Fibonacci series Fibonacci series was invented by Italian mathematician Leonardo Fibonacci (born in 1 170 and died in 1240). His birthplace may be Pisa. He is called "Leonardo of Pisa". He wrote the book Freedom of Apache. 19860/68617 He was the first European to study India and * * * mathematical theory. His father was hired as a diplomatic consul by a business group in Pisa and was stationed in Algeria today, so Leonardo was able to study mathematics under the guidance of his teacher. He also studied mathematics in Egypt, Syria, Greece, Sicily and Provence.
Fibonacci series refers to such a series: 1, 1, 2, 3, 5, 8, 13, 2 1 ...
This series begins with the third item, each item is equal to the sum of the first two items. The general formula is: (1√ 5) * {[(1+√ 5)/2] n-[(1-√ 5)/2] n} √.
It is very interesting that such a series of completely natural numbers are actually represented by irrational numbers.
This series has many wonderful properties.
For example, when the number of items in a series increases, the ratio of the previous item to the latter item is close to the golden section point of 0.6 180339887. ...
There is another attribute. Starting from the second term, the square of each odd term is more than the product of the first two terms 1, and the square of each even term is less than the product of the first two terms 1.
If you see such a topic: someone cut an 8*8 square into four pieces and made a 5* 13 rectangle, pretending to be surprised and asking you: Why is 64 = 65? In fact, this property of Fibonacci sequence is used: 5, 8, 13 are three adjacent terms in the sequence. In fact, the area difference between the front and rear blocks is really 1, but there is a slender slit in the back picture, which is not easy for ordinary people to notice.
If you start with any two numbers, such as 5, -2.4, and then add them together to form 5, -2.4, 2.6, 0.2, 2.8, 3, 5.8, 8.8, 14.6, etc. You will find that with the development of the series, the ratio of the first two items is getting closer and closer to the golden section.
The nth term of Fibonacci sequence also means that * * {1, 2, ..., n} does not contain adjacent positive integers.
Fibonacci sequence alias
Fibonacci sequence was introduced by mathematician Leonardo Fibonacci, so it is also called rabbit sequence.
Fibonacci sequence
Generally speaking, rabbits can reproduce two months after birth, and a pair of rabbits can give birth to a pair of rabbits every month. If all rabbits don't die, how many pairs of rabbits can you breed in a year?
We might as well take a pair of newborn rabbits to analyze:
In the first month, rabbits can't reproduce, so they are still a pair;
Two months later, two couples gave birth to a pair of rabbits.
Three months later, the old rabbit gave birth to another pair, because the little rabbit has no reproductive ability, so one pair is three pairs;
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By analogy, the following table can be listed:
Several months have passed: 012345678910112.
Rabbit logarithm:1123581321345 589144233.
The numbers 1, 1, 2, 3, 5, 8- in the table form a series. A very obvious feature of this series is that the sum of the first two adjacent terms constitutes the latter term.
This series was put forward by the Italian mathematician Fibonacci in the Encyclopedia of Abacus in the Middle Ages. Except a(n+2)=an+a(n+ 1)/, the general formula of this series can also be proved as an = 1/√ [( 1+).
Derivation of general term formula of Fibonacci sequence
Fibonacci series: 1, 1, 2,3,5,8,13,21 ...
If F(n) is the nth term of the series (n∈N+), then this sentence can be written as:
F( 1)=F(2)= 1，F(n)=F(n- 1)+F(n-2) (n≥3)
Obviously this is a linear recursive sequence.
General formula derivation method 1: Using characteristic equation
The characteristic equation of linear recursive sequence is:
X^2=X+ 1
solve
X 1=( 1+√5)/2，X2=( 1-√5)/2。
Then f (n) = c1* x1n+C2 * x2n.
∫F( 1)= F(2)= 1
∴C 1*X 1 + C2*X2
C 1*X 1^2 + C2*X2^2
The solution is c1=1√ 5, C2 =-1√ 5.
∴ f (n) = (1/√ 5) * {[(1+√ 5)/2] n-[(1-√ 5)/2] n} √ 5 represents the radical number 5.
Derivation method of general formula 2: general method
Let constant r, s
Let f (n)-r * f (n-1) = s * [f (n-1)-r * f (n-2)]
Then r+s= 1, -rs= 1.
When n≥3, there are
F(n)-r * F(n- 1)= s *[F(n- 1)-r * F(n-2)]
F(n- 1)-r * F(n-2)= s *[F(n-2)-r * F(n-3)]
F(n-2)-r * F(n-3)= s *[F(n-3)-r * F(n-4)]
……
F(3)-r * F(2)= s *[F(2)-r * F( 1)]
Multiply the above n-2 expressions to get:
f(n)-r*f(n- 1)=[s^(n-2)]*[f(2)-r*f( 1)]
∫s = 1-r，F( 1)=F(2)= 1
The above formula can be simplified as:
f(n)=s^(n- 1)+r*f(n- 1)
So:
f(n)=s^(n- 1)+r*f(n- 1)
= s^(n- 1)+r*s^(n-2)+r^2*f(n-2)
= s^(n- 1)+r*s^(n-2)+r^2*s^(n-3)+r^3*f(n-3)
……
= s^(n- 1)+r*s^(n-2)+r^2*s^(n-3)+……+r^(n-2)*s+r^(n- 1)*f( 1)
= s^(n- 1)+r*s^(n-2)+r^2*s^(n-3)+……+r^(n-2)*s+r^(n- 1)
(This is the sum of terms of geometric series with S (n- 1) as the first term, R (n- 1) as the last term, and r/s as the tolerance).
=[s^(n- 1)-r^(n- 1)*r/s]/( 1-r/s)
=(s^n - r^n)/(s-r)
The solutions of R+S = 1 and -RS = 1 are S = (1+√ 5)/2, and R = (1-√ 5)/2.
Then f (n) = (1√ 5) * {[(1+√ 5)/2] n-[(1-√ 5)/2] n}
C language program
Master ()
{
Long fiber [40] = {1,1};
int I；
for(I = 2； I