1088:
interest per annum and I add $ 1 every year, what is the amount of money I get in 20 years?". If you can do this one then you can do the former one by multiplying by 300. A still simpler example is: "If I were to invest $ 1 in a bank account with 0% compound interest per annum and I add $ 1 every year, what is the amount of money I get in 0 years?" It seems that the result is $ 1, but there is no consensus on this one.
1087:
Always begin with the simplest examples. The problem posed is: "If I were to invest $ 300 in a bank account with 3% compound interest per annum and I add $ 300 every year, what is the amount of money I get in 20 years?". A simpler example is "If I were to invest $ 1 in a bank account with 3% compound
1009:
By the way, 211's "iterations" are nothing more than a very obfuscated way of calculating the logarithm of 1.03, and the "differential equation" is an attempt to reinvent the solution to the continuous problem (where the original problem is supposedly discrete). This is what I would call "stopping in
988:
majority of banks would quote x% per annum as, if you invest £100 now, then in 1 year's time there will be £10x in your account, so I would have thought the problem would be as we originally had it. But I agree that "16 iterations" and differential equations looks to be overcomplicating the problem.
1060:
Assuming that there are 20 payments at yearly intervals and the final one earns one year's interest, i.e. the expression to be evaluated is 300 x 1.03 + 300 x 1.03 + ..... + 300 x 1.03, as given in an earlier post, then the answer will come from summing a GP with first term 300 X 1.03 and common
187:
I actually did this manually. By multiplying the compound interest with the initial value, I worked my way through each year until I get year 20. But I found it extremely time-consuming so I was hoping that there was a simpler method to obtain a similar answer. I also do not understand what the
936:
I have no idea what you have tried to do here - I don't think our purpose is to confuse the OP. You did, however, remind me of another subtle issue: All of our responses are based on the assumption that the annual interest is actually 3%. If the interest is what some would call "3% per year,
1071:
Please do not insult recurrence relations. In this case, they are easier to understand intuitively and just as easy to solve mathematically. Also, please do not give end results for a question which might be homework. Providing a complete solution is easy; hinting at a solution is hard. --
108:
If I were to invest $ 300 in a bank account with 3% compound interest per annum and I add $ 300 every year, what is the amount of money I get in 20 years? I have used the arithmetic and geometric progression, but I found it lengthy. Is there a shorter method? Thanks.
227:
That would depend on a more precise specification of the problem. I find it more plausible that the first payment is at the beginning of 1st year, 20th payment at the beginning of 20th year, and at the end of 20th year, instead of an additional deposit, we make a
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Those commentaries were mine(should have said so), for x^x, I guess the simple answer would be it has real values, but it does not have real values for all real numbers. Unfortunately, I can't find the particular archive that goes into more
453:
Funny that you should ask this question because I have written some notes in my notebook two months ago about how to turn the problem you just described into a differential equation which can then be solved just once (to get the solution).
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I was fitting a second order 2D function to the peak sub-region, but references indicate that the correct function is a sinc. My problem is that while I know how to formulate the A matrix for the 2D 'bowl' in c = A\y:
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231:
As for the solution... See if you can follow
Richard's suggestion. If you are familiar with sequences and recursion, you can also try Leland's and my suggestion - assuming the problem is as I stated, you can denote by
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Well, if you treat each individual payment separately, the first one has 20 years' worth of interest, the 2nd payment has 19 year's worth of interest, 3rd one 18 years, etc. 20th payment just 1 year of
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I can't figure out how to formulate it for a sinc function. Is there perhaps a better way to do this via a transform in the
Fourier domain? Not having done Fourier undergrad, I'm in deep water here.
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There are many methods to solve this. Can you show us what you have tried to do? See if you can find a recurrence relation between the amount of money between successive years and solve it. --
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173:+ 300) and then solve the recurrence relation for a closed form expression. Explanations and examples of solving such recurrence relations (in this case an inhomogeneous one) are give on the
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The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the
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ratio 1.03, giving $ 8302.95, I believe. Surely the OP will prefer this to a lot of flannel about recurrence relations and the continuous case.…
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The trick of course is to convert discrete_rate into inst_rate and to convert discrete_deposit into inst_deposit
590:{\displaystyle Money=\left(300+{\frac {295.588}{0.0295588}}\right)e^{0.0295588\,t}-{\frac {295.588}{0.0295588}}}
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that doesn't exclude the negative reals, then you can get real values for select negative rational numbers
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After 20 years you have made 21 payments, the first having 20 years of interest, the last having zero.
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150:(After Edit Conflict) A relatively easy approach is to express annual the interest and deposit as a
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without relying on previous terms. There are several ways to do this - my favorite is letting
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I'm doing sub-pixel image registration via the cross-correlation method. Matlab code:
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1965:{\displaystyle \{a/b\in \mathbb {Q} ^{-}:a,b\in \mathbb {Z} ,b{\mbox{ is odd}}\}}
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Your derivative is correct, but you can't really make a meaningful definition of
206:, and see if there's anything on that page which will solve the above expression.
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1290:{\displaystyle {\frac {d}{dx}}(x^{x})=e^{x\ln x}(1+\ln x)=x^{x}(1+\ln x)}
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M = (principle + inst_deposit/inst_rate) × Exp - (inst_deposit/inst_rate)
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Factor out the 300 and you get 300(1.03 + 1.03 + 1.03 + 1.03 + ... +1.03)
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is real valued. There was some extensive discussion about the function
196:
You can write this out as; 300 x 1.03 + 300 x 1.03 + ..... + 300 x 1.03
1599:. It works for negative integers, sure, but what do you do with, say,
1138:% Take the inverse 2D fft of the multiplied matrix and fftshift
1868:(after edit conflict) Well, if you choose a branch of ln over
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not, when they are effectively re-arrangements of each other?
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There's a nice explanation of how to solve this problem here
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After the calculations, the instatanous rate and deposit are
1566:{\displaystyle f(x):\mathbb {R} \to \mathbb {R} ,f(x)=x^{x}}
1132:% Multiply the first by the complex conjugate of the second
79:
Welcome to the
Knowledge Mathematics Reference Desk Archives
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unknown Ax represents in the recurrence relation. Thanks.
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im1 = (im1 - mean(im1(:))); im2 = (im2 - mean(im2(:)));
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I find that 16 iterations is enough for most purposes.
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th year, before an additional deposit. Then you have
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1141:peak = real(ifft2(mult)); peak = fftshift(peak);
913:Of course this is an converging approximation,
431:{\displaystyle b_{n}={\frac {a_{n}}{1.03^{n}}}}
1477:{\displaystyle y=e^{x\ln x}(1+\ln x),x<0\,}
357:. Solving this means finding a way to express
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1851:ln(x) not defined in the reals for x<0--
1010:Japan on the way from France to Spain". --
1129:im1_fft = fft2(im1); im2_fft = fft2(im2);
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350:{\displaystyle a_{n+1}=1.03(a_{n}+300)}
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1170:Am I correct in workings to say that
1113:Fitting the 2D sinc funtion to 2D data
204:Geometric progression#geometric series
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1829:{\displaystyle y=e^{x\ln x},x<0\,}
1399:{\displaystyle y=e^{x\ln x},x<0\,}
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1972:. The discussion I referenced is at
104:Investment and interest calculations
959:{\displaystyle \mathrm {e} ^{0.03}}
1678:{\displaystyle y=(x^{x}),x<0\,}
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917:inst_rate = 2^ bignumber * (a - 1)
259:the amount of money at the end of
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1135:mult = im1_fft .* conj(im2_fft);
1773:{\displaystyle y=x^{x},x<0\,}
1342:{\displaystyle y=x^{x},x<0\,}
1126:% Take the 2D fft of each image
1780:have real negative values and
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1625:{\displaystyle -\pi ^{-\pi }}
880:{\displaystyle Money=1940.52}
832:{\displaystyle Money=1592.74}
784:{\displaystyle Money=1255.09}
438:and proceeding naturally. --
33:
1883:{\displaystyle \mathbb {C} }
736:{\displaystyle Money=927.27}
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1980:17:21, 20 July 2007 (UTC)
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688:{\displaystyle Money=609}
640:{\displaystyle Money=300}
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289:{\displaystyle a_{1}=309}
221:22:16, 21 July 2007 (UTC)
211:14:13, 21 July 2007 (UTC)
182:14:53, 20 July 2007 (UTC)
144:14:48, 20 July 2007 (UTC)
906:inst_rate = 2^16 * (a-1)
18:Knowledge:Reference desk
1040:should indeed be £(100+
909:inst_deposit = 2^16 * b
496:So the formula becomes
492:inst_deposit is 295.588
478:discrete_deposit is 300
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128:) 07:38, 20 July 2007
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456:211.28.131.44
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369:
365:
341:
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27:
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19:
1977:Donald Hosek
1721:
1714:Donald Hosek
1487:
1302:
1299:
1172:
1169:
1156:
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1137:
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1128:
1125:
1122:
1119:
1116:
1104:Donald Hosek
1086:
1059:
1041:
1037:
1003:
999:
985:
984:I think the
926:
923:
912:
889:
495:
484:
470:
467:
452:
260:
215:
202:Now look at
186:
169:
165:
159:
155:
118:124.82.78.95
107:
94:
78:
1488:thank you.
903:b = b/(a+1)
228:withdrawal.
112:—Preceding
26:Mathematics
1733:Why does
1090:Bo Jacoby
1046:Richard B
1002:be £(100+
991:Richard B
900:a = Sqrt]
583:0.0295588
565:0.0295588
550:0.0295588
218:Bo Jacoby
208:Richard B
193:interest.
177:page. --
50:<<
1853:Cronholm
1725:Cronholm
1723:depth.--
126:contribs
114:unsigned
24: |
22:Archives
20: |
875:1940.52
827:1592.74
779:1255.09
580:295.588
547:295.588
154:(as in
99:July 20
89:pages.
67:July 21
46:July 19
1955:is odd
1148:A = ;
731:927.27
164:= 1.03
1839:Philc
1491:Philc
69:: -->
63:: -->
62:: -->
44:<
16:<
1819:<
1763:<
1668:<
1584:<
1573:for
1467:<
1389:<
1349:yes
1332:<
1297:yes
1078:talk
1016:talk
986:vast
972:talk
952:0.03
444:talk
418:1.03
323:1.03
296:and
140:talk
122:talk
56:July
1484:no
1406:no
683:609
635:300
539:300
342:300
284:309
60:Aug
52:Jun
1939:∈
1922:−
1912:∈
1805:
1802:ln
1618:π
1615:−
1611:π
1607:−
1528:→
1452:
1449:ln
1432:
1429:ln
1375:
1372:ln
1279:
1276:ln
1245:
1242:ln
1225:
1222:ln
1102:.
1095:.
1080:)
1044:)
1018:)
974:)
575:−
446:)
223:.
162:+1
142:)
124:•
58:|
54:|
1960:}
1950:b
1947:,
1943:Z
1936:b
1933:,
1930:a
1927::
1917:Q
1909:b
1905:/
1901:a
1898:{
1877:C
1822:0
1816:x
1813:,
1808:x
1799:x
1795:e
1791:=
1788:y
1766:0
1760:x
1757:,
1752:x
1748:x
1744:=
1741:y
1698:x
1694:x
1671:0
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1659:)
1654:x
1650:x
1646:(
1643:=
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1551:=
1548:)
1545:x
1542:(
1539:f
1536:,
1532:R
1524:R
1520::
1517:)
1514:x
1511:(
1508:f
1470:0
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1461:,
1458:)
1455:x
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1443:1
1440:(
1435:x
1426:x
1422:e
1418:=
1415:y
1392:0
1386:x
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1378:x
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1365:e
1361:=
1358:y
1335:0
1329:x
1326:,
1321:x
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1313:=
1310:y
1285:)
1282:x
1273:+
1270:1
1267:(
1262:x
1258:x
1254:=
1251:)
1248:x
1239:+
1236:1
1233:(
1228:x
1219:x
1215:e
1211:=
1208:)
1203:x
1199:x
1195:(
1189:x
1186:d
1182:d
1076:(
1042:x
1038:x
1014:(
1004:x
1000:x
970:(
947:e
872:=
869:]
866:5
863:[
860:y
857:e
854:n
851:o
848:M
824:=
821:]
818:4
815:[
812:y
809:e
806:n
803:o
800:M
776:=
773:]
770:3
767:[
764:y
761:e
758:n
755:o
752:M
728:=
725:]
722:2
719:[
716:y
713:e
710:n
707:o
704:M
680:=
677:]
674:1
671:[
668:y
665:e
662:n
659:o
656:M
632:=
629:]
626:0
623:[
620:y
617:e
614:n
611:o
608:M
570:t
561:e
556:)
542:+
535:(
531:=
528:]
525:t
522:[
519:y
516:e
513:n
510:o
507:M
442:(
422:n
412:n
408:a
402:=
397:n
393:b
370:n
366:a
345:)
339:+
334:n
330:a
326:(
320:=
315:1
312:+
309:n
305:a
281:=
276:1
272:a
261:n
245:n
241:a
170:n
166:a
160:n
156:a
138:(
120:(
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