If $f(1) = 1$ And $f(n) = 2 F(n-1$\], Then Find The Value Of $f(6$\].

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Introduction

In this article, we will explore a simple recursive function and find the value of f(6)f(6) given the initial condition f(1)=1f(1) = 1 and the recursive formula f(n)=2f(n1)f(n) = 2 f(n-1). This type of problem is a classic example of a linear recurrence relation, which is a fundamental concept in mathematics and computer science.

Understanding the Recursive Formula

The recursive formula f(n)=2f(n1)f(n) = 2 f(n-1) indicates that the value of f(n)f(n) is twice the value of f(n1)f(n-1). This means that to find the value of f(n)f(n), we need to know the value of f(n1)f(n-1), and to find the value of f(n1)f(n-1), we need to know the value of f(n2)f(n-2), and so on.

Finding the Value of f(2)f(2)

To find the value of f(2)f(2), we can use the recursive formula f(n)=2f(n1)f(n) = 2 f(n-1). Since f(1)=1f(1) = 1, we have:

f(2)=2f(1)=21=2f(2) = 2 f(1) = 2 \cdot 1 = 2

Finding the Value of f(3)f(3)

To find the value of f(3)f(3), we can use the recursive formula f(n)=2f(n1)f(n) = 2 f(n-1). Since f(2)=2f(2) = 2, we have:

f(3)=2f(2)=22=4f(3) = 2 f(2) = 2 \cdot 2 = 4

Finding the Value of f(4)f(4)

To find the value of f(4)f(4), we can use the recursive formula f(n)=2f(n1)f(n) = 2 f(n-1). Since f(3)=4f(3) = 4, we have:

f(4)=2f(3)=24=8f(4) = 2 f(3) = 2 \cdot 4 = 8

Finding the Value of f(5)f(5)

To find the value of f(5)f(5), we can use the recursive formula f(n)=2f(n1)f(n) = 2 f(n-1). Since f(4)=8f(4) = 8, we have:

f(5)=2f(4)=28=16f(5) = 2 f(4) = 2 \cdot 8 = 16

Finding the Value of f(6)f(6)

To find the value of f(6)f(6), we can use the recursive formula f(n)=2f(n1)f(n) = 2 f(n-1). Since f(5)=16f(5) = 16, we have:

f(6)=2f(5)=216=32f(6) = 2 f(5) = 2 \cdot 16 = 32

Conclusion

In this article, we have found the value of f(6)f(6) given the initial condition f(1)=1f(1) = 1 and the recursive formula f(n)=2f(n1)f(n) = 2 f(n-1). The value of f(6)f(6) is 32. This type of problem is a classic example of a linear recurrence relation, which is a fundamental concept in mathematics and computer science.

Example Use Cases

Linear recurrence relations have many practical applications in computer science, such as:

  • Dynamic programming: Linear recurrence relations are used to solve dynamic programming problems, which involve finding the optimal solution to a problem by breaking it down into smaller subproblems.
  • Algorithms: Linear recurrence relations are used to analyze the time and space complexity of algorithms, which is crucial in computer science.
  • Modeling: Linear recurrence relations are used to model real-world phenomena, such as population growth, financial markets, and epidemiology.

Further Reading

For more information on linear recurrence relations and their applications, we recommend the following resources:

  • "Introduction to Algorithms" by Thomas H. Cormen: This book provides a comprehensive introduction to algorithms, including linear recurrence relations.
  • "Linear Algebra and Its Applications" by Gilbert Strang: This book provides a comprehensive introduction to linear algebra, including linear recurrence relations.
  • "Dynamic Programming" by Richard E. Bellman: This book provides a comprehensive introduction to dynamic programming, including linear recurrence relations.

References

  • Cormen, T. H. (2009). Introduction to Algorithms. MIT Press.
  • Strang, G. (2016). Linear Algebra and Its Applications. Wellesley-Cambridge Press.
  • Bellman, R. E. (1957). Dynamic Programming. Princeton University Press.

Introduction

In our previous article, we explored a simple recursive function and found the value of f(6)f(6) given the initial condition f(1)=1f(1) = 1 and the recursive formula f(n)=2f(n1)f(n) = 2 f(n-1). In this article, we will answer some frequently asked questions about this problem.

Q: What is the recursive formula?

A: The recursive formula is f(n)=2f(n1)f(n) = 2 f(n-1). This means that the value of f(n)f(n) is twice the value of f(n1)f(n-1).

Q: How do I find the value of f(6)f(6)?

A: To find the value of f(6)f(6), you need to use the recursive formula f(n)=2f(n1)f(n) = 2 f(n-1) and start with the initial condition f(1)=1f(1) = 1. Then, you can find the value of f(2)f(2), f(3)f(3), f(4)f(4), f(5)f(5), and finally f(6)f(6).

Q: Can I use a calculator to find the value of f(6)f(6)?

A: Yes, you can use a calculator to find the value of f(6)f(6). Simply enter the recursive formula f(n)=2f(n1)f(n) = 2 f(n-1) and the initial condition f(1)=1f(1) = 1, and the calculator will give you the value of f(6)f(6).

Q: Is there a faster way to find the value of f(6)f(6)?

A: Yes, there is a faster way to find the value of f(6)f(6). You can use the formula for the nth term of a geometric sequence, which is an=a1rn1a_n = a_1 \cdot r^{n-1}. In this case, a1=1a_1 = 1 and r=2r = 2, so the formula becomes f(n)=12n1f(n) = 1 \cdot 2^{n-1}.

Q: How do I use the formula f(n)=12n1f(n) = 1 \cdot 2^{n-1} to find the value of f(6)f(6)?

A: To use the formula f(n)=12n1f(n) = 1 \cdot 2^{n-1} to find the value of f(6)f(6), simply plug in n=6n = 6 and calculate the result. You get f(6)=1261=125=32f(6) = 1 \cdot 2^{6-1} = 1 \cdot 2^5 = 32.

Q: What is the significance of the value of f(6)f(6)?

A: The value of f(6)f(6) is significant because it represents the 6th term of a geometric sequence with a common ratio of 2. This sequence has many practical applications in computer science, such as dynamic programming and algorithm analysis.

Q: Can I use the formula f(n)=12n1f(n) = 1 \cdot 2^{n-1} to find the value of f(n)f(n) for any positive integer nn?

A: Yes, you can use the formula f(n)=12n1f(n) = 1 \cdot 2^{n-1} to find the value of f(n)f(n) for any positive integer nn. Simply plug in the value of nn and calculate the result.

Conclusion

In this article, we have answered some frequently asked questions about the problem of finding the value of f(6)f(6) given the initial condition f(1)=1f(1) = 1 and the recursive formula f(n)=2f(n1)f(n) = 2 f(n-1). We have also provided a formula for the nth term of a geometric sequence, which can be used to find the value of f(n)f(n) for any positive integer nn.

Example Use Cases

The formula f(n)=12n1f(n) = 1 \cdot 2^{n-1} has many practical applications in computer science, such as:

  • Dynamic programming: The formula can be used to solve dynamic programming problems, which involve finding the optimal solution to a problem by breaking it down into smaller subproblems.
  • Algorithms: The formula can be used to analyze the time and space complexity of algorithms, which is crucial in computer science.
  • Modeling: The formula can be used to model real-world phenomena, such as population growth, financial markets, and epidemiology.

Further Reading

For more information on geometric sequences and their applications, we recommend the following resources:

  • "Introduction to Algorithms" by Thomas H. Cormen: This book provides a comprehensive introduction to algorithms, including geometric sequences.
  • "Linear Algebra and Its Applications" by Gilbert Strang: This book provides a comprehensive introduction to linear algebra, including geometric sequences.
  • "Dynamic Programming" by Richard E. Bellman: This book provides a comprehensive introduction to dynamic programming, including geometric sequences.

References

  • Cormen, T. H. (2009). Introduction to Algorithms. MIT Press.
  • Strang, G. (2016). Linear Algebra and Its Applications. Wellesley-Cambridge Press.
  • Bellman, R. E. (1957). Dynamic Programming. Princeton University Press.