f(x) is different of f(value of x)

Hi,

In my code, I have a function wich receive a lambda function f.

inside this function, I have a variable called newPoint

I stop with a breakpoint in the running, the value of newPoint is 1.0434

and if I write in the console f(newPoint), I get 0, but if I write f(1.0434) i get -5.2815e-05.

so in my function, I can not evaluate precisely f(newPoint)

(My function f is f = @(x) 2 - x^3 -sin(x))

I will very gratefull if you can help me.

[KSSV]

Can you elaborate....?

[undefined undefined]

Hi,

first of all thank you very much for you comment.

What do you mean ? what do you want me to tell you more ?

[Mathieu NOE]

hi

there shoud not be any difference between writting this in a script or execute it directly in the command window (waht you describe as "console" ?)

f(x)

ans =

  -5.2815e-05

>> f(1.0434)

ans =

  -5.2815e-05

[Rik]

The only reason I can think for this is that newPoint is not exactly 1.0434.

f=@(x) x.^2;
f(1.414)

ans = 1.9994

f(sqrt(2))

ans = 2.0000

[Walter Roberson]

give the command

format long g

and then display the value again

[undefined undefined]

This is what I get u=in the command window during the debugging

K>> newPoint

newPoint =

    1.0434

K>> f(newPoint)

ans =

     0

K>> f(1.0434)

ans =

  -5.2815e-05

Thank you very much for your help

[Walter Roberson]

fprintf('%.20g\n', newPoint)
newPoint - 1.0434

[undefined undefined]

Okay, I see the problem.

Here what I did : The difference is very small between a positive result and 0.

Thank you very much for your help, I hope I didn't bther you too much

K>> format long g
K>> newPoint

newPoint =

          1.04338598820955

K>> f(1.04338598820955)

ans =

        1.765254609154e-14

K>> f(newPoint)

ans =

     0

K>> fprintf('%.20g\n', newPoint)
1.0433859882095546379
K>> f(1.0433859882095546379)

ans =

     0

K>> 

Answer 1

[Steven Lord]

The value stored in an array is not necessarily the same value that is displayed.

Right now Forbes magazine lists the net worth of Jeff Bezos as $183.1 billion dollars. Is that his actual net worth down to the last penny? No. Is it close enough for purposes of this profile? Yes.

Similarly, when displaying pi in MATLAB in the default display format you don't see the full value of pi.

format
x = pi

x = 3.1416

y = x - 3.1416 % not exactly 0

y = -7.3464e-06

Using x or 3.1416 in a calculation could result in different answers if the answer changes based on the value of y (which is not itself exactly -7.3464e-06.)

z = y - (-7.3464e-06)

z = -1.0207e-11

[undefined undefined]

Yes exactly

Thank you !