sparrow2
  • sparrow2
need help
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
sparrow2
  • sparrow2
question is about compound interest. \[a=p(1+r/m)mt and a=p(1+r)t\]
sparrow2
  • sparrow2
symbols are ordinary, why this two result arn't the same?
sparrow2
  • sparrow2
a=p(1+r/m)^mt a=p(1+r)^t

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sparrow2
  • sparrow2
@Michele_Laino
0_0youscareme
  • 0_0youscareme
@.Gjallarhorn.
Michele_Laino
  • Michele_Laino
I'm sorry, I don't know your answer since I'm not good with financial mathematics
sparrow2
  • sparrow2
@Abhisar
sparrow2
  • sparrow2
@amistre64
sparrow2
  • sparrow2
@empty
anonymous
  • anonymous
\[a=p \left( 1+r \right)^t\] here interest is compounded annually \[a=p \left( 1+\frac{ r }{ m } \right)^{mt}\] here interest is compounded monthly. or we can say interest of first month is also principal for next month and so on.
sparrow2
  • sparrow2
i know that ,but they must be equal, i think so
anonymous
  • anonymous
no, for first year principal is same during whole year. but in second case principal is different for every month.
sparrow2
  • sparrow2
so if i have r as anual rate, and they tell me to fing future value of smt,but intereset must be monthly ,what i will do?
sparrow2
  • sparrow2
@Palmo4ka esli mojesh pomogi :)
anonymous
  • anonymous
let us take the case principal=100 rate=12% annually then amount after first month \[a=100\left( 1+\frac{ .12 }{ 12 } \right)^1=101\] amount after second month \[a=101\left( 1+\frac{ .12 }{ 12 } \right)^1=101*1.01=102.01\] .......... this becomes principal for third month and so on. we see principal increases every month. but in the first case principal is same whole year.
sparrow2
  • sparrow2
i understand that man, but why did you diveded anual rate by 12,is it ok?
anonymous
  • anonymous
@sparrow2 Do you speak Russian?
anonymous
  • anonymous
no
sparrow2
  • sparrow2
kaneshna gavariu, no po angliski luchshe :)
anonymous
  • anonymous
а я, наоборот, говорю лучше по-русски:) @sparrow2
anonymous
  • anonymous
rate is annual ,but we are calculating every month so we divide by 12 if we have to calculate after every 6 months,we divide by 2 if we have to calculate every 3 months ,we divide by 4
sparrow2
  • sparrow2
so the answers will be different when calculating anualy and monthly?
anonymous
  • anonymous
yes
sparrow2
  • sparrow2
and dividing by periods is ok? so do they mean so when asking for periods(we know only annualy)
sparrow2
  • sparrow2
@Palmo4ka potomushto ti iz rossi( no ia net)
anonymous
  • anonymous
i am sorry i don't understand your language fully,i am an indian.
sparrow2
  • sparrow2
i'm also not native speaker :D
sparrow2
  • sparrow2
so you divided it by 12 or by 6 and so on(anual rate) is it ok?
sparrow2
  • sparrow2
so if i have anual rate like 12%, if they askme me to calcualte anualy i use 12%,but by monthly i will use 1% and the answers will be different
anonymous
  • anonymous
correct
anonymous
  • anonymous
but at the sametime multiply t by 12
sparrow2
  • sparrow2
i thought that this was just 2 different ways,but the answers must be the same, i thougt so
anonymous
  • anonymous
no ,as you see principal goes on changing for compound interest.
sparrow2
  • sparrow2
on earth :)
sparrow2
  • sparrow2
okay thanks man :)
anonymous
  • anonymous
yw
ganeshie8
  • ganeshie8
It might be enlightening to see what happens if you calculate the interest every day, every hour, every second..
sparrow2
  • sparrow2
@ganeshie it's true. you see, if you increase periods the amount is increasing too(it's more beneficila for banks to use more periods :D )
sparrow2
  • sparrow2
pe^rt will be if its continous
ganeshie8
  • ganeshie8
Haha these days most banks calculate interest continuously(every nano second or so) The interest wont increase much by increasing the periods, it saturates after 12 periods or so
ganeshie8
  • ganeshie8
Notice that we get that continuous formula by letting \(m\to \infty\) in the discrete version : \[\large \lim\limits_{m\to\infty}~p\left(1+\frac{r}{m}\right)^{mt} = pe^{rt} \]
sparrow2
  • sparrow2
yeah i see, it easy to prove :)
ganeshie8
  • ganeshie8
Letting \(p=r=t=1\), we get the definition of euler constant \(e\) : \[\large \lim\limits_{m\to\infty}~\left(1+\frac{1}{m}\right)^{m} = e\]
sparrow2
  • sparrow2
yeah that is cool way to define e :) creative
sparrow2
  • sparrow2
okay thanks @ganeshie8
ganeshie8
  • ganeshie8
that is indeed one of the most useful definitions of \(e\) https://en.wikipedia.org/wiki/E_(mathematical_constant)#History
sparrow2
  • sparrow2
so when you have like 12% anualy can you always say that monthly will be like 1 % @ganeshie8
sparrow2
  • sparrow2
oh i closed the question

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