using inverse hyperbolic functions cosh in matlab

Hi every one

I have vector of complex data that has real and imaginary part. I want to use cosh of this data in MATLAB. then I want to use the invers of cosh to return me to the data input. I used in MATLAB this code

A= [  10.9837193883460 + 26.5347537674488i       9.08578029925812 + 27.2688648124603i    7.13892338965110 + 27.8674662179979i              5.15283024456107 + 28.3273169598239i            3.13739622908723 + 28.6458664819948i      1.10268121205304 + 28.8212686991090i -0.941140621386700 + 28.8523924881452i      -2.98383161797377 + 28.7388286111658i        -5.01514191135404 + 28.4808930279443i            -7.02486082745132 + 28.0796265756387i       -9.00286818246618 + 27.5367910108141i        -10.9391852159635 + 26.8548614273679i     -12.8240249028140 + 26.0370150821188i        -14.6478413905078 + 25.0871166778933i            -16.4013783122172 + 24.0097001718231i     -18.0757157312878 + 22.8099471941067i       -19.6623154792203 + 21.4936621796417i           -21.1530646568884 + 20.0672443316137i      -22.5403170775323 + 18.5376565522323i            -23.8169324400296 + 16.9123914912512i     -24.9763130319849 + 15.1994348776115i    -26.0124377741751 + 13.4072263135169i       -26.9198934309609 + 11.5446177232455i          -27.6939028251654 + 9.62082966110862i            -28.3303499106663 + 7.64540569408595i       -28.8258015714740 + 5.62816508468216i          -29.1775260322726 + 3.57915400845697i     -29.3835077821777 + 1.50859554847569i      -29.4424589308279 - 0.573161284572463i          -29.3538269336532 - 2.65569325247631i      -29.1177986412982 - 4.72855505165752i       -28.7353006465233 - 6.78133167652808i            -28.2079959204595 - 8.80369072161829i      -27.5382767486756 - 10.7854343599834i        -26.7292539961117 - 12.7165507365551i            -25.7847427483981 - 14.5872645175724i       -24.7092443953497 - 16.3880863410096i            -23.5079252403681 - 18.1098609180221i      -22.1865917370961 - 19.7438135417332i    -20.7516624717032 - 21.2815947673119i          -19.2101370258051 - 22.7153230359575i         -17.5695618707992 - 24.0376250254637i            -15.8379934596274 - 25.2416735209560i      -14.0239586962942 - 26.3212226115406i             -12.1364129769286 - 27.2706400316673i      -10.1846960087276 - 28.0849364800239i       -8.17848562456214 - 28.7597917636580i            -6.12774982152663 - 29.2915776306869i      -4.04269726096728 - 29.6773771713292i        -1.93372647568119 - 29.9150006840109i];
B= cosh(A);
C=acosh(B);

In theoretically, I should get C=A but I couldn’t find in MATLAB. Can you help me with this problem?

Answer 1

[Voss]

"theoretically, I should get C=A"

Not necessarily.

x = -1.5:0.1:1.5;
plot(x,cosh(x))

line([1 1],[1 cosh(1)],'Color','r','LineStyle','--')
text(-1,1,'x = -1','HorizontalAlignment','right','VerticalAlignment','bottom')
line([-2 1],cosh([1 1]),'Color','r','LineStyle','--')
text(-2,cosh(1),sprintf('y = cosh(x) = %.2f',cosh(1)),'VerticalAlignment','bottom')
line([-1 1],cosh([1 1]),'LineStyle','none','Marker','o','Color','r')
line([-1 -1],[1 cosh(1)],'Color','r','LineStyle','--')
text(1,1,'x = 1','VerticalAlignment','bottom')

cosh(1) == cosh(-1), and they are both approximately 1.54.

So acosh(1.54) returns 1, but what if you started with x = -1? You could not say that acosh(cosh(x)) = x.

A= [  10.9837193883460 + 26.5347537674488i       9.08578029925812 + 27.2688648124603i    7.13892338965110 + 27.8674662179979i              5.15283024456107 + 28.3273169598239i            3.13739622908723 + 28.6458664819948i      1.10268121205304 + 28.8212686991090i -0.941140621386700 + 28.8523924881452i      -2.98383161797377 + 28.7388286111658i        -5.01514191135404 + 28.4808930279443i            -7.02486082745132 + 28.0796265756387i       -9.00286818246618 + 27.5367910108141i        -10.9391852159635 + 26.8548614273679i     -12.8240249028140 + 26.0370150821188i        -14.6478413905078 + 25.0871166778933i            -16.4013783122172 + 24.0097001718231i     -18.0757157312878 + 22.8099471941067i       -19.6623154792203 + 21.4936621796417i           -21.1530646568884 + 20.0672443316137i      -22.5403170775323 + 18.5376565522323i            -23.8169324400296 + 16.9123914912512i     -24.9763130319849 + 15.1994348776115i    -26.0124377741751 + 13.4072263135169i       -26.9198934309609 + 11.5446177232455i          -27.6939028251654 + 9.62082966110862i            -28.3303499106663 + 7.64540569408595i       -28.8258015714740 + 5.62816508468216i          -29.1775260322726 + 3.57915400845697i     -29.3835077821777 + 1.50859554847569i      -29.4424589308279 - 0.573161284572463i          -29.3538269336532 - 2.65569325247631i      -29.1177986412982 - 4.72855505165752i       -28.7353006465233 - 6.78133167652808i            -28.2079959204595 - 8.80369072161829i      -27.5382767486756 - 10.7854343599834i        -26.7292539961117 - 12.7165507365551i            -25.7847427483981 - 14.5872645175724i       -24.7092443953497 - 16.3880863410096i            -23.5079252403681 - 18.1098609180221i      -22.1865917370961 - 19.7438135417332i    -20.7516624717032 - 21.2815947673119i          -19.2101370258051 - 22.7153230359575i         -17.5695618707992 - 24.0376250254637i            -15.8379934596274 - 25.2416735209560i      -14.0239586962942 - 26.3212226115406i             -12.1364129769286 - 27.2706400316673i      -10.1846960087276 - 28.0849364800239i       -8.17848562456214 - 28.7597917636580i            -6.12774982152663 - 29.2915776306869i      -4.04269726096728 - 29.6773771713292i        -1.93372647568119 - 29.9150006840109i]

A = 
  10.9837 +26.5348i   9.0858 +27.2689i   7.1389 +27.8675i   5.1528 +28.3273i   3.1374 +28.6459i   1.1027 +28.8213i  -0.9411 +28.8524i  -2.9838 +28.7388i  -5.0151 +28.4809i  -7.0249 +28.0796i  -9.0029 +27.5368i -10.9392 +26.8549i -12.8240 +26.0370i -14.6478 +25.0871i -16.4014 +24.0097i -18.0757 +22.8099i -19.6623 +21.4937i -21.1531 +20.0672i -22.5403 +18.5377i -23.8169 +16.9124i -24.9763 +15.1994i -26.0124 +13.4072i -26.9199 +11.5446i -27.6939 + 9.6208i -28.3303 + 7.6454i -28.8258 + 5.6282i -29.1775 + 3.5792i -29.3835 + 1.5086i -29.4425 - 0.5732i -29.3538 - 2.6557i -29.1178 - 4.7286i -28.7353 - 6.7813i -28.2080 - 8.8037i -27.5383 -10.7854i -26.7293 -12.7166i -25.7847 -14.5873i -24.7092 -16.3881i -23.5079 -18.1099i -22.1866 -19.7438i -20.7517 -21.2816i -19.2101 -22.7153i -17.5696 -24.0376i -15.8380 -25.2417i -14.0240 -26.3212i -12.1364 -27.2706i -10.1847 -28.0849i  -8.1785 -28.7598i  -6.1277 -29.2916i  -4.0427 -29.6774i  -1.9337 -29.9150i

B = cosh(A);
C = acosh(B)

C = 
  10.9837 + 1.4020i   9.0858 + 2.1361i   7.1389 + 2.7347i   5.1528 - 3.0886i   3.1374 - 2.7701i   1.1027 - 2.5947i   0.9411 + 2.5635i   2.9838 + 2.6771i   5.0151 + 2.9350i   7.0249 - 2.9469i   9.0029 - 2.4040i  10.9392 - 1.7221i  12.8240 - 0.9043i  14.6478 + 0.0456i  16.4014 + 1.1230i  18.0757 + 2.3228i  19.6623 - 2.6441i  21.1531 - 1.2177i  22.5403 + 0.3119i  23.8169 + 1.9372i  24.9763 - 2.6331i  26.0124 - 0.8409i  26.9199 + 1.0218i  27.6939 + 2.9455i  28.3303 - 1.3622i  28.8258 + 0.6550i  29.1775 + 2.7040i  29.3835 - 1.5086i  29.4425 + 0.5732i  29.3538 + 2.6557i  29.1178 - 1.5546i  28.7353 + 0.4981i  28.2080 + 2.5205i  27.5383 - 1.7809i  26.7293 + 0.1502i  25.7847 + 2.0209i  24.7092 - 2.4615i  23.5079 - 0.7397i  22.1866 + 0.8943i  20.7517 + 2.4320i  19.2101 - 2.4174i  17.5696 - 1.0951i  15.8380 + 0.1089i  14.0240 + 1.1885i  12.1364 + 2.1379i  10.1847 + 2.9522i   8.1785 - 2.6561i   6.1277 - 2.1243i   4.0427 - 1.7385i   1.9337 - 1.5009i

isequal(A,C)

ans = logical
   0

[mohammed hussein]

thank you very much for your answer. i would like to ask , if i neglect the negative part in all data ( all data is positive part )

for example A=[A= [ 10.9837193883460 + 26.5347537674488i 9.08578029925812 + 27.2688648124603i 7.13892338965110 + 27.8674662179979i 5.15283024456107 + 28.3273169598239i 3.13739622908723 + 28.6458664819948i 1.10268121205304 + 28.8212686991090i];

is there any way in mtlab to get C=A

[Voss]

That's a nice idea, and that would work if the inputs to cosh were all real. But I was showing real inputs just because it's easier to depict graphically, and to show that even in the real case, there's ambiguity with acosh.

Of course, the A you have are complex. Consider the definition of cosh(x):

For purely imaginary x, exp(x) is periodic with period 2*pi; therefore, for complex x, there's that periodic component in there, with a varying magnitude coming from the real part of x.

% all elements of A have positive real and imaginary parts:
A= [  10.9837193883460 + 26.5347537674488i       9.08578029925812 + 27.2688648124603i    7.13892338965110 + 27.8674662179979i              5.15283024456107 + 28.3273169598239i            3.13739622908723 + 28.6458664819948i      1.10268121205304 + 28.8212686991090i]

A = 
  10.9837 +26.5348i   9.0858 +27.2689i   7.1389 +27.8675i   5.1528 +28.3273i   3.1374 +28.6459i   1.1027 +28.8213i

B = cosh(A);
C = acosh(B)

C = 
  10.9837 + 1.4020i   9.0858 + 2.1361i   7.1389 + 2.7347i   5.1528 - 3.0886i   3.1374 - 2.7701i   1.1027 - 2.5947i

isequal(A,C) % still not the same

ans = logical
   0

Consider the fact that the (non-hyperbolic) cosine function is also periodic. If you were given cos(x) = 0.4 and asked "What's the value of x?" what could you say? There are infinitely many values of x such that cos(x) = 0.4

x = -8:0.01:8;
plot(x,cos(x))
y = 0.4;
acy = acos(y);
xd = [acy+2*pi*(-1:1) -acy+2*pi*(-1:1)];
xd = [xd; xd; NaN(1,numel(xd))];
line(xd(:),repmat([y; -1; NaN],size(xd,2),1),'LineStyle','--','Color','r')
line([-8 8],[y y],'LineStyle','--','Color','r')
line(xd(1,:),y*ones(1,size(xd,2)),'LineStyle','none','Color','r','Marker','o')

So the only way you'll get C = A is to change A to something else. It's like if I said, can you make 1 = 2? Yeah, you can, if you change what "2" means.

It's math, and there's nothing MATLAB or anything else in the universe (or any possible universe) can do to make it work differently.

[mohammed hussein]

thank you for your answer

[Voss]

You're welcome!