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Some of my algorithm ideas to deal with advanced mathematics with cas
12-25-2018, 04:26 PM (This post was last modified: 01-01-2019 01:20 PM by yangyongkang.)
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Some of my algorithm ideas to deal with advanced mathematics with cas
The articles I recently published in the forum are all in CAS. This article is no exception, but this is a discussion of some of my ideas for using hp prime or XCAS to handle advanced math problems. The difference between advanced mathematics and elementary mathematics is that elementary mathematics is a finite order operation, while higher mathematics is an infinite number of operations, that is, limit operations. The well-known Weierstrass approximation theorem states that continuous functions on closed intervals can be uniformly approximated by algebraic polynomials. This means that the derivatives and integrals of all functions can be converted into derivatives and integrals of polynomials, so sometimes complex calculus problems can be transformed into squirrel (sum) functional operations of hp prime or XCAS. This requires the sum function to be very strong. In order to solve some of the problem of summing limit, xcas adds sum_riemann(), but I don't think it is enough, such as solving limit(sum((k)^(1/k)/n,k,1,n),n,+infinity ), riemann sum doesn't work, but if we can solve it with Stolz's theorem, enter limit(x^(1/x), x, +infinity) on hp prime to get the answer to 1 to solve the problem. If the student does not have the stolz theorem, it will be difficult to solve. Stolz solves the problem of limit(sum(f(k)/g(n),k,1,n), n,+infinity). There are also some anomalous integrals that XCAS can't seem to solve. The integrand is transformed into a polynomial sum by the Taylor expansion expansion, and then the integral of the polynomial sum is obtained. Finally, the sum is used to find the answer, such as int(ln(1- x)/x,x,0,1), XCAS cannot be solved directly, but using Taylor expansion, ln(1-x)/x=sum(x^(n-1)/n,1,infinity), int(ln(1-x)/x,x,0,1)=sum((int(x^(n-1)/n,x,0,1)|n>0),n,1,infinity)= Pi^2/6

Above we talked about expanding the function Taylor into a polynomial. Below I try to convert the polynomial into a function and see the performance of XCAS. Since XCAS did not define (2n-1)!! as all odd numbers, this forced me to redefine the expression. 1/sqrt(1+x)=Σ((((-1))(2*k-1,k,1,n))/(product(2*k,k,1,n)) ) *x^n,n,0,∞), but the left side of the equation is not on the right side of the XCAS or hp prime input equation, which means that the polynomial and function cannot be fully transformed. Enter \!\( in Wolfram mathematica 11.3)
\*UnderoverscriptBox[\(\[Sum]\), \(n = 0\), \(\[Infinity]\)]
\*SuperscriptBox[\(E\), \(I\ n\ \[Pi]\)]\
\*SuperscriptBox[\(x\), \(n\)]\ \(\((
\*FractionBox[\(1\), \(2\)]\ \((\(-1\) + 2\ n)\))\)!\)\), \(
\*SqrtBox[\(\[Pi]\)]\ \(n!\)\)]\), you can get 1/Sqrt[1 + x], the answer we want.

Finally, let me say my opinion: I guess the first step of the sum of sum is to use the summation formula to find the sum and then the limit, or use sum_Riemann to solve, my opinion is that when the above two can not be solved, you can Add Stolz theorem to solve. Solving the definite integral is also the reason. The preferred method is to use the Newton Leibniz formula to find the original function. When the original function cannot be found, the Taylor formula can be used to convert into a polynomial summation.
sorry my poor english

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Some of my algorithm ideas to deal with advanced mathematics with cas - yangyongkang - 12-25-2018 04:26 PM

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