error in multiplication of array by a number

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Martin Kavka
Martin Kavka el 28 de Jul. de 2020
Editada: James Tursa el 28 de Jul. de 2020
Hello, I have a question, I have array X that has 3352 numbers in it and i am trying to multiply every number by 3017.4. And its not going well.
These are the last 2 numbers of array 0.2204 0.1897. After I write x1=x*3017.4 it returns new array x1 and in it the last 2 numbers look like this 0.6650 0.5723.
And I dont get why it is doing this, and it is happening to all numbers in that array, I even tried to multiply by 1000 and 3 in kinda works but not completely. If I multiply it by 1000 it calculates correctly those last numbers 220.3920 189.6600 but then i multiply it by 3 it jumps back to 0.6650 0.5723.
I really want to know how to multiply it please help me, if thats my mistake or some matlab smart thingie.
  6 comentarios
Bruno Luong
Bruno Luong el 28 de Jul. de 2020
Editada: Bruno Luong el 28 de Jul. de 2020
Look at the exponent displayed on top of the result when you display the result.
>> a=rand(1,10)
a =
0.0975 0.2785 0.5469 0.9575 0.9649 0.1576 0.9706 0.9572 0.4854 0.8003
>> a*1e6
ans =
1.0e+05 * % <====== THIS THING, is a common exponent of on numbers bellow
0.9754 2.7850 5.4688 9.5751 9.6489 1.5761 9.7059 9.5717 4.8538 8.0028
>>
Martin Kavka
Martin Kavka el 28 de Jul. de 2020
Editada: James Tursa el 28 de Jul. de 2020
my data is so long that i cant even see the beginning, but I have been looking for this but didint see it so I asked here we when i plotted it those numbers are in thousands so thanks a lot guys, love you

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Respuestas (1)

the cyclist
the cyclist el 28 de Jul. de 2020
Just a guess here. If you have
x = [0.2204 0.1897]
and then type
1000*x
in the command window, it will display [220.4 189.7], but it does not change the value of x. So, if you were to now do
3*x
you will get [0.6612 0.5691].
If you wanted to multiply x by 1000, you need
x = 1000*x
  5 comentarios
the cyclist
the cyclist el 28 de Jul. de 2020
Oh, I just noticed Bruno mentioned the same thing in a comment above. I will still take credit for independent discovery. :-)
Bruno Luong
Bruno Luong el 28 de Jul. de 2020
No problem I give you all credit you want Mr cyclist.

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