Decimal places for value stored

2020 12 1
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更新时间：2020 12 1
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Hello, im wondering if there any way i could perform calculation using values up to 4 decimal places only.
For example,i write this in command windows:
x=4.21;
h=0.001;
f=@(x) x^3 + 8*x^2 + 23*x +3;
f2_prime = (f(x+h)-2*f(x)+f(x-h))/(h^2);
how can i make the value of f(x), f(x+h) and f(x-h) to be only up to 4 decimal places when calculating for f2_prime?

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Ameer Hamza
Ameer Hamza 2020-12-1

0 个投票

You can use round()
x=4.21;
h=0.001;
f=@(x) x^3 + 8*x^2 + 23*x +3;
f2_prime = (round(f(x+h),4)-2*f(x)+round(f(x-h),4))/(h^2);

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fikri osman
fikri osman 2020-12-1
thank you, i will try and see if it works
fikri osman
fikri osman 2020-12-1
f2_prime = (round(f(x+h),4)-2*f(x)+round(f(x-h),4))/(h^2);
does this round off the f(x) too?
fikri osman
fikri osman 2020-12-1
f2_prime = (round(f(x+h),4)-(2*round(f(x),4)+round(f(x-h),4)/(h^2);
i use this and got error
fikri osman
fikri osman 2020-12-1
f2_prime = (round(f(x+h),4)-2*round(f(x),4)+round(f(x-h),4))/(h^2);
i used this and got the answer i wanted, thank you sir

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Walter Roberson
Walter Roberson 2020-12-1

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how can i make the value of f(x), f(x+h) and f(x-h) to be only up to 4 decimal places when calculating for f2_prime?
You would need to use fixed-point arithemetic, such as from the Fixed Point Toolbox.
A close second would be to use the Symbolic Toolbox with digits(4), but it turns out that the Symbolic Toolbox uses "guard digits" ... I think I read once that it uses 5 guard digits -- and the minimum permitted is digits(2)

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