Give upper and lower bounds for $left|frac{sin^3{na}-2}{left(1+frac{cos b}{2}right)^n}right|$
Give upper and lower bounds for $left|frac{sin^3{na}-2}{left(1+frac{cos b}{2}right)^n}right|$
The only way that book explained up to this point in other examples uses this kind of technique:
$|sin^3{na}|le1$; $-3lesin^3{na}-2le-1$
$-1lecos{b}le1$; $-1/2le(cos{b})/2le1/2$;$1/2le1+(cos{b})/2le3/2$;
$2/3le1/(1+(cos{b})/2)le2$; $(2/3)^nle(1/(1+(cos{b})/2))^nle(2)^n$
So, $-1(2^n)lefrac{sin^3{na}-2}{left(1+frac{cos b}{2}right)^n}le-3left(frac23right)^n$
However, this doesn't make sense.
analysis upper-lower-bounds
edited Dec 2 '18 at 12:39
José Carlos Santos
151k22123224
asked Dec 2 '18 at 11:14
Sargis Iskandaryan
560112
add a comment |
Give upper and lower bounds for $left|frac{sin^3{na}-2}{left(1+frac{cos b}{2}right)^n}right|$
The only way that book explained up to this point in other examples uses this kind of technique:
$|sin^3{na}|le1$; $-3lesin^3{na}-2le-1$
$-1lecos{b}le1$; $-1/2le(cos{b})/2le1/2$;$1/2le1+(cos{b})/2le3/2$;
$2/3le1/(1+(cos{b})/2)le2$; $(2/3)^nle(1/(1+(cos{b})/2))^nle(2)^n$
So, $-1(2^n)lefrac{sin^3{na}-2}{left(1+frac{cos b}{2}right)^n}le-3left(frac23right)^n$
However, this doesn't make sense.
analysis upper-lower-bounds
edited Dec 2 '18 at 12:39
José Carlos Santos
151k22123224
asked Dec 2 '18 at 11:14
Sargis Iskandaryan
560112
add a comment |
Give upper and lower bounds for $left|frac{sin^3{na}-2}{left(1+frac{cos b}{2}right)^n}right|$
The only way that book explained up to this point in other examples uses this kind of technique:
$|sin^3{na}|le1$; $-3lesin^3{na}-2le-1$
$-1lecos{b}le1$; $-1/2le(cos{b})/2le1/2$;$1/2le1+(cos{b})/2le3/2$;
$2/3le1/(1+(cos{b})/2)le2$; $(2/3)^nle(1/(1+(cos{b})/2))^nle(2)^n$
So, $-1(2^n)lefrac{sin^3{na}-2}{left(1+frac{cos b}{2}right)^n}le-3left(frac23right)^n$
However, this doesn't make sense.
analysis upper-lower-bounds
edited Dec 2 '18 at 12:39
José Carlos Santos
151k22123224
asked Dec 2 '18 at 11:14
Sargis Iskandaryan
560112
Give upper and lower bounds for $left|frac{sin^3{na}-2}{left(1+frac{cos b}{2}right)^n}right|$
The only way that book explained up to this point in other examples uses this kind of technique:
$|sin^3{na}|le1$; $-3lesin^3{na}-2le-1$
$-1lecos{b}le1$; $-1/2le(cos{b})/2le1/2$;$1/2le1+(cos{b})/2le3/2$;
$2/3le1/(1+(cos{b})/2)le2$; $(2/3)^nle(1/(1+(cos{b})/2))^nle(2)^n$
So, $-1(2^n)lefrac{sin^3{na}-2}{left(1+frac{cos b}{2}right)^n}le-3left(frac23right)^n$
However, this doesn't make sense.
analysis upper-lower-bounds
analysis upper-lower-bounds
edited Dec 2 '18 at 12:39
José Carlos Santos
151k22123224
asked Dec 2 '18 at 11:14
Sargis Iskandaryan
560112
edited Dec 2 '18 at 12:39
José Carlos Santos
151k22123224
asked Dec 2 '18 at 11:14
Sargis Iskandaryan
560112
edited Dec 2 '18 at 12:39
José Carlos Santos
151k22123224
edited Dec 2 '18 at 12:39
José Carlos Santos
151k22123224
edited Dec 2 '18 at 12:39
José Carlos Santos
151k22123224
151k22123224
asked Dec 2 '18 at 11:14
Sargis Iskandaryan
560112
asked Dec 2 '18 at 11:14
Sargis Iskandaryan
560112
asked Dec 2 '18 at 11:14
Sargis Iskandaryan
560112
560112
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
There is n error in the last inequality.Let
$$
A=sin^3(n,a)-2,quad B=frac{1}{Bigl(1+dfrac{cos b}{2}Bigr)^n}.
$$
You have shown that
$$-3le Ale-1quadtext{and}quad Bigl(frac{2}{3}Bigr)^nle Ble2^n.$$
Since $B>0$, we have $-3,Ble A,Ble -B$ and
$$
-3times2^nle A,Ble-Bigl(frac{2}{3}Bigr)^n.
$$
This makes perfectly good sense. Taking absolute values we get
$$
Bigl(frac{2}{3}Bigr)^nle| A,B|le3times2^n.
$$
answered Dec 2 '18 at 12:38
Julián Aguirre
67.6k24094
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
There is n error in the last inequality.Let
$$
A=sin^3(n,a)-2,quad B=frac{1}{Bigl(1+dfrac{cos b}{2}Bigr)^n}.
$$
You have shown that
$$-3le Ale-1quadtext{and}quad Bigl(frac{2}{3}Bigr)^nle Ble2^n.$$
Since $B>0$, we have $-3,Ble A,Ble -B$ and
$$
-3times2^nle A,Ble-Bigl(frac{2}{3}Bigr)^n.
$$
This makes perfectly good sense. Taking absolute values we get
$$
Bigl(frac{2}{3}Bigr)^nle| A,B|le3times2^n.
$$
answered Dec 2 '18 at 12:38
Julián Aguirre
67.6k24094
add a comment |
There is n error in the last inequality.Let
$$
A=sin^3(n,a)-2,quad B=frac{1}{Bigl(1+dfrac{cos b}{2}Bigr)^n}.
$$
You have shown that
$$-3le Ale-1quadtext{and}quad Bigl(frac{2}{3}Bigr)^nle Ble2^n.$$
Since $B>0$, we have $-3,Ble A,Ble -B$ and
$$
-3times2^nle A,Ble-Bigl(frac{2}{3}Bigr)^n.
$$
This makes perfectly good sense. Taking absolute values we get
$$
Bigl(frac{2}{3}Bigr)^nle| A,B|le3times2^n.
$$
answered Dec 2 '18 at 12:38
Julián Aguirre
67.6k24094
add a comment |
There is n error in the last inequality.Let
$$
A=sin^3(n,a)-2,quad B=frac{1}{Bigl(1+dfrac{cos b}{2}Bigr)^n}.
$$
You have shown that
$$-3le Ale-1quadtext{and}quad Bigl(frac{2}{3}Bigr)^nle Ble2^n.$$
Since $B>0$, we have $-3,Ble A,Ble -B$ and
$$
-3times2^nle A,Ble-Bigl(frac{2}{3}Bigr)^n.
$$
This makes perfectly good sense. Taking absolute values we get
$$
Bigl(frac{2}{3}Bigr)^nle| A,B|le3times2^n.
$$
answered Dec 2 '18 at 12:38
Julián Aguirre
67.6k24094
There is n error in the last inequality.Let
$$
A=sin^3(n,a)-2,quad B=frac{1}{Bigl(1+dfrac{cos b}{2}Bigr)^n}.
$$
You have shown that
$$-3le Ale-1quadtext{and}quad Bigl(frac{2}{3}Bigr)^nle Ble2^n.$$
Since $B>0$, we have $-3,Ble A,Ble -B$ and
$$
-3times2^nle A,Ble-Bigl(frac{2}{3}Bigr)^n.
$$
This makes perfectly good sense. Taking absolute values we get
$$
Bigl(frac{2}{3}Bigr)^nle| A,B|le3times2^n.
$$
answered Dec 2 '18 at 12:38
Julián Aguirre
67.6k24094
answered Dec 2 '18 at 12:38
Julián Aguirre
67.6k24094
answered Dec 2 '18 at 12:38
Julián Aguirre
67.6k24094
answered Dec 2 '18 at 12:38
Julián Aguirre
67.6k24094
67.6k24094
add a comment |
add a comment |
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