I don't know if matlab is calculating this properly
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Is this calculating correctly? It returns my answer as 1. It is meant to represent this series 1 + 1⁄r + 1⁄r^2 + 1⁄r^3 + … + 1⁄r^n.
function sum = mysum(r,n)
sum = 1;
for i = 1:n;
sum = sum + 1/(r^n);
end
Please help.
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Risposte (2)
John D'Errico
il 11 Set 2019
Modificato: John D'Errico
il 11 Set 2019
MATLAB is calculating what it calculated properly. The issue is, you told it to calculate the wrong thing. Computers are sooooo picky. :)
You don't tell us what values of r that you used. Or what n was when you ran this code. But if you used a large value for n, then yes, it SHOULD return 1. Or it might return inf. Really? What did you write, and why is that?
In fact, you wrote:
1 + 1⁄r^n + 1⁄r^n + 1⁄r^n + … + 1⁄r^n
Your exponent was fixed, at n.
The exponent in what you wrote was CONSTANT, at n. So if n was large, and abs(r ) was relatively large so that 1/abs(r ) is small, then you would see 1 as a result, because the powers will underflow. Or if r is itself small, then you would just get overflows.
The fix is easy, of course. Change n to i in the expression. That is, the exponent of r needs to be the index variable, not the number n itself.
function sum = mysum(r,n)
sum = 1;
for i = 1:n;
sum = sum + 1/(r^i);
end
And using the name sum for a variable is a really bad idea, as it will cause bugs in your code sometime, when you actually want to use the FUNCTION named sum.
Always avoid using existing function names as variable names.
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madhan ravi
il 11 Set 2019
Modificato: madhan ravi
il 11 Set 2019
Result = mysum(2,10) % just call it like this
doc function % to know how to use functions
function SUM = mysum(r,n)
SUM = 1; % don’t name variable sum because it will shadow the MATLAB’s inbuilt function
for ii = 1:n;
SUM = SUM + 1/(r^ii);
% ^^-- have a look here
end
end
11 Commenti
madhan ravi
il 11 Set 2019
function SUM = mysum(r,n)
SUM = 1;
for ii = 1:n;
SUM = SUM + 1/(r^ii);
% ^^-- have a look here
end
end
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