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Is there a simpler way to determine m, n, p, such that the following holds for all reals?
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I am given the following equation and I am asked to find $m$, $n$ and $p$, so that the equation holds for all reals:
$$ sin^4x + cos^4x + m(sin^6x + cos^6x) + n(sin^8x + cos^8x) + p(sin^10x + cos^10x) = 1, space forall x in mathbb R $$
I have managed to solve it by, in advance, calculating the following power reduction formulas, and then applying them to the equation:
$$
sin^4x + cos^4x = 1 - frac12sin^22x \
sin^6x + cos^6x = 1 - frac34sin^22x \
sin^8x + cos^8x = 1 - sin^22x + frac18sin^42x \
sin^10x + cos^10x = 1 - frac54sin^22x + frac516sin^42x
$$
As a result, I managed to simplify it to the following, which is only in terms of powers of $sin2x$:
$$ left( 1+m+n+p right) - left( frac12 + frac3m4 + n + frac5p4 right) sin^22x + left( fracn8 + frac5p16 right) sin^42x = 1, space forall x in mathbb R $$
I then came to the conclusion that the only possible way for which this can be true is if:
$1+m+n+p=1$, $ frac12 + frac3m4 + n + frac5p4 = 0 $ and $ fracn8 + frac5p16 = 0 $.
By solving the system of equations below I arrive at the solutions $m=6$, $n=-10$, $p=4$.
$$
left{
beginaligned
1 + m + n + p = 1 \
frac12 + frac3m4 + n + frac5p4 = 0 \
fracn8 + frac5p16 = 0
endaligned
right.
$$
My question is if my reasoning is correct and if there exists any simpler way to solve the problem.
trigonometry
asked Jul 21 at 16:58
Cheez
283
add a comment |Â
up vote
5
down vote
favorite
I am given the following equation and I am asked to find $m$, $n$ and $p$, so that the equation holds for all reals:
$$ sin^4x + cos^4x + m(sin^6x + cos^6x) + n(sin^8x + cos^8x) + p(sin^10x + cos^10x) = 1, space forall x in mathbb R $$
I have managed to solve it by, in advance, calculating the following power reduction formulas, and then applying them to the equation:
$$
sin^4x + cos^4x = 1 - frac12sin^22x \
sin^6x + cos^6x = 1 - frac34sin^22x \
sin^8x + cos^8x = 1 - sin^22x + frac18sin^42x \
sin^10x + cos^10x = 1 - frac54sin^22x + frac516sin^42x
$$
As a result, I managed to simplify it to the following, which is only in terms of powers of $sin2x$:
$$ left( 1+m+n+p right) - left( frac12 + frac3m4 + n + frac5p4 right) sin^22x + left( fracn8 + frac5p16 right) sin^42x = 1, space forall x in mathbb R $$
I then came to the conclusion that the only possible way for which this can be true is if:
$1+m+n+p=1$, $ frac12 + frac3m4 + n + frac5p4 = 0 $ and $ fracn8 + frac5p16 = 0 $.
By solving the system of equations below I arrive at the solutions $m=6$, $n=-10$, $p=4$.
$$
left{
beginaligned
1 + m + n + p = 1 \
frac12 + frac3m4 + n + frac5p4 = 0 \
fracn8 + frac5p16 = 0
endaligned
right.
$$
My question is if my reasoning is correct and if there exists any simpler way to solve the problem.
trigonometry
asked Jul 21 at 16:58
Cheez
283
1
Your reasoning is correct. My method is less complicated, but not by much.
– Batominovski
Jul 21 at 17:41
add a comment |Â
up vote
5
down vote
favorite
up vote
5
down vote
favorite
I am given the following equation and I am asked to find $m$, $n$ and $p$, so that the equation holds for all reals:
$$ sin^4x + cos^4x + m(sin^6x + cos^6x) + n(sin^8x + cos^8x) + p(sin^10x + cos^10x) = 1, space forall x in mathbb R $$
I have managed to solve it by, in advance, calculating the following power reduction formulas, and then applying them to the equation:
$$
sin^4x + cos^4x = 1 - frac12sin^22x \
sin^6x + cos^6x = 1 - frac34sin^22x \
sin^8x + cos^8x = 1 - sin^22x + frac18sin^42x \
sin^10x + cos^10x = 1 - frac54sin^22x + frac516sin^42x
$$
As a result, I managed to simplify it to the following, which is only in terms of powers of $sin2x$:
$$ left( 1+m+n+p right) - left( frac12 + frac3m4 + n + frac5p4 right) sin^22x + left( fracn8 + frac5p16 right) sin^42x = 1, space forall x in mathbb R $$
I then came to the conclusion that the only possible way for which this can be true is if:
$1+m+n+p=1$, $ frac12 + frac3m4 + n + frac5p4 = 0 $ and $ fracn8 + frac5p16 = 0 $.
By solving the system of equations below I arrive at the solutions $m=6$, $n=-10$, $p=4$.
$$
left{
beginaligned
1 + m + n + p = 1 \
frac12 + frac3m4 + n + frac5p4 = 0 \
fracn8 + frac5p16 = 0
endaligned
right.
$$
My question is if my reasoning is correct and if there exists any simpler way to solve the problem.
trigonometry
asked Jul 21 at 16:58
Cheez
283
I am given the following equation and I am asked to find $m$, $n$ and $p$, so that the equation holds for all reals:
$$ sin^4x + cos^4x + m(sin^6x + cos^6x) + n(sin^8x + cos^8x) + p(sin^10x + cos^10x) = 1, space forall x in mathbb R $$
I have managed to solve it by, in advance, calculating the following power reduction formulas, and then applying them to the equation:
$$
sin^4x + cos^4x = 1 - frac12sin^22x \
sin^6x + cos^6x = 1 - frac34sin^22x \
sin^8x + cos^8x = 1 - sin^22x + frac18sin^42x \
sin^10x + cos^10x = 1 - frac54sin^22x + frac516sin^42x
$$
As a result, I managed to simplify it to the following, which is only in terms of powers of $sin2x$:
$$ left( 1+m+n+p right) - left( frac12 + frac3m4 + n + frac5p4 right) sin^22x + left( fracn8 + frac5p16 right) sin^42x = 1, space forall x in mathbb R $$
I then came to the conclusion that the only possible way for which this can be true is if:
$1+m+n+p=1$, $ frac12 + frac3m4 + n + frac5p4 = 0 $ and $ fracn8 + frac5p16 = 0 $.
By solving the system of equations below I arrive at the solutions $m=6$, $n=-10$, $p=4$.
$$
left{
beginaligned
1 + m + n + p = 1 \
frac12 + frac3m4 + n + frac5p4 = 0 \
fracn8 + frac5p16 = 0
endaligned
right.
$$
My question is if my reasoning is correct and if there exists any simpler way to solve the problem.
trigonometry
asked Jul 21 at 16:58
Cheez
283
asked Jul 21 at 16:58
Cheez
283
asked Jul 21 at 16:58
Cheez
283
asked Jul 21 at 16:58
Cheez
283
283
1
Your reasoning is correct. My method is less complicated, but not by much.
– Batominovski
Jul 21 at 17:41
add a comment |Â
1
Your reasoning is correct. My method is less complicated, but not by much.
– Batominovski
Jul 21 at 17:41
1
1
Your reasoning is correct. My method is less complicated, but not by much.
– Batominovski
Jul 21 at 17:41
Your reasoning is correct. My method is less complicated, but not by much.
– Batominovski
Jul 21 at 17:41
add a comment |Â
2 Answers
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up vote
3
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accepted
Consider the polynomial
$$P(t):=-1+t^2+(1-t)^2+mt^3+m(1-t)^3+nt^4+n(1-t)^4+pt^5+p(1-t)^5,.$$
If $t=sin^2(x)$ is a root for every $xinmathbbR$, then $Pequiv 0$ identically. In particular,
$$0=[t^4],P(t)=2n+5p,,$$
$$0=[t^2],P(t)=2+3m+6n+10p,,$$
and
$$0=[t^0],P(t)=P(0)=m+n+p,,$$
where $[t^k],P(t)$ denotes the coefficient of the term $t^k$ in $P(t)$ for every $k=0,1,2,ldots$. (Here, the coefficients are found via the Binomial Theorem.) It follows immediately that $m=6$, $n=-10$, and $p=4$.
We still need to show that $Pequiv 0$ is indeed the case. First, $deg(P)le4$ can be easily seen. Therefore, we need to check that $[t],P(t)=0$ and $[t^3],P(t)=0$, but
$$[t],P(t)=-2-3m-4n-5p=-big([t^2],P(t)big)+big([t^4],P(t)big)=0$$
and
$$[t^3],P(t)=-4n-10p=-2,big([t^4],P(t)big)=0,.$$
Thus, $P$ is indeed the zero polynomial.
answered Jul 21 at 17:38
Batominovski
23.3k22777
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up vote
0
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Your reasoning is correct. You may save a little algebraa by formulating it as follows. Define $, f(t) := t^2 + m t^3 + n t^4 + p t^5. ,$ We want the equation
$, f(t) + f(1-t) - 1 = 0,$ to be true for all $, t = sin^2(x). ,$ If you expand the left side, it is a fourth degreee polynomial in $,t.,$ All of its coefficients must be zero. This gives five linear equations in the four unknowns and the solution is as you found. The advantage of this approach is that it avoids the need to express $, sin^2n(x) + cos^2n(x) ,$ in terms of even powers of $, sin(2x). ,$ All that is needed is to know the identity $, sin^2 x + cos^2 x = 1. $
answered Jul 21 at 17:56
Somos
11.5k1933
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Consider the polynomial
$$P(t):=-1+t^2+(1-t)^2+mt^3+m(1-t)^3+nt^4+n(1-t)^4+pt^5+p(1-t)^5,.$$
If $t=sin^2(x)$ is a root for every $xinmathbbR$, then $Pequiv 0$ identically. In particular,
$$0=[t^4],P(t)=2n+5p,,$$
$$0=[t^2],P(t)=2+3m+6n+10p,,$$
and
$$0=[t^0],P(t)=P(0)=m+n+p,,$$
where $[t^k],P(t)$ denotes the coefficient of the term $t^k$ in $P(t)$ for every $k=0,1,2,ldots$. (Here, the coefficients are found via the Binomial Theorem.) It follows immediately that $m=6$, $n=-10$, and $p=4$.
We still need to show that $Pequiv 0$ is indeed the case. First, $deg(P)le4$ can be easily seen. Therefore, we need to check that $[t],P(t)=0$ and $[t^3],P(t)=0$, but
$$[t],P(t)=-2-3m-4n-5p=-big([t^2],P(t)big)+big([t^4],P(t)big)=0$$
and
$$[t^3],P(t)=-4n-10p=-2,big([t^4],P(t)big)=0,.$$
Thus, $P$ is indeed the zero polynomial.
answered Jul 21 at 17:38
Batominovski
23.3k22777
add a comment |Â
up vote
3
down vote
accepted
Consider the polynomial
$$P(t):=-1+t^2+(1-t)^2+mt^3+m(1-t)^3+nt^4+n(1-t)^4+pt^5+p(1-t)^5,.$$
If $t=sin^2(x)$ is a root for every $xinmathbbR$, then $Pequiv 0$ identically. In particular,
$$0=[t^4],P(t)=2n+5p,,$$
$$0=[t^2],P(t)=2+3m+6n+10p,,$$
and
$$0=[t^0],P(t)=P(0)=m+n+p,,$$
where $[t^k],P(t)$ denotes the coefficient of the term $t^k$ in $P(t)$ for every $k=0,1,2,ldots$. (Here, the coefficients are found via the Binomial Theorem.) It follows immediately that $m=6$, $n=-10$, and $p=4$.
We still need to show that $Pequiv 0$ is indeed the case. First, $deg(P)le4$ can be easily seen. Therefore, we need to check that $[t],P(t)=0$ and $[t^3],P(t)=0$, but
$$[t],P(t)=-2-3m-4n-5p=-big([t^2],P(t)big)+big([t^4],P(t)big)=0$$
and
$$[t^3],P(t)=-4n-10p=-2,big([t^4],P(t)big)=0,.$$
Thus, $P$ is indeed the zero polynomial.
answered Jul 21 at 17:38
Batominovski
23.3k22777
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Consider the polynomial
$$P(t):=-1+t^2+(1-t)^2+mt^3+m(1-t)^3+nt^4+n(1-t)^4+pt^5+p(1-t)^5,.$$
If $t=sin^2(x)$ is a root for every $xinmathbbR$, then $Pequiv 0$ identically. In particular,
$$0=[t^4],P(t)=2n+5p,,$$
$$0=[t^2],P(t)=2+3m+6n+10p,,$$
and
$$0=[t^0],P(t)=P(0)=m+n+p,,$$
where $[t^k],P(t)$ denotes the coefficient of the term $t^k$ in $P(t)$ for every $k=0,1,2,ldots$. (Here, the coefficients are found via the Binomial Theorem.) It follows immediately that $m=6$, $n=-10$, and $p=4$.
We still need to show that $Pequiv 0$ is indeed the case. First, $deg(P)le4$ can be easily seen. Therefore, we need to check that $[t],P(t)=0$ and $[t^3],P(t)=0$, but
$$[t],P(t)=-2-3m-4n-5p=-big([t^2],P(t)big)+big([t^4],P(t)big)=0$$
and
$$[t^3],P(t)=-4n-10p=-2,big([t^4],P(t)big)=0,.$$
Thus, $P$ is indeed the zero polynomial.
answered Jul 21 at 17:38
Batominovski
23.3k22777
Consider the polynomial
$$P(t):=-1+t^2+(1-t)^2+mt^3+m(1-t)^3+nt^4+n(1-t)^4+pt^5+p(1-t)^5,.$$
If $t=sin^2(x)$ is a root for every $xinmathbbR$, then $Pequiv 0$ identically. In particular,
$$0=[t^4],P(t)=2n+5p,,$$
$$0=[t^2],P(t)=2+3m+6n+10p,,$$
and
$$0=[t^0],P(t)=P(0)=m+n+p,,$$
where $[t^k],P(t)$ denotes the coefficient of the term $t^k$ in $P(t)$ for every $k=0,1,2,ldots$. (Here, the coefficients are found via the Binomial Theorem.) It follows immediately that $m=6$, $n=-10$, and $p=4$.
We still need to show that $Pequiv 0$ is indeed the case. First, $deg(P)le4$ can be easily seen. Therefore, we need to check that $[t],P(t)=0$ and $[t^3],P(t)=0$, but
$$[t],P(t)=-2-3m-4n-5p=-big([t^2],P(t)big)+big([t^4],P(t)big)=0$$
and
$$[t^3],P(t)=-4n-10p=-2,big([t^4],P(t)big)=0,.$$
Thus, $P$ is indeed the zero polynomial.
answered Jul 21 at 17:38
Batominovski
23.3k22777
edited Jul 22 at 6:04
answered Jul 21 at 17:38
Batominovski
23.3k22777
answered Jul 21 at 17:38
Batominovski
23.3k22777
answered Jul 21 at 17:38
Batominovski
23.3k22777
23.3k22777
add a comment |Â
add a comment |Â
up vote
0
down vote
Your reasoning is correct. You may save a little algebraa by formulating it as follows. Define $, f(t) := t^2 + m t^3 + n t^4 + p t^5. ,$ We want the equation
$, f(t) + f(1-t) - 1 = 0,$ to be true for all $, t = sin^2(x). ,$ If you expand the left side, it is a fourth degreee polynomial in $,t.,$ All of its coefficients must be zero. This gives five linear equations in the four unknowns and the solution is as you found. The advantage of this approach is that it avoids the need to express $, sin^2n(x) + cos^2n(x) ,$ in terms of even powers of $, sin(2x). ,$ All that is needed is to know the identity $, sin^2 x + cos^2 x = 1. $
answered Jul 21 at 17:56
Somos
11.5k1933
add a comment |Â
up vote
0
down vote
Your reasoning is correct. You may save a little algebraa by formulating it as follows. Define $, f(t) := t^2 + m t^3 + n t^4 + p t^5. ,$ We want the equation
$, f(t) + f(1-t) - 1 = 0,$ to be true for all $, t = sin^2(x). ,$ If you expand the left side, it is a fourth degreee polynomial in $,t.,$ All of its coefficients must be zero. This gives five linear equations in the four unknowns and the solution is as you found. The advantage of this approach is that it avoids the need to express $, sin^2n(x) + cos^2n(x) ,$ in terms of even powers of $, sin(2x). ,$ All that is needed is to know the identity $, sin^2 x + cos^2 x = 1. $
answered Jul 21 at 17:56
Somos
11.5k1933
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Your reasoning is correct. You may save a little algebraa by formulating it as follows. Define $, f(t) := t^2 + m t^3 + n t^4 + p t^5. ,$ We want the equation
$, f(t) + f(1-t) - 1 = 0,$ to be true for all $, t = sin^2(x). ,$ If you expand the left side, it is a fourth degreee polynomial in $,t.,$ All of its coefficients must be zero. This gives five linear equations in the four unknowns and the solution is as you found. The advantage of this approach is that it avoids the need to express $, sin^2n(x) + cos^2n(x) ,$ in terms of even powers of $, sin(2x). ,$ All that is needed is to know the identity $, sin^2 x + cos^2 x = 1. $
answered Jul 21 at 17:56
Somos
11.5k1933
Your reasoning is correct. You may save a little algebraa by formulating it as follows. Define $, f(t) := t^2 + m t^3 + n t^4 + p t^5. ,$ We want the equation
$, f(t) + f(1-t) - 1 = 0,$ to be true for all $, t = sin^2(x). ,$ If you expand the left side, it is a fourth degreee polynomial in $,t.,$ All of its coefficients must be zero. This gives five linear equations in the four unknowns and the solution is as you found. The advantage of this approach is that it avoids the need to express $, sin^2n(x) + cos^2n(x) ,$ in terms of even powers of $, sin(2x). ,$ All that is needed is to know the identity $, sin^2 x + cos^2 x = 1. $
answered Jul 21 at 17:56
Somos
11.5k1933
edited Jul 23 at 20:46
answered Jul 21 at 17:56
Somos
11.5k1933
answered Jul 21 at 17:56
Somos
11.5k1933
answered Jul 21 at 17:56
Somos
11.5k1933
11.5k1933
add a comment |Â
add a comment |Â
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1
Your reasoning is correct. My method is less complicated, but not by much.
– Batominovski
Jul 21 at 17:41