Who can help me solve this matlab ques?

How to solve this problem by matlab?
Start with number 5, get unknown(I) as the equally space, then I want 20 terms of number. I tried this before>> [5:3:(5+19*3)] Take 3 as the unknown(I)

1 Comment

Can you have a native English speaker look this over. I can't figure out what you want. Let's use x for the spacing instead of 3. So you want to solve for [5, 5+s, 5+2s, 5+19s], but I don't know what to solve for . Nothing is unknown if you specify s. If you specify the value of the last element, 5+19s, then you can solve for s. But as it is worded right now everyone (not just me) is confused.

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Answers (2)

Have you tried [5:3:(5+19*3)] in matlab?
If I understand your question correct, it will give you the whanted answer.
[5:3:(5+19*3)]
ans =
5 8 11 14 17 20 23 26 29 32 35 38 41 44 47 50 53 56 59 62

7 Comments

Kent Lam
Kent Lam on 25 Jun 2015
Edited: Kent Lam on 25 Jun 2015
Yes I had tried this in Matlab before. But my teacher want the 3 be the unknown(I), this equation isn't what he want. He want the answer be more creative
You mean creative like
5:unknown(i):(5+19*unknown(i))
or creative like
5 + unknown(i) * (0:19)
shouldn't the creativity come from you?
This answer is creative but it cant present in Matlab right? I also tried it
Which answer are you referring to?
i = 1;
unknown = 3;
5 + unknown(i) * (0:19)
ans =
5 8 11 14 17 20 23 26 29 32 35 38 41 44 47 50 53 56 59 62
Except this kind of answer, I want the another answer
i = 1;
unknown = 3;
5:unknown(i):(5+19*unknown(i))
ans =
5 8 11 14 17 20 23 26 29 32 35 38 41 44 47 50 53 56 59 62
Nothing remarkable there?

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Is there any one to help me in MATLAB?

3 Comments

You should create Question asking about your interests.
How to implement the mu law in matlab, it means this formula x= (_1,1) and mu= 255 Y(n)= ln(1+mu)x(n)/ln(1+mu).sgn(x(n) but it is not complet formulla if you can solve it please send me your email I should send you the whole file
https://www.mathworks.com/matlabcentral/answers/425337-mu-law-compressor-function

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