How does Horner method evaluate the derivative of a function
up vote
1
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From my understanding, Horner method is mainly used to evaluate polynomial functions by altering the equation into a simpler recursive relation with lesser number of operations.
Say for example, I was given $f (x) = 4x^4 + 3x^3 +2x^2+x+5$
This can be rewritten as $5 +x (1+x (2+x (3+x (4)))$
Were we can evaluate the function as a recurrent relation of simpler terms starting from:
$b_n=4 $
$b_{n-1} = 3 + b_n* x$
And $b_0$ would be the whole term evaluated and therefore the image of the function.
What I want to understand how is running horner method to the $b_n$ values result in the derivative?
derivatives numerical-methods
add a comment |
up vote
1
down vote
favorite
From my understanding, Horner method is mainly used to evaluate polynomial functions by altering the equation into a simpler recursive relation with lesser number of operations.
Say for example, I was given $f (x) = 4x^4 + 3x^3 +2x^2+x+5$
This can be rewritten as $5 +x (1+x (2+x (3+x (4)))$
Were we can evaluate the function as a recurrent relation of simpler terms starting from:
$b_n=4 $
$b_{n-1} = 3 + b_n* x$
And $b_0$ would be the whole term evaluated and therefore the image of the function.
What I want to understand how is running horner method to the $b_n$ values result in the derivative?
derivatives numerical-methods
@JeanMarie: Links are overlapped therefore clicking doesn't lead to any of them.
– kub0x
Feb 11 '17 at 9:23
Here is the first link, the most interesting :(cut-the-knot.org/Curriculum/Calculus/HornerMethod.shtml)
– Jean Marie
Feb 11 '17 at 9:28
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
From my understanding, Horner method is mainly used to evaluate polynomial functions by altering the equation into a simpler recursive relation with lesser number of operations.
Say for example, I was given $f (x) = 4x^4 + 3x^3 +2x^2+x+5$
This can be rewritten as $5 +x (1+x (2+x (3+x (4)))$
Were we can evaluate the function as a recurrent relation of simpler terms starting from:
$b_n=4 $
$b_{n-1} = 3 + b_n* x$
And $b_0$ would be the whole term evaluated and therefore the image of the function.
What I want to understand how is running horner method to the $b_n$ values result in the derivative?
derivatives numerical-methods
From my understanding, Horner method is mainly used to evaluate polynomial functions by altering the equation into a simpler recursive relation with lesser number of operations.
Say for example, I was given $f (x) = 4x^4 + 3x^3 +2x^2+x+5$
This can be rewritten as $5 +x (1+x (2+x (3+x (4)))$
Were we can evaluate the function as a recurrent relation of simpler terms starting from:
$b_n=4 $
$b_{n-1} = 3 + b_n* x$
And $b_0$ would be the whole term evaluated and therefore the image of the function.
What I want to understand how is running horner method to the $b_n$ values result in the derivative?
derivatives numerical-methods
derivatives numerical-methods
edited Feb 11 '17 at 9:33
kub0x
1,7331821
1,7331821
asked Feb 11 '17 at 7:25
Raed Tabani
7218
7218
@JeanMarie: Links are overlapped therefore clicking doesn't lead to any of them.
– kub0x
Feb 11 '17 at 9:23
Here is the first link, the most interesting :(cut-the-knot.org/Curriculum/Calculus/HornerMethod.shtml)
– Jean Marie
Feb 11 '17 at 9:28
add a comment |
@JeanMarie: Links are overlapped therefore clicking doesn't lead to any of them.
– kub0x
Feb 11 '17 at 9:23
Here is the first link, the most interesting :(cut-the-knot.org/Curriculum/Calculus/HornerMethod.shtml)
– Jean Marie
Feb 11 '17 at 9:28
@JeanMarie: Links are overlapped therefore clicking doesn't lead to any of them.
– kub0x
Feb 11 '17 at 9:23
@JeanMarie: Links are overlapped therefore clicking doesn't lead to any of them.
– kub0x
Feb 11 '17 at 9:23
Here is the first link, the most interesting :(cut-the-knot.org/Curriculum/Calculus/HornerMethod.shtml)
– Jean Marie
Feb 11 '17 at 9:28
Here is the first link, the most interesting :(cut-the-knot.org/Curriculum/Calculus/HornerMethod.shtml)
– Jean Marie
Feb 11 '17 at 9:28
add a comment |
3 Answers
3
active
oldest
votes
up vote
3
down vote
I do not know if this will answer the question; if it does not, please forgive me.
Consider the polynomial to be $$p=sum_{i=1}^n c_i, x^{i-1}implies p'=sum_{i=1}^n (i-1), c_i, x^{i-2}$$ As a pseudo code, you would have
p = c(n)
dp = 0
do j = n-1 , 1 , -1
dp = dp * x + p
p = p * x + c(j)
enddo
which computes at the same time the polynomial and its derivative with a minimum number of basic operations.
The first summation (the polynomial $p$, not the derivative $p'$) should have $i=0$. I assume that code is Fortran "pseudocode" - as Fortran isn't all that mainstream, a comment explaining the loop might help. It's not hard to work out that the loop body runs for $j=n-1$ down to $j=0$ inclusive, but thatdo
syntax doesn't clearly express that IMO (actually it seems actively misleading, and makes me wonder if the, 1 ,
should be a, 0 ,
).
– Steve314
Nov 22 at 11:25
@Steve314. You are right ! Typo that I shall fix.
– Claude Leibovici
Nov 22 at 11:50
add a comment |
up vote
2
down vote
If you apply the Horner scheme, or perhaps the Ruffini method (both originally published papers on variants of the Newton method that use a trick for fast polynomial evaluation that was previously well-known), then you perform a polynomial division with remainder by a linear term. From the Horner table, you can read off the coefficients for a polynomial $q$ so that
$$
p(x)=p(a)+(x-a)q(x)iff q(x)=frac{p(x)-p(a)}{x-a}
$$
which tells you that $q(a)=p'(a)$. A second Horner evaluation below the first table will thus evaluate the derivative value.
You can also see this as algorithmic differentiation, if the original algorithm is
val = 0
for k=0 to deg
val = val*x + a[deg-k]
end for
then the derivative by using the chain rule in every step gives
dval = 0; val = 0
for k=0 to deg
dval = dval*x + val
val = val*x + a[deg-k]
end for
The derivative steps needs to be computed first as it uses the last value of val
.
add a comment |
up vote
-1
down vote
The key is to make the indices the same and use original Horner's rule in the same for loop to evaluate both polynomial and its derivative.
Let's say the original problem is to compute f(x)=sum[ c[n] x^n, for {n,0 ,k }]
.
The derivative is of the form f'(x)=sum[ m c[m] x^(m-1), for {m,1 ,k }]
.
A change of index n=m-1
make the base of iterator the same, i.e. f'(x)=sum[ (n+1) c[n+1] x^(n), for {n,0 ,k-1 }]
Now we have same iterator which can be used except the case n=k
which only applies to evaluation of function and not the derivative. This case can be computed outside the loop easily.
Here is the way to do it in Python:
def horner_eval(x, a):
'''Uses Horner's rule to return the polynomial
a[n] + a[n+1] x^1 + a[n+2] x^2 + ... + a[0] x^(n)
AND its derivative evaluated at x.
NOTICE: coefficient vector is given backward for ease of indexing
'''
dist, speed = 0, 0
for i in range(len(a) - 1):
dist = a[i] + (x * dist)
speed = i * a[i] + (x * speed)
dist = a[-1] + (x * dist) # since function has one term more that the derivative
return dist, speed
dist
stores the values of the polynomial at x
.
speed
stores the values of derivative of the polynomial at x
.
1
You might want to include a little about why this evaluates the derivative
– eepperly16
Dec 22 '17 at 17:18
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
I do not know if this will answer the question; if it does not, please forgive me.
Consider the polynomial to be $$p=sum_{i=1}^n c_i, x^{i-1}implies p'=sum_{i=1}^n (i-1), c_i, x^{i-2}$$ As a pseudo code, you would have
p = c(n)
dp = 0
do j = n-1 , 1 , -1
dp = dp * x + p
p = p * x + c(j)
enddo
which computes at the same time the polynomial and its derivative with a minimum number of basic operations.
The first summation (the polynomial $p$, not the derivative $p'$) should have $i=0$. I assume that code is Fortran "pseudocode" - as Fortran isn't all that mainstream, a comment explaining the loop might help. It's not hard to work out that the loop body runs for $j=n-1$ down to $j=0$ inclusive, but thatdo
syntax doesn't clearly express that IMO (actually it seems actively misleading, and makes me wonder if the, 1 ,
should be a, 0 ,
).
– Steve314
Nov 22 at 11:25
@Steve314. You are right ! Typo that I shall fix.
– Claude Leibovici
Nov 22 at 11:50
add a comment |
up vote
3
down vote
I do not know if this will answer the question; if it does not, please forgive me.
Consider the polynomial to be $$p=sum_{i=1}^n c_i, x^{i-1}implies p'=sum_{i=1}^n (i-1), c_i, x^{i-2}$$ As a pseudo code, you would have
p = c(n)
dp = 0
do j = n-1 , 1 , -1
dp = dp * x + p
p = p * x + c(j)
enddo
which computes at the same time the polynomial and its derivative with a minimum number of basic operations.
The first summation (the polynomial $p$, not the derivative $p'$) should have $i=0$. I assume that code is Fortran "pseudocode" - as Fortran isn't all that mainstream, a comment explaining the loop might help. It's not hard to work out that the loop body runs for $j=n-1$ down to $j=0$ inclusive, but thatdo
syntax doesn't clearly express that IMO (actually it seems actively misleading, and makes me wonder if the, 1 ,
should be a, 0 ,
).
– Steve314
Nov 22 at 11:25
@Steve314. You are right ! Typo that I shall fix.
– Claude Leibovici
Nov 22 at 11:50
add a comment |
up vote
3
down vote
up vote
3
down vote
I do not know if this will answer the question; if it does not, please forgive me.
Consider the polynomial to be $$p=sum_{i=1}^n c_i, x^{i-1}implies p'=sum_{i=1}^n (i-1), c_i, x^{i-2}$$ As a pseudo code, you would have
p = c(n)
dp = 0
do j = n-1 , 1 , -1
dp = dp * x + p
p = p * x + c(j)
enddo
which computes at the same time the polynomial and its derivative with a minimum number of basic operations.
I do not know if this will answer the question; if it does not, please forgive me.
Consider the polynomial to be $$p=sum_{i=1}^n c_i, x^{i-1}implies p'=sum_{i=1}^n (i-1), c_i, x^{i-2}$$ As a pseudo code, you would have
p = c(n)
dp = 0
do j = n-1 , 1 , -1
dp = dp * x + p
p = p * x + c(j)
enddo
which computes at the same time the polynomial and its derivative with a minimum number of basic operations.
edited Nov 22 at 11:53
answered Feb 11 '17 at 8:55
Claude Leibovici
118k1156131
118k1156131
The first summation (the polynomial $p$, not the derivative $p'$) should have $i=0$. I assume that code is Fortran "pseudocode" - as Fortran isn't all that mainstream, a comment explaining the loop might help. It's not hard to work out that the loop body runs for $j=n-1$ down to $j=0$ inclusive, but thatdo
syntax doesn't clearly express that IMO (actually it seems actively misleading, and makes me wonder if the, 1 ,
should be a, 0 ,
).
– Steve314
Nov 22 at 11:25
@Steve314. You are right ! Typo that I shall fix.
– Claude Leibovici
Nov 22 at 11:50
add a comment |
The first summation (the polynomial $p$, not the derivative $p'$) should have $i=0$. I assume that code is Fortran "pseudocode" - as Fortran isn't all that mainstream, a comment explaining the loop might help. It's not hard to work out that the loop body runs for $j=n-1$ down to $j=0$ inclusive, but thatdo
syntax doesn't clearly express that IMO (actually it seems actively misleading, and makes me wonder if the, 1 ,
should be a, 0 ,
).
– Steve314
Nov 22 at 11:25
@Steve314. You are right ! Typo that I shall fix.
– Claude Leibovici
Nov 22 at 11:50
The first summation (the polynomial $p$, not the derivative $p'$) should have $i=0$. I assume that code is Fortran "pseudocode" - as Fortran isn't all that mainstream, a comment explaining the loop might help. It's not hard to work out that the loop body runs for $j=n-1$ down to $j=0$ inclusive, but that
do
syntax doesn't clearly express that IMO (actually it seems actively misleading, and makes me wonder if the , 1 ,
should be a , 0 ,
).– Steve314
Nov 22 at 11:25
The first summation (the polynomial $p$, not the derivative $p'$) should have $i=0$. I assume that code is Fortran "pseudocode" - as Fortran isn't all that mainstream, a comment explaining the loop might help. It's not hard to work out that the loop body runs for $j=n-1$ down to $j=0$ inclusive, but that
do
syntax doesn't clearly express that IMO (actually it seems actively misleading, and makes me wonder if the , 1 ,
should be a , 0 ,
).– Steve314
Nov 22 at 11:25
@Steve314. You are right ! Typo that I shall fix.
– Claude Leibovici
Nov 22 at 11:50
@Steve314. You are right ! Typo that I shall fix.
– Claude Leibovici
Nov 22 at 11:50
add a comment |
up vote
2
down vote
If you apply the Horner scheme, or perhaps the Ruffini method (both originally published papers on variants of the Newton method that use a trick for fast polynomial evaluation that was previously well-known), then you perform a polynomial division with remainder by a linear term. From the Horner table, you can read off the coefficients for a polynomial $q$ so that
$$
p(x)=p(a)+(x-a)q(x)iff q(x)=frac{p(x)-p(a)}{x-a}
$$
which tells you that $q(a)=p'(a)$. A second Horner evaluation below the first table will thus evaluate the derivative value.
You can also see this as algorithmic differentiation, if the original algorithm is
val = 0
for k=0 to deg
val = val*x + a[deg-k]
end for
then the derivative by using the chain rule in every step gives
dval = 0; val = 0
for k=0 to deg
dval = dval*x + val
val = val*x + a[deg-k]
end for
The derivative steps needs to be computed first as it uses the last value of val
.
add a comment |
up vote
2
down vote
If you apply the Horner scheme, or perhaps the Ruffini method (both originally published papers on variants of the Newton method that use a trick for fast polynomial evaluation that was previously well-known), then you perform a polynomial division with remainder by a linear term. From the Horner table, you can read off the coefficients for a polynomial $q$ so that
$$
p(x)=p(a)+(x-a)q(x)iff q(x)=frac{p(x)-p(a)}{x-a}
$$
which tells you that $q(a)=p'(a)$. A second Horner evaluation below the first table will thus evaluate the derivative value.
You can also see this as algorithmic differentiation, if the original algorithm is
val = 0
for k=0 to deg
val = val*x + a[deg-k]
end for
then the derivative by using the chain rule in every step gives
dval = 0; val = 0
for k=0 to deg
dval = dval*x + val
val = val*x + a[deg-k]
end for
The derivative steps needs to be computed first as it uses the last value of val
.
add a comment |
up vote
2
down vote
up vote
2
down vote
If you apply the Horner scheme, or perhaps the Ruffini method (both originally published papers on variants of the Newton method that use a trick for fast polynomial evaluation that was previously well-known), then you perform a polynomial division with remainder by a linear term. From the Horner table, you can read off the coefficients for a polynomial $q$ so that
$$
p(x)=p(a)+(x-a)q(x)iff q(x)=frac{p(x)-p(a)}{x-a}
$$
which tells you that $q(a)=p'(a)$. A second Horner evaluation below the first table will thus evaluate the derivative value.
You can also see this as algorithmic differentiation, if the original algorithm is
val = 0
for k=0 to deg
val = val*x + a[deg-k]
end for
then the derivative by using the chain rule in every step gives
dval = 0; val = 0
for k=0 to deg
dval = dval*x + val
val = val*x + a[deg-k]
end for
The derivative steps needs to be computed first as it uses the last value of val
.
If you apply the Horner scheme, or perhaps the Ruffini method (both originally published papers on variants of the Newton method that use a trick for fast polynomial evaluation that was previously well-known), then you perform a polynomial division with remainder by a linear term. From the Horner table, you can read off the coefficients for a polynomial $q$ so that
$$
p(x)=p(a)+(x-a)q(x)iff q(x)=frac{p(x)-p(a)}{x-a}
$$
which tells you that $q(a)=p'(a)$. A second Horner evaluation below the first table will thus evaluate the derivative value.
You can also see this as algorithmic differentiation, if the original algorithm is
val = 0
for k=0 to deg
val = val*x + a[deg-k]
end for
then the derivative by using the chain rule in every step gives
dval = 0; val = 0
for k=0 to deg
dval = dval*x + val
val = val*x + a[deg-k]
end for
The derivative steps needs to be computed first as it uses the last value of val
.
edited Dec 22 '17 at 16:59
answered Feb 11 '17 at 9:00
LutzL
55k42053
55k42053
add a comment |
add a comment |
up vote
-1
down vote
The key is to make the indices the same and use original Horner's rule in the same for loop to evaluate both polynomial and its derivative.
Let's say the original problem is to compute f(x)=sum[ c[n] x^n, for {n,0 ,k }]
.
The derivative is of the form f'(x)=sum[ m c[m] x^(m-1), for {m,1 ,k }]
.
A change of index n=m-1
make the base of iterator the same, i.e. f'(x)=sum[ (n+1) c[n+1] x^(n), for {n,0 ,k-1 }]
Now we have same iterator which can be used except the case n=k
which only applies to evaluation of function and not the derivative. This case can be computed outside the loop easily.
Here is the way to do it in Python:
def horner_eval(x, a):
'''Uses Horner's rule to return the polynomial
a[n] + a[n+1] x^1 + a[n+2] x^2 + ... + a[0] x^(n)
AND its derivative evaluated at x.
NOTICE: coefficient vector is given backward for ease of indexing
'''
dist, speed = 0, 0
for i in range(len(a) - 1):
dist = a[i] + (x * dist)
speed = i * a[i] + (x * speed)
dist = a[-1] + (x * dist) # since function has one term more that the derivative
return dist, speed
dist
stores the values of the polynomial at x
.
speed
stores the values of derivative of the polynomial at x
.
1
You might want to include a little about why this evaluates the derivative
– eepperly16
Dec 22 '17 at 17:18
add a comment |
up vote
-1
down vote
The key is to make the indices the same and use original Horner's rule in the same for loop to evaluate both polynomial and its derivative.
Let's say the original problem is to compute f(x)=sum[ c[n] x^n, for {n,0 ,k }]
.
The derivative is of the form f'(x)=sum[ m c[m] x^(m-1), for {m,1 ,k }]
.
A change of index n=m-1
make the base of iterator the same, i.e. f'(x)=sum[ (n+1) c[n+1] x^(n), for {n,0 ,k-1 }]
Now we have same iterator which can be used except the case n=k
which only applies to evaluation of function and not the derivative. This case can be computed outside the loop easily.
Here is the way to do it in Python:
def horner_eval(x, a):
'''Uses Horner's rule to return the polynomial
a[n] + a[n+1] x^1 + a[n+2] x^2 + ... + a[0] x^(n)
AND its derivative evaluated at x.
NOTICE: coefficient vector is given backward for ease of indexing
'''
dist, speed = 0, 0
for i in range(len(a) - 1):
dist = a[i] + (x * dist)
speed = i * a[i] + (x * speed)
dist = a[-1] + (x * dist) # since function has one term more that the derivative
return dist, speed
dist
stores the values of the polynomial at x
.
speed
stores the values of derivative of the polynomial at x
.
1
You might want to include a little about why this evaluates the derivative
– eepperly16
Dec 22 '17 at 17:18
add a comment |
up vote
-1
down vote
up vote
-1
down vote
The key is to make the indices the same and use original Horner's rule in the same for loop to evaluate both polynomial and its derivative.
Let's say the original problem is to compute f(x)=sum[ c[n] x^n, for {n,0 ,k }]
.
The derivative is of the form f'(x)=sum[ m c[m] x^(m-1), for {m,1 ,k }]
.
A change of index n=m-1
make the base of iterator the same, i.e. f'(x)=sum[ (n+1) c[n+1] x^(n), for {n,0 ,k-1 }]
Now we have same iterator which can be used except the case n=k
which only applies to evaluation of function and not the derivative. This case can be computed outside the loop easily.
Here is the way to do it in Python:
def horner_eval(x, a):
'''Uses Horner's rule to return the polynomial
a[n] + a[n+1] x^1 + a[n+2] x^2 + ... + a[0] x^(n)
AND its derivative evaluated at x.
NOTICE: coefficient vector is given backward for ease of indexing
'''
dist, speed = 0, 0
for i in range(len(a) - 1):
dist = a[i] + (x * dist)
speed = i * a[i] + (x * speed)
dist = a[-1] + (x * dist) # since function has one term more that the derivative
return dist, speed
dist
stores the values of the polynomial at x
.
speed
stores the values of derivative of the polynomial at x
.
The key is to make the indices the same and use original Horner's rule in the same for loop to evaluate both polynomial and its derivative.
Let's say the original problem is to compute f(x)=sum[ c[n] x^n, for {n,0 ,k }]
.
The derivative is of the form f'(x)=sum[ m c[m] x^(m-1), for {m,1 ,k }]
.
A change of index n=m-1
make the base of iterator the same, i.e. f'(x)=sum[ (n+1) c[n+1] x^(n), for {n,0 ,k-1 }]
Now we have same iterator which can be used except the case n=k
which only applies to evaluation of function and not the derivative. This case can be computed outside the loop easily.
Here is the way to do it in Python:
def horner_eval(x, a):
'''Uses Horner's rule to return the polynomial
a[n] + a[n+1] x^1 + a[n+2] x^2 + ... + a[0] x^(n)
AND its derivative evaluated at x.
NOTICE: coefficient vector is given backward for ease of indexing
'''
dist, speed = 0, 0
for i in range(len(a) - 1):
dist = a[i] + (x * dist)
speed = i * a[i] + (x * speed)
dist = a[-1] + (x * dist) # since function has one term more that the derivative
return dist, speed
dist
stores the values of the polynomial at x
.
speed
stores the values of derivative of the polynomial at x
.
edited Dec 31 '17 at 23:06
answered Dec 22 '17 at 16:48
Pourmehrab
11
11
1
You might want to include a little about why this evaluates the derivative
– eepperly16
Dec 22 '17 at 17:18
add a comment |
1
You might want to include a little about why this evaluates the derivative
– eepperly16
Dec 22 '17 at 17:18
1
1
You might want to include a little about why this evaluates the derivative
– eepperly16
Dec 22 '17 at 17:18
You might want to include a little about why this evaluates the derivative
– eepperly16
Dec 22 '17 at 17:18
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@JeanMarie: Links are overlapped therefore clicking doesn't lead to any of them.
– kub0x
Feb 11 '17 at 9:23
Here is the first link, the most interesting :(cut-the-knot.org/Curriculum/Calculus/HornerMethod.shtml)
– Jean Marie
Feb 11 '17 at 9:28