dividing defined variable by named variable

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Liam Bates
Liam Bates 2022 年 9 月 28 日
コメント済み: Image Analyst 2022 年 9 月 28 日
everytime i try to divide a defined variable by a calculated variable, it only outputs one number multiple times. What syntax am i missing?
d=1:.2:3;
a=sqrt((d.^2)*(3^2));
Y=d./a
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Stephen23
Stephen23 2022 年 9 月 28 日
編集済み: Stephen23 2022 年 9 月 28 日
" it only outputs one number multiple times."
Because that is the correct result for your code, which is equivalent to d./(3*d)
"What syntax am i missing?"
Because you did not state the expected output nor what you are trying to achieve, we cannot guess the syntax that is required to achieve an unstated goal.

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回答 (2 件)

Image Analyst
Image Analyst 2022 年 9 月 28 日
Well yeah. Just look at the numbers:
d = 1 : 0.2 : 3
d = 1×11
1.0000 1.2000 1.4000 1.6000 1.8000 2.0000 2.2000 2.4000 2.6000 2.8000 3.0000
a = sqrt((d.^2)*(3^2))
a = 1×11
3.0000 3.6000 4.2000 4.8000 5.4000 6.0000 6.6000 7.2000 7.8000 8.4000 9.0000
Y = d ./ a
Y = 1×11
0.3333 0.3333 0.3333 0.3333 0.3333 0.3333 0.3333 0.3333 0.3333 0.3333 0.3333
They all look right to me. Exactly what output numbers did you expect? Tell me which element of Y it got wrong and should be a different number.
  3 件のコメント
Image Analyst
Image Analyst 2022 年 9 月 28 日
You can define "a" differently then, like perhaps add noise
d = 1 : 0.2 : 3
d = 1×11
1.0000 1.2000 1.4000 1.6000 1.8000 2.0000 2.2000 2.4000 2.6000 2.8000 3.0000
a = sqrt((d.^2)*(3^2)) + rand(size(d))
a = 1×11
3.3071 4.0240 4.6150 5.2096 5.6304 6.3306 7.5081 8.0647 7.8299 8.5545 9.7844
Y = d ./ a
Y = 1×11
0.3024 0.2982 0.3034 0.3071 0.3197 0.3159 0.2930 0.2976 0.3321 0.3273 0.3066

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Walter Roberson
Walter Roberson 2022 年 9 月 28 日
a=sqrt((d.^2)*(3^2))
3^2 is 9 so inside the sqrt() you have 9*d^2. sqrt() of that is sqrt(9)*sqrt(d^2). For positive d that is 3*d
Y = d ./ a
So you have d divided by 3*d. For non-zero d that is going to be 1/3 no matter what the positive value of d is. For negative d you would get -1/3. For 0 you get nan.

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