# How to implement a formula involving several elements of the same vector?

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FW on 23 Sep 2019
Commented: Shubham Gupta on 23 Sep 2019
Suppose we have two row vectors
Time = [ 1 2 3 4 5 6 7 8 9 10];
width= [0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0]; % width of peaks
There is a formula for peak resolution = (t2-t1)/(2*(w2+w1)), where t2 is element of Time(1,2) and t1 is element of Time(1,1). Similarly, w1 and w2 correspond to elements width(1,1) and width(1,2) respectively. With that one should be able to generate a row vector with 9 elements instead of individually calculating those nine values. In such cases how should we implement a formula involving several elements of the same vector?
Could someone also point out an intermediate level reference which discusses the sytax for operating on several elements of the same row or column vector to produce another vector of different dimension such as in this case? Thanks.

Shubham Gupta on 23 Sep 2019
Edited: Shubham Gupta on 23 Sep 2019
There are several way to do this, but most basic would be to use 'for' loop to get the feeling of how Matrix indexing works:
PeakResolution = zeros(1,length(Time)-1)
for i = 1:length(Time)-1
PeakResolution(i) = (Time(i+1)-Time(i))/(2*(width(i+1)+width(i)));
end
Once you get the feel of it, you can use functions or inline indexing to make your work easy. For e.g. to calculate difference between consecutive elements can be achieved using 'diff'. So, numerators for peak resolution can be written as:
PeakResolution_num = diff(Time); % diff function
% or
PeakResolution_num = Time(2:end)-Time(1:end-1); % inline indexing
and to take mean of consecutive numbers you can use movmean (only for matlab version above 2016a), so denominators become :
PeakResolution_den = movmean(width,[0,1])*2; % movmean function
PeakResolution_den(end) = [];
% or
PeakResolution_den = (width(2:end) + width(1:end-1)) % inline indexing
Now, you can simply use division to calculate PeakResolution:
PeakResolution = PeakResolution_num./(2*PeakResolution_den);
I hope it helps !
Shubham Gupta on 23 Sep 2019
Yes, this should work as you have mentioned !

madhan ravi on 23 Sep 2019
Edited: madhan ravi on 23 Sep 2019
peak_resolution = (Time(2:end)-Time(1:end-1))./...
(2*(Width(2:end)+Width(1:end-1))) % if understood correctly
doc colon
doc end
FW on 23 Sep 2019
Thanks Ravi.

R2019a

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