Mixing
Find the value of all $x$ satisfying $(fcirc gcirc gcirc f)(x)=(gcirc gcirc f)(x)$, where $(fcirc g)(x)=f(g(x))$.
Clash Royale CLAN TAG#URR8PPP
up vote
2
down vote
favorite
Let $f(x)=x^2$ and $g(x)=sin(x) forall xin mathbb R$.
Then find the value of all $x$ satisfying $(fcirc gcirc gcirc f)(x)=(gcirc gcirc f)(x)$, where $(fcirc g)(x)=f(g(x))$.
Solution. From the given data, we have
$(fcirc gcirc gcirc f)(x)=(sin(sin(x^2)))^2$ and $(gcirc gcirc f)(x)=sin(sin(x^2))$.
Now we have to find the values of $x$ satisfying
$$(sin(sin(x^2)))^2=sin(sin(x^2))tag1$$
Let $sin(sin(x^2))=t$, then $(1)$ becomes $t^2=timplies t=0 text or 1$
Case 1: $sin(sin(x^2))=0implies sin(x^2)=npi, nin mathbb Z$. Need help in proceeding further
Case 2: $sin(sin(x^2))=1implies sin(x^2)=npi+(-1)^n left (fracpi2 right)$. Need help in proceeding further
trigonometry systems-of-equations roots
edited 13 hours ago
Rócherz
2,1811417
asked 13 hours ago
PK Styles
1,392524
add a comment |Â
up vote
2
down vote
favorite
Let $f(x)=x^2$ and $g(x)=sin(x) forall xin mathbb R$.
Then find the value of all $x$ satisfying $(fcirc gcirc gcirc f)(x)=(gcirc gcirc f)(x)$, where $(fcirc g)(x)=f(g(x))$.
Solution. From the given data, we have
$(fcirc gcirc gcirc f)(x)=(sin(sin(x^2)))^2$ and $(gcirc gcirc f)(x)=sin(sin(x^2))$.
Now we have to find the values of $x$ satisfying
$$(sin(sin(x^2)))^2=sin(sin(x^2))tag1$$
Let $sin(sin(x^2))=t$, then $(1)$ becomes $t^2=timplies t=0 text or 1$
Case 1: $sin(sin(x^2))=0implies sin(x^2)=npi, nin mathbb Z$. Need help in proceeding further
Case 2: $sin(sin(x^2))=1implies sin(x^2)=npi+(-1)^n left (fracpi2 right)$. Need help in proceeding further
trigonometry systems-of-equations roots
edited 13 hours ago
Rócherz
2,1811417
asked 13 hours ago
PK Styles
1,392524
Please edit the text in RED"Need help in proceeding further"
– PK Styles
13 hours ago
do you mean $f(g(g(f(x))))=g(g(f(x)))$
– Henry Lee
13 hours ago
@HenryLee:Yeah,you got it right
– PK Styles
13 hours ago
1
Your solution for Case 2, while accurate, is not optimal. For example, the solution is the same for $n=0$ and $n=1.$ Ideally, we'd like a different solution for each $n,$ such as by $fracpi2+2npi$ for $ninBbb Z.$
– Cameron Buie
13 hours ago
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let $f(x)=x^2$ and $g(x)=sin(x) forall xin mathbb R$.
Then find the value of all $x$ satisfying $(fcirc gcirc gcirc f)(x)=(gcirc gcirc f)(x)$, where $(fcirc g)(x)=f(g(x))$.
Solution. From the given data, we have
$(fcirc gcirc gcirc f)(x)=(sin(sin(x^2)))^2$ and $(gcirc gcirc f)(x)=sin(sin(x^2))$.
Now we have to find the values of $x$ satisfying
$$(sin(sin(x^2)))^2=sin(sin(x^2))tag1$$
Let $sin(sin(x^2))=t$, then $(1)$ becomes $t^2=timplies t=0 text or 1$
Case 1: $sin(sin(x^2))=0implies sin(x^2)=npi, nin mathbb Z$. Need help in proceeding further
Case 2: $sin(sin(x^2))=1implies sin(x^2)=npi+(-1)^n left (fracpi2 right)$. Need help in proceeding further
trigonometry systems-of-equations roots
edited 13 hours ago
Rócherz
2,1811417
asked 13 hours ago
PK Styles
1,392524
Let $f(x)=x^2$ and $g(x)=sin(x) forall xin mathbb R$.
Then find the value of all $x$ satisfying $(fcirc gcirc gcirc f)(x)=(gcirc gcirc f)(x)$, where $(fcirc g)(x)=f(g(x))$.
Solution. From the given data, we have
$(fcirc gcirc gcirc f)(x)=(sin(sin(x^2)))^2$ and $(gcirc gcirc f)(x)=sin(sin(x^2))$.
Now we have to find the values of $x$ satisfying
$$(sin(sin(x^2)))^2=sin(sin(x^2))tag1$$
Let $sin(sin(x^2))=t$, then $(1)$ becomes $t^2=timplies t=0 text or 1$
Case 1: $sin(sin(x^2))=0implies sin(x^2)=npi, nin mathbb Z$. Need help in proceeding further
Case 2: $sin(sin(x^2))=1implies sin(x^2)=npi+(-1)^n left (fracpi2 right)$. Need help in proceeding further
trigonometry systems-of-equations roots
edited 13 hours ago
Rócherz
2,1811417
asked 13 hours ago
PK Styles
1,392524
edited 13 hours ago
Rócherz
2,1811417
edited 13 hours ago
Rócherz
2,1811417
edited 13 hours ago
Rócherz
2,1811417
2,1811417
asked 13 hours ago
PK Styles
1,392524
asked 13 hours ago
PK Styles
1,392524
asked 13 hours ago
PK Styles
1,392524
1,392524
Please edit the text in RED"Need help in proceeding further"
– PK Styles
13 hours ago
do you mean $f(g(g(f(x))))=g(g(f(x)))$
– Henry Lee
13 hours ago
@HenryLee:Yeah,you got it right
– PK Styles
13 hours ago
1
Your solution for Case 2, while accurate, is not optimal. For example, the solution is the same for $n=0$ and $n=1.$ Ideally, we'd like a different solution for each $n,$ such as by $fracpi2+2npi$ for $ninBbb Z.$
– Cameron Buie
13 hours ago
add a comment |Â
Please edit the text in RED"Need help in proceeding further"
– PK Styles
13 hours ago
do you mean $f(g(g(f(x))))=g(g(f(x)))$
– Henry Lee
13 hours ago
@HenryLee:Yeah,you got it right
– PK Styles
13 hours ago
1
Your solution for Case 2, while accurate, is not optimal. For example, the solution is the same for $n=0$ and $n=1.$ Ideally, we'd like a different solution for each $n,$ such as by $fracpi2+2npi$ for $ninBbb Z.$
– Cameron Buie
13 hours ago
Please edit the text in RED"Need help in proceeding further"
– PK Styles
13 hours ago
Please edit the text in RED"Need help in proceeding further"
– PK Styles
13 hours ago
do you mean $f(g(g(f(x))))=g(g(f(x)))$
– Henry Lee
13 hours ago
do you mean $f(g(g(f(x))))=g(g(f(x)))$
– Henry Lee
13 hours ago
@HenryLee:Yeah,you got it right
– PK Styles
13 hours ago
@HenryLee:Yeah,you got it right
– PK Styles
13 hours ago
1
1
Your solution for Case 2, while accurate, is not optimal. For example, the solution is the same for $n=0$ and $n=1.$ Ideally, we'd like a different solution for each $n,$ such as by $fracpi2+2npi$ for $ninBbb Z.$
– Cameron Buie
13 hours ago
Your solution for Case 2, while accurate, is not optimal. For example, the solution is the same for $n=0$ and $n=1.$ Ideally, we'd like a different solution for each $n,$ such as by $fracpi2+2npi$ for $ninBbb Z.$
– Cameron Buie
13 hours ago
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
3
down vote
Recall that $sin$ is a bounded function between $-1$ and $1$. So the only way that $sin(x^2)$ is of the form $npi$ for some $nin mathbbZ$ is when $n=0$. You can then solve this easily for $x$.
Can you solve case $2$ in a similar manner?
EDIT: As for case $2$, as @CameronBuie remarked in the comment, the usual writing of the solutions of $sin(z)=0$ is $z=fracpi2+2npi$. Note that no real number of this form lie between $1$ and $-1$ (the least such positive real number is $fracpi2=1.57ldots$ and the greatest negative real number of this form is $fracpi2-2pi=-4.71ldots$). Hence, there is no solution in the case $2$ where $sin(x^2)=fracpi2+2npi$.
answered 13 hours ago
Suzet
2,003324
:Following your comment ,$sin(x^2)=0implies$ $x^2=npi implies x=pm sqrt npi$.Am i correct?
– PK Styles
12 hours ago
Exactly ! Note that necessarily, $n$ is non-negative.
– Suzet
12 hours ago
:will you please clarify case 2 also?
– PK Styles
12 hours ago
@PKStyles I added the case $2$ in my answer.
– Suzet
3 hours ago
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
Recall that $sin$ is a bounded function between $-1$ and $1$. So the only way that $sin(x^2)$ is of the form $npi$ for some $nin mathbbZ$ is when $n=0$. You can then solve this easily for $x$.
Can you solve case $2$ in a similar manner?
EDIT: As for case $2$, as @CameronBuie remarked in the comment, the usual writing of the solutions of $sin(z)=0$ is $z=fracpi2+2npi$. Note that no real number of this form lie between $1$ and $-1$ (the least such positive real number is $fracpi2=1.57ldots$ and the greatest negative real number of this form is $fracpi2-2pi=-4.71ldots$). Hence, there is no solution in the case $2$ where $sin(x^2)=fracpi2+2npi$.
answered 13 hours ago
Suzet
2,003324
:Following your comment ,$sin(x^2)=0implies$ $x^2=npi implies x=pm sqrt npi$.Am i correct?
– PK Styles
12 hours ago
Exactly ! Note that necessarily, $n$ is non-negative.
– Suzet
12 hours ago
:will you please clarify case 2 also?
– PK Styles
12 hours ago
@PKStyles I added the case $2$ in my answer.
– Suzet
3 hours ago
add a comment |Â
up vote
3
down vote
Recall that $sin$ is a bounded function between $-1$ and $1$. So the only way that $sin(x^2)$ is of the form $npi$ for some $nin mathbbZ$ is when $n=0$. You can then solve this easily for $x$.
Can you solve case $2$ in a similar manner?
EDIT: As for case $2$, as @CameronBuie remarked in the comment, the usual writing of the solutions of $sin(z)=0$ is $z=fracpi2+2npi$. Note that no real number of this form lie between $1$ and $-1$ (the least such positive real number is $fracpi2=1.57ldots$ and the greatest negative real number of this form is $fracpi2-2pi=-4.71ldots$). Hence, there is no solution in the case $2$ where $sin(x^2)=fracpi2+2npi$.
answered 13 hours ago
Suzet
2,003324
:Following your comment ,$sin(x^2)=0implies$ $x^2=npi implies x=pm sqrt npi$.Am i correct?
– PK Styles
12 hours ago
Exactly ! Note that necessarily, $n$ is non-negative.
– Suzet
12 hours ago
:will you please clarify case 2 also?
– PK Styles
12 hours ago
@PKStyles I added the case $2$ in my answer.
– Suzet
3 hours ago
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Recall that $sin$ is a bounded function between $-1$ and $1$. So the only way that $sin(x^2)$ is of the form $npi$ for some $nin mathbbZ$ is when $n=0$. You can then solve this easily for $x$.
Can you solve case $2$ in a similar manner?
EDIT: As for case $2$, as @CameronBuie remarked in the comment, the usual writing of the solutions of $sin(z)=0$ is $z=fracpi2+2npi$. Note that no real number of this form lie between $1$ and $-1$ (the least such positive real number is $fracpi2=1.57ldots$ and the greatest negative real number of this form is $fracpi2-2pi=-4.71ldots$). Hence, there is no solution in the case $2$ where $sin(x^2)=fracpi2+2npi$.
answered 13 hours ago
Suzet
2,003324
Recall that $sin$ is a bounded function between $-1$ and $1$. So the only way that $sin(x^2)$ is of the form $npi$ for some $nin mathbbZ$ is when $n=0$. You can then solve this easily for $x$.
Can you solve case $2$ in a similar manner?
EDIT: As for case $2$, as @CameronBuie remarked in the comment, the usual writing of the solutions of $sin(z)=0$ is $z=fracpi2+2npi$. Note that no real number of this form lie between $1$ and $-1$ (the least such positive real number is $fracpi2=1.57ldots$ and the greatest negative real number of this form is $fracpi2-2pi=-4.71ldots$). Hence, there is no solution in the case $2$ where $sin(x^2)=fracpi2+2npi$.
answered 13 hours ago
Suzet
2,003324
edited 3 hours ago
answered 13 hours ago
Suzet
2,003324
answered 13 hours ago
Suzet
2,003324
answered 13 hours ago
Suzet
2,003324
2,003324
:Following your comment ,$sin(x^2)=0implies$ $x^2=npi implies x=pm sqrt npi$.Am i correct?
– PK Styles
12 hours ago
Exactly ! Note that necessarily, $n$ is non-negative.
– Suzet
12 hours ago
:will you please clarify case 2 also?
– PK Styles
12 hours ago
@PKStyles I added the case $2$ in my answer.
– Suzet
3 hours ago
add a comment |Â
:Following your comment ,$sin(x^2)=0implies$ $x^2=npi implies x=pm sqrt npi$.Am i correct?
– PK Styles
12 hours ago
Exactly ! Note that necessarily, $n$ is non-negative.
– Suzet
12 hours ago
:will you please clarify case 2 also?
– PK Styles
12 hours ago
@PKStyles I added the case $2$ in my answer.
– Suzet
3 hours ago
:Following your comment ,$sin(x^2)=0implies$ $x^2=npi implies x=pm sqrt npi$.Am i correct?
– PK Styles
12 hours ago
:Following your comment ,$sin(x^2)=0implies$ $x^2=npi implies x=pm sqrt npi$.Am i correct?
– PK Styles
12 hours ago
Exactly ! Note that necessarily, $n$ is non-negative.
– Suzet
12 hours ago
Exactly ! Note that necessarily, $n$ is non-negative.
– Suzet
12 hours ago
:will you please clarify case 2 also?
– PK Styles
12 hours ago
:will you please clarify case 2 also?
– PK Styles
12 hours ago
@PKStyles I added the case $2$ in my answer.
– Suzet
3 hours ago
@PKStyles I added the case $2$ in my answer.
– Suzet
3 hours ago
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2873094%2ffind-the-value-of-all-x-satisfying-f-circ-g-circ-g-circ-fx-g-circ-g-circ%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Please edit the text in RED"Need help in proceeding further"
– PK Styles
13 hours ago
do you mean $f(g(g(f(x))))=g(g(f(x)))$
– Henry Lee
13 hours ago
@HenryLee:Yeah,you got it right
– PK Styles
13 hours ago
1
Your solution for Case 2, while accurate, is not optimal. For example, the solution is the same for $n=0$ and $n=1.$ Ideally, we'd like a different solution for each $n,$ such as by $fracpi2+2npi$ for $ninBbb Z.$
– Cameron Buie
13 hours ago