How to find an appropriate constant value to which the function will be defined
$begingroup$
I have this function $y=frac{-ln(2cos{t}+e^{2k}-2)}{2}$, and i'm trying to find a value for k for which $y$ is defined for all $t$.
So i need the thing in the $ln$ to be larger than zero, i see that this inequality should hold $2cos(t) +e^{2k}>2$.
How do i continue form here to reach that if $k$ is larger than some constant then $y$ is defined for every $t$?
functions inequality
edited Dec 1 '18 at 15:04
Key Flex
7,77441232
asked Dec 1 '18 at 15:02
user3133165user3133165
1838
$endgroup$
add a comment |
$begingroup$
I have this function $y=frac{-ln(2cos{t}+e^{2k}-2)}{2}$, and i'm trying to find a value for k for which $y$ is defined for all $t$.
So i need the thing in the $ln$ to be larger than zero, i see that this inequality should hold $2cos(t) +e^{2k}>2$.
How do i continue form here to reach that if $k$ is larger than some constant then $y$ is defined for every $t$?
functions inequality
edited Dec 1 '18 at 15:04
Key Flex
7,77441232
asked Dec 1 '18 at 15:02
user3133165user3133165
1838
$endgroup$
add a comment |
$begingroup$
I have this function $y=frac{-ln(2cos{t}+e^{2k}-2)}{2}$, and i'm trying to find a value for k for which $y$ is defined for all $t$.
So i need the thing in the $ln$ to be larger than zero, i see that this inequality should hold $2cos(t) +e^{2k}>2$.
How do i continue form here to reach that if $k$ is larger than some constant then $y$ is defined for every $t$?
functions inequality
edited Dec 1 '18 at 15:04
Key Flex
7,77441232
asked Dec 1 '18 at 15:02
user3133165user3133165
1838
$endgroup$
I have this function $y=frac{-ln(2cos{t}+e^{2k}-2)}{2}$, and i'm trying to find a value for k for which $y$ is defined for all $t$.
So i need the thing in the $ln$ to be larger than zero, i see that this inequality should hold $2cos(t) +e^{2k}>2$.
How do i continue form here to reach that if $k$ is larger than some constant then $y$ is defined for every $t$?
functions inequality
functions inequality
edited Dec 1 '18 at 15:04
Key Flex
7,77441232
asked Dec 1 '18 at 15:02
user3133165user3133165
1838
edited Dec 1 '18 at 15:04
Key Flex
7,77441232
asked Dec 1 '18 at 15:02
user3133165user3133165
1838
edited Dec 1 '18 at 15:04
Key Flex
7,77441232
edited Dec 1 '18 at 15:04
Key Flex
7,77441232
edited Dec 1 '18 at 15:04
Key Flex
7,77441232
7,77441232
asked Dec 1 '18 at 15:02
user3133165user3133165
1838
asked Dec 1 '18 at 15:02
user3133165user3133165
1838
asked Dec 1 '18 at 15:02
user3133165user3133165
1838
1838
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
You should recall that minimum value of cos is -1. So for "worst" scenario you have $$2*(-1)+e^{2k}>2$$ or $$e^{2k}>4$$
Then, ln both sides: $2k>ln4$, so $k>0.5 ln4$
edited Dec 1 '18 at 16:09
Key Flex
7,77441232
answered Dec 1 '18 at 15:07
Kelly ShepphardKelly Shepphard
2298
$endgroup$
add a comment |
$begingroup$
Hint $cos t$ is bounded, so all you need is to find a $k$ s.t. $e^{2k}>4$… Taking logs seem the next obvious step...
answered Dec 1 '18 at 15:07
MacavityMacavity
35.2k52554
$endgroup$
add a comment |
$begingroup$
Hint: Recall that $$-1 leq cos (t) leq 1$$
What is the least possible value for $2cos (t)$ then?
answered Dec 1 '18 at 15:10
KM101KM101
5,9611523
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add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You should recall that minimum value of cos is -1. So for "worst" scenario you have $$2*(-1)+e^{2k}>2$$ or $$e^{2k}>4$$
Then, ln both sides: $2k>ln4$, so $k>0.5 ln4$
edited Dec 1 '18 at 16:09
Key Flex
7,77441232
answered Dec 1 '18 at 15:07
Kelly ShepphardKelly Shepphard
2298
$endgroup$
add a comment |
$begingroup$
You should recall that minimum value of cos is -1. So for "worst" scenario you have $$2*(-1)+e^{2k}>2$$ or $$e^{2k}>4$$
Then, ln both sides: $2k>ln4$, so $k>0.5 ln4$
edited Dec 1 '18 at 16:09
Key Flex
7,77441232
answered Dec 1 '18 at 15:07
Kelly ShepphardKelly Shepphard
2298
$endgroup$
add a comment |
$begingroup$
You should recall that minimum value of cos is -1. So for "worst" scenario you have $$2*(-1)+e^{2k}>2$$ or $$e^{2k}>4$$
Then, ln both sides: $2k>ln4$, so $k>0.5 ln4$
edited Dec 1 '18 at 16:09
Key Flex
7,77441232
answered Dec 1 '18 at 15:07
Kelly ShepphardKelly Shepphard
2298
$endgroup$
You should recall that minimum value of cos is -1. So for "worst" scenario you have $$2*(-1)+e^{2k}>2$$ or $$e^{2k}>4$$
Then, ln both sides: $2k>ln4$, so $k>0.5 ln4$
edited Dec 1 '18 at 16:09
Key Flex
7,77441232
answered Dec 1 '18 at 15:07
Kelly ShepphardKelly Shepphard
2298
edited Dec 1 '18 at 16:09
Key Flex
7,77441232
edited Dec 1 '18 at 16:09
Key Flex
7,77441232
edited Dec 1 '18 at 16:09
Key Flex
7,77441232
7,77441232
answered Dec 1 '18 at 15:07
Kelly ShepphardKelly Shepphard
2298
answered Dec 1 '18 at 15:07
Kelly ShepphardKelly Shepphard
2298
answered Dec 1 '18 at 15:07
Kelly ShepphardKelly Shepphard
2298
2298
add a comment |
add a comment |
$begingroup$
Hint $cos t$ is bounded, so all you need is to find a $k$ s.t. $e^{2k}>4$… Taking logs seem the next obvious step...
answered Dec 1 '18 at 15:07
MacavityMacavity
35.2k52554
$endgroup$
add a comment |
$begingroup$
Hint $cos t$ is bounded, so all you need is to find a $k$ s.t. $e^{2k}>4$… Taking logs seem the next obvious step...
answered Dec 1 '18 at 15:07
MacavityMacavity
35.2k52554
$endgroup$
add a comment |
$begingroup$
Hint $cos t$ is bounded, so all you need is to find a $k$ s.t. $e^{2k}>4$… Taking logs seem the next obvious step...
answered Dec 1 '18 at 15:07
MacavityMacavity
35.2k52554
$endgroup$
Hint $cos t$ is bounded, so all you need is to find a $k$ s.t. $e^{2k}>4$… Taking logs seem the next obvious step...
answered Dec 1 '18 at 15:07
MacavityMacavity
35.2k52554
answered Dec 1 '18 at 15:07
MacavityMacavity
35.2k52554
answered Dec 1 '18 at 15:07
MacavityMacavity
35.2k52554
answered Dec 1 '18 at 15:07
MacavityMacavity
35.2k52554
35.2k52554
add a comment |
add a comment |
$begingroup$
Hint: Recall that $$-1 leq cos (t) leq 1$$
What is the least possible value for $2cos (t)$ then?
answered Dec 1 '18 at 15:10
KM101KM101
5,9611523
$endgroup$
add a comment |
$begingroup$
Hint: Recall that $$-1 leq cos (t) leq 1$$
What is the least possible value for $2cos (t)$ then?
answered Dec 1 '18 at 15:10
KM101KM101
5,9611523
$endgroup$
add a comment |
$begingroup$
Hint: Recall that $$-1 leq cos (t) leq 1$$
What is the least possible value for $2cos (t)$ then?
answered Dec 1 '18 at 15:10
KM101KM101
5,9611523
$endgroup$
Hint: Recall that $$-1 leq cos (t) leq 1$$
What is the least possible value for $2cos (t)$ then?
answered Dec 1 '18 at 15:10
KM101KM101
5,9611523
answered Dec 1 '18 at 15:10
KM101KM101
5,9611523
answered Dec 1 '18 at 15:10
KM101KM101
5,9611523
answered Dec 1 '18 at 15:10
KM101KM101
5,9611523
5,9611523
add a comment |
add a comment |
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