a regular saving of $500 is made into a sinking fund at the start of each year for 10 years. Determine the value of the fund at the end of the tenth year on the assumption that the rate of interest is A) 11% compounded annually b) 10% compounded continously
How far have you gotten? Does it matter that it's a "Sinking Fund" or is any old accumulation sufficient?
the chapter is geometric series
Okay, how do we find a geometric series?
a(r^n-2)/(r-1) right?
If you built it, you will understand it. Let's try 11% annual i = 0.11 r = 1+i = 111 t = 10: 500 t = 9: 500*r t = 8: 500*r^2 ... t = 1: 500*r^9 Add those all up and see what you get.
for annual 500(1+11/100)^10=1419.7
so i do 500(1.11^10-1/1.11-1)
?? You just accumulated 10 payments of $500 and only had $1419 at the end? I hope not. Did you add up the payments? 500 + 500*r + 500*r^2 + ... + 500*r^9 = What?
500(1.11^10-1/1.11-1
500(1.11^10-1/1.11-1=8361
Now we're talking! What's your plan for the continuous interest?
it is correct?^^
11% Annual - 8361.0044821
11%=8361 right? no -
it 11%annual=8361 right? not 11%-8361
Not sure what you are asking. I'll be horribly clear. Determine the value of the fund at the end of the tenth year on the assumption that the rate of interest is A) 11% compounded annually Answer: $8,361.00
Off to the continuous interest. How shall we tackle that one?
i am sorry yes but you 11- 8361 i am tellyouin for 11% annual =8631
Okay, now what is different about continuous interest?
how we can do conitnuous ?
Barely different. Rather that \(500(1.11)^{n}\), we get to use \(500e^{0.10\cdot n}\).
ahhh kk
500e^0.10-10?
Divided by....
wait it is not pe^r%T
pe^r/100time t
\(500\cdot\dfrac{e^{0.10\cdot n} - 1}{e^{0.10} - 1}\) You must add up 10 of them.
it same way we didnt learn that that why i dont know that kk
2113 the answer
It will be of great value to you to learn to add up a geometric series. 500 + 500*r + 500*r^2 + ... + 500*r^9 You will NEVER struggle with finding the "right" formula, again, if you concentrate on this fundamental principle.
No, no. Again, $500*10 = $5,000! It MUST be greater than that? Never be satisfied with a value less than $5,000.
i found sorry 8591
Where did "-10" come from? \(e^{0.01\cdot 10} = e^{0.10} = 1.1051709180756476152426726271225\)
8591 is no good. Something went wrong. Give it another go.
kk
first stepe 11/100 =0.11 so e^0.11*10 -1=2.00
I wondered if you had done that. Read Part B of the problem statement, again.
kk
10% so 10/100=0.1 so e^0.1*10-1=
i am doing same annually but continuously
so e^0.1*10-1=1.718
(1.718/0.10)*500
I would certainly recommend using more decimal places.
mmm
The Numerator \(\dfrac{1.718}{1.718281828459} = 0.9998359\) and is only a small error, 0.02%. \(\dfrac{0.10}{0.105170918} = 0.950833\) and is a whopping 5% error! Are you sure you're okay with a deliberate error that size?
can you explain me how
1.718.1718.92 ?
How did you get 1.718? The value of e-1 is 1.718281828459045... You must DECIDE what level of precision is sufficient for your needs. Generally, if you are dealing in thousands of dollars, you will need 5 ir 6 decimal places to have any reliablility at all in the pennies. You MUST look at the first decimal place you lop off and decide how important it is and what difference it will make. Let's look at the denominator. You used 0.10 ==> 1/0.10 = 10 One more decimal place 0.105 ==> 1/0.105 = 9.524 That's quite a difference! One more decimal place 0.1051 ==> 1/0.1051 = 9.5147 One more decimal place 0.10517 ==> 1/0.10517 = 9.50842 You can't just go about it willy nilly. You must make choices.
denomerator it is e^0.10-1
=0.1051809181
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