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I would like to measure the first column vector as defined here using Mathematica. The list formed of the sum of the adults and infants at each period is a Fibonacci sequence.

$\qquad %Translator MathMagic Pro for InDesign Mac v9.14, LaTeX converter, 2016.9.30 12:45 \mathrm{\left({\begin{array}{c}{{adults}_{t}}\\{{\mathrm{infants}}_{t}}\end{array}}\right)}\mathrm{{=}}\mathrm{\left({\begin{array}{cc}{1}&{1}\\{1}&{0}\end{array}}\right)}\mathrm{\left({\begin{array}{c}{{adults}_{{t}{-}{1}}}\\{{\mathrm{infants}}_{{t}{-}{1}}}\end{array}}\right)} %MathMagic MMF.7h_S5*00]EQ]Km|f4?h5n*on:0)Z`N);B?FKhgZY|DP^K3OM|?B3U[RMDML)E:E[EoRoSdMAd|Vb;AC3ZR;QLlnA?3hlWZSLcm9OiX]T_9[=dj|OEoNCaGBlVRn292_nB[:?VhO1Vg`on;3?1k?MZoFGcLOM8(TN1UoS4LSQeOeZ][ZM1|nkHU=|ehoFL3MM;)e*P*h740jH]K`J[jI170h42`NL*F*9oU:ZUd8=LKY0gH2noPBojmLF_i^U]o=9X7D8XE8j=03F)_g]cNehUPK3Zl?EeOeb^U[=dQ^(M[U(alTdB;kWVncAS_iDF)l_AL12Kl;PIhD=9?A?bD(?cg]h4BI9=[S9en]?bfF7UAMIeC=fA?P)ZBnA9Yc|Wo?=)^l`LIS|MmU3]`_d:]DW5OAY1IO5P|]Z*ImLL54_^2PHW5L(cT[6nbCS_MWE9aVo;1Vo;1W_ThaOU8aOU8bOUhbOUL`NoBBI[7O5LohmGOoMiJ67icfl2(=STnbJE=MTUDRVaD=gA7f)(2N9Ofc9hB7gKL3ZiM_L]UfY|feQfb0mT1IXkjB*h9j8T?1]CDHbj1Ai45]PZ^VH1FG_Y`3K[6aR650f)O70:=C8U02S4=hOXhRm4`H1_Zg]d]Q8U|0h84Z0DD0e6j]SOL:bdHAQQJWR09`lZPQ99P6LWO|fcUkI]G)Z^SPAF8ER6QUW91Y)0^2hn[P2PTc:J*ALTBhA7D|ghG2ooPYB1**=*02A*cSm:b3Z_G_2?2Em(08a4|;;:hP)*Q?UQ:63a`A8eWBAA0?9BE^*=UE0U_?[2^;d;j[PI9T6XTXL3(5DWS5IWf9TB0G4CF4*_[nRnj0TY|RX*XY^|W9KDNLXA^7CA1WBSY|TRaQY4`dRW=hWBb*JWB8gNnFT:8S8B;P1XXXS(^i4EBQ^I]7d76RJ1IXgLfZb1eXjIKE7RWJ9b?WD90neJMJRGA9j8*a);oaHQTi_bRB|T:2D9?]WU0_7CfBR9U2SJD8H*aQj3V:BPc7A?nITbUPdd|JBi7J|j50AkJ99b|B60YYo=W)`PUH;00I]B4h2(986`2*5R[[ALP1(]i2[21DPJ*S0Z93*ZXW0BK45EaIOR9ZD[BX7=1B8:=2QK00ZlZ8N|RF9ZhgegX)[SUFaMNFa1Z9iJ*2WP[Sbb6]4=`ILQC*e8XT1W1`=4?C]i2YTWDdPb)T0Fb=IhnRb|h[*5DUIZbYJFb(d3J]E9|7EBEVoNeSmHTH4I79IEn^KlI|0go5^NJ6`_d]DaXH|)=AVkN=IkSQG3lU_l:b`_K1OfKolJEamA5IJ1QmTW1XUmZaZ[S;dlFcTXSPGUJiIkESMH|g5__5U=G[4:]Mio(BFG|cWZh2=V=dFNf84o[Ag7GcO(?c9g2inoQ2dKi:U3DkHb8ea_WedEnFB42NLiCUWNi=)aWVnmgihFkcNk;;lNhWaTYR^?fK5^|Ah=eb|]eVafNm:RkdB9gNI_Dgo^Eec7bd[Ki0NHO3CKdFNNLa9Sk6g2F:kmSI9K1=_DlCfb]^RDh;YDdIcCPGl5^X:c(jihoL?E*doMeZb^LlKZY_kX6T;aeEG)AiAiKQ^:lM=EcTNMiEc7b=7b[T?T2?USShj_?5DnXRcnB?D:OOX[;]^:nLn=ZQb8SiBC[:)LQ:jbTU)UI)R[IbDGNFTjRXWXjibDQlYUfg]ZL4[[Td*hoiK5`m|VkVf*njFVnD]MhUVEaC*8;fWa46|lnAfH*]0^|loIm_1J_f]l8FaO634G6U5;g`7DKnbo;Km[WmehmWFlVdBX5:cM1EPD(_GlgOGhnGDTGOCbGPb2K17)Wlk^A^k?_R7Z(TlGDgC5OhQj_Z?|?X7kgoLfe9V:oF?[]UMX]fK`;:h6)QjCH;|lGUKO1UJOhQ)lLG`L3Qdc0Pe8QVn3cJk3mV^M``Li|hcEVl[^WU_5NSJi?1`_k*U*Pmke|KkePK3K^AX??aT5cKl^NSombfan_0HQWheogekS|KkVLDNK7[n2o;=Z3d.mmf $

With $ %Translator MathMagic Pro for InDesign Mac v9.14, LaTeX converter, 2016.9.30 12:52 \begin{array}{l} {{\mathrm{adults}}_{0}\mathrm{{=}}{1}} , {{\mathrm{infants}}_{0}\mathrm{{=}}{0}} \end{array} %MathMagic MMF.7h_74`00SEQ]LmXh4?h5n*ml=3(^XmF;9Mlg*[VD^APjh:IgLoB3;b4iYQ*b[]=[[i?oOUYI]]LHl(DIf6MgYEdoFZeUe[?i[h]U(TiWRoWEcj_eI3TMYh]UT6C5gdWf];dO_(l?PlM3?YS]gfjnKYofPbBk7gb;Ab27En]dU]i)PiMm|BefV`N[^9|^EgJZ*8L320O(J]j)dfT*Q`?1`P5W45T3odGQoa331NX6m?EWn4)o|oSSK7jkV0AJQa0ZYD(3H;GCgmoOSVOcH7SeNWFeGTgCM3Jo`FaGZoThV*K9SgbK?MSIW`_[oKD8F)QEV?b||8V4oR[]d6?W?GHA9TTf^(TgVln[ELLZ;eYEcm`A|GN(nY;AQ9?3BkkMi1e;72J7OGKO7*:mC?EA1GeL`FFbh39Kd4LGG)*;;Q86iaV3|iCa?|YhKgGeDLH_DlH_DlKk:)(G:N(G:N?W:N=W:K=K?dTVVggaT_nHKokYfZ77cW_|8Pb?EK:[DUfEIB:I5_OM6ODiPcUYn=Nf71ib;`=f;bmc:`]Eb|;:83f*5VS_Y=30_B52PiLeVLVPDnA1K86Y`S4;b]7?0LZ|534=:4E)?30;=C8U`2b4mlL|H^n4BH2G]Kde=Y8U(0j84V0FD4ESMJk?f3JJ=2`aEAj0`J?:84T*`)SLbaRmdV_WE0ea9;0:aC*ccTPfW2C0lNkS2PPBU=((^2938SZGK];Qo_h[B1T*=041Q0kQn:n0Z=O^6N^DS(4(a4P8CjlP?0Q=V1)6CQhC85TcA18)92Nb839U*9KaM*Dao9|Z)EVFPJP:1e(`UFM(kTla(ZD2hZH`2CmNdGE*4T]TE2555eViYJQ[5;?`IJ8(TN)Vb296I(91Q)5m|DBRhBUbdB|WAD54I|8545DNTG4kZT9a4dGCOJ1Y5FSNa=AT3KAdc6Z?51dBTOfYBAeZdmb;MTGXRC0HG_Ri30e_bR:|T:0VBMK?:9N)3fBR9U6SJD4H*bad7lBT1V?2Ola9b5Pde(JBe7J|j5*A7J99bLB60UYo]W:`PeHg00cJT)`4H:*(P4T:57FSk*2HKR7G4BY0bQ20DB:QeA)1TfH;[Rfn4KEA][XLd5*PXT27|P78b9]jbQHU[SOFJ`n^)eK=e[G76XSVX*6L4^;J8jlAGAQ`7M;DR1*6L;8e*=2WTn^*MCF18;|3K8mTSJ)[cRY3ebAUcJYX;Hg*=:eFV`CG9fGmk67ePaTAT)2bk]HghoL1?^?MkHG2OYNXc0f]h53KJRm_iLkVnR7i1VlEMQB):lNGWlKeAkA:Jl4;;Hj=4W^[JXhbm?;Fb6Ea;R]MFkFcjYKEG1`KGfJSQjcb?Xn_f9ZGRdDJ|16cbf9gS(1?NmK1i`g3CnIFlL]Sd3i9USXhXB(Wa|G^`AfEBh(hhBc?)M^CM3;)lh?g`m?RmGJOiCm:S8O4nNHY:cHUa[?QL[?;R^eQGf[|TCRibnaYnZoMQ_]|FGV2m0RCWgh_l|aSCTJ(_Dh*gKGGBJ:KN9dR^[MN5idRC9mBVW(|h;]*Uf1fcQgOObQ[n;[CX|fmgU3Ng0]=VcR^^|caR3;7MI|iK[[(lKS;W7|I)F;)_H0L(GOdd^6EYlY7W:dOXDjiAfOMMI|imk91VA?a4G)BMIRCd6E)L|ZL56gVY)`b9eFG)AUeVI?jR;U|IgL=7W5]PASgKedl|39c|T?^U9_U;GN9J]LDD26mYlA9[??TMVTK`?b*OlUfPgCc_O2=|KaPa5a[_Ok=nEUYmB49T87I?0d`f)[Mh^?eN3Eea[_YI3bI13QR__P`^A^k(OP3df*aCjOc57mPB[3m3gn^[ol(Zcohm7(m2K:7UegaMKQ)k*hkIFO3emOebVhZ?EaST=N)2d:=B8JOP^gn(M_oooT*fECo0dUoYDD.mmf $

However I would like a way that returns the number of adults and infants at each period.

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  • $\begingroup$ You need to provide initial conditions! $\endgroup$ – yarchik Sep 30 '16 at 16:49
  • $\begingroup$ You're right, I just added them $\endgroup$ – EBassal Sep 30 '16 at 16:54
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f[{ad_, in_}] := {ad + in, ad};
NestList[f, {1, 0}, 8]

{{1, 0}, {1, 1}, {2, 1}, {3, 2}, {5, 3}, {8, 5}, {13, 8}, {21, 13}, {34, 21}}

Or directly using the Fibonnaci command:

{Fibonacci[#], Fibonacci[#-1]} & /@ Range[8]

which give the same output.

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  • $\begingroup$ Very simple answer, thanks! $\endgroup$ – EBassal Sep 30 '16 at 17:01
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r = RecurrenceTable[{{a[t], i[t]} == {{1, 1}, {1, 0}}.{a[t - 1], i[t - 1]},
 a[0] == 1, i[0] == 0}, {a, i}, {t, 1, 10}]

{{1, 1}, {2, 1}, {3, 2}, {5, 3}, {8, 5}, {13, 8}, {21, 13}, {34, 21}, {55, 34}, {89, 55}}

Both a and i are sequences of Fibonacci numbers, but with an offset of one iteration.

One can also obtain the explicit formula for a and i:

RSolve[{{a[t], i[t]} == {{1, 1}, {1, 0}}.{a[t - 1], i[t - 1]}, a[0] == 1, i[0] == 0},
 {a[t], i[t]}, t][[1]] // Simplify

enter image description here


EDIT: The formulae for sequence r might be guessed with FindSequenceFunction:

FindSequenceFunction[#, t] & @ r[[All, 1]]
FindSequenceFunction[#, t] & @ r[[All, 2]]

1/2 (Fibonacci[t] + LucasL[t])

Fibonacci[t]

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  • $\begingroup$ Thank you, I didn't know about the RecurrenceTable function. $\endgroup$ – EBassal Sep 30 '16 at 17:01
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It's worth knowing that you can also use

MatrixPower[{{1, 1}, {1, 0}}, t, {1, 0}]
(* {(2^(-1 - t) (1 - Sqrt[5])^t (-1 + Sqrt[5]))/Sqrt[5] + (
  2^(-1 - t) (1 + Sqrt[5])^(1 + t))/Sqrt[
  5], -((1/2 (1 - Sqrt[5]))^t/Sqrt[5]) + (1/2 (1 + Sqrt[5]))^t/Sqrt[5]
  } *)
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