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The importance of financial mathematics

A few words about mathematics financial to start.

Undoubtedly, everyone makes use of mathematical to do basic calculations in the life of every day, either with consciousness or not. For a financial tool that is invaluable is essential for proper management of flows of money.

One of the interests of mathematics in finance is to simplify the complexity of the calculations and financial problems. Take a small example to understand the meaning:

Mr. Y is 50000 Euro bank at 6%. We want to know how much it will ave the first year, second year, third year ...

By process of spontaneous calculations:

• First year: It will be the 50,000 Euro plus interest. To calculate interest by multiplying the percentage of 6% interest (ie 0, O6) by the amount deposited in the bank 50000 Euro, which will be: 50 000x0, 06 = 3000 Euro. So in total there will be at the end of the first year: 50,000 + 6% = 53,000 Euro x50000

• Second year: He will have the 53,000 plus interest calculated on the 53000, ie: 53000 + 6% = 56,180 Euro x53000

• The third year: 56,180 + 6% = 59,550.8 Euro x56180

So you see it's very tiring make these calculations, especially if you want to know how Mr. Y will be after 10 or 20 years for example. Well that's where financial mathematics will intervene to simplify any calculations of this kind.

could well have a mathematical formula that allows us to calculate a time and at any time the principal amount plus interest earned.

Noting in general, C 0 is the amount of capital deposited at time 0 (today) so C n and the amount of capital available to the bank after the n th year and also note i = the interest rate.

• The first year we have: C = C 1 0 + C 0 x i = 0 C (1 + i)

• The second year we have: C 2 = C + C 1 1 x i = C 1 (1 + i) = C 0 (1 + i) (1 + i) = C 0 (1 + i) ²

....

• By inference, the n th years we have: C n = 0 C (1 + i) n

So it suffices to replace in this formula, n the number of years. For example, after the 20 th year, Mr. Y will ave 50000 (1 +6%) 20 = Euro 160,356.77.

This case seems more or less simple because it can easily happen to calculate, with a little patience, that Mr. Y strike at the end of the 20 th year. But if one asks a question such as: how can we calculate, for example, when Mr. Y reaches 100,000 Euro. mentally, it is impossible to calculate, but mathematically speaking, it is very simple. Just solve the equation C = C n 0 (1 + i) n No stranger to one that represents the number of years on the capital deposited in the bank. Here we asked when C n = 100000 So by replacing in formula (C n = 0 C (1 + i) n ) i by 6%, C No. 100000 and by C 0 by 50,000 we would have:

100000 = 50000 (1 +0.06) n

==> 1.06 ⁿ 100000/50000 =

==> 1.06 n = 2 To determine the value of n, we must use a function called the sign logarithm "ln" (because one of the characteristics and that this function lnx n = n . lnx)

==> LN1, 06 n = ln2

==> x n LN1, 06 = ln2

==> No = ln2/ln1, 06 = 11.895660996576

That is Mr. Y will be Euro 100,000 in the bank in 11 years , 0.895660996576x12m = 10 months and 22 days 0.747931958912x30j =

To check the answer, just replace n 11.895660996576 by the formula 50000 (1 +0.06) n and see if it gives good Euro 100,000.

This is a simple example to show you how important points is to use the formulas mathematics to solve practical problems .

In this blog, I would put more emphasis on calculations with bank loan explanations and practical examples.