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How to apply a polynomial to a sequence


How to apply a polynomial to a sequence

By : Knuhl
Date : November 22 2020, 10:48 AM
I think the issue was by ths following , The notation [i] means taking the coefficient of z^i. Notice that i runs only from 0 to l-d in the definition; so, the result of computation is empty if d is greater than l. Your example with d=4 and l=3 is not a valid input for this computation. In the actual algorithm in the paper, the degree of g is less than the length of the sequence.
Although g(z^(-1)) is not a polynomial (it has negative powers of z), it is closely related to the reciprocal polynomial, which is what Ben Voigt mentioned in a comment. Precisely, g(z^(-1)) = z^(-d)h(z) where h is the polynomial with coefficients given by g(end:-1:1).
code :
g = [1 2 0 0 2];
s = [3 1 4 1 5 9 2 6 5];
d = length(g)-1;
c = conv(g(end:-1:1),s);
result = c(d+1:end-d);


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Prolog Common Differences, Next Number in Polynomial Sequence

Prolog Common Differences, Next Number in Polynomial Sequence


By : user3055771
Date : March 29 2020, 07:55 AM
this one helps. You have a typo in your nextitem rule, NextI should be case-sensitive and you have it as nextI in the definition. Also, do you need the list brackets around [Z], rowbelow results in a list for that operand, so they are not needed. So you end up with:
code :
nextitem([A|B], NextI):-lastitem([A|B],X),
rowbelow([A|B],Z),lastitem(Z,Y), NextI is X+Y.
nextitem([],0).
nextitem([A|B], NextI):-lastitem([A|B],X), rowbelow([A|B],Z), nextitem(Z,Y), NextI is X+Y.
nextitem([A],A).
nextitem([],0).
How to use apply to create multiple polynomial variables?

How to use apply to create multiple polynomial variables?


By : Imogen.ds
Date : March 29 2020, 07:55 AM
hope this fix your issue I have several variables, , We can make use of a named vector to change the values to numeric
code :
newdata <- combi[qual_cols]
newdata[] <- lapply(combi[qual_cols], function(x) 
         setNames(c(1, 2, 4, 6, 11), grades)[x])
nm1 <- grep("(Cond|Qual)$", names(newdata), value = TRUE)
nm2 <- sub("[A-Z][a-z]+$", "", nm1)
nm3 <- paste0(unique(nm2), 'Grade')
newdata[nm3] <- lapply(split.default(newdata[nm1], nm2), function(x) Reduce(`*`, x))
set.seed(24)
combi <- as.data.frame(matrix(sample(grades, 10 * 5, replace = TRUE), 
   ncol = 10, dimnames = list(NULL, qual_cols)), stringsAsFactors = FALSE)
CRC implementing a specific polynomial. How does the polynomial relate to the polynomial used in code?

CRC implementing a specific polynomial. How does the polynomial relate to the polynomial used in code?


By : hildegard
Date : March 29 2020, 07:55 AM
may help you . I have the following CRC function: ,
How does 0x18 relate to the polynomial X^8+X^5+X^4+X^0?
code :
typedef unsigned char uint8_t;
typedef unsigned short uint16_t;
#define CRC8INIT 0x00
#define CRC8POLY 0x8c   // 0x119 >> 1

uint8_t crc8 (uint8_t *data, uint16_t number_of_bytes_in_data)
{
    uint8_t  crc;
    uint16_t loop_count;
    uint8_t  bit_counter;
    uint8_t  b;
    uint8_t  feedback_bit;

    crc = CRC8INIT;

    for (loop_count = 0; loop_count != number_of_bytes_in_data; loop_count++) {
        b = data[loop_count];
        bit_counter = 8;
        do {
            feedback_bit = (crc ^ b) & 0x01;
            crc = (crc >> 1) ^ ((0-feedback_bit) & CRC8POLY);
            b = b >> 1;
            bit_counter--;
        } while (bit_counter > 0);
    }

    return crc;
}
I expected a `numpy.lib.polynomial.poly1d` object, I've found a sequence of integers

I expected a `numpy.lib.polynomial.poly1d` object, I've found a sequence of integers


By : Grant
Date : March 29 2020, 07:55 AM
I wish this helpful for you , A poly1d object is iterable
code :
In [1]: np.poly1d((3,2))                                                        
Out[1]: poly1d([3, 2])
In [2]: list(_)                                                                 
Out[2]: [3, 2]
In [12]: arr = np.empty(2, object)                                              
In [13]: arr[:] = [np.poly1d((3,2)), np.poly1d((4,2))]                          
In [14]: arr                                                                    
Out[14]: array([poly1d([3, 2]), poly1d([4, 2])], dtype=object)
In [17]: np.array([np.poly1d((3,2)), np.poly1d((3,2,1))])                       
Out[17]: array([poly1d([3, 2]), poly1d([3, 2, 1])], dtype=object)
In [18]: np.array([np.poly1d((3,2,1)), np.poly1d((3,2,1))])                     
---------------------------------------------------------------------------
ValueError                                Traceback (most recent call last)
<ipython-input-18-00b20d14a2b6> in <module>
----> 1 np.array([np.poly1d((3,2,1)), np.poly1d((3,2,1))])

ValueError: cannot copy sequence with size 2 to array axis with dimension 3
Finding a polynomial formula for sequence of numbers

Finding a polynomial formula for sequence of numbers


By : user2930301
Date : March 29 2020, 07:55 AM
Hope this helps You can code a routine yourself using the sympy module in Python. (This is a popular third-party module for Python.) This code uses the base formula for the Lagrange polynomial, the polynomial of smallest degree that yields a given sequence. This code allow you to define your own x-values in addition to the y-values: if you do not define the x-values, this routine will use 1, 2, .... Note that there are other ways to get this polynomial--I used the formula used in Wikipedia in the link.
code :
import sympy

x = sympy.symbols('x')
zeropoly = x - x
onepoly = zeropoly + 1


def lagrangepoly(yseq, xseq=None):
    """Build a Lagrange polynomial from a sequence of `y` values.
    If no sequence of `x`s is given, use x = 1, 2, ..."""
    if xseq is None:
        xseq = list(range(1, len(yseq) + 1))
    assert len(yseq) == len(xseq)

    result = zeropoly
    for j, (xj, yj) in enumerate(zip(xseq, yseq)):
        # Build the j'th base polynomial
        polyj = onepoly
        for m, xm in enumerate(xseq):
            if m != j:
                polyj *= (x - xm) / (xj - xm)
        # Add in the j'th polynomial
        result += yj * polyj
    return sympy.expand(result)
x**2
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