Good morning,I was hoping to get some assistance on question number 4. With the function defined I need to prove f(n). 4) Define f : N-R as follows:f(1) = 3, f(2)and for n 2 3 f(n)2'f(n- 1) + f(n-2)Prove that f(n) = 2+ (-)"-1 for all n E N11+ - + •..41.<2 Vn e N5) Prove that 1+2n-1x + y6) Let x > 0 and y > 0, prove that ry <

(4) Given data are f:N→Rf(1)=3, f(2)=32 and for…

Answered: 4) Define f: N→R as follows: f(1) = 3,…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

See similar textbooks

Related questions

Question

Good morning, I was hoping to get some assistance on question number 4. With the function defined I need to prove f(n).

4) Define f : N-R as follows:
f(1) = 3, f(2)
and for n 2 3 f(n)
2'
f(n- 1) + f(n-2)
Prove that f(n) = 2+ (-)"-1 for all n E N
1
1
+ - + •..
4
1.
<2 Vn e N
5) Prove that 1+
2n-1
x + y
6) Let x > 0 and y > 0, prove that ry <

Expert Solution

Step 1

(4) Given data are

$f : N \to R f (1) = 3, f (2) = \frac{3}{2} a n d f o r n \geq 3 f (n) = \frac{f (n - 1) + f (n - 2)}{2}$

$\begin{array}{rcl} f (1) & = & 3 = 2 + 1 = 2 + {(- \frac{1}{2})}^{0} = 2 + {(- \frac{1}{2})}^{1 - 1} \\ f (2) & = & \frac{3}{2} = 2 + {(- \frac{1}{2})}^{2 - 1} \\ f (3) & = & \frac{f (2) + f (1)}{2} = \frac{\frac{3}{2} + 3}{2} = \frac{9}{4} = 2 + {(- \frac{1}{2})}^{3 - 1} \\ f (4) & = & \frac{f (3) + f (2)}{2} = \frac{\frac{9}{4} + \frac{3}{2}}{2} = \frac{\frac{15}{4}}{2} = \frac{15}{8} = 2 + {(- \frac{1}{2})}^{4 - 1} \\ . . . . . . . . . . . . . . . . . . . . . . \\ . . . . . . . . . . . . . . . . . . . . . . \\ f (n) & = & 2 + {(- \frac{1}{2})}^{n - 1} \end{array}$

Hence proved.

Step by step

Solved in 3 steps

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

$Power Series$

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, advanced-math and related others by exploring similar questions and additional content below.

Similar questions