# Bounded or unbounded.

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#1
A = {exp(-x)sinx , x– R}
How do i find lower and upper bounds of this set using their ranges?
I know the range of exp(-x) is (-∞,0) and range of sinx is [-1,1]
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2 years ago
#2
Presuming this is on domain , double check the range of function as the graph of this function is:

What can you say of the set A now where:

?
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2 years ago
#3
Exp(-x) can't be negative or 0 so clearly your range is wrong.
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#4
(Original post by simon0)
Presuming this is on domain , double check the range of function as the graph of this function is:

What can you say of the set A now where:

?
Range of exp(-x) is (0,∞).
Sinx is is an oscillating function so as n tends to infinity sinx still oscillates.
so as n tends to infinity , this function oscillates infinitely.
so range of the function is (-∞,∞) and the function is unbounded.
is this right approach to prove the function unbounded?
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#5
(Original post by Whistlejacket)
Exp(-x) can't be negative or 0 so clearly your range is wrong.
What is the range of exp(-x)? all positive real numbers i guess.
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2 years ago
#6
(Original post by Infinity...)
Range of exp(-x) is (0,∞).
Sinx is is an oscillating function so as n tends to infinity sinx still oscillates.
so as n tends to infinity , this function oscillates infinitely.
so range of the function is (-∞,∞) and the function is unbounded.
is this right approach to prove the function unbounded?
Assuming the domain is (-inf,inf). If it's (0,inf), just differentiate and find the first couple of turning points.
Insights seem about right (apart from the exp(-x) typo in post 1). consider sequences of
x_k = pi/2 + 2*pi*k
and you're just left with the exponential part. Similarly
x_k = -pi/2 + 2*pi*k
and you're left with the -exponential part (for an integer k). They tend to inf, -inf and the function is continuous so the range is (-inf,inf). You're almost treating the sin() as a "soft" sign() function and the exp() as the magnitude.
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2 years ago
#7
(Original post by Infinity...)
A = {exp(-x)sinx , x– R}
How do i find lower and upper bounds of this set using their ranges?
I know the range of exp(-x) is (-∞,0) and range of sinx is [-1,1]
To show it's unbounded you just need to show it can be greater/smaller than an arbitrary large positive/negative number, N.
1
#8
(Original post by mqb2766)
Assuming the domain is (-inf,inf). If it's (0,inf), just differentiate and find the first couple of turning points.
Insights seem about right (apart from the exp(-x) typo in post 1). consider sequences of
x_k = pi/2 + 2*pi*k
and you're just left with the exponential part. Similarly
x_k = -pi/2 + 2*pi*k
and you're left with the -exponential part (for an integer k). They tend to inf, -inf and the function is continuous so the range is (-inf,inf). You're almost treating the sin() as a "soft" sign() function and the exp() as the magnitude.
consider sequences of
x_k = pi/2 + 2*pi*k
and you're just left with the exponential part. Similarly
x_k = -pi/2 + 2*pi*k
and you're left with the -exponential part (for an integer k).

I didn't understand this. where do these sequences come from?
0
2 years ago
#9
(Original post by Infinity...)
consider sequences of
x_k = pi/2 + 2*pi*k
and you're just left with the exponential part. Similarly
x_k = -pi/2 + 2*pi*k
and you're left with the -exponential part (for an integer k).

I didn't understand this. where do these sequences come from?
where sin(x_k) = 1 or -1. k -> -inf (integer) will give sequences of +/- exp(x_k)
0
2 years ago
#10
Given the first post isn't 100% clear, can the OP confirm we're allowing x to range over all of , not just ?

It radically alters the solution (and the method needed) depending on which is the case.
1
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