I am having trouble with the if cycle.

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Muhendisleksi
Muhendisleksi 2017-8-10
评论： Jan 2017-8-10
kontrl = 3.197442310920451e-14
if kontrl>=-10^-20 & kontrl<=10^-20
a = 4
else
a = 10
end
% "a = 4" should be.
  2 个评论
Stephen23
Stephen23 2017-8-10
编辑：Stephen23 2017-8-10
@Muhendisleksi: your upper bound is much smaller than kontrl:
>> 10^-20 <= 3.197442310920451e-14
ans = 1
Why do you think that a should be 4 ?
Jan
Jan 2017-8-10
Note: While 10^-20 is an expensive power operation, 1e-20 is a cheap constant.

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Walter Roberson
Walter Roberson 2017-8-10
No, that is not correct. "a=4" is given only for values that are in the range +/- 1/10^20 but your value is more than 3000000 times larger than the upper end of that, at roughly +3/10^14
  2 个评论
Muhendisleksi
Muhendisleksi 2017-8-10
编辑：Muhendisleksi 2017-8-10
"kontrl" is constantly changing. How can I make this dynamic?
Walter Roberson
Walter Roberson 2017-8-10
a = zeros(1, 100);
for N = 1 : 100
kontrl = randn(1,1) * 1E-20;
if kontrl>=-10^-20 & kontrl<=10^-20
a(N) = 4;
else
a(N) = 10;
end
end
a

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