Mixing
Dual Number for derivation of tangens to power of arbitrary number
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I have recently stumbled across dual numbers and thought they may help me to solve a problem. What I ultimately want to do is root finding of a polynomial of an arbitrary degree. The polynomial is of form
$f(x) = sum_i=0^n k_i (g(x))^i$
where $g(x)$ can be just $x$ or a trigonometric function (including $tan(x)$) and $g(x)^i$ is the $g$ to the power of $i$ (rather than $g$ applied to itself $i$ times).
Currently I'm assuming $g(x)=x$ and use Newton's method which works fine, but obviously this won't work as easily with $g(x)=tan(x)$. Now I could use regula falsi to avoid having to calculate the derivative of $f$, but since I need two points $x_1$ and $x_2$ to start the algorithm with I have problem, because if my $x_2$ is too great then I end up with $f(x_1) <0$ and $f(x_2) < 0$.
Now come the dual numbers into play, because I have a rough estimate of where the root is, if it exists. I want to use them for Newton's method to calculate the root I'm looking for but I'm uncertain of how to actually apply this.
In my research so far I couldn't find how to "extend" my function $f$ to dual numbers in order to solve for both the value at $x$ and the derivative $x$. In a blog and a few other resources they "just use" the value of x and add $1epsilon$ (or at least that's how far my understanding of the texts goes).
Is this correct in my case, when using $g(x)=tan(x)$ and thus
$f(x) = sum_i=0^n k_i tan(x+1epsilon)^i$
? This just seems way to easy, but I have no feeling for this method yet.
derivatives hypercomplex-numbers
asked Aug 6 at 7:31
Tare
1011
add a comment |Â
up vote
0
down vote
favorite
I have recently stumbled across dual numbers and thought they may help me to solve a problem. What I ultimately want to do is root finding of a polynomial of an arbitrary degree. The polynomial is of form
$f(x) = sum_i=0^n k_i (g(x))^i$
where $g(x)$ can be just $x$ or a trigonometric function (including $tan(x)$) and $g(x)^i$ is the $g$ to the power of $i$ (rather than $g$ applied to itself $i$ times).
Currently I'm assuming $g(x)=x$ and use Newton's method which works fine, but obviously this won't work as easily with $g(x)=tan(x)$. Now I could use regula falsi to avoid having to calculate the derivative of $f$, but since I need two points $x_1$ and $x_2$ to start the algorithm with I have problem, because if my $x_2$ is too great then I end up with $f(x_1) <0$ and $f(x_2) < 0$.
Now come the dual numbers into play, because I have a rough estimate of where the root is, if it exists. I want to use them for Newton's method to calculate the root I'm looking for but I'm uncertain of how to actually apply this.
In my research so far I couldn't find how to "extend" my function $f$ to dual numbers in order to solve for both the value at $x$ and the derivative $x$. In a blog and a few other resources they "just use" the value of x and add $1epsilon$ (or at least that's how far my understanding of the texts goes).
Is this correct in my case, when using $g(x)=tan(x)$ and thus
$f(x) = sum_i=0^n k_i tan(x+1epsilon)^i$
? This just seems way to easy, but I have no feeling for this method yet.
derivatives hypercomplex-numbers
asked Aug 6 at 7:31
Tare
1011
You are correct. It is that easy. In general, since $epsilon^2=0$, we have $f(x+epsilon) = f(x)+f'(x)epsilon.$ Or else, as you did, just substitute $x+epsilon$ for $x$.
– Somos
Aug 6 at 14:52
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I have recently stumbled across dual numbers and thought they may help me to solve a problem. What I ultimately want to do is root finding of a polynomial of an arbitrary degree. The polynomial is of form
$f(x) = sum_i=0^n k_i (g(x))^i$
where $g(x)$ can be just $x$ or a trigonometric function (including $tan(x)$) and $g(x)^i$ is the $g$ to the power of $i$ (rather than $g$ applied to itself $i$ times).
Currently I'm assuming $g(x)=x$ and use Newton's method which works fine, but obviously this won't work as easily with $g(x)=tan(x)$. Now I could use regula falsi to avoid having to calculate the derivative of $f$, but since I need two points $x_1$ and $x_2$ to start the algorithm with I have problem, because if my $x_2$ is too great then I end up with $f(x_1) <0$ and $f(x_2) < 0$.
Now come the dual numbers into play, because I have a rough estimate of where the root is, if it exists. I want to use them for Newton's method to calculate the root I'm looking for but I'm uncertain of how to actually apply this.
In my research so far I couldn't find how to "extend" my function $f$ to dual numbers in order to solve for both the value at $x$ and the derivative $x$. In a blog and a few other resources they "just use" the value of x and add $1epsilon$ (or at least that's how far my understanding of the texts goes).
Is this correct in my case, when using $g(x)=tan(x)$ and thus
$f(x) = sum_i=0^n k_i tan(x+1epsilon)^i$
? This just seems way to easy, but I have no feeling for this method yet.
derivatives hypercomplex-numbers
asked Aug 6 at 7:31
Tare
1011
I have recently stumbled across dual numbers and thought they may help me to solve a problem. What I ultimately want to do is root finding of a polynomial of an arbitrary degree. The polynomial is of form
$f(x) = sum_i=0^n k_i (g(x))^i$
where $g(x)$ can be just $x$ or a trigonometric function (including $tan(x)$) and $g(x)^i$ is the $g$ to the power of $i$ (rather than $g$ applied to itself $i$ times).
Currently I'm assuming $g(x)=x$ and use Newton's method which works fine, but obviously this won't work as easily with $g(x)=tan(x)$. Now I could use regula falsi to avoid having to calculate the derivative of $f$, but since I need two points $x_1$ and $x_2$ to start the algorithm with I have problem, because if my $x_2$ is too great then I end up with $f(x_1) <0$ and $f(x_2) < 0$.
Now come the dual numbers into play, because I have a rough estimate of where the root is, if it exists. I want to use them for Newton's method to calculate the root I'm looking for but I'm uncertain of how to actually apply this.
In my research so far I couldn't find how to "extend" my function $f$ to dual numbers in order to solve for both the value at $x$ and the derivative $x$. In a blog and a few other resources they "just use" the value of x and add $1epsilon$ (or at least that's how far my understanding of the texts goes).
Is this correct in my case, when using $g(x)=tan(x)$ and thus
$f(x) = sum_i=0^n k_i tan(x+1epsilon)^i$
? This just seems way to easy, but I have no feeling for this method yet.
derivatives hypercomplex-numbers
asked Aug 6 at 7:31
Tare
1011
asked Aug 6 at 7:31
Tare
1011
asked Aug 6 at 7:31
Tare
1011
asked Aug 6 at 7:31
Tare
1011
1011
You are correct. It is that easy. In general, since $epsilon^2=0$, we have $f(x+epsilon) = f(x)+f'(x)epsilon.$ Or else, as you did, just substitute $x+epsilon$ for $x$.
– Somos
Aug 6 at 14:52
add a comment |Â
You are correct. It is that easy. In general, since $epsilon^2=0$, we have $f(x+epsilon) = f(x)+f'(x)epsilon.$ Or else, as you did, just substitute $x+epsilon$ for $x$.
– Somos
Aug 6 at 14:52
You are correct. It is that easy. In general, since $epsilon^2=0$, we have $f(x+epsilon) = f(x)+f'(x)epsilon.$ Or else, as you did, just substitute $x+epsilon$ for $x$.
– Somos
Aug 6 at 14:52
You are correct. It is that easy. In general, since $epsilon^2=0$, we have $f(x+epsilon) = f(x)+f'(x)epsilon.$ Or else, as you did, just substitute $x+epsilon$ for $x$.
– Somos
Aug 6 at 14:52
add a comment |Â
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You are correct. It is that easy. In general, since $epsilon^2=0$, we have $f(x+epsilon) = f(x)+f'(x)epsilon.$ Or else, as you did, just substitute $x+epsilon$ for $x$.
– Somos
Aug 6 at 14:52