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Trying to investigate and find the sensitivity of function

akerman

New member
Jan 25, 2014
26
I have a question about function F(x) = sqrt(x)

I found that it has Kappa value equal to 1/2

I am not too sure what happens if when x = 0 is it just a minimum?

But now I am trying to investigate and find the sensitivity of F(x) to errors in x when we use x+ϵ, where ϵ is small.
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,888
Re: trying to investigate and find the sensitivity of function

I have a question about function F(x) = sqrt(x)

I found that it has Kappa value equal to 1/2

I am not too sure what happens if when x = 0 is it just a minimum?

But now I am trying to investigate and find the sensitivity of F(x) to errors in x when we use x+ϵ, where ϵ is small.
Welcome to MHB, akerman! :)

I am assuming that with a Kappa value you mean the condition number.
If so then we have:
$$\frac{|\Delta y|}{|y|} = \kappa \frac{|\Delta x|}{|x|}$$
where $y=F(x)$, where $\Delta x$ is the error in $x$, and $\Delta y$ is the error in $y$.
In other words: $\kappa$ gives the amplification of the relative error.

At $x=0$, $x$ has an infinite relative error (we're dividing by zero), so $\kappa$ is not defined there.
Since $F(0)=0$, the relative error of $y$ is also infinite.

We can calculate $\kappa$ with:
$$\kappa = \left| \frac{x \cdot F'(x)}{F(x)} \right|$$
This stems from the first order Taylor approximation:
$$F(x + \Delta x) \approx F(x) + \Delta x \cdot F'(x)$$

Since we are using an approximation, the end result is also an approximation.
In particular for points where $F'(x)$ or $F''(x)$ are not defined, the relationship will break down.
This is in particular the case for your function at $x=0$.
 

akerman

New member
Jan 25, 2014
26
Re: trying to investigate and find the sensitivity of function

Welcome to MHB, akerman! :)

I am assuming that with a Kappa value you mean the condition number.
If so then we have:
$$\frac{|\Delta y|}{|y|} = \kappa \frac{|\Delta x|}{|x|}$$
where $y=F(x)$, where $\Delta x$ is the error in $x$, and $\Delta y$ is the error in $y$.
In other words: $\kappa$ gives the amplification of the relative error.

At $x=0$, $x$ has an infinite relative error (we're dividing by zero), so $\kappa$ is not defined there.
Since $F(0)=0$, the relative error of $y$ is also infinite.

We can calculate $\kappa$ with:
$$\kappa = \left| \frac{x \cdot F'(x)}{F(x)} \right|$$
This stems from the first order Taylor approximation:
$$F(x + \Delta x) \approx F(x) + \Delta x \cdot F'(x)$$

Since we are using an approximation, the end result is also an approximation.
In particular for points where $F'(x)$ or $F''(x)$ are not defined, the relationship will break down.
This is in particular the case for your function at $x=0$.
I still don't get it...
So what is the sensitivity of f(x) to errors in x?
And if we consider limit x→0, how many digits can one compute x√ when x is known to an error of 10^−16?

Can you give more detailed explanation.
thanks
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,888
Re: trying to investigate and find the sensitivity of function

I still don't get it...
So what is the sensitivity of f(x) to errors in x?
Since the derivative of $\sqrt x$ is $\frac 1 {2\sqrt x}$, Taylor's approximation gives us:
$$\sqrt{x+ε} \approx \sqrt{x} + ε \cdot \frac{1}{2\sqrt x}$$
So if the error in $x$ is $ε$, then the error in $√x$ is approximately $ε \cdot \frac{1}{2\sqrt x}$.

The so called absolute sensitivity to errors in x is $\frac{1}{2\sqrt x}$, since an error gets multiplied by this amount.

The relative sensitivity is $\frac 1 2$, since relative errors get multiplied by this amount.
A relative error is the error relative to the value measured. For $x$ this is $ε / x$.


And if we consider limit x→0, how many digits can one compute x√ when x is known to an error of 10^−16?
I do not understand your question.
In the limit x→0, √x is simply 0.

However, if x is a regular non-zero value known with an error of $10^{−16}$, then the resultant √x will have an absolute error of $10^{−16} \cdot \frac{1}{2\sqrt x}$ and a relative error of $0.5 \cdot 10^{−16}$.

Can you give more detailed explanation.
thanks
Where would you like more details?
 

akerman

New member
Jan 25, 2014
26
The lastest answer is just something I was looking for.
So having a quesiton such as "how many digits can one compute x√ when x is known to an error of 10^−16?" Can I simply say that absolute error of 10−16⋅12x√ and a relative error of 0.5⋅10^−16?
Also if we have x= x+ ε
Can I specify exactly what the ε is for y =x√ ? Or is it just an assumption that is it a small number?
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,888
The lastest answer is just something I was looking for.
So having a quesiton such as "how many digits can one compute x√ when x is known to an error of 10^−16?" Can I simply say that absolute error of 10−16⋅12x√ and a relative error of 0.5⋅10^−16?
The term "how many digits" is somewhat confusing.
It can typically mean either how many digits behind the decimal point, or it can mean how many significant digits.
Can you clarify which one is intended?
Similarly, when you say "error" do you mean an absolute error or a relative error?

If you have 16 significant digits, that means that your relative error is $10^{-16}$.
In this case the resultant relative error is $0.5 \cdot 10^{-16}$, meaning you have slightly over 16 significant digits (usually treated as just 16).

What you write about the errors is correct, assuming your initial error is an absolute error.
However, that is apparently not what is being asked, since the question asks "how many digits".


Also if we have x= x+ ε
Can I specify exactly what the ε is for y =x√ ? Or is it just an assumption that is it a small number?
"Exactly" is a strong word.
If you want to have the "exact" error in y, you need to calculate $\sqrt{x+ε}-\sqrt x$.
If you are satisfied with the approximate error, you can use the formulas I gave.
 

akerman

New member
Jan 25, 2014
26
Now I got it. Thanks for help. I believe you the only person in number of forum who could explain and answer it.