lost with the term $x_0$. I am using this function in a MILP problem and I do solve for $x$ and other thousands of variables while there are several inequality constraints including $x$. If you may put your solution approach down, I will be thrilled to read it.
– user8028576
Jul 16 at 1:53
 |Â
show 2 more comments
Linear Approximation? You want to approximate the function using a straight line?
– W. mu
Jul 15 at 2:50
Yes. Does it sound impossible?
– user8028576
Jul 15 at 2:51
Of course, it's possible, but the error may be large. en.wikipedia.org/wiki/Approximation_theory
– W. mu
Jul 15 at 3:04
If you are looking for the usual calculus approximation, then it will be $h(x)=x$. This can be found by the formula $f(x)≈f'(x_0)(x-x_0)+f(x_0)$ when $x≈x_0$.
– user496634
Jul 15 at 9:53
@user496634 thanks for your comment. In my search for solutions, I confronted with the calculus approximation. Yet, I could not quite understand how it can make the function linear because the first derivative of the function still remains to be non-linear with squared $x$ terms. Additionally, I am lost with the term $x_0$. I am using this function in a MILP problem and I do solve for $x$ and other thousands of variables while there are several inequality constraints including $x$. If you may put your solution approach down, I will be thrilled to read it.
– user8028576
Jul 16 at 1:53
Linear Approximation? You want to approximate the function using a straight line?
– W. mu
Jul 15 at 2:50
Linear Approximation? You want to approximate the function using a straight line?
– W. mu
Jul 15 at 2:50
Yes. Does it sound impossible?
– user8028576
Jul 15 at 2:51
Yes. Does it sound impossible?
– user8028576
Jul 15 at 2:51
Of course, it's possible, but the error may be large. en.wikipedia.org/wiki/Approximation_theory
– W. mu
Jul 15 at 3:04
Of course, it's possible, but the error may be large. en.wikipedia.org/wiki/Approximation_theory
– W. mu
Jul 15 at 3:04
If you are looking for the usual calculus approximation, then it will be $h(x)=x$. This can be found by the formula $f(x)≈f'(x_0)(x-x_0)+f(x_0)$ when $x≈x_0$.
– user496634
Jul 15 at 9:53
If you are looking for the usual calculus approximation, then it will be $h(x)=x$. This can be found by the formula $f(x)≈f'(x_0)(x-x_0)+f(x_0)$ when $x≈x_0$.
– user496634
Jul 15 at 9:53
@user496634 thanks for your comment. In my search for solutions, I confronted with the calculus approximation. Yet, I could not quite understand how it can make the function linear because the first derivative of the function still remains to be non-linear with squared $x$ terms. Additionally, I am lost with the term $x_0$. I am using this function in a MILP problem and I do solve for $x$ and other thousands of variables while there are several inequality constraints including $x$. If you may put your solution approach down, I will be thrilled to read it.
– user8028576
Jul 16 at 1:53
@user496634 thanks for your comment. In my search for solutions, I confronted with the calculus approximation. Yet, I could not quite understand how it can make the function linear because the first derivative of the function still remains to be non-linear with squared $x$ terms. Additionally, I am lost with the term $x_0$. I am using this function in a MILP problem and I do solve for $x$ and other thousands of variables while there are several inequality constraints including $x$. If you may put your solution approach down, I will be thrilled to read it.
– user8028576
Jul 16 at 1:53
 |Â
show 2 more comments
3 Answers
3
active
oldest
votes
up vote
2
down vote
accepted
If, over a range $aleq x leq b$, you want the best linear approximation $A+Bx$ of
$$f(x) = fracx2m(1-x)$$ the solution is to minimize the norm
$$F=int_a^b left(A+Bx-fracx2m(1-x) right)^2$$ with respect to parameters $A$ and $B$.
Integrating and then computing the partial derivatives $fracpartial Fpartial A$ and $fracpartial Fpartial B$ and setting them equal to $0$ leads to two linear equations in $(A,B)$
$$2m(b-a) A + m (b^2-a^2)B+(b-a)+log left(frac1-b1-aright)=0$$
$$6m(b^2-a^2)A+4m(b^3-a^3)B+3 (b-a) (a+b+2)+6log left(frac1-b1-aright)=0$$ which you can simplify factoring $(b-a)$ in several places.
I let to you the pleasure of finding $A,B$ (but this is simple).
Around $x=0$, for $a=-b$, using Taylor series, we have
$$A=fracb^26 m+fracb^410 m+Oleft(b^6right)qquad B=frac12 m+frac3 b^210 m+frac3 b^414 m+Oleft(b^6right)qquad F=frac2 b^545 m^2+Oleft(b^7right)$$
Taking an example using $a=-0.1$, $b=0.1$, $m=1$, this would lead to
$$A=frac12 left(-1+5 log left(frac1110right)+5 log
left(frac109right)right)approx 0.00167674$$
$$B=-150 (1+10 log (3)-5 log (11))approx 0.503022$$
answered Jul 15 at 4:22
Claude Leibovici
112k1055126
Thanks a lot! This is what I was looking for. Yet, I need some verbal explanation to digest all this and make it useful for my case. First of all, I use the function $f(x)$ in a mixed integer (non)linear programming model and $f(x)$ is the term that makes it non. Again, $0leq x<1$. So, I guess, I need to set $a=0$ and possibly $b=0.999$ as close as to $1$. Then, instead of having my $x$ as a variable in my problem, I would have $A$ and $B$. But, I still need $x$ as a variable because I have to use in some other equality constraints. So, what would you recommend in this case?
– user8028576
Jul 16 at 0:55
Dr. Leibovici, if I understand your solution correctly with my limited math, I guess, you mean, I probably need to find $A$ and $B$ given that $a=0$ and (say) $b=0.999$. Then, plugging these in $f(x)=A+Bx$ will yield me an approximation of my original function, right? If I get it right until here, either this approximation does not really work or I miss something. I found $A=36.9472, B=-68.0547$ given that $a=0, b=0.999, m=1$. When I test the approximation with $x=0.6$, I get $f(x)_approx = −3.88562$ while $f(x)=0.75$. Furthermore, $f(x)geq 0 $ must be held in any case.
– user8028576
Jul 16 at 1:46
@user8028576. Linearization of any function is local and valid over a small range.
– Claude Leibovici
Jul 16 at 3:05
Dr. Leibovici, so, do you mean $a=0$ and $b=0.999$ is a large range? If yes, how can I find the optimal range to make the approximation valid? I guess, I may create multiple equations of the above form with e.g., $a=0, b=0.1$, $a=0.1, b=0.2$ $...$ $a=0.9, b=0.999$. Right? But, is $0.1$ step size okay?
– user8028576
Jul 16 at 3:38
@user8028576. For each ("small") range, you need to compute $A$ and $B$.
– Claude Leibovici
Jul 16 at 3:51
 |Â
show 2 more comments
up vote
1
down vote
I will present an elementary calculus solution. I will first rename your $g(x)$ into $f(x)$ because I can and because it looks nicer to the eye. It is apparent that the best linear approximation to any function at $x=0$ must be the tangent line to the function at $x=0$. In our case, since $f(0)=0$, our tangent line must pass through $(0,0)$ too and so it must be of the form $y=mx$. Since it is the tangent line, the gradient $m$ of this line must have the same "gradient" (i.e. rate of change) as $f$ at $0$. To find the gradient, thus, we set
$$ m = f'(0) $$
Where the RHS is the derivative at $0$; so the rate of change as previously mentioned. So now we just have to take the derivative of $f$ and evaluate it at $0$ to find $m$. I'm sure you know how to do this, but I'll put it here for completeness's sake:
$$fracddx fracx1-x = frac(1-x)-(-1)x(1-x)^2$$
Evaluated at $0$ this is $1$. So $m=1$ and the closest linear approximation would be $y=mx=x$.
answered Jul 16 at 5:51
user496634
30518
you basically propose $F = fracx2m$ as a n approximation to the original $f(x)$ given in the post. If I get it correctly, this approximation is not any close to $f(x)$. Let $x=0.9, m=1$, $f(x)=4.5$ and $F=0.45$. If I interpret it wrongly, please let me know.
– user8028576
Jul 16 at 15:40
@user8028576 You are absolutely right that it is not a very good approximation. However, it is the best linear one, locally around $x=0$. This is unless, of course, you want to approximate it around a range around $x=0$, not just best at that point and it's "immediate vicinity". In that case, you can consider Leibovici's answer. You will see that in a small range immediately around $0$, my approximation is better; while in the range $a$ and $b$ in his answer his is generally better.
– user496634
Jul 16 at 22:34
@user8028576 A helpful way to consider this is that my answer is the limit of Lebivoci's answer as $a$ and $b$ both approach $0$ (this is quite easy to prove). Around points extremely close to $0$, my approximation is extremely good; while his answer minimises the sum of the squared of the difference in values over a whole range $[a,b]$. One is an approximation at a point, the other is the approximation over a range.
– user496634
Jul 16 at 22:40
@user8028576 I totally get the logic of both approximations. Your approximation draws a linear line starting from the origin and it does not really worry about minimizing the difference in the approximation and the actual. Your approximation is the easiest with a single $F=fracx2m function. On the other hand, Leibovici's Taylor series draws a line for a "small" range of the whole plot that is as close as to the originial nonlinear line in that segmentation. Once I create "a lot of"of those lines with the given function, I will be able to approximate much more precisely.
– user8028576
Jul 17 at 13:42
1
But, when I get much closer to the tails of the approximation, his approximation yields an infeasible (negative) solution. For example, $a=0, b=0.1, m=1$, then, when I search for $x=0.0001$, it will give me $Fapprox -0.00088$. Yet, this situation is understandable. To deal with it, I will adjust the ranges according to my expectations of the results. For instance, if I know $x=0.0001$ is an expected result, then I will add shorter ranges, which is equivalent of saying a decimal adjustment. I think, I got it! I am happy. Thank you very much to all, as you thought me something very useful!
– user8028576
Jul 17 at 13:47
add a comment |Â
up vote
0
down vote
$$frac ddx frac1 1-x = frac1(1-x)^2$$ Additionally,
since $frac11-x = sum_k=0^inftyx^k$ clearly $$frac1(1-x)^2 = sum_k=1^inftykx^k-1$$Unfortunately, it does not really make sense to approximate linearly something like this in any global sense. Locally, evaluating $m = frac1(1-x_0)^2$ would give you the slope of a line tangent to $frac1(1-x)$ at point $x_0$ which locally approximates it linearly. I believe this is what you're asking.
answered Jul 15 at 3:47
BelowAverageIntelligence
339212
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
If, over a range $aleq x leq b$, you want the best linear approximation $A+Bx$ of
$$f(x) = fracx2m(1-x)$$ the solution is to minimize the norm
$$F=int_a^b left(A+Bx-fracx2m(1-x) right)^2$$ with respect to parameters $A$ and $B$.
Integrating and then computing the partial derivatives $fracpartial Fpartial A$ and $fracpartial Fpartial B$ and setting them equal to $0$ leads to two linear equations in $(A,B)$
$$2m(b-a) A + m (b^2-a^2)B+(b-a)+log left(frac1-b1-aright)=0$$
$$6m(b^2-a^2)A+4m(b^3-a^3)B+3 (b-a) (a+b+2)+6log left(frac1-b1-aright)=0$$ which you can simplify factoring $(b-a)$ in several places.
I let to you the pleasure of finding $A,B$ (but this is simple).
Around $x=0$, for $a=-b$, using Taylor series, we have
$$A=fracb^26 m+fracb^410 m+Oleft(b^6right)qquad B=frac12 m+frac3 b^210 m+frac3 b^414 m+Oleft(b^6right)qquad F=frac2 b^545 m^2+Oleft(b^7right)$$
Taking an example using $a=-0.1$, $b=0.1$, $m=1$, this would lead to
$$A=frac12 left(-1+5 log left(frac1110right)+5 log
left(frac109right)right)approx 0.00167674$$
$$B=-150 (1+10 log (3)-5 log (11))approx 0.503022$$
answered Jul 15 at 4:22
Claude Leibovici
112k1055126
Thanks a lot! This is what I was looking for. Yet, I need some verbal explanation to digest all this and make it useful for my case. First of all, I use the function $f(x)$ in a mixed integer (non)linear programming model and $f(x)$ is the term that makes it non. Again, $0leq x<1$. So, I guess, I need to set $a=0$ and possibly $b=0.999$ as close as to $1$. Then, instead of having my $x$ as a variable in my problem, I would have $A$ and $B$. But, I still need $x$ as a variable because I have to use in some other equality constraints. So, what would you recommend in this case?
– user8028576
Jul 16 at 0:55
Dr. Leibovici, if I understand your solution correctly with my limited math, I guess, you mean, I probably need to find $A$ and $B$ given that $a=0$ and (say) $b=0.999$. Then, plugging these in $f(x)=A+Bx$ will yield me an approximation of my original function, right? If I get it right until here, either this approximation does not really work or I miss something. I found $A=36.9472, B=-68.0547$ given that $a=0, b=0.999, m=1$. When I test the approximation with $x=0.6$, I get $f(x)_approx = −3.88562$ while $f(x)=0.75$. Furthermore, $f(x)geq 0 $ must be held in any case.
– user8028576
Jul 16 at 1:46
@user8028576. Linearization of any function is local and valid over a small range.
– Claude Leibovici
Jul 16 at 3:05
Dr. Leibovici, so, do you mean $a=0$ and $b=0.999$ is a large range? If yes, how can I find the optimal range to make the approximation valid? I guess, I may create multiple equations of the above form with e.g., $a=0, b=0.1$, $a=0.1, b=0.2$ $...$ $a=0.9, b=0.999$. Right? But, is $0.1$ step size okay?
– user8028576
Jul 16 at 3:38
@user8028576. For each ("small") range, you need to compute $A$ and $B$.
– Claude Leibovici
Jul 16 at 3:51
 |Â
show 2 more comments
up vote
2
down vote
accepted
If, over a range $aleq x leq b$, you want the best linear approximation $A+Bx$ of
$$f(x) = fracx2m(1-x)$$ the solution is to minimize the norm
$$F=int_a^b left(A+Bx-fracx2m(1-x) right)^2$$ with respect to parameters $A$ and $B$.
Integrating and then computing the partial derivatives $fracpartial Fpartial A$ and $fracpartial Fpartial B$ and setting them equal to $0$ leads to two linear equations in $(A,B)$
$$2m(b-a) A + m (b^2-a^2)B+(b-a)+log left(frac1-b1-aright)=0$$
$$6m(b^2-a^2)A+4m(b^3-a^3)B+3 (b-a) (a+b+2)+6log left(frac1-b1-aright)=0$$ which you can simplify factoring $(b-a)$ in several places.
I let to you the pleasure of finding $A,B$ (but this is simple).
Around $x=0$, for $a=-b$, using Taylor series, we have
$$A=fracb^26 m+fracb^410 m+Oleft(b^6right)qquad B=frac12 m+frac3 b^210 m+frac3 b^414 m+Oleft(b^6right)qquad F=frac2 b^545 m^2+Oleft(b^7right)$$
Taking an example using $a=-0.1$, $b=0.1$, $m=1$, this would lead to
$$A=frac12 left(-1+5 log left(frac1110right)+5 log
left(frac109right)right)approx 0.00167674$$
$$B=-150 (1+10 log (3)-5 log (11))approx 0.503022$$
answered Jul 15 at 4:22
Claude Leibovici
112k1055126
Thanks a lot! This is what I was looking for. Yet, I need some verbal explanation to digest all this and make it useful for my case. First of all, I use the function $f(x)$ in a mixed integer (non)linear programming model and $f(x)$ is the term that makes it non. Again, $0leq x<1$. So, I guess, I need to set $a=0$ and possibly $b=0.999$ as close as to $1$. Then, instead of having my $x$ as a variable in my problem, I would have $A$ and $B$. But, I still need $x$ as a variable because I have to use in some other equality constraints. So, what would you recommend in this case?
– user8028576
Jul 16 at 0:55
Dr. Leibovici, if I understand your solution correctly with my limited math, I guess, you mean, I probably need to find $A$ and $B$ given that $a=0$ and (say) $b=0.999$. Then, plugging these in $f(x)=A+Bx$ will yield me an approximation of my original function, right? If I get it right until here, either this approximation does not really work or I miss something. I found $A=36.9472, B=-68.0547$ given that $a=0, b=0.999, m=1$. When I test the approximation with $x=0.6$, I get $f(x)_approx = −3.88562$ while $f(x)=0.75$. Furthermore, $f(x)geq 0 $ must be held in any case.
– user8028576
Jul 16 at 1:46
@user8028576. Linearization of any function is local and valid over a small range.
– Claude Leibovici
Jul 16 at 3:05
Dr. Leibovici, so, do you mean $a=0$ and $b=0.999$ is a large range? If yes, how can I find the optimal range to make the approximation valid? I guess, I may create multiple equations of the above form with e.g., $a=0, b=0.1$, $a=0.1, b=0.2$ $...$ $a=0.9, b=0.999$. Right? But, is $0.1$ step size okay?
– user8028576
Jul 16 at 3:38
@user8028576. For each ("small") range, you need to compute $A$ and $B$.
– Claude Leibovici
Jul 16 at 3:51
 |Â
show 2 more comments
up vote
2
down vote
accepted
up vote
2
down vote
accepted
If, over a range $aleq x leq b$, you want the best linear approximation $A+Bx$ of
$$f(x) = fracx2m(1-x)$$ the solution is to minimize the norm
$$F=int_a^b left(A+Bx-fracx2m(1-x) right)^2$$ with respect to parameters $A$ and $B$.
Integrating and then computing the partial derivatives $fracpartial Fpartial A$ and $fracpartial Fpartial B$ and setting them equal to $0$ leads to two linear equations in $(A,B)$
$$2m(b-a) A + m (b^2-a^2)B+(b-a)+log left(frac1-b1-aright)=0$$
$$6m(b^2-a^2)A+4m(b^3-a^3)B+3 (b-a) (a+b+2)+6log left(frac1-b1-aright)=0$$ which you can simplify factoring $(b-a)$ in several places.
I let to you the pleasure of finding $A,B$ (but this is simple).
Around $x=0$, for $a=-b$, using Taylor series, we have
$$A=fracb^26 m+fracb^410 m+Oleft(b^6right)qquad B=frac12 m+frac3 b^210 m+frac3 b^414 m+Oleft(b^6right)qquad F=frac2 b^545 m^2+Oleft(b^7right)$$
Taking an example using $a=-0.1$, $b=0.1$, $m=1$, this would lead to
$$A=frac12 left(-1+5 log left(frac1110right)+5 log
left(frac109right)right)approx 0.00167674$$
$$B=-150 (1+10 log (3)-5 log (11))approx 0.503022$$
answered Jul 15 at 4:22
Claude Leibovici
112k1055126
If, over a range $aleq x leq b$, you want the best linear approximation $A+Bx$ of
$$f(x) = fracx2m(1-x)$$ the solution is to minimize the norm
$$F=int_a^b left(A+Bx-fracx2m(1-x) right)^2$$ with respect to parameters $A$ and $B$.
Integrating and then computing the partial derivatives $fracpartial Fpartial A$ and $fracpartial Fpartial B$ and setting them equal to $0$ leads to two linear equations in $(A,B)$
$$2m(b-a) A + m (b^2-a^2)B+(b-a)+log left(frac1-b1-aright)=0$$
$$6m(b^2-a^2)A+4m(b^3-a^3)B+3 (b-a) (a+b+2)+6log left(frac1-b1-aright)=0$$ which you can simplify factoring $(b-a)$ in several places.
I let to you the pleasure of finding $A,B$ (but this is simple).
Around $x=0$, for $a=-b$, using Taylor series, we have
$$A=fracb^26 m+fracb^410 m+Oleft(b^6right)qquad B=frac12 m+frac3 b^210 m+frac3 b^414 m+Oleft(b^6right)qquad F=frac2 b^545 m^2+Oleft(b^7right)$$
Taking an example using $a=-0.1$, $b=0.1$, $m=1$, this would lead to
$$A=frac12 left(-1+5 log left(frac1110right)+5 log
left(frac109right)right)approx 0.00167674$$
$$B=-150 (1+10 log (3)-5 log (11))approx 0.503022$$
answered Jul 15 at 4:22
Claude Leibovici
112k1055126
edited Jul 17 at 7:47
answered Jul 15 at 4:22
Claude Leibovici
112k1055126
answered Jul 15 at 4:22
Claude Leibovici
112k1055126
answered Jul 15 at 4:22
Claude Leibovici
112k1055126
112k1055126
Thanks a lot! This is what I was looking for. Yet, I need some verbal explanation to digest all this and make it useful for my case. First of all, I use the function $f(x)$ in a mixed integer (non)linear programming model and $f(x)$ is the term that makes it non. Again, $0leq x<1$. So, I guess, I need to set $a=0$ and possibly $b=0.999$ as close as to $1$. Then, instead of having my $x$ as a variable in my problem, I would have $A$ and $B$. But, I still need $x$ as a variable because I have to use in some other equality constraints. So, what would you recommend in this case?
– user8028576
Jul 16 at 0:55
Dr. Leibovici, if I understand your solution correctly with my limited math, I guess, you mean, I probably need to find $A$ and $B$ given that $a=0$ and (say) $b=0.999$. Then, plugging these in $f(x)=A+Bx$ will yield me an approximation of my original function, right? If I get it right until here, either this approximation does not really work or I miss something. I found $A=36.9472, B=-68.0547$ given that $a=0, b=0.999, m=1$. When I test the approximation with $x=0.6$, I get $f(x)_approx = −3.88562$ while $f(x)=0.75$. Furthermore, $f(x)geq 0 $ must be held in any case.
– user8028576
Jul 16 at 1:46
@user8028576. Linearization of any function is local and valid over a small range.
– Claude Leibovici
Jul 16 at 3:05
Dr. Leibovici, so, do you mean $a=0$ and $b=0.999$ is a large range? If yes, how can I find the optimal range to make the approximation valid? I guess, I may create multiple equations of the above form with e.g., $a=0, b=0.1$, $a=0.1, b=0.2$ $...$ $a=0.9, b=0.999$. Right? But, is $0.1$ step size okay?
– user8028576
Jul 16 at 3:38
@user8028576. For each ("small") range, you need to compute $A$ and $B$.
– Claude Leibovici
Jul 16 at 3:51
 |Â
show 2 more comments
Thanks a lot! This is what I was looking for. Yet, I need some verbal explanation to digest all this and make it useful for my case. First of all, I use the function $f(x)$ in a mixed integer (non)linear programming model and $f(x)$ is the term that makes it non. Again, $0leq x<1$. So, I guess, I need to set $a=0$ and possibly $b=0.999$ as close as to $1$. Then, instead of having my $x$ as a variable in my problem, I would have $A$ and $B$. But, I still need $x$ as a variable because I have to use in some other equality constraints. So, what would you recommend in this case?
– user8028576
Jul 16 at 0:55
Dr. Leibovici, if I understand your solution correctly with my limited math, I guess, you mean, I probably need to find $A$ and $B$ given that $a=0$ and (say) $b=0.999$. Then, plugging these in $f(x)=A+Bx$ will yield me an approximation of my original function, right? If I get it right until here, either this approximation does not really work or I miss something. I found $A=36.9472, B=-68.0547$ given that $a=0, b=0.999, m=1$. When I test the approximation with $x=0.6$, I get $f(x)_approx = −3.88562$ while $f(x)=0.75$. Furthermore, $f(x)geq 0 $ must be held in any case.
– user8028576
Jul 16 at 1:46
@user8028576. Linearization of any function is local and valid over a small range.
– Claude Leibovici
Jul 16 at 3:05
Dr. Leibovici, so, do you mean $a=0$ and $b=0.999$ is a large range? If yes, how can I find the optimal range to make the approximation valid? I guess, I may create multiple equations of the above form with e.g., $a=0, b=0.1$, $a=0.1, b=0.2$ $...$ $a=0.9, b=0.999$. Right? But, is $0.1$ step size okay?
– user8028576
Jul 16 at 3:38
@user8028576. For each ("small") range, you need to compute $A$ and $B$.
– Claude Leibovici
Jul 16 at 3:51
Thanks a lot! This is what I was looking for. Yet, I need some verbal explanation to digest all this and make it useful for my case. First of all, I use the function $f(x)$ in a mixed integer (non)linear programming model and $f(x)$ is the term that makes it non. Again, $0leq x<1$. So, I guess, I need to set $a=0$ and possibly $b=0.999$ as close as to $1$. Then, instead of having my $x$ as a variable in my problem, I would have $A$ and $B$. But, I still need $x$ as a variable because I have to use in some other equality constraints. So, what would you recommend in this case?
– user8028576
Jul 16 at 0:55
Thanks a lot! This is what I was looking for. Yet, I need some verbal explanation to digest all this and make it useful for my case. First of all, I use the function $f(x)$ in a mixed integer (non)linear programming model and $f(x)$ is the term that makes it non. Again, $0leq x<1$. So, I guess, I need to set $a=0$ and possibly $b=0.999$ as close as to $1$. Then, instead of having my $x$ as a variable in my problem, I would have $A$ and $B$. But, I still need $x$ as a variable because I have to use in some other equality constraints. So, what would you recommend in this case?
– user8028576
Jul 16 at 0:55
Dr. Leibovici, if I understand your solution correctly with my limited math, I guess, you mean, I probably need to find $A$ and $B$ given that $a=0$ and (say) $b=0.999$. Then, plugging these in $f(x)=A+Bx$ will yield me an approximation of my original function, right? If I get it right until here, either this approximation does not really work or I miss something. I found $A=36.9472, B=-68.0547$ given that $a=0, b=0.999, m=1$. When I test the approximation with $x=0.6$, I get $f(x)_approx = −3.88562$ while $f(x)=0.75$. Furthermore, $f(x)geq 0 $ must be held in any case.
– user8028576
Jul 16 at 1:46
Dr. Leibovici, if I understand your solution correctly with my limited math, I guess, you mean, I probably need to find $A$ and $B$ given that $a=0$ and (say) $b=0.999$. Then, plugging these in $f(x)=A+Bx$ will yield me an approximation of my original function, right? If I get it right until here, either this approximation does not really work or I miss something. I found $A=36.9472, B=-68.0547$ given that $a=0, b=0.999, m=1$. When I test the approximation with $x=0.6$, I get $f(x)_approx = −3.88562$ while $f(x)=0.75$. Furthermore, $f(x)geq 0 $ must be held in any case.
– user8028576
Jul 16 at 1:46
@user8028576. Linearization of any function is local and valid over a small range.
– Claude Leibovici
Jul 16 at 3:05
@user8028576. Linearization of any function is local and valid over a small range.
– Claude Leibovici
Jul 16 at 3:05
Dr. Leibovici, so, do you mean $a=0$ and $b=0.999$ is a large range? If yes, how can I find the optimal range to make the approximation valid? I guess, I may create multiple equations of the above form with e.g., $a=0, b=0.1$, $a=0.1, b=0.2$ $...$ $a=0.9, b=0.999$. Right? But, is $0.1$ step size okay?
– user8028576
Jul 16 at 3:38
Dr. Leibovici, so, do you mean $a=0$ and $b=0.999$ is a large range? If yes, how can I find the optimal range to make the approximation valid? I guess, I may create multiple equations of the above form with e.g., $a=0, b=0.1$, $a=0.1, b=0.2$ $...$ $a=0.9, b=0.999$. Right? But, is $0.1$ step size okay?
– user8028576
Jul 16 at 3:38
@user8028576. For each ("small") range, you need to compute $A$ and $B$.
– Claude Leibovici
Jul 16 at 3:51
@user8028576. For each ("small") range, you need to compute $A$ and $B$.
– Claude Leibovici
Jul 16 at 3:51
 |Â
show 2 more comments
up vote
1
down vote
I will present an elementary calculus solution. I will first rename your $g(x)$ into $f(x)$ because I can and because it looks nicer to the eye. It is apparent that the best linear approximation to any function at $x=0$ must be the tangent line to the function at $x=0$. In our case, since $f(0)=0$, our tangent line must pass through $(0,0)$ too and so it must be of the form $y=mx$. Since it is the tangent line, the gradient $m$ of this line must have the same "gradient" (i.e. rate of change) as $f$ at $0$. To find the gradient, thus, we set
$$ m = f'(0) $$
Where the RHS is the derivative at $0$; so the rate of change as previously mentioned. So now we just have to take the derivative of $f$ and evaluate it at $0$ to find $m$. I'm sure you know how to do this, but I'll put it here for completeness's sake:
$$fracddx fracx1-x = frac(1-x)-(-1)x(1-x)^2$$
Evaluated at $0$ this is $1$. So $m=1$ and the closest linear approximation would be $y=mx=x$.
answered Jul 16 at 5:51
user496634
30518
you basically propose $F = fracx2m$ as a n approximation to the original $f(x)$ given in the post. If I get it correctly, this approximation is not any close to $f(x)$. Let $x=0.9, m=1$, $f(x)=4.5$ and $F=0.45$. If I interpret it wrongly, please let me know.
– user8028576
Jul 16 at 15:40
@user8028576 You are absolutely right that it is not a very good approximation. However, it is the best linear one, locally around $x=0$. This is unless, of course, you want to approximate it around a range around $x=0$, not just best at that point and it's "immediate vicinity". In that case, you can consider Leibovici's answer. You will see that in a small range immediately around $0$, my approximation is better; while in the range $a$ and $b$ in his answer his is generally better.
– user496634
Jul 16 at 22:34
@user8028576 A helpful way to consider this is that my answer is the limit of Lebivoci's answer as $a$ and $b$ both approach $0$ (this is quite easy to prove). Around points extremely close to $0$, my approximation is extremely good; while his answer minimises the sum of the squared of the difference in values over a whole range $[a,b]$. One is an approximation at a point, the other is the approximation over a range.
– user496634
Jul 16 at 22:40
@user8028576 I totally get the logic of both approximations. Your approximation draws a linear line starting from the origin and it does not really worry about minimizing the difference in the approximation and the actual. Your approximation is the easiest with a single $F=fracx2m function. On the other hand, Leibovici's Taylor series draws a line for a "small" range of the whole plot that is as close as to the originial nonlinear line in that segmentation. Once I create "a lot of"of those lines with the given function, I will be able to approximate much more precisely.
– user8028576
Jul 17 at 13:42
1
But, when I get much closer to the tails of the approximation, his approximation yields an infeasible (negative) solution. For example, $a=0, b=0.1, m=1$, then, when I search for $x=0.0001$, it will give me $Fapprox -0.00088$. Yet, this situation is understandable. To deal with it, I will adjust the ranges according to my expectations of the results. For instance, if I know $x=0.0001$ is an expected result, then I will add shorter ranges, which is equivalent of saying a decimal adjustment. I think, I got it! I am happy. Thank you very much to all, as you thought me something very useful!
– user8028576
Jul 17 at 13:47
add a comment |Â
up vote
1
down vote
I will present an elementary calculus solution. I will first rename your $g(x)$ into $f(x)$ because I can and because it looks nicer to the eye. It is apparent that the best linear approximation to any function at $x=0$ must be the tangent line to the function at $x=0$. In our case, since $f(0)=0$, our tangent line must pass through $(0,0)$ too and so it must be of the form $y=mx$. Since it is the tangent line, the gradient $m$ of this line must have the same "gradient" (i.e. rate of change) as $f$ at $0$. To find the gradient, thus, we set
$$ m = f'(0) $$
Where the RHS is the derivative at $0$; so the rate of change as previously mentioned. So now we just have to take the derivative of $f$ and evaluate it at $0$ to find $m$. I'm sure you know how to do this, but I'll put it here for completeness's sake:
$$fracddx fracx1-x = frac(1-x)-(-1)x(1-x)^2$$
Evaluated at $0$ this is $1$. So $m=1$ and the closest linear approximation would be $y=mx=x$.
answered Jul 16 at 5:51
user496634
30518
you basically propose $F = fracx2m$ as a n approximation to the original $f(x)$ given in the post. If I get it correctly, this approximation is not any close to $f(x)$. Let $x=0.9, m=1$, $f(x)=4.5$ and $F=0.45$. If I interpret it wrongly, please let me know.
– user8028576
Jul 16 at 15:40
@user8028576 You are absolutely right that it is not a very good approximation. However, it is the best linear one, locally around $x=0$. This is unless, of course, you want to approximate it around a range around $x=0$, not just best at that point and it's "immediate vicinity". In that case, you can consider Leibovici's answer. You will see that in a small range immediately around $0$, my approximation is better; while in the range $a$ and $b$ in his answer his is generally better.
– user496634
Jul 16 at 22:34
@user8028576 A helpful way to consider this is that my answer is the limit of Lebivoci's answer as $a$ and $b$ both approach $0$ (this is quite easy to prove). Around points extremely close to $0$, my approximation is extremely good; while his answer minimises the sum of the squared of the difference in values over a whole range $[a,b]$. One is an approximation at a point, the other is the approximation over a range.
– user496634
Jul 16 at 22:40
@user8028576 I totally get the logic of both approximations. Your approximation draws a linear line starting from the origin and it does not really worry about minimizing the difference in the approximation and the actual. Your approximation is the easiest with a single $F=fracx2m function. On the other hand, Leibovici's Taylor series draws a line for a "small" range of the whole plot that is as close as to the originial nonlinear line in that segmentation. Once I create "a lot of"of those lines with the given function, I will be able to approximate much more precisely.
– user8028576
Jul 17 at 13:42
1
But, when I get much closer to the tails of the approximation, his approximation yields an infeasible (negative) solution. For example, $a=0, b=0.1, m=1$, then, when I search for $x=0.0001$, it will give me $Fapprox -0.00088$. Yet, this situation is understandable. To deal with it, I will adjust the ranges according to my expectations of the results. For instance, if I know $x=0.0001$ is an expected result, then I will add shorter ranges, which is equivalent of saying a decimal adjustment. I think, I got it! I am happy. Thank you very much to all, as you thought me something very useful!
– user8028576
Jul 17 at 13:47
add a comment |Â
up vote
1
down vote
up vote
1
down vote
I will present an elementary calculus solution. I will first rename your $g(x)$ into $f(x)$ because I can and because it looks nicer to the eye. It is apparent that the best linear approximation to any function at $x=0$ must be the tangent line to the function at $x=0$. In our case, since $f(0)=0$, our tangent line must pass through $(0,0)$ too and so it must be of the form $y=mx$. Since it is the tangent line, the gradient $m$ of this line must have the same "gradient" (i.e. rate of change) as $f$ at $0$. To find the gradient, thus, we set
$$ m = f'(0) $$
Where the RHS is the derivative at $0$; so the rate of change as previously mentioned. So now we just have to take the derivative of $f$ and evaluate it at $0$ to find $m$. I'm sure you know how to do this, but I'll put it here for completeness's sake:
$$fracddx fracx1-x = frac(1-x)-(-1)x(1-x)^2$$
Evaluated at $0$ this is $1$. So $m=1$ and the closest linear approximation would be $y=mx=x$.
answered Jul 16 at 5:51
user496634
30518
I will present an elementary calculus solution. I will first rename your $g(x)$ into $f(x)$ because I can and because it looks nicer to the eye. It is apparent that the best linear approximation to any function at $x=0$ must be the tangent line to the function at $x=0$. In our case, since $f(0)=0$, our tangent line must pass through $(0,0)$ too and so it must be of the form $y=mx$. Since it is the tangent line, the gradient $m$ of this line must have the same "gradient" (i.e. rate of change) as $f$ at $0$. To find the gradient, thus, we set
$$ m = f'(0) $$
Where the RHS is the derivative at $0$; so the rate of change as previously mentioned. So now we just have to take the derivative of $f$ and evaluate it at $0$ to find $m$. I'm sure you know how to do this, but I'll put it here for completeness's sake:
$$fracddx fracx1-x = frac(1-x)-(-1)x(1-x)^2$$
Evaluated at $0$ this is $1$. So $m=1$ and the closest linear approximation would be $y=mx=x$.
answered Jul 16 at 5:51
user496634
30518
answered Jul 16 at 5:51
user496634
30518
answered Jul 16 at 5:51
user496634
30518
answered Jul 16 at 5:51
user496634
30518
30518
you basically propose $F = fracx2m$ as a n approximation to the original $f(x)$ given in the post. If I get it correctly, this approximation is not any close to $f(x)$. Let $x=0.9, m=1$, $f(x)=4.5$ and $F=0.45$. If I interpret it wrongly, please let me know.
– user8028576
Jul 16 at 15:40
@user8028576 You are absolutely right that it is not a very good approximation. However, it is the best linear one, locally around $x=0$. This is unless, of course, you want to approximate it around a range around $x=0$, not just best at that point and it's "immediate vicinity". In that case, you can consider Leibovici's answer. You will see that in a small range immediately around $0$, my approximation is better; while in the range $a$ and $b$ in his answer his is generally better.
– user496634
Jul 16 at 22:34
@user8028576 A helpful way to consider this is that my answer is the limit of Lebivoci's answer as $a$ and $b$ both approach $0$ (this is quite easy to prove). Around points extremely close to $0$, my approximation is extremely good; while his answer minimises the sum of the squared of the difference in values over a whole range $[a,b]$. One is an approximation at a point, the other is the approximation over a range.
– user496634
Jul 16 at 22:40
@user8028576 I totally get the logic of both approximations. Your approximation draws a linear line starting from the origin and it does not really worry about minimizing the difference in the approximation and the actual. Your approximation is the easiest with a single $F=fracx2m function. On the other hand, Leibovici's Taylor series draws a line for a "small" range of the whole plot that is as close as to the originial nonlinear line in that segmentation. Once I create "a lot of"of those lines with the given function, I will be able to approximate much more precisely.
– user8028576
Jul 17 at 13:42
1
But, when I get much closer to the tails of the approximation, his approximation yields an infeasible (negative) solution. For example, $a=0, b=0.1, m=1$, then, when I search for $x=0.0001$, it will give me $Fapprox -0.00088$. Yet, this situation is understandable. To deal with it, I will adjust the ranges according to my expectations of the results. For instance, if I know $x=0.0001$ is an expected result, then I will add shorter ranges, which is equivalent of saying a decimal adjustment. I think, I got it! I am happy. Thank you very much to all, as you thought me something very useful!
– user8028576
Jul 17 at 13:47
add a comment |Â
you basically propose $F = fracx2m$ as a n approximation to the original $f(x)$ given in the post. If I get it correctly, this approximation is not any close to $f(x)$. Let $x=0.9, m=1$, $f(x)=4.5$ and $F=0.45$. If I interpret it wrongly, please let me know.
– user8028576
Jul 16 at 15:40
@user8028576 You are absolutely right that it is not a very good approximation. However, it is the best linear one, locally around $x=0$. This is unless, of course, you want to approximate it around a range around $x=0$, not just best at that point and it's "immediate vicinity". In that case, you can consider Leibovici's answer. You will see that in a small range immediately around $0$, my approximation is better; while in the range $a$ and $b$ in his answer his is generally better.
– user496634
Jul 16 at 22:34
@user8028576 A helpful way to consider this is that my answer is the limit of Lebivoci's answer as $a$ and $b$ both approach $0$ (this is quite easy to prove). Around points extremely close to $0$, my approximation is extremely good; while his answer minimises the sum of the squared of the difference in values over a whole range $[a,b]$. One is an approximation at a point, the other is the approximation over a range.
– user496634
Jul 16 at 22:40
@user8028576 I totally get the logic of both approximations. Your approximation draws a linear line starting from the origin and it does not really worry about minimizing the difference in the approximation and the actual. Your approximation is the easiest with a single $F=fracx2m function. On the other hand, Leibovici's Taylor series draws a line for a "small" range of the whole plot that is as close as to the originial nonlinear line in that segmentation. Once I create "a lot of"of those lines with the given function, I will be able to approximate much more precisely.
– user8028576
Jul 17 at 13:42
1
But, when I get much closer to the tails of the approximation, his approximation yields an infeasible (negative) solution. For example, $a=0, b=0.1, m=1$, then, when I search for $x=0.0001$, it will give me $Fapprox -0.00088$. Yet, this situation is understandable. To deal with it, I will adjust the ranges according to my expectations of the results. For instance, if I know $x=0.0001$ is an expected result, then I will add shorter ranges, which is equivalent of saying a decimal adjustment. I think, I got it! I am happy. Thank you very much to all, as you thought me something very useful!
– user8028576
Jul 17 at 13:47
you basically propose $F = fracx2m$ as a n approximation to the original $f(x)$ given in the post. If I get it correctly, this approximation is not any close to $f(x)$. Let $x=0.9, m=1$, $f(x)=4.5$ and $F=0.45$. If I interpret it wrongly, please let me know.
– user8028576
Jul 16 at 15:40
you basically propose $F = fracx2m$ as a n approximation to the original $f(x)$ given in the post. If I get it correctly, this approximation is not any close to $f(x)$. Let $x=0.9, m=1$, $f(x)=4.5$ and $F=0.45$. If I interpret it wrongly, please let me know.
– user8028576
Jul 16 at 15:40
@user8028576 You are absolutely right that it is not a very good approximation. However, it is the best linear one, locally around $x=0$. This is unless, of course, you want to approximate it around a range around $x=0$, not just best at that point and it's "immediate vicinity". In that case, you can consider Leibovici's answer. You will see that in a small range immediately around $0$, my approximation is better; while in the range $a$ and $b$ in his answer his is generally better.
– user496634
Jul 16 at 22:34
@user8028576 You are absolutely right that it is not a very good approximation. However, it is the best linear one, locally around $x=0$. This is unless, of course, you want to approximate it around a range around $x=0$, not just best at that point and it's "immediate vicinity". In that case, you can consider Leibovici's answer. You will see that in a small range immediately around $0$, my approximation is better; while in the range $a$ and $b$ in his answer his is generally better.
– user496634
Jul 16 at 22:34
@user8028576 A helpful way to consider this is that my answer is the limit of Lebivoci's answer as $a$ and $b$ both approach $0$ (this is quite easy to prove). Around points extremely close to $0$, my approximation is extremely good; while his answer minimises the sum of the squared of the difference in values over a whole range $[a,b]$. One is an approximation at a point, the other is the approximation over a range.
– user496634
Jul 16 at 22:40
@user8028576 A helpful way to consider this is that my answer is the limit of Lebivoci's answer as $a$ and $b$ both approach $0$ (this is quite easy to prove). Around points extremely close to $0$, my approximation is extremely good; while his answer minimises the sum of the squared of the difference in values over a whole range $[a,b]$. One is an approximation at a point, the other is the approximation over a range.
– user496634
Jul 16 at 22:40
@user8028576 I totally get the logic of both approximations. Your approximation draws a linear line starting from the origin and it does not really worry about minimizing the difference in the approximation and the actual. Your approximation is the easiest with a single $F=fracx2m function. On the other hand, Leibovici's Taylor series draws a line for a "small" range of the whole plot that is as close as to the originial nonlinear line in that segmentation. Once I create "a lot of"of those lines with the given function, I will be able to approximate much more precisely.
– user8028576
Jul 17 at 13:42
@user8028576 I totally get the logic of both approximations. Your approximation draws a linear line starting from the origin and it does not really worry about minimizing the difference in the approximation and the actual. Your approximation is the easiest with a single $F=fracx2m function. On the other hand, Leibovici's Taylor series draws a line for a "small" range of the whole plot that is as close as to the originial nonlinear line in that segmentation. Once I create "a lot of"of those lines with the given function, I will be able to approximate much more precisely.
– user8028576
Jul 17 at 13:42
1
1
But, when I get much closer to the tails of the approximation, his approximation yields an infeasible (negative) solution. For example, $a=0, b=0.1, m=1$, then, when I search for $x=0.0001$, it will give me $Fapprox -0.00088$. Yet, this situation is understandable. To deal with it, I will adjust the ranges according to my expectations of the results. For instance, if I know $x=0.0001$ is an expected result, then I will add shorter ranges, which is equivalent of saying a decimal adjustment. I think, I got it! I am happy. Thank you very much to all, as you thought me something very useful!
– user8028576
Jul 17 at 13:47
But, when I get much closer to the tails of the approximation, his approximation yields an infeasible (negative) solution. For example, $a=0, b=0.1, m=1$, then, when I search for $x=0.0001$, it will give me $Fapprox -0.00088$. Yet, this situation is understandable. To deal with it, I will adjust the ranges according to my expectations of the results. For instance, if I know $x=0.0001$ is an expected result, then I will add shorter ranges, which is equivalent of saying a decimal adjustment. I think, I got it! I am happy. Thank you very much to all, as you thought me something very useful!
– user8028576
Jul 17 at 13:47
add a comment |Â
up vote
0
down vote
$$frac ddx frac1 1-x = frac1(1-x)^2$$ Additionally,
since $frac11-x = sum_k=0^inftyx^k$ clearly $$frac1(1-x)^2 = sum_k=1^inftykx^k-1$$Unfortunately, it does not really make sense to approximate linearly something like this in any global sense. Locally, evaluating $m = frac1(1-x_0)^2$ would give you the slope of a line tangent to $frac1(1-x)$ at point $x_0$ which locally approximates it linearly. I believe this is what you're asking.
answered Jul 15 at 3:47
BelowAverageIntelligence
339212
add a comment |Â
up vote
0
down vote
$$frac ddx frac1 1-x = frac1(1-x)^2$$ Additionally,
since $frac11-x = sum_k=0^inftyx^k$ clearly $$frac1(1-x)^2 = sum_k=1^inftykx^k-1$$Unfortunately, it does not really make sense to approximate linearly something like this in any global sense. Locally, evaluating $m = frac1(1-x_0)^2$ would give you the slope of a line tangent to $frac1(1-x)$ at point $x_0$ which locally approximates it linearly. I believe this is what you're asking.
answered Jul 15 at 3:47
BelowAverageIntelligence
339212
add a comment |Â
up vote
0
down vote
up vote
0
down vote
$$frac ddx frac1 1-x = frac1(1-x)^2$$ Additionally,
since $frac11-x = sum_k=0^inftyx^k$ clearly $$frac1(1-x)^2 = sum_k=1^inftykx^k-1$$Unfortunately, it does not really make sense to approximate linearly something like this in any global sense. Locally, evaluating $m = frac1(1-x_0)^2$ would give you the slope of a line tangent to $frac1(1-x)$ at point $x_0$ which locally approximates it linearly. I believe this is what you're asking.
answered Jul 15 at 3:47
BelowAverageIntelligence
339212
$$frac ddx frac1 1-x = frac1(1-x)^2$$ Additionally,
since $frac11-x = sum_k=0^inftyx^k$ clearly $$frac1(1-x)^2 = sum_k=1^inftykx^k-1$$Unfortunately, it does not really make sense to approximate linearly something like this in any global sense. Locally, evaluating $m = frac1(1-x_0)^2$ would give you the slope of a line tangent to $frac1(1-x)$ at point $x_0$ which locally approximates it linearly. I believe this is what you're asking.
answered Jul 15 at 3:47
BelowAverageIntelligence
339212
answered Jul 15 at 3:47
BelowAverageIntelligence
339212
answered Jul 15 at 3:47
BelowAverageIntelligence
339212
answered Jul 15 at 3:47
BelowAverageIntelligence
339212
339212
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2852141%2flinear-approximation-of-x-1-x%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Linear Approximation? You want to approximate the function using a straight line?
– W. mu
Jul 15 at 2:50
Yes. Does it sound impossible?
– user8028576
Jul 15 at 2:51
Of course, it's possible, but the error may be large. en.wikipedia.org/wiki/Approximation_theory
– W. mu
Jul 15 at 3:04
If you are looking for the usual calculus approximation, then it will be $h(x)=x$. This can be found by the formula $f(x)≈f'(x_0)(x-x_0)+f(x_0)$ when $x≈x_0$.
– user496634
Jul 15 at 9:53
@user496634 thanks for your comment. In my search for solutions, I confronted with the calculus approximation. Yet, I could not quite understand how it can make the function linear because the first derivative of the function still remains to be non-linear with squared $x$ terms. Additionally, I am lost with the term $x_0$. I am using this function in a MILP problem and I do solve for $x$ and other thousands of variables while there are several inequality constraints including $x$. If you may put your solution approach down, I will be thrilled to read it.
– user8028576
Jul 16 at 1:53