Let a n = 2 n + 5 ⋅ 3 n for n = 0 , 1 , 2 , , ... Find a 0 , a 1 , a 2 , a 3 , and a 4 . Show that a 2 = 5 a 1 − 6 a 0 , a 3 = 5 a 2 − 6 a 1 , and a 4 = 5 a 3 − 6 a 2 . Show that a n = 5 a n − 1 − 6 a n − 2 for all integers n with n ≥ 2 .
Let a n = 2 n + 5 ⋅ 3 n for n = 0 , 1 , 2 , , ... Find a 0 , a 1 , a 2 , a 3 , and a 4 . Show that a 2 = 5 a 1 − 6 a 0 , a 3 = 5 a 2 − 6 a 1 , and a 4 = 5 a 3 − 6 a 2 . Show that a n = 5 a n − 1 − 6 a n − 2 for all integers n with n ≥ 2 .
Solution Summary: The author explains how to find the value of the term a_0.
A function is defined on the interval (-π/2,π/2) by this multipart rule:
if -π/2 < x < 0
f(x) =
a
if x=0
31-tan x
+31-cot x
if 0 < x < π/2
Here, a and b are constants. Find a and b so that the function f(x) is continuous at x=0.
a=
b= 3
Use the definition of continuity and the properties of limits to show that the function is continuous at the given number a.
f(x) = (x + 4x4) 5,
a = -1
lim f(x)
X--1
=
lim
x+4x
X--1
lim
X-1
4
x+4x
5
))"
5
))
by the power law
by the sum law
lim (x) + lim
X--1
4
4x
X-1
-(0,00+(
Find f(-1).
f(-1)=243
lim (x) +
-1 +4
35
4 ([
)
lim (x4)
5
x-1
Thus, by the definition of continuity, f is continuous at a = -1.
by the multiple constant law
by the direct substitution property
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MFCS unit-1 || Part:1 || JNTU || Well formed formula || propositional calculus || truth tables; Author: Learn with Smily;https://www.youtube.com/watch?v=XV15Q4mCcHc;License: Standard YouTube License, CC-BY