How can I simplify this expression using "abs" function?

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Sathish
Sathish 2023-11-20
编辑： Torsten 2023-11-20
  8 个评论
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Walter Roberson
Walter Roberson 2023-11-20
I think in the Wolfram output that the # stand in for the variable whose value has to be found to make the expression zero

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回答（2 个）

Star Strider
Star Strider 2023-11-20
Ran in:
This seems to work —
syms n k
Expr = 7/6 * symsum((2*n^3 + 3*n^2 + n - 2*k^3 - 3*k^3 - k)-(k*(k+1)*(2*k+1))/6, k, 1, n-1)
Expr = 
Expr = simplify(Expr, 500)
Expr = 
.
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Star Strider
Star Strider 2023-11-20
编辑：Star Strider 2023-11-20
Ran in:
Edited —
syms n k
Expr = symsum(abs((2*n^3 + 3*n^2 + n - 2*k^3 - 3*k^2 - k)/6-(k*(k+1)*(2*k+1))/6), k, 1, n-1)
Expr = 
Expr = simplify(Expr, 400)
Expr = 
.

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Torsten
Torsten 2023-11-20
编辑：Torsten 2023-11-20
Ran in:
You must determine the value for k0 where the expression
2*n^3 + 3*n^2 + n - 2*k^3 - 3*k^2 - k - k*(k+1)*(2*k+1)
changes sign from positive to negative. Then you can add
1/6*(2*n^3 + 3*n^2 + n - 2*k^3 - 3*k^2 - k - k*(k+1)*(2*k+1))
from k = 1 to k = floor(k0) and add
-1/6*(2*n^3 + 3*n^2 + n - 2*k^3 - 3*k^2 - k - k*(k+1)*(2*k+1)))
from k = floor(k0)+1 to k = n-1.
syms n k
p = simplify(2*n^3 + 3*n^2 + n - 2*k^3 - 3*k^2 - k - k*(k+1)*(2*k+1))
p = 
s = solve(p,k,'MaxDegree',3)
s = 
result = simplify(1/6*symsum(2*n^3 + 3*n^2 + n - 2*k^3 - 3*k^2 - k - k*(k+1)*(2*k+1),k,1,floor(s(1)))-...
1/6*symsum(2*n^3 + 3*n^2 + n - 2*k^3 - 3*k^2 - k - k*(k+1)*(2*k+1),k,floor(s(1))+1,n-1))
result = 
subs(result,n,13)
ans = 
6444
k = 1:12;
n = 13;
expr = 1/6*abs(2*n^3 + 3*n^2 + n - 2*k.^3 - 3*k.^2 - k - k.*(k+1).*(2*k+1))
expr = 1×12
817 809 791 759 709 637 539 411 249 49 193 481
sum(expr)
ans = 6444

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