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Find asymptotically equivalent function $fsim g$
Find a majorizing functionSign of a functionTaylor expansion of a complex function on a discHow do we find a Taylor series for complex numbers?Fourier Series for $f(x)=sin(x)+cos(2x)$Find any rational approximation of $a = fracsin(sqrt2)sqrt2$Complex analysis problem with laurent series and singular pointsAlternative expansion for the sine functionFind function with this propertiesTaylor series for $sqrt1+sqrtx$ around $x=0$
$begingroup$
I'm trying to find a asymptotically equivalent function $g$ for $f(x)= sin(fracpi6^x) - dfrac12$, ie $f sim g$.
Should I use $sinx$ Taylor expansion or something else?
Here, $x$ approaches to $1$.
analysis
edited Mar 12 at 19:12
dmtri
1,6612521
asked Mar 12 at 18:53
MetsoMetso
15310
$endgroup$
add a comment |
$begingroup$
I'm trying to find a asymptotically equivalent function $g$ for $f(x)= sin(fracpi6^x) - dfrac12$, ie $f sim g$.
Should I use $sinx$ Taylor expansion or something else?
Here, $x$ approaches to $1$.
analysis
edited Mar 12 at 19:12
dmtri
1,6612521
asked Mar 12 at 18:53
MetsoMetso
15310
$endgroup$
$begingroup$
@ Jakobian : sorry, edited, approached to 1
$endgroup$
– Metso
Mar 12 at 18:57
$begingroup$
Yes, I think Taylor series would be the best in here.
$endgroup$
– Jakobian
Mar 12 at 18:58
$begingroup$
Is it in a neighbour hood of $0$ or of $infty$ of of $1$?
$endgroup$
– Bernard
Mar 12 at 19:04
$begingroup$
@Bernard: a neighbourhood of 1
$endgroup$
– Metso
Mar 12 at 19:06
add a comment |
$begingroup$
I'm trying to find a asymptotically equivalent function $g$ for $f(x)= sin(fracpi6^x) - dfrac12$, ie $f sim g$.
Should I use $sinx$ Taylor expansion or something else?
Here, $x$ approaches to $1$.
analysis
edited Mar 12 at 19:12
dmtri
1,6612521
asked Mar 12 at 18:53
MetsoMetso
15310
$endgroup$
I'm trying to find a asymptotically equivalent function $g$ for $f(x)= sin(fracpi6^x) - dfrac12$, ie $f sim g$.
Should I use $sinx$ Taylor expansion or something else?
Here, $x$ approaches to $1$.
analysis
analysis
edited Mar 12 at 19:12
dmtri
1,6612521
asked Mar 12 at 18:53
MetsoMetso
15310
edited Mar 12 at 19:12
dmtri
1,6612521
asked Mar 12 at 18:53
MetsoMetso
15310
edited Mar 12 at 19:12
dmtri
1,6612521
edited Mar 12 at 19:12
dmtri
1,6612521
edited Mar 12 at 19:12
dmtri
1,6612521
1,6612521
asked Mar 12 at 18:53
MetsoMetso
15310
asked Mar 12 at 18:53
MetsoMetso
15310
asked Mar 12 at 18:53
MetsoMetso
15310
15310
$begingroup$
@ Jakobian : sorry, edited, approached to 1
$endgroup$
– Metso
Mar 12 at 18:57
$begingroup$
Yes, I think Taylor series would be the best in here.
$endgroup$
– Jakobian
Mar 12 at 18:58
$begingroup$
Is it in a neighbour hood of $0$ or of $infty$ of of $1$?
$endgroup$
– Bernard
Mar 12 at 19:04
$begingroup$
@Bernard: a neighbourhood of 1
$endgroup$
– Metso
Mar 12 at 19:06
add a comment |
$begingroup$
@ Jakobian : sorry, edited, approached to 1
$endgroup$
– Metso
Mar 12 at 18:57
$begingroup$
Yes, I think Taylor series would be the best in here.
$endgroup$
– Jakobian
Mar 12 at 18:58
$begingroup$
Is it in a neighbour hood of $0$ or of $infty$ of of $1$?
$endgroup$
– Bernard
Mar 12 at 19:04
$begingroup$
@Bernard: a neighbourhood of 1
$endgroup$
– Metso
Mar 12 at 19:06
$begingroup$
@ Jakobian : sorry, edited, approached to 1
$endgroup$
– Metso
Mar 12 at 18:57
$begingroup$
@ Jakobian : sorry, edited, approached to 1
$endgroup$
– Metso
Mar 12 at 18:57
$begingroup$
Yes, I think Taylor series would be the best in here.
$endgroup$
– Jakobian
Mar 12 at 18:58
$begingroup$
Yes, I think Taylor series would be the best in here.
$endgroup$
– Jakobian
Mar 12 at 18:58
$begingroup$
Is it in a neighbour hood of $0$ or of $infty$ of of $1$?
$endgroup$
– Bernard
Mar 12 at 19:04
$begingroup$
Is it in a neighbour hood of $0$ or of $infty$ of of $1$?
$endgroup$
– Bernard
Mar 12 at 19:04
$begingroup$
@Bernard: a neighbourhood of 1
$endgroup$
– Metso
Mar 12 at 19:06
$begingroup$
@Bernard: a neighbourhood of 1
$endgroup$
– Metso
Mar 12 at 19:06
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Since
$$(sin(pi/6^x)-1/2)' = -ln 6 cdotfracpi cos(pi/6^x)6^x $$
Taylor formula gives us
$$sin(pi/6^x)-1/2 = -ln 6 cdotfracpisqrt312(x-1)+o(x-1) $$
so
$$sin(pi/6^x)-1/2 sim -ln 6 cdotfracpisqrt312(x-1)$$
as $xto 1$.
answered Mar 12 at 19:03
JakobianJakobian
2,695721
$endgroup$
$begingroup$
@Jacobian: Thank you for yor answer. It's interesting that $1/2=sin(pi/6)$. Does it make sense?
$endgroup$
– Metso
Mar 12 at 19:10
$begingroup$
You forgot a factor $pi$ in your derivative.
$endgroup$
– Bernard
Mar 12 at 19:12
$begingroup$
@Bernard Thank you kindly
$endgroup$
– Jakobian
Mar 12 at 19:14
$begingroup$
@Metso What do you mean?
$endgroup$
– Jakobian
Mar 12 at 19:16
$begingroup$
@Jacobian: I mean in this exercise we can rewrite $sin(pi/6^x)-sin(pi/6)$. But it does not help. Thank you.
$endgroup$
– Metso
Mar 12 at 19:18
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Since
$$(sin(pi/6^x)-1/2)' = -ln 6 cdotfracpi cos(pi/6^x)6^x $$
Taylor formula gives us
$$sin(pi/6^x)-1/2 = -ln 6 cdotfracpisqrt312(x-1)+o(x-1) $$
so
$$sin(pi/6^x)-1/2 sim -ln 6 cdotfracpisqrt312(x-1)$$
as $xto 1$.
answered Mar 12 at 19:03
JakobianJakobian
2,695721
$endgroup$
$begingroup$
@Jacobian: Thank you for yor answer. It's interesting that $1/2=sin(pi/6)$. Does it make sense?
$endgroup$
– Metso
Mar 12 at 19:10
$begingroup$
You forgot a factor $pi$ in your derivative.
$endgroup$
– Bernard
Mar 12 at 19:12
$begingroup$
@Bernard Thank you kindly
$endgroup$
– Jakobian
Mar 12 at 19:14
$begingroup$
@Metso What do you mean?
$endgroup$
– Jakobian
Mar 12 at 19:16
$begingroup$
@Jacobian: I mean in this exercise we can rewrite $sin(pi/6^x)-sin(pi/6)$. But it does not help. Thank you.
$endgroup$
– Metso
Mar 12 at 19:18
add a comment |
$begingroup$
Since
$$(sin(pi/6^x)-1/2)' = -ln 6 cdotfracpi cos(pi/6^x)6^x $$
Taylor formula gives us
$$sin(pi/6^x)-1/2 = -ln 6 cdotfracpisqrt312(x-1)+o(x-1) $$
so
$$sin(pi/6^x)-1/2 sim -ln 6 cdotfracpisqrt312(x-1)$$
as $xto 1$.
answered Mar 12 at 19:03
JakobianJakobian
2,695721
$endgroup$
$begingroup$
@Jacobian: Thank you for yor answer. It's interesting that $1/2=sin(pi/6)$. Does it make sense?
$endgroup$
– Metso
Mar 12 at 19:10
$begingroup$
You forgot a factor $pi$ in your derivative.
$endgroup$
– Bernard
Mar 12 at 19:12
$begingroup$
@Bernard Thank you kindly
$endgroup$
– Jakobian
Mar 12 at 19:14
$begingroup$
@Metso What do you mean?
$endgroup$
– Jakobian
Mar 12 at 19:16
$begingroup$
@Jacobian: I mean in this exercise we can rewrite $sin(pi/6^x)-sin(pi/6)$. But it does not help. Thank you.
$endgroup$
– Metso
Mar 12 at 19:18
add a comment |
$begingroup$
Since
$$(sin(pi/6^x)-1/2)' = -ln 6 cdotfracpi cos(pi/6^x)6^x $$
Taylor formula gives us
$$sin(pi/6^x)-1/2 = -ln 6 cdotfracpisqrt312(x-1)+o(x-1) $$
so
$$sin(pi/6^x)-1/2 sim -ln 6 cdotfracpisqrt312(x-1)$$
as $xto 1$.
answered Mar 12 at 19:03
JakobianJakobian
2,695721
$endgroup$
Since
$$(sin(pi/6^x)-1/2)' = -ln 6 cdotfracpi cos(pi/6^x)6^x $$
Taylor formula gives us
$$sin(pi/6^x)-1/2 = -ln 6 cdotfracpisqrt312(x-1)+o(x-1) $$
so
$$sin(pi/6^x)-1/2 sim -ln 6 cdotfracpisqrt312(x-1)$$
as $xto 1$.
answered Mar 12 at 19:03
JakobianJakobian
2,695721
edited Mar 12 at 19:14
answered Mar 12 at 19:03
JakobianJakobian
2,695721
answered Mar 12 at 19:03
JakobianJakobian
2,695721
answered Mar 12 at 19:03
JakobianJakobian
2,695721
2,695721
$begingroup$
@Jacobian: Thank you for yor answer. It's interesting that $1/2=sin(pi/6)$. Does it make sense?
$endgroup$
– Metso
Mar 12 at 19:10
$begingroup$
You forgot a factor $pi$ in your derivative.
$endgroup$
– Bernard
Mar 12 at 19:12
$begingroup$
@Bernard Thank you kindly
$endgroup$
– Jakobian
Mar 12 at 19:14
$begingroup$
@Metso What do you mean?
$endgroup$
– Jakobian
Mar 12 at 19:16
$begingroup$
@Jacobian: I mean in this exercise we can rewrite $sin(pi/6^x)-sin(pi/6)$. But it does not help. Thank you.
$endgroup$
– Metso
Mar 12 at 19:18
add a comment |
$begingroup$
@Jacobian: Thank you for yor answer. It's interesting that $1/2=sin(pi/6)$. Does it make sense?
$endgroup$
– Metso
Mar 12 at 19:10
$begingroup$
You forgot a factor $pi$ in your derivative.
$endgroup$
– Bernard
Mar 12 at 19:12
$begingroup$
@Bernard Thank you kindly
$endgroup$
– Jakobian
Mar 12 at 19:14
$begingroup$
@Metso What do you mean?
$endgroup$
– Jakobian
Mar 12 at 19:16
$begingroup$
@Jacobian: I mean in this exercise we can rewrite $sin(pi/6^x)-sin(pi/6)$. But it does not help. Thank you.
$endgroup$
– Metso
Mar 12 at 19:18
$begingroup$
@Jacobian: Thank you for yor answer. It's interesting that $1/2=sin(pi/6)$. Does it make sense?
$endgroup$
– Metso
Mar 12 at 19:10
$begingroup$
@Jacobian: Thank you for yor answer. It's interesting that $1/2=sin(pi/6)$. Does it make sense?
$endgroup$
– Metso
Mar 12 at 19:10
$begingroup$
You forgot a factor $pi$ in your derivative.
$endgroup$
– Bernard
Mar 12 at 19:12
$begingroup$
You forgot a factor $pi$ in your derivative.
$endgroup$
– Bernard
Mar 12 at 19:12
$begingroup$
@Bernard Thank you kindly
$endgroup$
– Jakobian
Mar 12 at 19:14
$begingroup$
@Bernard Thank you kindly
$endgroup$
– Jakobian
Mar 12 at 19:14
$begingroup$
@Metso What do you mean?
$endgroup$
– Jakobian
Mar 12 at 19:16
$begingroup$
@Metso What do you mean?
$endgroup$
– Jakobian
Mar 12 at 19:16
$begingroup$
@Jacobian: I mean in this exercise we can rewrite $sin(pi/6^x)-sin(pi/6)$. But it does not help. Thank you.
$endgroup$
– Metso
Mar 12 at 19:18
$begingroup$
@Jacobian: I mean in this exercise we can rewrite $sin(pi/6^x)-sin(pi/6)$. But it does not help. Thank you.
$endgroup$
– Metso
Mar 12 at 19:18
add a comment |
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$begingroup$
@ Jakobian : sorry, edited, approached to 1
$endgroup$
– Metso
Mar 12 at 18:57
$begingroup$
Yes, I think Taylor series would be the best in here.
$endgroup$
– Jakobian
Mar 12 at 18:58
$begingroup$
Is it in a neighbour hood of $0$ or of $infty$ of of $1$?
$endgroup$
– Bernard
Mar 12 at 19:04
$begingroup$
@Bernard: a neighbourhood of 1
$endgroup$
– Metso
Mar 12 at 19:06