Solve 1-(0.955.^​n)-(0.005.​^n)*(0.995​.^(n-1))*n​-(0.005.^2​)*(0.995.^​(n-2))*n*(​(n-1)/2)= 1/2 numerically with matlab

fh = @(n) 1-(0.955.^n)-(0.005.^n)*(0.995.^(n-1))*n-(0.005.^2)*(0.995.^(n-2))*n*((n-1)/2)- 0.5;

n_solved = fsolve(@fh, n_guess);

when I run it, it doesnot work! I need to find n which is the number of projectiles Any help please? Thank you

Answer 1

[Sean de Wolski]

fh(1.2)

Gives an error because:

n((n-1)/2)

is an indexing assignment. I don't know what you want here, but n should not be indexing into itself. Perhaps

n.*((n-1)/2)

?

If so, it works fine:

    n_solved = fsolve(fh, pi);

    Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the default value of the function tolerance, and
the problem appears regular as measured by the gradient.

<stopping criteria details>

n_solved =
   15.1631

[Abdel Lootfun]

the answer in the book shows that n = 535

[Torsten]

Then I guess that you made a mistake while copying the equation for fh from the book.

Plot your fh and you'll see that the zero at 15.1631 is correct.

Best wishes

Torsten-

[dpb]

The answer is simply wrong, then, or perhaps you've mistyped the problem definition. The function is monotonically increasing and positive in that area...

 >> [x;fh(x)].'
ans =
  530.0000    0.2516
  530.5000    0.2517
  531.0000    0.2519
  531.5000    0.2520
  532.0000    0.2522
  532.5000    0.2523
  533.0000    0.2525
  533.5000    0.2526
  534.0000    0.2528
  534.5000    0.2530
  535.0000    0.2531
  535.5000    0.2533
  536.0000    0.2534
  536.5000    0.2536
  537.0000    0.2537
  537.5000    0.2539
  538.0000    0.2541
  538.5000    0.2542
  539.0000    0.2544
  539.5000    0.2545
  540.0000    0.2547
>>

Answer 2

[dpb]

fh=@(n) 1-0.955.^n-0.005.^n.*0.995.^(n-1).*n-0.000025.*0.995.^(n-2).*n.*(n-1)/2- 0.5;

NB: n.*(n-1)/2, not n(n-1)/2