I know I can make a sequence of numbers in matlab / octave
using x=1:1:5 and it will give me 1,2,3,4,5 but how can I get a sequence to be
1,1/2,1/4,1/6,1/8,1/10...
[Math] making a sequence of numbers in octave and matlab
MATLABoctavesequences-and-series
Related Solutions
Answered in a comment by Arturo Magidin:
It seems to me that you are confusing series and sequences. Given a series $\sum a_n$, you get two sequences:
- the sequence of the terms, $a_1,a_2,a_3,\dots$
- the sequence of partial sums, $a_1,a_1+a_2,a_1+a_2+a_3,\dots$
Convergence of the series is convergence of the sequence of partial sums. By the Divergence Test, if the series converges then the sequence of terms must converge to zero. So if the terms don't converge to zero (either they diverge or they converge to something else), then the series diverges.
(quick comment regarding @Hayoung.Chung 's answer since I can't comment on it yet. While that code might prove to be a valid solution for Matlab, but he mentioned Octave and Matlab. Use of Matlab File Exchange code in Octave appears to be against The Mathworks's restrictive Terms of Use for the File Exchange. Specifically Section 2.a.iii, even though it appears to conflict with Section 2.c and Section 5. Octave has been avoiding using File Exchange code internally for that reason. Obviously, you are in control of your own actions in that regard.)
what you're looking for is a "zero" crossing of the interpolated line between points after you've shifted it by -4. if your data had many more data points between zero crossings you could do a polyline fit and use the matlab/octave tools to find zeros of a polyline. Here, you may just want to make a loop that generates the equation of a line between successive points, does a (shifted) zero-find with that line, and then has some clean up functions to account for points that hit y=0 exactly (hence both segments will report the same zero location giving you duplicate points), or don't hit it at all. It would then be simple to loop this through all of your collected points.
As your comments indicated that 'always crossing 4' was arbitrary, and wouldn't always even cross at a fixed value, the script below will take care of the situation your provided in the your question, as well as the case where they randomly cross. it takes the two curves, subtracts them, and finds where that equals zero. it then returns the intersection frequencies and the amplitudes at those points.
note that the code below could be vectorized to run much faster without a for loop, but i think the for loop is more instructive at this point. what I'm doing is really repeated linear interpolation, and there are built in functions to do that for an array of points. I believe even the direct method here could likely be vectorized as well. But this appears sufficient for your purpose.
clear all;
freq=[20,30,40,50,60,70,80];
#amp_orig1 = floor(10*rand(1,7))
#amp_orig2 = floor(10*rand(1,7))
amp_orig1 = [8 8 9 0 5 1 2]
amp_orig2 = [8 8 4 2 4 5 6]
numpoints = numel(freq);
crossings_xvals = NaN(numpoints-1,2); #ones with no crossing will stay as NaN
crossings_yvals = NaN(numpoints-1,2);
#crossing_amp_value = 4; #selected since your inversion is about amp=4.
amp_shift = amp_orig1-amp_orig2; #solve for zero crossing in rest of code
#first find the zero crossing for each segment
for fval = 1 : numpoints-1
xs = freq(fval:fval+1);
ys = amp_shift(fval:fval+1);
slope = (ys(2)-ys(1))/(xs(2)-xs(1));
yintercept = ys(1) - slope*xs(1);
xintercept = -yintercept/slope;
#exclude cases where it doesn't intersect in that segment
if (xintercept>=xs(1) && xintercept<=xs(2))
#get actual y-value at crossing
ys_orig = amp_orig1(fval:fval+1);
yval_at_crossing= ((ys_orig(2)-ys_orig(1))/(xs(2)-xs(1)))*(xintercept-xs(1)) + ys_orig(1);
crossings_xvals(fval) = xintercept;
crossings_yvals(fval) = yval_at_crossing;
end
end
#drop NaNs to account for it not always oscillating/crossing amp=4;
crossings_xvals = crossings_xvals(~isnan(crossings_xvals));
crossings_yvals = crossings_yvals(~isnan(crossings_yvals));
#remove duplicates. floating point arithmetic, so check if they're closer than
#numerical error eps
dups = find(diff(crossings_xvals)<=eps);
#final xvalues where it crosses/equals '4'
crossings_xvals(dups)=[];
crossings_yvals(dups)=[];
crossings = [crossings_xvals,crossings_yvals]
plot(freq,amp_orig1,freq,amp_orig2,freq,amp_shift,crossings_xvals,crossings_yvals,'ok',[freq(1),freq(end)],[0,0],'--');
xlabel('frequency');ylabel('amplitude');
legend('orig1','orig2','orig_{diff}','intersect');
if they're sampled at different frequency values, you'll have to do a lot more interpolating freq values before calculating the rest.
Best Answer
Try
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