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

[Dyuman Joshi]

What have you tried yet?

[Sathish]

>> syms n k

>> A = (1/6)*(2*n^3 + 3*n^2 + n - 2*k^3 - 3*k^2 - k)

A =

- k^3/3 - k^2/2 - k/6 + n^3/3 + n^2/2 + n/6

>> B = (1/6)*(k*(k+1)*(2*k+1))

B =

(k*(2*k + 1)*(k + 1))/6

>> C = symsum((abs(A-B)),k,1,n-1)

C =

symsum(abs(k/6 - n/6 + k^2/2 + k^3/3 - n^2/2 - n^3/3 + (k*(2*k + 1)*(k + 1))/6), k, 1, n - 1)

[Sathish]

I want this expression to be simplified.

[Star Strider]

I am not certain what you want to do. Using symsum and simplify (allowing 500 iterations) worked when I tried it.

[Sathish]

It is working when we give values of 'n' and 'k'.

But I am trying to get simplified expression with variable 'n' only.

For example:

>> syms n k

>> A = k^2

A =

k^2

>> B = n^2

B =

n^2

>> C = symsum((abs(A-B)),k,1,n-1)

C =

n^2*(n - 1) - (n*(2*n - 1)*(n - 1))/6

[Dyuman Joshi]

Unfortunately (for you), MATLAB does not seem to be able to simplify this expression significantly beyond a particular point, see below.

WolframAlpha gives a weird-ish output (I can't recognize what the symbols similar to # are supposed to be) - https://www.wolframalpha.com/input?i=sum+abs%28k%2F6+-+n%2F6+%2B+k%5E2%2F2+%2B+k%5E3%2F3+-+n%5E2%2F2+-+n%5E3%2F3+%2B+%28k*%282*k+%2B+1%29*%28k+%2B+1%29%29%2F6%29%2C+k%3D1+to+n-1

You could try using MAPLE and see what output it gives.

It is also worth noting that not all symbolic expression can be simplified.

syms n k
A = (1/6)*(2*n^3 + 3*n^2 + n - 2*k^3 - 3*k^2 - k);
B = (1/6)*(k*(k+1)*(2*k+1));
C = symsum((abs(A-B)),k,1,n-1)

C = 

simplify(C, 'Steps', 100)

ans = 

simplify(C, 'Steps', 200)

ans = 

[Sathish]

Thank you for your suggestions.

[Walter Roberson]

I think in the Wolfram output that the # stand in for the variable whose value has to be found to make the expression zero

Answer 1

[Star Strider]

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 = 

.

[Dyuman Joshi]

@Star Strider, there is an absolute sign missing in your code.

[Sathish]

Yes. abs( ) is missing.

[Star Strider]

I thought the ‘|’ were 1.

Edited version —

syms n k 
Expr = abs(1/6 * 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 = 

Added abs call.

.

[Dyuman Joshi]

The abs() call is inside the symsum() call.

[Star Strider]

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 = 

.

Answer 2

[Torsten]

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