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{SECT 0 {EXCHG {PARA 18 "" 0 "" {TEXT -1 21 "Financial mathematics" }}
{PARA 18 "" 0 "" {TEXT 257 23 "Demonstrative exercises" }}}{SECT 1 
{PARA 3 "" 0 "" {TEXT -1 14 "Relative value" }}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 71 "i-th commodity, i=1,2,3 is being traded at time t=1,2,3,4
,5 and course " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 55 "restart;
\nkappa:=(i,t)->abs(2*i^3+(5*i^2)*sin(i*t/6))+1;" }}{PARA 0 "> " 0 "" 
{MPLTEXT 1 0 111 "plot(\{kappa(1,t),kappa(2,t),kappa(3,t)\},t=0..6, co
lor=[GREEN,NAVY,RED],title=`Courses of commodities in time`);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&kappaGf*6$%\"iG%\"tG6\"6$%)operator
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{GLPLOT2D 671 671 671 {PLOTDATA 2 "6(-%'CURVESG6$7S7$$\"\"!F)$\"\"$F)7
$$\"3%*******\\#HyI\"!#=$\"3ltB*39x*3J!#<7$$\"33++]([kdW#F/$\"3))\\V8p
sv.KF27$$\"3++++v;\\DPF/$\"3hAb[!>e-J$F27$$\"3W+++D(>O7F8%F27$$\"36+++&>q0]\"F2
$\"3eIY`\"=![PUF27$$\"3'******\\U80j\"F2$\"3p4fE4*)4UVF27$$\"35+++0ytb
ev!\\ClRYF27$$\"3.+++!y?#>@F2$\"3Q)*\\#F2$\"3$=_h7(\\VB]F27$$\"3:++DEP/BEF2$\"3Yo]]!\\/p6&F27$$\"3=++](
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\"3#****\\7:xWC$F2$\"374T`=$zQd&F27$$\"37++]Zn%)oLF2$\"3#QMR.Ju@m&F27$
$\"3y******4FL(\\$F2$\"3IenoP+>_dF27$$\"3#)****\\d6.BOF2$\"3*pr47*=.Re
F27$$\"3(****\\(o3lWPF2$\"3lFAxX-(=#fF27$$\"3!*****\\A))ozQF2$\"3A[\"p
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ph?L'F27$$\"3g***\\(3/3(\\%F2$\"3Z&H6I?8kS'F27$$\"33++vB4JBYF2$\"3!y'[
m_kl#['F27$$\"3u*****\\KCnu%F2$\"3?q)p7?3db'F27$$\"3s***\\(=n#f([F2$\"
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3c%G\\'=j^nOFh[l7$F[y$\"37pXGZ&)3OFh[l7$Fjy$\"39cG/\"*4%>e$Fh[l7$F_z$\"3I6Y(HgMIb$Fh[l7$Fdz$
\"3^j8l`[f=NFh[l-Fiz6&F[[l$\")!\\DP\"F^[lFhdl$\")viobF^[l-F$6$7S7$F($
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(Fh[l7$FM$\"3Y^$HU4b%*R(Fh[l7$FR$\"3g;4cY%o]l(Fh[l7$FW$\"3p@!3Cty6!zFh
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fi&*Fh[l7$Fcq$\"3g(eiz$ffp'*Fh[l7$Fhq$\"3'psc`\"*****Fh[l7$Ffs$\"3!*yHwEr/%***Fh[
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*Fh[l7$Fjt$\"3M$G*41d(pz*Fh[l7$F_u$\"3[))HAx,-(p*Fh[l7$Fdu$\"3Y='*G/o(
3f*Fh[l7$Fiu$\"3]'z4?zS4Y*Fh[l7$F^v$\"3z-U\"[jM)G$*Fh[l7$Fcv$\"3?`'=Ux
M$p\"*Fh[l7$Fhv$\"3r#QOo<_a+*Fh[l7$F]w$\"3c7OR=/]?))Fh[l7$Fbw$\"3]rd[+
r)oi)Fh[l7$Fgw$\"3#y&yxS)Fh[l7$Fax$\"3C_!=
A " 0 "" {MPLTEXT 1 0 115 "Digits:=3;\nj:='j';\nfor i f
rom 0 to 5 do\nprint(i,evalf(kappa('j',i)) $'j'=1..3);\n#print(i,(kapp
a(i,j)) $j=1..3);\nod;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%'DigitsG\"
\"$" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"jGF$" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6&\"\"!$\"\"$F#$\"#)F)" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 71 "C
ourse margin - increase (or decrease) in courses per unit of commodity
" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 254 "for i from 0 to 5. do\nxxx:=ev
alf(kappa(j,i)):yyy:=evalf(kappa(j,i+1)):\nzzz:=([xxx,yyy] $j=1..3);\n
zzzz:=evalf(kappa(j,i+1)-kappa(j,i));\nprint(i,evalf(kappa(j,i+1)-kapp
a(j,i)) \n$j=1..3);\nod:\nevalf([5, -3.70+5.*sin(1), -19.9+20.*sin(2),
 -26.9+45.*sin(3)]);\n" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6&\"\"!$\"#$)!\"#$\"#m!\"\"$\"$;#F)" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6&\"\"\"$\"#\")!\"#$\"#e!\"\"$\"$i\"F)" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6&\"\"#$\"#w!\"#$\"#W!\"\"$\"#qF)" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6&\"\"$$\"#q!\"#$\"#E!\"\"$!#RF)" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6&\"\"%$\"#g!\"#$\"\"&!\"\"$!$S\"F)" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6&\"\"&,&$\"$q$!\"#!\"\"*&$F#\"\"!\"\"\"
-%$sinG6#F,F,F,,&$\"$*>F(F(*&$\"#?F+F,-F.6#\"\"#F,F,,&$\"$p#F(F(*&$\"#
XF+F,-F.6#\"\"$F,F," }}{PARA 11 "" 1 "" {XPPMATH 20 "6#7&$\"\"&\"\"!$
\"#]!\"#$!# " 0 "" {MPLTEXT 1 0 108 "for i from 0.1 to 5.1 do\ni:=i-.1;
\nprint(i,evalf(kappa(j,i+1.)-kappa(j,i))/(kappa(j,i)) $j=1..3);\ni:=i
+.1\nod:" }}{PARA 11 "" 1 "" {XPPMATH 20 "6&$\"\"!F$$\"$x#!\"$$\"$)QF'
$\"$$RF'" }}{PARA 11 "" 1 "" {XPPMATH 20 "6&$\"#5!\"\"$\"$6#!\"$$\"$Y#
F(F&" }}{PARA 11 "" 1 "" {XPPMATH 20 "6&$\"#?!\"\"$\"$k\"!\"$$\"$]\"F(
$\"$a(!\"%" }}{PARA 11 "" 1 "" {XPPMATH 20 "6&$\"#I!\"\"$\"$I\"!\"$$\"
$p(!\"%$!$\"RF+" }}{PARA 11 "" 1 "" {XPPMATH 20 "6&$\"#S!\"\"$\"$%)*!
\"%$\"$P\"F($!$Y\"!\"$" }}{PARA 11 "" 1 "" {XPPMATH 20 "6&$\"#]!\"\"$
\"$Y(!\"%$!$h%F($!$]#!\"$" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
136 "for i from 1 to 3 do\nplot(\{kappa(i,t),kappa(i,t+1)-kappa(i,t),(
kappa(i,t+1)-kappa(i,t))/kappa(i,t)\},t=0..6, color=[GREEN,NAVY,RED]);
\nod;" }}{PARA 13 "" 1 "" {GLPLOT2D 671 671 671 {PLOTDATA 2 "6'-%'CURV
ESG6$7S7$$\"\"!F)$\"\"$F)7$$\"3%*******\\#HyI\"!#=$\"3ltB*39x*3J!#<7$$
\"33++]([kdW#F/$\"3))\\V8psv.KF27$$\"3++++v;\\DPF/$\"3hAb[!>e-J$F27$$
\"3W+++D(>O
7F8%F27$$\"36+++&>q0]\"F2$\"3eIY`\"=![PUF27$$\"3'******\\U80j\"F2$\"3p
4fE4*)4UVF27$$\"35+++0ytbev!\\ClRYF27$$\"3.+++!y?#>@F2$\"3Q)*\\#F2$\"3$=_h7(\\VB]F27$$\"3:++DEP/BEF2$\"
3Yo]]!\\/p6&F27$$\"3=++](o:;v#F2$\"3\"4GuhrxM@&F27$$\"3=++v$)[opGF2$\"
3]wc#H#zE,`F27$$\"3%*****\\i%Qq*HF2$\"3IBn/t:'\\R&F27$$\"3&****\\(QIKH
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*\\d6.BOF2$\"3*pr47*=.ReF27$$\"3(****\\(o3lWPF2$\"3lFAxX-(=#fF27$$\"3!
*****\\A))ozQF2$\"3A[\"pq@SC,'F27$$\"3e******Hk-,SF2$\"3I&p?`=@D4'F27$
$\"36+++D-eITF2$\"3a$puY&yiwhF27$$\"3u***\\(=_(zC%F2$\"37+o=NIc^iF27$$
\"3M+++b*=jP%F2$\"3C&fGsph?L'F27$$\"3g***\\(3/3(\\%F2$\"3Z&H6I?8kS'F27
$$\"33++vB4JBYF2$\"3!y'[m_kl#['F27$$\"3u*****\\KCnu%F2$\"3?q)p7?3db'F2
7$$\"3s***\\(=n#f([F2$\"3jFK#*3VdImF27$$\"3P+++!)RO+]F2$\"3+g3,8#)3,nF
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$\"3G+++5jyp`F2$\"3tZW69S$4!pF27$$\"3<++]Ujp-bF2$\"3usiv:eDppF27$$\"3+
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3=(\\sua2Ta\"F/7$Fcq$\"3wY4id7a+:F/7$Fhq$\"3%G;3Vx`TX\"F/7$F]r$\"3IzR7
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al7$F\\x$\"3M^fW%)zVdvFeal7$Fax$\"3'\\M!)>=H-F(Feal7$Ffx$\"3_Ww1Ajv*)p
Feal7$F[y$\"3IL;c\"yuct'Feal7$F`y$\"3CQcu_Kb[kFeal7$Fey$\"3om1m,(o_>'F
eal7$Fjy$\"3e*pT[R'pGfFeal7$F_z$\"3m\"oMcHdnn&Feal7$Fdz$\"3$oY$
\"3*fWL2gO!3#)F/7$FC$\"3UB;hA%=m<)F/7$FH$\"3Jj!f1IaT9)F/7$FM$\"3\"=\"
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Fjo$\"3)R5m:WRBu(F/7$F_p$\"3%)f?sd/.$o(F/7$Fdp$\"30d'[sO;*3wF/7$Fip$\"
39i+?`q5VvF/7$F^q$\"3Q_
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(F(-%+AXESLABELSG6$Q\"t6\"Q!Fh^m-%%VIEWG6$;F(Fdz%(DEFAULTG" 1 2 0 1 
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$\"3W+++D#F27$$\"3i****\\()fB:()F/$\"3s$**fe
pxGF#F27$$\"39++](Q=\"))**F/$\"3AFU![$3k`BF27$$\"3(****\\P'=pD6!#<$\"3
pg%\\$GU(HV#F27$$\"33+++lN?c7FY$\"3frK>l#4K^#F27$$\"3-++]U$e6P\"FY$\"3
c()>@R1h#e#F27$$\"36+++&>q0]\"FY$\"3/VlMbY=fEF27$$\"3'******\\U80j\"FY
$\"36s#))RHxUt#F27$$\"35+++0ytbnGF27$$\"3.+++XDn/?FY$\"3P:(pu;'=RHF27$$\"3.+++!y?#>@FY$
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DEFAULTG" 1 2 0 1 10 0 2 9 1 4 2 1.000000 45.000000 45.000000 0 0 "Cur
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se derivation according to time is " }{XPPEDIT 18 0 "limit((kappa(i,t)
-kappa(i,t+delta(t)))/delta(t),delta(t)=0)" "6#-%&limitG6$*&,&-%&kappa
G6$%\"iG%\"tG\"\"\"-F)6$F+,&F,F--%&deltaG6#F,F-!\"\"F--F26#F,F4/-F26#F
,\"\"!" }{TEXT -1 50 " which shows instantaneous total change in cours
e." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 141 "\nplot(\{diff(kappa(1,t),t),
diff(kappa(2,t),t),diff(kappa(3,t),t)\},t=0..6, color=[GREEN,NAVY,RED]
,title=`Course derivation according to time`);" }}{PARA 13 "" 1 "" 
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l7$F^q$\"3_Av\"=w;kn*FX7$Fcq$\"3uoJZw@Pi%)FX7$Fhq$\"33/(zPL%o'4(FX7$F]
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=+<#Fbel7$F_z$!3Hj7ME,2-AFbel7$Fdz$!3J-5N " 0 "" {MPLTEXT 1 0 62 "diff(Kappa(i,t)/Kappa(i,t0),t)=diff
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\"DG6#\"\"#6#%&KappaG6$\"\"%%\"tG\"\"\"-F,6$F.%#t0G!\"\"F$" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 35 "Relative total change of course is " }
{XPPEDIT 18 0 "limit((kappa(i,t)-kappa(i,t+delta(t)))/(kappa(i,t)*delt
a(t)),delta(t)=0)=diff(kappa(i,t),t)/kappa(i,t)" "6#/-%&limitG6$*&,&-%
&kappaG6$%\"iG%\"tG\"\"\"-F*6$F,,&F-F.-%&deltaG6#F-F.!\"\"F.*&-F*6$F,F
-F.-F36#F-F.F5/-F36#F-\"\"!*&-%%diffG6$-F*6$F,F-F-F.-F*6$F,F-F5" }
{TEXT -1 1 "." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 186 "plot(\n\{
diff(kappa(1,t),t)/kappa(1,t)\n,diff(kappa(2,t),t)/kappa(2,t)\n,diff(k
appa(3,t),t)/kappa(3,t)\n\},t=0..6, color=[GREEN,NAVY,RED],title=`Rela
tive derivation commodity values in time`);" }}{PARA 13 "" 1 "" 
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,7$Fiv$!3yeW9YPkm:F,7$F^w$!3K02H)QU;s\"F,7$Fcw$!3#y)4!Hd5c(=F,7$Fhw$!3
(G%=StwiR?F,7$F]x$!3&G5Er\"*z0?#F,7$Fbx$!3=6>Gv'o%oBF,7$Fgx$!3;6<_'3@&
QDF,7$F\\y$!3w\"[KosL\")p#F,7$Fay$!3KP<[#Q&H&)GF,7$Ffy$!3og'pI$owcIF,7
$F[z$!3#)oH)Q\\%3WKF,7$F`z$!3(e+(4zc%zU$F,7$Fez$!3 " 0 "" {MPLTEXT 1 0 11 "Digits:=12;" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#>%'DigitsG\"#7" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 120 "Supposing that we could trade commodities continually, t
he most profitable is to buy and keep commodity #3 till time T1," }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 72 "T1:=fsolve(diff(kappa(2,t),t
)/kappa(2,t)=diff(kappa(3,t),t)/kappa(3,t));" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%#T1G$\"-a!e200(!#7" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 71 "then sell commodity #3 and buy commodity #2. Keep commodity #2 \+
till T2," }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 73 "T2:=fsolve(diff
(kappa(1,t),t)/kappa(1,t)=diff(kappa(2,t),t)/kappa(2,t));\n" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#>%#T2G$\"-9fVQ " 0 "" {MPLTEXT 1 0 
140 "NumberOfTitles:=3;\nCapital[0]:=100;\nRateOfProfit:=(i,t)->kappa(
i,t+1)/kappa(i,t);\nValue:=t->Sum(kappa(i,t)*Number[i],i=1..NumberOfTi
tles);\n\n" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%/NumberOfTitlesG\"\"$
" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>&%(CapitalG6#\"\"!\"$+\"" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%-RateOfProfitGf*6$%\"iG%\"tG6\"6$%)o
peratorG%&arrowGF)*&-%&kappaG6$9$,&\"\"\"F39%F3F3-F/6$F1F4!\"\"F)F)F)
" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&ValueGf*6#%\"tG6\"6$%)operatorG
%&arrowGF(-%$SumG6$*&-%&kappaG6$%\"iG9$\"\"\"&%'NumberG6#F3F5/F3;F5%/N
umberOfTitlesGF(F(F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 52 "Max
:=t->max(RateOfProfit(i,t) $i=1..NumberOfTitles);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%$MaxGf*6#%\"tG6\"6$%)operatorG%&arrowGF(-%$maxG6#-%\"
$G6$-%-RateOfProfitG6$%\"iG9$/F5;\"\"\"%/NumberOfTitlesGF(F(F(" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 131 "Vyber:=if(t)\nglobal Number
OfTitle;\nfor i from 1 to NumberOfTitles do\nif Max(t)=RateOfProfit(i,
t)\nthen NumberOfTitle:=i;\nfi \nod\nend;" }}{PARA 8 "" 1 "" {TEXT -1 
37 "Error, reserved word `if` unexpected\n" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 401 "for t from  0 to 4 do \ni:='i';\nVyber(t); \n  for
 i from 1 to NumberOfTitles do\n   Number[i]:=0;\n  od:\nNumber[Number
OfTitle]:=Capital[t]/kappa(NumberOfTitle,t);\nCapital[t+1]:=Number[Num
berOfTitle]*kappa(NumberOfTitle,t+1);\nprint(`at time `,t, ` I purchas
e `, evalf(Number[NumberOfTitle]),\n ` items of commodity `,   NumberO
fTitle, `at time`, t+1, `will I have the capital of `, evalf(Capital[t
+1]));\nod:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}{PARA 12 "" 1 "" 
{XPPMATH 20 "6,%)at~time~G\"\"!%-~I~purchase~G,$*&$\"$+\"F$\"\"\",&*&$
\"\"#F$F*)-%$absG6#%.NumberOfTitleG\"\"$F*F*$F*F$F*!\"\"F*%5~items~of~
commodity~GF3%(at~timeGF*% " 0 "" {MPLTEXT 1 0 163 "RateOfProfit := if (i, t1, t2) options oper
ator, arrow; kappa(i,t2)/kappa(i,t1) end if; evalf(100*RateOfProfit(3,
0,T1)*RateOfProfit(2,T1,T2)*RateOfProfit(1,T2,5));" }}{PARA 0 "> " 0 "
" {MPLTEXT 1 0 0 "" }}{PARA 8 "" 1 "" {TEXT -1 37 "Error, reserved wor
d `if` unexpected\n" }}}}}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 18 "Coumpo
und interest" }}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 14 "Common formula" }
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 147 "restart;\nPhi[1]:=(x0,xi,t
)->x0*(1+xi)^(floor(t));\n\nPhi[2]:=(x0,xi,t)->x0*(1+xi)^(floor(t))*(1
+xi*(t-floor(t)));\n\nPhi[3]:=(x0,xi,t)->x0*(1+xi)^(t);\n" }}{PARA 11 
"" 1 "" {XPPMATH 20 "6#>&%$PhiG6#\"\"\"f*6%%#x0G%#xiG%\"tG6\"6$%)opera
torG%&arrowGF-*&9$F'),&F'F'9%F'-%&floorG6#9&F'F-F-F-" }}{PARA 11 "" 1 
"" {XPPMATH 20 "6#>&%$PhiG6#\"\"#f*6%%#x0G%#xiG%\"tG6\"6$%)operatorG%&
arrowGF-*(9$\"\"\"),&F3F39%F3-%&floorG6#9&F3,&F3F3*&F6F3,&F:F3F7!\"\"F
3F3F3F-F-F-" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>&%$PhiG6#\"\"$f*6%%#x0
G%#xiG%\"tG6\"6$%)operatorG%&arrowGF-*&9$\"\"\"),&F3F39%F39&F3F-F-F-" 
}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 36 "Let's assign values into variab
les: " }{XPPEDIT 18 0 "x0=10,xi=1.9" "6$/%#x0G\"#5/%#xiG-%&FloatG6$\"#
>!\"\"" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 125 "Arg:=(10,1.9,t);\nplot(
\{\nPhi[1](Arg),\nPhi[2](Arg),\nPhi[3](Arg)\},\nt=0..5,\nthickness=3,
\ncolor=[coral,gold,maroon],discont=true);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%$ArgG6%\"#5$\"#>!\"\"%\"tG" }}{PARA 13 "" 1 "" 
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Fjbp7$F]jm$\"3G97SZau%*>Fjbp7$F`jm$\"3B7&p5?;>-#Fjbp7$Fcjm$\"3[h;c')[6
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%oig[-c\"Fg^l7$$\"3emm\"z%4\\Y_F>$\"3=/7;UCA[$\"
3t&R\"[]?nU>Fg^l7$$\"3]***\\il'pisF>$\"3'yR(**))R$o;#Fg^l7$$\"3>MLe*)>
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\"37++]i^Z]7Fjt$\"3W19+%4>jy$Fg^l7$$\"33++](=h(e8Fjt$\"3)))f\\R%o,\\UF
g^l7$$\"3/++]P[6j9Fjt$\"3k\"pV:AN$[ZFg^l7$$\"3UL$e*[z(yb\"Fjt$\"3q`3@(
RACD&Fg^l7$$\"3wmm;a/cq;Fjt$\"3y_SOht$>#fFg^l7$$\"3%ommmJFjt$
\"3G+u')*\\.@>)Fg^l7$$\"3K+]i!f#=$3#Fjt$\"3uSb1pR\"))=*Fg^l7$$\"3?+](=
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3um;zpSS\"R#Fjt$\"3U^%zBq$zv7F^hl7$$\"3GLL3_?`(\\#Fjt$\"3+:)3s38%G9F^h
l7$$\"3fL$e*)>pxg#Fjt$\"3`40\"*3AI1;F^hl7$$\"33+]Pf4t.FFjt$\"3#*\\!eaj
(4zF^hl7$$\"30+++DRW9HFjt$\"3M
\\W_#*HbEAF^hl7$$\"3:++DJE>>IFjt$\"3?rU(Gk]#*[#F^hl7$$\"3F+]i!RU07$Fjt
$\"3OkOnR%*)Gx#F^hl7$$\"3+++v=S2LKFjt$\"3)\\DIT]Te7$F^hl7$$\"3Jmmm\"p)
=MLFjt$\"3CNhi*\\X6[$F^hl7$$\"3B++](=]@W$Fjt$\"3l!*[;q1?0RF^hl7$$\"35L
$e*[$z*RNFjt$\"3ZDlQjZ!RL%F^hl7$$\"3e++]iC$pk$Fjt$\"3yuG7B;ic[F^hl7$$
\"3[m;H2qcZPFjt$\"3='o?cW$*eS&F^hl7$$\"3O+]7.\"fF&QFjt$\"3Y>@4EibYgF^h
l7$$\"3Ymm;/OgbRFjt$\"39)H!>V#[Z%F
jt$\"3[E+!\\y'fs6Fjbp7$$\"3SnmT&G!e&e%Fjt$\"3Q%zLP[f$>8Fjbp7$$\"3#RLLL
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++DM\"3%[Fjt$\"3/XSD\"HS8t\"Fjbp7$$\"3a+]P40O\"*[Fjt$\"3/(\\#G/02F=Fjb
p7$$\"3s+voa-oX\\Fjt$\"3Wx$zJ@ae$>Fjbp7$$\"\"&F-$\"3e********[6^?Fjbp-
Ffjm6&Fhjm$\")viobF[[n$\")!\\DP\"F[[n$\")%yg>%F[[nF_[n-%'POINTSG6)Fcjp
7$$\"\"\"F-$\"$!H!\"\"7$$\"\"#F-$\"%5%)!\"#7$$Fb[nF-$\"'!*QC!\"$7$$\"
\"%F-$\"(5G2(!\"%7$Fjiq$\"*!\\6^?!\"&Fejm-Fgjq6)Fcjp7$Fjjq$\"%+HFd[r7$
F`[r$\"&+T)Fi[r7$Ff[r$\"(+*QCF_\\r7$F[\\r$\")+\"G2(Fc\\r7$Fjiq$\"++\\6
^?!\"'Fjip-%+AXESLABELSG6%Q\"t6\"Q!Fj]r-%%FONTG6#%(DEFAULTG-%%VIEWG6$;
F^[nFjiqF_^r" 1 2 0 1 10 0 2 9 1 4 2 1.000000 45.000000 45.000000 0 0 
"Curve 1" "Curve 2" "Curve 3" "Curve 4" "Curve 5" }}}}{EXCHG {PARA 0 "
" 0 "" {TEXT -1 165 "Distinction between the values of continual and c
ompound interest; meaning the distinction between the values of expone
ntial function and the values on it's secants." }}{PARA 0 "> " 0 "" 
{MPLTEXT 1 0 37 "plot(Phi[2](Arg)-Phi[3](Arg),t=0..5);" }}{PARA 13 "" 
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7$$\"3Y\\7`p2_#)RF5$\"3$3HC5p>\\F57$$\"3K&Rs31]#*)RF5$\"3W[cQM?mnIF5
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*fDVF5$\"3xyl3jh!\\W\"Fdhl7$$\"3Cn;HdO=yVF5$\"3w%3k>o8bd\"Fdhl7$$\"3MM
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++DM\"3%[F5$\"3CK1OGHae5Fdhl7$$\"3)3](=np3m[F5$\"3SxZde\"f'f#*Fgu7$$\"
3a+]P40O\"*[F5$\"3%zp*oLl40yFgu7$$\"3#e7.dWS\\!\\F5$\"3uu5nTq6ppFgu7$$
\"3>]7.#Q?&=\\F5$\"3c76+T\\Q%4'Fgu7$$\"3Yv$f$=.5K\\F5$\"3s8:*)[eL!=&Fg
u7$$\"3s+voa-oX\\F5$\"3TQ?.jrREUFgu7$$\"35Dc,\">g#f\\F5$\"3<%*H)=%z)>B
$Fgu7$$\"3O]PMF,%G(\\F5$\"3wP`A$y=l>#Fgu7$$\"3iv=nj+U')\\F5$\"3\\#3T4q
\"R>6Fgu7$$\"\"&F)F(-%'COLOURG6&%$RGBG$\"#5!\"\"F(F(-%+AXESLABELSG6$Q
\"t6\"Q!Fdbm-%%VIEWG6$;F(Fgam%(DEFAULTG" 1 2 0 1 10 0 2 9 1 4 2 
1.000000 45.000000 45.000000 0 0 "Curve 1" }}}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 89 "We'll find the moment, when the distinction is equal to t
he half of the original deposit." }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 38 "T:=solve(Phi[2](Arg)-Phi[3](Arg)=5,t);" }}{PARA 11 "
" 1 "" {XPPMATH 20 "6#>%\"TG$\"+`Ypa7!\"*" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 59 "evalf(subs(t=T,Phi[2](Arg)));\nevalf(subs(t=T,Phi[3
](Arg)));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+RvO.V!\")" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#$\"+RvO.Q!\")" }}}}{SECT 1 {PARA 4 "" 0 "" 
{TEXT -1 7 "Samples" }}{SECT 1 {PARA 0 "" 0 "" {TEXT -1 200 "1.  1. On
 my current account there are 10 golden coins(GC). On 1. March I depos
it another 10GC and on 1. September 20GC more. What will be my portfol
io on 1. 1. next year if the interest rate is 0.05?" }}{SECT 1 {PARA 
5 "" 0 "" {TEXT -1 8 "Solution" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 407 "NumberOfDaysInM[1]:=31:\nNumberOfDaysInM[2]:=28:\nNumberOfDaysI
nM[3]:=31:\nNumberOfDaysInM[4]:=30:\nNumberOfDaysInM[5]:=31:\nNumberOf
DaysInM[6]:=30:\nNumberOfDaysInM[7]:=31:\nNumberOfDaysInM[8]:=31:\nNum
berOfDaysInM[9]:=30:\nNumberOfDaysInM[10]:=31:\nNumberOfDaysInM[11]:=3
0:\nNumberOfDaysInM[12]:=31:\nNumberOfDaysTillEndOfY:=(d,m)->\nNumberO
fDaysInM[m]-d+sum(NumberOfDaysInM[i],i=m+1..12);\nNumberOfDaysTillEndO
fY(1,1);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%7NumberOfDaysTillEndOfYG
f*6$%\"dG%\"mG6\"6$%)operatorG%&arrowGF),(&%0NumberOfDaysInMG6#9%\"\"
\"9$!\"\"-%$sumG6$&F/6#%\"iG/F:;,&F2F2F1F2\"#7F2F)F)F)" }}{PARA 11 "" 
1 "" {XPPMATH 20 "6#\"$k$" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
159 "zeta := (1+5/100)^(1/365)-1;  evalf(10*(1+zeta)^NumberOfDaysTillE
ndOfY(21,1)+10*(1+zeta)^NumberOfDaysTillEndOfY(16,3)+20*(1+zeta)^Numbe
rOfDaysTillEndOfY(7,9));" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}
{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#
>%%zetaG,&*(\"#?!\"\"\"#@#\"\"\"\"$l$F'#\"$k$F,F+F+F(" }}{PARA 11 "" 
1 "" {XPPMATH 20 "6#$\"+FYc " 0 "" 
{MPLTEXT 1 0 88 "NumberOfDaysTillEndOfY(21,1);\nNumberOfDaysTillEndOfY
(16,3);\nNumberOfDaysTillEndOfY(7,9);" }}{PARA 11 "" 1 "" {XPPMATH 20 
"6#\"$W$" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"$!H" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#\"$:\"" }}}}}{PARA 4 "" 0 "" {TEXT -1 0 "" }}{SECT 1 
{PARA 0 "" 0 "" {TEXT -1 271 "When your current account (CA) reaches t
he nominal value of 120 giggles(G), if you deposit 90G at time t=0 and
 the interest rate is 0.05 during the time when the nominal value of t
he CA is below 100G and 0.07 during the time when the nominal value of
 the CA is above 100?" }}{SECT 1 {PARA 5 "" 0 "" {TEXT -1 8 "Solution
" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 60 "We have to calculate the time \+
before we reach 100G on our CA" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 39 "restart;  T[1]:=solve(90*1.05^t=100,t);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>&%\"TG6#\"\"\"$\"+3AYf@!\"*" }}}{EXCHG {PARA 0 "" 0 "
" {TEXT -1 72 "then we'll count the time when the value of the CA grow
s from 100 to 120" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "T[2]:=
solve(100*1.07^t=120,t);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>&%\"TG6#
\"\"#$\"+cls%p#!\"*" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 29 "finally we
 add up T1 and T2. " }{MPLTEXT 1 0 14 "Tt:=T[1]+T[2];" }}{PARA 11 "" 
1 "" {XPPMATH 20 "6#>%#TtG$\"+k()=a[!\"*" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 146 "Considering that the time unit is the same in which the \+
interest rate is being accredited. For example if the interest is accr
edited only at time " }{XPPEDIT 18 0 "t=1/365,2/365,3/365" "6%/%\"tG*&
\"\"\"F&\"$l$!\"\"*&\"\"#F&F'F(*&\"\"$F&F'F(" }{TEXT -1 130 ",... we c
an say, that the CA will never reach the nominal value of 120G. Yet th
e firs moment when we can expect 120G an our CA is:" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 171 "T[1]:=ceil(solve(T[1]=x*1/365,x));\nT[2]
 := solve((90*1.05^(T[1]/365))*1.07^t = 120,t);\n#(90*1.07^(T[1]/365))
*1.07^T[2];\nT[2]:=ceil(solve(T[2]=x*1/365,x));\nTt:=T[1]+T[2];\n\n" }
}{PARA 11 "" 1 "" {XPPMATH 20 "6#>&%\"TG6#\"\"\"\"$*y" }}{PARA 11 "" 
1 "" {XPPMATH 20 "6#>&%\"TG6#\"\"#$\"+XL:$p#!\"*" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>&%\"TG6#\"\"#\"$%)*" }}{PARA 11 "" 1 "" {XPPMATH 20 "6
#>%#TtG\"%t<" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 11 "t-th moment" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "evalf(Tt/365);" }}{PARA 11 "
" 1 "" {XPPMATH 20 "6#$\"+ZU`d[!\"*" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 35 "the value of the CA at t-th moment:" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 37 "(90*1.05^(T[1]/365))*1.07^(T[2]/365);" }}{PARA 11 "
" 1 "" {XPPMATH 20 "6#$\"+XAA+7!\"(" }}}}}{SECT 1 {PARA 0 "" 0 "" 
{TEXT -1 45 "Think of two CAs. On the first one there are " }{XPPEDIT 
18 0 "x1:=1234;" "6#>%#x1G\"%M7" }{TEXT -1 42 "GC deposited. On the se
cond one there are " }{XPPEDIT 18 0 "x2 := 1230" "6#>%#x2G\"%I7" }
{TEXT -1 81 "GK. What is the interest rate the CA must have been charg
ed with for the period  " }{XPPEDIT 18 0 "T:=2;" "6#>%\"TG\"\"#" }
{TEXT -1 12 " to achieve " }{XPPEDIT 18 0 "x1" "6#%#x1G" }{TEXT -1 3 "
 = " }{XPPEDIT 18 0 "x2" "6#%#x2G" }{TEXT -1 9 ", if the " }{XPPEDIT 
18 0 "x1" "6#%#x1G" }{TEXT -1 42 " is charged with simple interest rat
e and " }{XPPEDIT 18 0 "x2" "6#%#x2G" }{TEXT -1 21 " with compound one
?  " }}{SECT 1 {PARA 5 "" 0 "" {TEXT -1 8 "solution" }}{EXCHG {PARA 0 
"> " 0 "" {MPLTEXT 1 0 37 "formula:=x[1]*(1+xi*T)=x[2]*(1+xi)^T;" }}
{PARA 0 "> " 0 "" {MPLTEXT 1 0 56 "solution:=solve(\{formula,xi>0\},xi
);\nallvalues(solution);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "zeta:=o
p(2,op(evalf(solution)));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%(formul
aG/*&&%\"xG6#\"\"\"F*,&F*F**&%#xiGF*%\"TGF*F*F**&&F(6#\"\"#F*),&F*F*F-
F*F.F*" }}{PARA 8 "" 1 "" {TEXT -1 60 "Error, (in index/FillInitVals) \+
too many levels of recursion\n" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#%)so
lutionG" }}{PARA 8 "" 1 "" {TEXT -1 41 "Error, improper op or subscrip
t selector\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 189 "plot(x[2]*
(1+xi)^T-x[1]*(1+xi*T),xi=0..zeta+0.1,title=`The distinction between t
he current value of simple and compond interrest in relation to the in
terest rate.`,titlefont=[HELVETICA,7]);" }}{PARA 0 "> " 0 "" {MPLTEXT 
1 0 0 "" }}{PARA 8 "" 1 "" {TEXT -1 53 "Error, (in plot) range values \+
must be real constants\n" }}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{PARA 
4 "" 0 "" {TEXT -1 0 "" }}}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 33 "Depos
iting in equidistant moments" }}{SECT 1 {PARA 0 "" 0 "" {TEXT -1 173 "
 Suppose a year of 360 days with 12 months of 30 days. Your deposit is
 allways 100CZK. The interest rate is 0.05 p. a. How much is there goi
ng to be at the end of a year if:" }}{PARA 15 "" 0 "" {TEXT -1 16 "You
 deposit once" }}{PARA 16 "" 0 "" {TEXT -1 28 "at the beginning of the
 year" }}{PARA 16 "" 0 "" {TEXT -1 23 "at the half of the year" }}
{PARA 15 "" 0 "" {TEXT -1 18 "You deposit twice " }}{PARA 16 "" 0 "" 
{TEXT -1 45 "at the beginning and at the half of the yearu" }}{PARA 
16 "" 0 "" {TEXT -1 38 "at the half and at the end of the year" }}
{PARA 15 "" 0 "" {TEXT -1 13 "Twelve times " }}{PARA 16 "" 0 "" {TEXT 
-1 30 "at the beginning of each month" }}{PARA 16 "" 0 "" {TEXT -1 25 
"at the end of each month." }}{SECT 1 {PARA 20 "" 0 "" {TEXT -1 0 "" }
}{EXCHG {PARA 0 "" 0 "" {TEXT -1 69 "Trying to calculate Present value
 (PV) of the CA. Time unit is 1 day." }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 54 "restart;\nPV:=sum(Z(t)*(1+xi)^((360-t)/360), t=1..360
):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 48 "PVdem:=Sum(Z(t)*(1+xi)^((360-
t)/360), t=1..360):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "xi:=.05;" }
{TEXT -1 24 ", t is a time in years, " }{XPPEDIT 18 0 "z(t);" "6#-%\"z
G6#%\"tG" }{TEXT -1 36 " is the value of a deposit at time t" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#>%#xiG$\"\"&!\"#" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 201 "z[1]:=t->piecewise(t=1,100,t<>0,0);\nz[1](t);\nz[2
]:=t->piecewise(t=360/2,100,t<>0,0);\nz[2](t);\nz[3]:=t->piecewise(t=1
,100,t=360/2,100,0);\nz[3](t);\nz[4]:=t->piecewise(t=360/2,100,t=360,1
00,0);\nz[4](t);\n\n" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>&%\"zG6#\"\"
\"f*6#%\"tG6\"6$%)operatorG%&arrowGF+-%*piecewiseG6&/9$F'\"$+\"0F3\"\"
!F6F+F+F+" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%*PIECEWISEG6$7$\"$+\"/%
\"tG\"\"\"7$\"\"!0F)F," }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>&%\"zG6#\"
\"#f*6#%\"tG6\"6$%)operatorG%&arrowGF+-%*piecewiseG6&/9$\"$!=\"$+\"0F3
\"\"!F7F+F+F+" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%*PIECEWISEG6$7$\"$+
\"/%\"tG\"$!=7$\"\"!0F)F," }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>&%\"zG6#
\"\"$f*6#%\"tG6\"6$%)operatorG%&arrowGF+-%*piecewiseG6'/9$\"\"\"\"$+\"
/F3\"$!=F5\"\"!F+F+F+" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%*PIECEWISEG
6%7$\"$+\"/%\"tG\"\"\"7$F'/F)\"$!=7$\"\"!%*otherwiseG" }}{PARA 11 "" 
1 "" {XPPMATH 20 "6#>&%\"zG6#\"\"%f*6#%\"tG6\"6$%)operatorG%&arrowGF+-
%*piecewiseG6'/9$\"$!=\"$+\"/F3\"$g$F5\"\"!F+F+F+" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#-%*PIECEWISEG6%7$\"$+\"/%\"tG\"$!=7$F'/F)\"$g$7$\"\"!%*
otherwiseG" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 60 "for i from 1 \+
to 4 do\nZ:=t->z[i](t);\nPresentValue[i]:=PV;\nod;" }}{PARA 11 "" 1 "
" {XPPMATH 20 "6#>%\"ZG&%\"zG6#\"\"\"" }}}{EXCHG {PARA 0 "" 0 "" 
{XPPEDIT 18 0 "PresentValue[3]=PresentValue[2]+PresentValue[1]" "6#/&%
-PresentValueG6#\"\"$,&&F%6#\"\"#\"\"\"&F%6#F,F," }{TEXT -1 2 ", " }
{XPPEDIT 18 0 "PresentValue[4]=Presentvalue[2]+100" "6#/&%-PresentValu
eG6#\"\"%,&&%-PresentvalueG6#\"\"#\"\"\"\"$+\"F-" }{TEXT -1 64 ", Beca
use the last deposit isn't being charged with an interest." }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "with(plots):" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 102 "\nz[5]:=t->piecewise(type((t-1)/(30),integer)
,100,0);\nz[5](31);\npointplot(\{seq([n,z[5](n)],n=0..95)\});\n" }}
{PARA 0 "> " 0 "" {MPLTEXT 1 0 64 "# z[6]:=t->piecewise(type((t)/(30),
integer),100,0);\n# z[6](31);\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 36 "Z:=t->z[5](t);\nPresentValue[5]:=PV;\n" }}}{EXCHG {PARA 0 "" 
0 "" {TEXT -1 44 "In fact we're summing the geometrical series" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "#restart;" }}{PARA 0 "> " 0 "
" {MPLTEXT 1 0 119 "i:='i';Xi:=xi;Time:='Time':\nxi:='xi';\nxxx:=Sum(1
00*(1+xi)^(Time(i)/360),i=1..12)=sum(100*(1+xi)^(Time(i)/360),i=1..12)
;" }}{PARA 0 "" 0 "" {TEXT -1 5 "where" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#>%\"iGF$" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#XiG$\"\"&!\"#" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#xiGF$" }}{PARA 12 "" 1 "" {XPPMATH 
20 "6#>%$xxxG/-%$SumG6$,$*&\"$+\"\"\"\"),&F,F,%#xiGF,,$*&#F,\"$g$F,-%%
DobaG6#%\"iGF,F,F,F,/F7;F,\"#7,:*&F+F,)F.,$*&F2F,-F56#F,F,F,F,F,*&F+F,
)F.,$*&F2F,-F56#\"\"#F,F,F,F,*&F+F,)F.,$*&F2F,-F56#\"\"$F,F,F,F,*&F+F,
)F.,$*&F2F,-F56#\"\"%F,F,F,F,*&F+F,)F.,$*&F2F,-F56#\"\"&F,F,F,F,*&F+F,
)F.,$*&F2F,-F56#\"\"'F,F,F,F,*&F+F,)F.,$*&F2F,-F56#\"\"(F,F,F,F,*&F+F,
)F.,$*&F2F,-F56#\"\")F,F,F,F,*&F+F,)F.,$*&F2F,-F56#\"\"*F,F,F,F,*&F+F,
)F.,$*&F2F,-F56#\"#5F,F,F,F,*&F+F,)F.,$*&F2F,-F56#\"#6F,F,F,F,*&F+F,)F
.,$*&F2F,-F56#F:F,F,F,F," }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 62 
"Time := if (i) options operator, arrow; 360-(i-1)*30-1 end if;" }
{TEXT -1 61 " is time from depositing i-th deposit till the end of a y
ear." }}{PARA 8 "" 1 "" {TEXT -1 37 "Error, reserved word `if` unexpec
ted\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "xi:=Xi;\nxxx;" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 32 "geometrical series with quotient" 
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 59 "q:=(100*(1+xi)^(Time(i+1)
/360))/(100*(1+xi)^(Time(i)/360));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 55 "q:=simplify(q);\n
a1:=subs(i=1,100*(1+xi)^(Time(i)/360));" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 56 "We could sum this series according to the common formula
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "PresentValue[5]:=a1*(1-
q^12)/(1-q);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 206 "the case we depo
sit at the end of a month is different just in function Time and in pr
esent value of the first deposit. Quotient is the same like in the pre
vious case and the number of series members also: " }}}{EXCHG {PARA 0 
"> " 0 "" {MPLTEXT 1 0 124 "Time := i->360-(i-1)*30-30;\na1:=subs(i=1,
100*(1+xi)^(Time(i)/360));\nPresentValue[6]:=sum(100*(1+xi)^(Time(i)/3
60),i=1..12);\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 "\n" }
{TEXT 256 7 "valid:\n" }{MPLTEXT 1 0 51 "(PresentValue[5]/PresentValue
[6]=(1+xi)^(29/360));\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "
" }}}}}{PARA 3 "" 0 "" {TEXT -1 0 "" }}}{SECT 1 {PARA 3 "" 0 "" {TEXT 
-1 7 "Savings" }}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 56 "Saving on compou
nd interest through equidistant deposits" }}{SECT 1 {PARA 4 "" 0 "" 
{TEXT -1 15 "General formula" }}{PARA 0 "" 0 "" {TEXT -1 74 "Let's ass
ume we're depositing in the equidistant moments, with proportion " }
{XPPEDIT 18 0 "tau" "6#%$tauG" }{TEXT -1 2 ". " }{XPPEDIT 18 0 "epsilo
n" "6#%(epsilonG" }{TEXT -1 37 " is the moment of the first deposit, \+
" }{XPPEDIT 18 0 "epsilon+tau" "6#,&%(epsilonG\"\"\"%$tauGF%" }{TEXT 
-1 16 " of the second, " }{XPPEDIT 18 0 "epsilon+2*tau" "6#,&%(epsilon
G\"\"\"*&\"\"#F%%$tauGF%F%" }{TEXT -1 18 " of the third and " }
{XPPEDIT 18 0 "epsilon+T" "6#,&%(epsilonG\"\"\"%\"TGF%" }{TEXT -1 23 "
 of the last one.      " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "#
restart;\nWholePart:=x->floor(x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>
%*WholePartG%&floorG" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 31 "We're dep
ositing at the moment " }{XPPEDIT 18 0 "n*tau+epsilon,n = 0 .. T;" "6$
,&*&%\"nG\"\"\"%$tauGF&F&%(epsilonGF&/F%;\"\"!%\"TG" }{TEXT -1 36 "\nH
ow much deposits do we do in time " }{XPPEDIT 18 0 "t < T+epsilon;" "6
#2%\"tG,&%\"TG\"\"\"%(epsilonGF'" }{TEXT -1 1 "?" }}{PARA 0 "> " 0 "" 
{MPLTEXT 1 0 40 "numberA := WholePart((t-epsilon)/tau)+1:" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 13 "and in time  " }{XPPEDIT 18 0 "t > T+epsi
lon;" "6#2,&%\"TG\"\"\"%(epsilonGF&%\"tG" }{TEXT -1 2 "? " }{MPLTEXT 
1 0 19 "numberB := T/tau+1;" }{TEXT -1 26 ", which is integer number.
" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#>%(numberBG,&*&%\"TG\"\"\"%$tauG!\"\"F(F(F(" }}}{EXCHG {PARA 0 "
" 0 "" {TEXT -1 39 "Generally: If we're depositing at time " }
{XPPEDIT 18 0 "n*tau+epsilon,n = 0 .. T;" "6$,&*&%\"nG\"\"\"%$tauGF&F&
%(epsilonGF&/F%;\"\"!%\"TG" }{TEXT -1 20 "\nwe receive in time " }
{XPPEDIT 18 0 "t" "6#%\"tG" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "numbe
r := min(NumberA,NumberB);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%'numbe
rG-%$minG6$%(NumberAG%(NumberBG" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 
75 "the number of deposits. At which moment is the i-th deposit being \+
deposited" }{XPPEDIT 18 0 "%?;" "6#%#%?G" }}{PARA 0 "> " 0 "" 
{MPLTEXT 1 0 63 "moment := if (i) options operator, arrow; i*tau+epsil
on end if;" }}{PARA 8 "" 1 "" {TEXT -1 37 "Error, reserved word `if` u
nexpected\n" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 60 "How long is the in
terest paid on i-th deposit till the time " }{XPPEDIT 18 0 "t" "6#%\"t
G" }{TEXT -1 8 " if the " }{XPPEDIT 19 1 "moment(i) <= t;" "6#1-%'mome
ntG6#%\"iG%\"tG" }{TEXT -1 1 "?" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 68 "
time := if (i) options operator, arrow; t-moment(i) end if; time(i);" 
}}{PARA 8 "" 1 "" {TEXT -1 37 "Error, reserved word `if` unexpected\n
" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 46 "What is the value of the i-th
 deposit at time " }{XPPEDIT 18 0 "t" "6#%\"tG" }{TEXT -1 26 " at(on) \+
the interest rate " }{XPPEDIT 18 0 "xi" "6#%#xiG" }{TEXT -1 5 ", if " 
}{XPPEDIT 19 1 "moment(i) <= t;" "6#1-%'momentG6#%\"iG%\"tG" }{TEXT 
-1 48 " and if its value at the time of depositing was " }{XPPEDIT 18 
0 "z" "6#%\"zG" }{TEXT -1 1 "?" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 75 "v
alue := if (i) options operator, arrow; z*(1+xi)^time(i) end if; value
(i);" }{TEXT -1 25 " is on the condition that" }}{PARA 8 "" 1 "" 
{TEXT -1 37 "Error, reserved word `if` unexpected\n" }}}{EXCHG {PARA 
0 "" 0 "" {TEXT -1 69 "If we intent to sum current values of all of de
posits done at moment " }{XPPEDIT 18 0 "t" "6#%\"tG" }{TEXT -1 33 " or
 before we became the series  " }{XPPEDIT 18 0 "Sum(value(i),i = 0 .. \+
number);" "6#-%$SumG6$-%&valueG6#%\"iG/F);\"\"!%'numberG" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 112 "I-th + 1st divided by i-th member of the
 series is independent on i. Afterwards we can call it quotient, becau
se" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 64 "quotient := value(i+1)/value(
i); quotient := simplify(quotient);" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }
}{PARA 0 "" 0 "" {TEXT -1 74 "Thus we're daling  with a geometrical se
ries determined by this quotient. " }}}{EXCHG {PARA 16 "" 0 "" {TEXT 
-1 24 "The number of members is" }}{PARA 16 "> " 0 "" {MPLTEXT 1 0 25 
"NumberOfMembers = number;" }}}{EXCHG {PARA 16 "" 0 "" {TEXT -1 13 "0t
h member is" }}{PARA 16 "> " 0 "" {MPLTEXT 1 0 25 "ZerothMember := val
ue(0);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 22 "Thus the sum equals to
" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 75 "Formula:=Sum(value(i),i=0..numb
er-1)=simplify(sum(value(i),i=0..number-1));" }}{PARA 0 "" 0 "" {TEXT 
-1 43 "Using this formula for defining functions:\n" }}{PARA 0 "> " 0 
"" {MPLTEXT 1 0 92 "Formula:=simplify(op(2,Formula)):\n#Formula:=subs(
tau=1/k,Formula);\n#limit(subs(k=K,\"),K=k);\n" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 75 "SUMGeometricalSeries:=Sum(a[0]*q^i,i=0..n)=sim
plify(sum(a[0]*q^i,i=0..n));\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 74 "Formula2:=subs(a[0]=NultyClen,q=quotient,n=number-1,SUMGeometr
icalSeries):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "Formula2:=op(2,Form
ula2);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 180 "psi2:=(Tau,Epsil
on,Xi,Z,TT,time)->subs(tau=Tau,epsilon=Epsilon,xi=Xi,z=Z,T=TT,t=time,F
ormula2);\nprint(`----------`);\nsimplify(psi2(tau,epsilon,xi,z,T,t)-p
si(tau,epsilon,xi,z,T,t));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 90 "The
 first function determines the value of the saved up amount, from the \+
parameters given." }}{PARA 14 "" 0 "" {TEXT -1 61 "Cross distance of d
eposits in time (same for all neighboring)" }}{PARA 14 "" 0 "" {TEXT 
-1 50 "The moment at which the first deposit is deposited" }}{PARA 14 
"" 0 "" {TEXT -1 22 "Constant interest rate" }}{PARA 14 "" 0 "" {TEXT 
-1 34 "The constant value of each deposit" }}{PARA 14 "" 0 "" {TEXT 
-1 61 "The period of time for which the deposits are being deposited" 
}}{PARA 14 "" 0 "" {TEXT -1 0 "" }}{PARA 14 "" 0 "" {TEXT -1 71 "The m
oment at which we want to know the nominal balance on the account." }}
{PARA 0 "> " 0 "" {MPLTEXT 1 0 93 "psi:=(Tau,Epsilon,Xi,Z,TT,time)->su
bs(tau=Tau,epsilon=Epsilon,xi=Xi,z=Z,T=TT,t=time,Formula);" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 97 "If we intend to know the nominal residue \+
on the account while we're still depositing, it is also " }{XPPEDIT 
18 0 "trunc(t-epsilon/tau) <= T+epsilon+1;" "6#1-%&truncG6#,&%\"tG\"\"
\"*&%(epsilonGF)%$tauG!\"\"F-,(%\"TGF)F+F)F)F)" }{TEXT -1 10 " and the
n " }{XPPEDIT 18 0 "min(trunc((t-epsilon)/tau),T/tau)=trunc((t-epsilon
)/tau)" "6#/-%$minG6$-%&truncG6#*&,&%\"tG\"\"\"%(epsilonG!\"\"F-%$tauG
F/*&%\"TGF-F0F/-F(6#*&,&F,F-F.F/F-F0F/" }{TEXT -1 2 ". " }}{PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 321 "Formula_:=subs(min(trunc((t-epsilon)/tau),T/t
au)=trunc((t-epsilon)/tau),Formula):\nFormula_:=subs(min((T/tau,trunc(
(t-epsilon+trunc(abs((t-epsilon)/tau))*tau+tau)/tau)-trunc(abs((t-epsi
lon)/tau))))=\ntrunc(T/tau,trunc((t-epsilon+trunc(abs((t-epsilon)/tau)
)*tau+tau)/tau)-trunc(abs((t-epsilon)/tau)))\n,Formula_):\nFormula_;\n
\n" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#%)VzorecekG" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 17 "And the func
tion " }{XPPEDIT 18 0 "psi" "6#%$psiG" }{TEXT -1 73 " determines the v
alue of the saved up amount from the parameters given:  " }}{PARA 14 "
" 0 "" {TEXT -1 61 "Cross distance of deposits in time (same for all n
eighboring)" }}{PARA 14 "" 0 "" {TEXT -1 50 "The moment at which the f
irst deposit is deposited" }}{PARA 14 "" 0 "" {TEXT -1 22 "Constant in
terest rate" }}{PARA 14 "" 0 "" {TEXT -1 34 "The constant value of eac
h deposit" }}{PARA 14 "" 0 "" {TEXT -1 71 "The moment at which we want
 to know the nominal balance on the account." }}{PARA 0 "" 0 "" {TEXT 
-1 3 "If " }{XPPEDIT 18 0 "trunc((t-epsilon)/tau) <= T/tau;" "6#1-%&tr
uncG6#*&,&%\"tG\"\"\"%(epsilonG!\"\"F*%$tauGF,*&%\"TGF*F-F," }{TEXT 
-1 4 ", is" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 86 "psi_:=(Tau,Epsilon,Xi
,Z,T)->subs(tau=Tau,epsilon=Epsilon,xi=Xi,z=Z,N=TT,t=T,Formula_);" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 9 "If also  " }{XPPEDIT 18 0 "t-epsilo
n)/tau" "6#*&,&%\"tG\"\"\"%(epsilonG!\"\"F&%$tauGF(" }{TEXT -1 96 " is
 an integer number, meaning just at the moments of depositing, the pre
vious function will be:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 64 "FormulaC
:=subs(trunc((t-epsilon)/tau)=(t-epsilon)/tau,Formula_);" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 85 "Phi:=(Tau,Epsilon,Xi,Z,T)->subs(tau
=Tau,epsilon=Epsilon,xi=Xi,z=Z,N=TT,t=T,FormulaC);" }}}{SECT 0 {PARA 
20 "" 0 "" {TEXT -1 6 "Sample" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" 
}{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 62 "p2:=pl
ot(psi2(1/3,1/6,3/4,5,2-1/3,t),t=0..2.75,discont=false):" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 165 "p3:=plot(psi2(1/3,1/6,3/4,5,2-1/3,
1.4)*(1+3/4)^(t-1.4),t=1.4..2.75#,title=`the interest is paid from the
 balance on the acount, though we don't deposit any more.`\n):" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 155 "l := [[n/3+1/6,(n+1)*5] $n=
0..5]:\np1:=plot(l, x=0..2.5, style=point,symbol=circle#,title=`The po
ints illustrate the sum of the nominal values deposited`\n):" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "with(plots):\ndisplay(p1,p2,
p3);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 54 "`plotsetup/devices`
[jpeg]:=[jpeg,`plot.jpg`,[],[],``]:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 
0 "" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 43 "#plotsetup(jpeg, plotoutput=
`sporeni.jpg`);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 49 "#plotsetup(ploto
ptions=`height=1200,width=1200`);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 
17 "#eval(plotsetup):" }}}{EXCHG {PARA 12 "> " 1 "" {MPLTEXT 1 0 34 " \+
evalf(psi2(1/3,1/6,1/3,5,2,0.1));" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 202 "ll := [trunc(100*(n/3+1/6))/100 $n=0..5];\nplot(eval
f(subs(tau=1/3,epsilon=1/6,T=2-1/3,t=t,number)),t=0..4,y=0..7,xtickmar
ks=ll, title=`graph of the relation of the munber of amounts deposited
 on time`);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 49 "evalf(subs(t
au=1/3,epsilon=1/6,T=2,t=.1,number));" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 49 "evalf(subs(T=2,tau=1,t=10/4,epsilon=1/2,number));" }}
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}}{PARA 4 "" 0 "" 
{TEXT -1 0 "" }}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 55 " Saving on simpl
e interest through equidistant deposits" }}{SECT 1 {PARA 4 "" 0 "" 
{TEXT -1 15 "General formula" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
28 "#restart; \nWholePart:=floor;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 
74 "Let's assume we're depositing in the equidistant moments, with pro
portion " }{XPPEDIT 18 0 "tau" "6#%$tauG" }{TEXT -1 2 ". " }{XPPEDIT 
18 0 "epsilon" "6#%(epsilonG" }{TEXT -1 37 " is the moment of the firs
t deposit, " }{XPPEDIT 18 0 "epsilon+tau" "6#,&%(epsilonG\"\"\"%$tauGF
%" }{TEXT -1 16 " of the second, " }{XPPEDIT 18 0 "epsilon+2*tau" "6#,
&%(epsilonG\"\"\"*&\"\"#F%%$tauGF%F%" }{TEXT -1 18 " of the third and \+
" }{XPPEDIT 18 0 "epsilon+T" "6#,&%(epsilonG\"\"\"%\"TGF%" }{TEXT -1 
19 " of the last one.  " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 8 "at time " }{XPPEDIT 18 0 "t < T+epsilon;" 
"6#2%\"tG,&%\"TG\"\"\"%(epsilonGF'" }{TEXT -1 13 " we receive  " }
{MPLTEXT 1 0 40 "numberA := WholePart((t-epsilon)/tau)+1:" }{TEXT -1 
14 " and at time  " }{XPPEDIT 18 0 "t > T+epsilon;" "6#2,&%\"TG\"\"\"%
(epsilonGF&%\"tG" }{TEXT -1 13 " we receive  " }{MPLTEXT 1 0 17 "Numbe
rB:=T/tau+1:" }{TEXT -1 38 " the number of deposits, thus at time " }
{XPPEDIT 18 0 "t>0" "6#2\"\"!%\"tG" }{TEXT -1 13 " we receive  " }
{MPLTEXT 1 0 31 "number := min(numberA,numberB):" }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 56 "deposits, i-th de
posit is being deposited at the moment " }{MPLTEXT 1 0 25 "moment:=i->
i*tau+epsilon;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 60 "How long is the
 interest paid on i-th deposit till the time " }{XPPEDIT 18 0 "t" "6#%
\"tG" }{TEXT -1 8 " if the " }{MPLTEXT 1 0 31 "time:=i->t-okamzik(i): \+
time(i);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 20 "its value at moment \+
" }{XPPEDIT 18 0 "t" "6#%\"tG" }{TEXT -1 4 " is " }{MPLTEXT 1 0 37 "va
lue:=i->z*(1+xi*time(i)): value(i);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 95 "We must sum present values (original values + interest) of all \+
of the deposits till the moment " }{XPPEDIT 18 0 "t" "6#%\"tG" }{TEXT 
-1 59 ". The difference between i-th and i-th + 1st deposit is    " }}
{PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "diference:=simplify(value(i+1)-valu
e(i));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 37 "So that the series is i
ndependent on " }{XPPEDIT 18 0 "t" "6#%\"tG" }{TEXT -1 87 ", thus the \+
series is arithmetic one. The value of the 0th member of the series at
 time " }{XPPEDIT 18 0 "t" "6#%\"tG" }{TEXT -1 4 " is " }{XPPEDIT 19 
1 "value(0);" "6#-%&valueG6#\"\"!" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 
42 "and the value of the last deposit at time " }{XPPEDIT 18 0 "t" "6#
%\"tG" }{TEXT -1 4 " is " }{MPLTEXT 1 0 16 "value(number-1);" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 28 "Consequently the formula is:" }}
{PARA 0 "> " 0 "" {MPLTEXT 1 0 76 "Formula:=Sum(value(i),i=0..number-1
)\n=simplify(sum(value(i),i=0..number-1));" }}{PARA 0 "> " 0 "" 
{MPLTEXT 1 0 23 "Formula:=op(2,Formula);" }}{PARA 0 "> " 0 "" 
{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 58 "Formula
2:=simplify(1/2*(value(0)+value(number-1))*number);" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 87 "subs(min(T/tau,-trunc((-t+epsilon)/tau))=
K,\n2*simplify(xxx/sum(value(i),i=0..number)));" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 70 "simplify(subs(min(trunc((t-epsilon)/tau),T/tau
)=K, Formula-Formula2));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
382 "W:=(Tau,Epsilon,Xi,Z,TT,time)->\n[evalf(subs(tau=Tau,epsilon=Epsi
lon,xi=Xi,z=Z,T=TT,t=time,Formula)),evalf(subs(tau=Tau,epsilon=Epsilon
,xi=Xi,z=Z,T=TT,t=time,Formula2)),\nevalf(subs(tau=Tau,epsilon=Epsilon
,xi=Xi,z=Z,T=TT,t=time,value(0))),\nevalf(subs(tau=Tau,epsilon=Epsilon
,xi=Xi,z=Z,T=TT,t=time,value(number-1))),\nevalf(subs(tau=Tau,epsilon=
Epsilon,xi=Xi,z=Z,T=TT,t=time,number))\n];" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 17 "W(1,0,1,5,10,1);\n" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 196 "psi:=(Tau,Epsilon,Xi,Z,TT,time)->subs(tau=Tau,epsilo
n=Epsilon,xi=Xi,z=Z,T=TT,t=time,op(2,Formula));\npsi2:=(Tau,Epsilon,Xi
,Z,TT,time)->subs(tau=Tau,epsilon=Epsilon,xi=Xi,z=Z,T=TT,t=time,Formul
a2);\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 50 "evalf(psi(1,0,1,5
,2,1));\nevalf(psi2(1,0,1,5,2,1));" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}}
{SECT 1 {PARA 0 "" 0 "" {TEXT -1 268 "At the beginning of the year we \+
begin to deposit on an account. At the beginning of each month we depo
sit 114,7 Oobols. At the end of the year we have to pay the tax 1/15 o
f the interest rate. The interest rate itself is 1/50 p. m. How much d
o you save up in 11 years?" }}{SECT 1 {PARA 5 "" 0 "" {TEXT -1 8 "Solu
tion" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 40 "restart;\nz:=114.7;x
i:=1/50;delta:=1/15;\n" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 12 "In one \+
year:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 58 "xxx := 12.*z+(sum(z*(1+xi)
^t,t = 1 .. 12)-12*z)*(1-delta);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 
12 "In 11 years:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "\nsum(xxx*(1+xi
*(1-delta))^(12*t),t=0..10);" }}}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}
{SECT 1 {PARA 3 "" 0 "" {TEXT -1 23 "Internal rate of return" }}{SECT 
1 {PARA 0 "" 0 "" {TEXT -1 363 "Dan Has bought a car via financial lea
sing. He should pay remaining 130 000 CZK in 36 monthy part payments o
f the value of 4592.5 CZK. By the peripeties of life he was forced to \+
repay the part payments from his savings. What interest rate must the \+
savings be charged with to be able to repay the debt preferable to pay
 immediately the whole amount of 130 000 CZK?" }}{SECT 1 {PARA 5 "" 0 
"" {TEXT -1 8 "Solution" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "re
start;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 40 "Resolving the equation \+
36 with variable " }{XPPEDIT 18 0 "xi" "6#%#xiG" }{TEXT -1 2 ". " }
{XPPEDIT 18 0 "xi" "6#%#xiG" }{TEXT -1 45 " is interest rate per month
 in this case.    " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 51 "rce:=
sum(4592.5*(1+xi)^i,i=0..35)=130000*(1+xi)^36;" }}}{EXCHG {PARA 0 "" 
0 "" {TEXT -1 78 "This equation must be solved numerically. Maple disp
oses of built-in function " }{HYPERLNK 17 "fsolve:" 2 "fsolve" "" }
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "> " 0 "" 
{MPLTEXT 1 0 20 "xxx:=fsolve(rce,xi);" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 174 "We're interested just in the non-negative solution out o
f two solutions of the equation. The interest rate per month must be r
ecalculated into the interest rate per a year.  " }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 109 "`It is worth repaying the depth from savings
 if the interest rate per year is bigger than:`, (1+xxx[2])^12-1;" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 80 "This equation can also be rewritte
n with help of the sries-summing formula into:" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 47 "rce2:=4592.5*(1-(1+xi)^36)/(1-(1+xi))=rhs(rce)
;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 82 "and be solved directly. Havi
ng to analyse the equation and find out equation root:" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "rce3:=factor((lhs(rce2)-rhs(rce2)))
=0;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 67 "The only one positive equa
tion root is the interest rate per month:" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 19 "solve(\{rce3,xi>0\});" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 57 "Surprisingly the same one as from the previous procedure.
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 
0 "" {TEXT -1 108 "The interest rate is the internal rate of return of
 the cash flow, -130 000, 4592.5, 4592.5, . . . , 4592.5." }}}}{PARA 
0 "" 0 "" {TEXT -1 127 "Let's assume that Dan would rather like to pay
 the last 13 part payments 'en block' altogether. He would like to pay
 less than " }{XPPEDIT 18 0 "13*4592.5" "6#*&\"#8\"\"\"-%&FloatG6$\"&D
f%!\"\"F%" }{TEXT -1 451 " CZK indeed, 'cause he repays the money imme
diately en block. Consider that the demanded for the financial leasing
 services higher than supply, thus the leasing company can immediately
 re-invest the money on the same conditions (meaning the same interest
 rate), then the question can also be formulated: what is the marginal
 value of the part payment (1 instead of 13) to keep the profit of the
 leasing company? (while naglecting the transaction cost)" }}{SECT 1 
{PARA 5 "" 0 "" {TEXT -1 8 "Solution" }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 9 "restart; " }{TEXT -1 162 "Also in this case we have to
 compute the internal rate of return, which means the interest rate on
 which the company experiences the profit out of money invested." }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "rce:=sum(45925/10*(1+xi)^(-i
),i=1..36)=130000:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "Xi:=f
solve(rce,xi=0.013..0.014); " }{TEXT -1 76 "We have to discont the wal
ue of 13 part payments with this rate of interest." }}{PARA 0 "> " 0 "
" {MPLTEXT 1 0 75 "xi := Xi; `the reasonable price is`, sum(45925/10*(
1+xi)^(-i),i = 0 .. 13);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}
{PARA 8 "" 1 "" {TEXT -1 65 "Error, (in fsolve) rce is in the equation
, and is not solved for\n" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#xiG%#X
iG" }}{PARA 12 "" 1 "" {XPPMATH 20 "6$%4spravedliv|\\y~cena~jeG,>#\"%&
=*\"\"#\"\"\"*(F&F(F'!\"\",&F(F(%#XiGF(F*F(*(F&F(F'F*F+!\"#F(*(F&F(F'F
*F+!\"$F(*(F&F(F'F*F+!\"%F(*(F&F(F'F*F+!\"&F(*(F&F(F'F*F+!\"'F(*(F&F(F
'F*F+!\"(F(*(F&F(F'F*F+!\")F(*(F&F(F'F*F+!\"*F(*(F&F(F'F*F+!#5F(*(F&F(
F'F*F+!#6F(*(F&F(F'F*F+!#7F(*(F&F(F'F*F+!#8F(" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 0 "" }}}}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 9 "Exactn
ess" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 22 "Assuming the cash flow" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "CF:=[-1000,3600,-4310,1716];
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "PV:=sum(CF[i]*(1+xi)^(-
i),i=1..4);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "solve(PV);" 
}}{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 0 "" }}}}{PARA 3 "" 0 "" {TEXT -1 0 "" }}}{SECT 1 {PARA 
3 "" 0 "" {TEXT -1 37 "Speculation demand for money (sample)" }}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 539 "According to John Maynard Keynes \+
one of the incentive for the demand for money is the speculation motiv
e. Keynes asks the question why people demand higher amount of money t
han the amount demanded because of transactional and cautiousness moti
ve. He induced, that the money is held because of uncertainty about th
e changes in the rate of interest in the future and  in the consequenc
e of the relations between the interest rate and the bond prices. The \+
consequence of the grow of the interest rate is the loss experienced b
y holding bonds." }}}{SECT 1 {PARA 0 "" 0 "" {TEXT -1 147 "Let's assum
e a bond which will never be repaied. Meaning that it's principal will
 never be repaied whereas it brings fixed coupon revenue per year." }}
{PARA 0 "" 0 "" {TEXT -1 193 "Supposing that the bond asset is bought \+
for the price of 1000 CZK, while the annual fixed coupon revenue is 10
0 CZK. What is the real value of the bond if the interest rate falls t
o it's half? " }}{SECT 1 {PARA 5 "" 0 "" {TEXT -1 8 "Solution" }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "restart; " }}{PARA 0 "" 0 "" 
{TEXT -1 243 "Holding this kind of bond is similar to receiving a ethe
rnal (everlasting) rent of 100 CZK p.a. out of 1000 CZK locked. The in
terest rate corresponding to the annual interest of 100 CZK out of 100
0 CZK at the moment when the bond was bought is" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 26 "xi[0]:=solve(1000*xi=100);" }}{PARA 11 "" 1 "
" {XPPMATH 20 "6#>&%#xiG6#\"\"!#\"\"\"\"#5" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 40 "yet the interest rate nowadays went half" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "xi[1]:=xi[0]/2;" }}{PARA 11 "" 1 "
" {XPPMATH 20 "6#>&%#xiG6#\"\"\"#F'\"#?" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 38 "consequently the value of the bond is " }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 22 "Z:=solve(z*xi[1]=100);" }}{PARA 11 "" 1 "
" {XPPMATH 20 "6#>%\"ZG\"%+?" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 28 "T
hus the profit is 1000 CZK." }}}}}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 
23 "Two examples - revision" }}{SECT 1 {PARA 0 "" 0 "" {TEXT -1 60 "wh
at is the interest rate on the 3rd month in the period if:" }}{PARA 
15 "" 0 "" {TEXT -1 104 "the annual interest rate is 0.05, and the int
erest rate per month is the same on all months in the year?" }}{PARA 
15 "" 0 "" {TEXT -1 134 "the interest rate during the first month is 0
.01, on the second month it's 0.02 and on the first quarter the rate o
f interest is 0.06?" }}{PARA 15 "" 0 "" {TEXT -1 307 "the annual inter
est rate is 0.1 and  \non the 4th quarter the rate of interest equals \+
to 0.02 ,\non the 4th quarter the rate of interest is 0.03,\non the 4t
h quarter the rate of interest is 0.02 again,\nand on the 1st month th
e interest rate is 0.003,\nfinally on the 2nd month the interest rate \+
equals to 0.004?" }}}{SECT 1 {PARA 0 "" 0 "" {TEXT -1 138 "How long wi
ll it take to repay the debt of 10 000 CZK by monthly afterdate (reali
zed at the end of each month) part payments of 1000 CZK. " }}{PARA 15 
"" 0 "" {TEXT -1 32 "if the interest rate is 0 p. m.," }}{PARA 15 "" 
0 "" {TEXT -1 130 "if the interest rate is 0.05 p. m; of what high the
 monthly part payments must have been to shorten the repay period by o
ne month?" }}{PARA 15 "" 0 "" {TEXT -1 33 "if the interest rate is 0.1
 p. m." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 75 "pom:=Sum(1000*(1+x
i)^(t-i),i=1..t)=simplify(sum(1000*(1+xi)^(t-i),i=1..t));" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "pom:=op(2,pom)=10000*(1+xi)^t;" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "solve(\{pom,xi=.1\});" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "simplify(sum(a*q^i,i=1..n));
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "10*(1-1.1)^(20)/0.1;" }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{PARA 3 "" 0 "" 
{TEXT -1 0 "" }}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 44 "Deposits denomin
ated in different currenceis" }}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 25 "T
he course-change problem" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 178 "The c
ourse of  jen-min-piao (juan) is 36:37 to rupee. The expected course a
t following period is 34:39. Which of these currenceis is preferable t
o keep? Quantify the destinction." }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 34 "restart;\nE[0]:=36/37;\nE[1]:=34/39;" }}}{EXCHG 
{PARA 0 "" 0 "" {XPPEDIT 18 0 "E[t]" "6#&%\"EG6#%\"tG" }{TEXT -1 112 "
 is the value of rupee denominated in juan at moment t. The value of j
uan denominated in rupee is reciprocal of " }{XPPEDIT 18 0 "E[t]" "6#&
%\"EG6#%\"tG" }}}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "if E[1] < E[0
] then `The rupee course falls` else `The rupee course rises` end if;
" "6#@%2&%\"EG6#\"\"\"&F&6#\"\"!%7The~rupee~course~fallsG%7The~rupee~c
ourse~risesG" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 155 "Profit measuring
: You will neither gain nor lose anything if we keep rupee. If we keep
 juan instead of rupee, your relative profit denominated in rupee is:
" }}}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "x*E[0]/E[1]*`rupi\355`;" 
"6#**%\"xG\"\"\"&%\"EG6#\"\"!F%&F'6#F%!\"\"%&rupi|hyGF%" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 28 "x rupee invested gives back " }{XPPEDIT 
18 0 "x*E[0]/E[1]" "6#*(%\"xG\"\"\"&%\"EG6#\"\"!F%&F'6#F%!\"\"" }
{TEXT -1 30 " rupee. So that the profit is:" }}}{EXCHG {PARA 0 "> " 0 
"" {XPPEDIT 19 1 "E[0]/E[1]-1;" "6#,&*&&%\"EG6#\"\"!\"\"\"&F&6#F)!\"\"
F)F)F," }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 123 "If we calculate the pr
ofit denominated in juan, we will neither gain nor lose anything, but \+
keeping rupee would mean loss. " }}{PARA 0 "" 0 "" {TEXT -1 74 "if we \+
invest x juan into rupee, we'll have (in the following period) just:" 
}}}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "x*E[0]^(-1)*E[1]*juan;" "6#
**%\"xG\"\"\")&%\"EG6#\"\"!,$F%!\"\"F%&F(6#F%F%%%juanGF%" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 21 "and the lose would be" }}}{EXCHG {PARA 0 
"> " 0 "" {MPLTEXT 1 0 17 "E[0]^(-1)*E[1]-1;" }}}{EXCHG {PARA 0 "" 0 "
" {TEXT -1 280 "How is it possible that we didn't get a number with th
e same absolute value? That's because we have chosen different (grater
 in this case) unit. (Lose one juan is grater loss than lose one rupee
.) Therefore choosing one fixed currency for this kind of computation \+
seems suitable. " }}}{SECT 0 {PARA 0 "" 0 "" {TEXT -1 147 "The courses
 of currencies (1) jen-min-piao (2) ngultrum (Bhutan) a (3) kyata (Bar
ma) is to the Danish crown (DKK) at time, t=1,2,3,4,5 is equal to " }
{XPPEDIT 18 0 "kappa(i,t)=9/i+abs(sin(i*t*Pi/6))" "6#/-%&kappaG6$%\"iG
%\"tG,&*&\"\"*\"\"\"F'!\"\"F,-%$absG6#-%$sinG6#**F'F,F(F,%#PiGF,\"\"'F
-F," }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "
" {TEXT -1 196 "Supposing that we have 100 DKK disposable and we know \+
all the courses in advance. We can also change the money without trans
action cost in actual courses at time i=1,2,3,4,5. How much we can mak
e?" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "restart;\nkappa:=(i,t)
->9/i+abs(sin(i*t*Pi/6));" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 78 "plot(
\{kappa(1,t),kappa(2,t),kappa(3,t)\},t=0..6, color=[navy,blue,aquamari
ne]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "kappa(1,1);" }}}
{SECT 0 {PARA 5 "" 0 "" {TEXT -1 8 "Solution" }}{EXCHG {PARA 0 "> " 0 
"" {MPLTEXT 1 0 137 "NumberOfTitlValueCapital[1]:=100;\nRateOfProfit:=
(i,t)->kappa(i,t+1)/kappa(i,t);\nNumber:=t->Sum(kappa(i)*Number[iCapit
alNumberOfTitles);\n\n" }}{PARA 8 "" 1 "" {TEXT -1 22 "Error, `)` unex
pected\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 52 "Max:=t->max(Rat
eOfProfit(i,t) $i=1..NumberOfTitles);" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 277 "Vyber:=if(t)\nglobal NumberOfTitle;\ni:='i';\n#print
(evalf(RateOfProfit(i,t)) $i=1..NumberOfTitles);\nfor i from 1 to Numb
erOfTitles do\nif Max(t)=RateOfProfit(i,t)\nthen NumberOfTitle:=i;\n#p
rint(`The biggest profit makes the currency`,i,`a to`,evalf(RateOfProf
it(i,t)));\nfi \nod\nend;" }}{PARA 8 "" 1 "" {TEXT -1 37 "Error, reser
ved word `if` unexpected\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
658 "for t from  1 to 6 do \ni:='i';\nVyber(t); MAX:=evalf(Max(t));#pr
int(MAX);\nif MAX>1 then\n  for i from 1 to NumberOfTitles do\n   Numb
er[i]:=0;\n  od:\nNumber[NumberOfTitle]:=Capital[t]/kappa(NumberOfTitl
e,t);\nCapital[t+1]:=Number[NumberOfTitle]*kappa(NumberOfTitle,t+1);\n
print(`At time `,t, ` I buy `, evalf(Number[NumberOfTitle]),\n ` units
 of `,   NumberOfTitle, `at time`, t+1, `I will have`, evalf(Capital[t
+1]),`of the original currency`);\nelse\nCapital[t+1]:=Capital[t];\npr
int(MAX,evalf(RateOfProfit(i,t)) $i=1..NumberOfTitles,`At time`,t,`I w
ill keep the original currency, at time`, t+1, `I will have`, evalf(Ca
pital[t+1]),`of the original currency`)\nfi\nod:" }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 0 "" }}}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
0 "" }}}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 88 "The Condition of uncover
ed interest parity condition, International interest arbitration" }}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 110 "Assuming two currencies CZK and U
SD for instance. Their real courses at time 0 and expected course at t
ime 1. " }{XPPEDIT 18 0 "E(0)" "6#-%\"EG6#\"\"!" }{TEXT -1 5 " and " }
{XPPEDIT 18 0 "E(1)" "6#-%\"EG6#\"\"\"" }{TEXT -1 84 " (meaning E(i) a
s the price of USD denominated in CZK at time i) and interest rates " 
}{XPPEDIT 18 0 "xi[CZK]" "6#&%#xiG6#%$CZKG" }{TEXT -1 5 " and " }
{XPPEDIT 18 0 "xi[USD]" "6#&%#xiG6#%$USDG" }{TEXT -1 4 ".   " }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 84 "If we invest CZK at time 0 into US
D deposits, the profit at time 1 will be equal to:" }}}{EXCHG {PARA 0 
"> " 0 "" {MPLTEXT 1 0 8 "restart;" }}}{EXCHG {PARA 0 "> " 0 "" 
{XPPEDIT 19 1 "Profit[US] := x*E(0)^(-1)*(1+xi[USD])*E(1)-x;" "6#>&%'P
rofitG6#%#USG,&**%\"xG\"\"\")-%\"EG6#\"\"!,$F+!\"\"F+,&F+F+&%#xiG6#%$U
SDGF+F+-F.6#F+F+F+F*F2" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 30 "What is the rate of interest (" }
{XPPEDIT 18 0 "xi[CZK]" "6#&%#xiG6#%$CZKG" }{TEXT -1 67 ") by which th
e profit will be the same as the profit of CZK chosens" }{XPPEDIT 18 
0 "%?;" "6#%#%?G" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "Profit[
CZ]:=x*xi[CZK];" }}}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "rce := Pro
fit[US] = Profit[CZ];" "6#>%$rceG/&%'ProfitG6#%#USG&F'6#%#CZG" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "xxx:=solve(rce,xi[CZK]);" }}
{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 75 "Modify. Let's define Q as expected rate of depreciation of CZK \+
against USD." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 56 "Q:=(E(1)-E(
0))/E(0);\nxxx2:=sort(expand(xxx/Q),xi[USD]);\n" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 129 "EquilibriumCondition:=xi[CZK]=sort(\n(simplif
y((op(1,xxx2)-xi[USD])*Q)+xi[USD]*Q)+\nsimplify(op(2,xxx2)*Q+op(3,xxx2
)*Q)\n,xi[USD]);\n" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 
"" 0 "" {TEXT -1 14 "The expression" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 
33 "op(2,op(2,EquilibriumCondition));" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 35 "is a neglectable one (Weather both " }{XPPEDIT 18 0 "xi[U
SD]" "6#&%#xiG6#%$USDG" }{TEXT -1 187 " and the rate of depreciation a
re low). Let's sum up: in the equilibrium the difference between the i
nterest rates is equal to the the interest payed on the expected rate \+
of depreciation." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 143 "lhs(Eq
uilibriumCondition)=op(1,rhs(EquilibriumCondition))+\nfactor(\nsimplif
y(op(2,rhs(EquilibriumCondition))+op(3,rhs(EquilibriumCondition)))\n);
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 0 "" }}}}}}{MARK "8" 0 }{VIEWOPTS 1 1 0 1 1 1803 
0 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }