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Evaluating definite integral of $p(x)$
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Let $p(x)$ be fifth degree polynomial such that $p(x)+1$ is divisible by $(x-1)^3$ and $p(x)-1 $is divisible by $(x+1)^3 $. Then find the value of the definite integral $$int _-10^10p(x)dx$$
Attempt:
$p(x)-1 = (x+1)^3 Q(x)$
$p(x)+1 = (x-1)^3 H(x)$
Where $Q(x)$ and $H(x)$ are unknown quadratics.
But there's not sufficient information to find $Q$ and $H$ thats why I am unable to proceed.
Please provide only a guiding hint, I want to solve it myself.
calculus
edited Jul 16 at 8:00
Martin Sleziak
43.5k6113259
asked Jul 15 at 22:34
Abcd
2,3831624
add a comment |Â
up vote
8
down vote
favorite
Let $p(x)$ be fifth degree polynomial such that $p(x)+1$ is divisible by $(x-1)^3$ and $p(x)-1 $is divisible by $(x+1)^3 $. Then find the value of the definite integral $$int _-10^10p(x)dx$$
Attempt:
$p(x)-1 = (x+1)^3 Q(x)$
$p(x)+1 = (x-1)^3 H(x)$
Where $Q(x)$ and $H(x)$ are unknown quadratics.
But there's not sufficient information to find $Q$ and $H$ thats why I am unable to proceed.
Please provide only a guiding hint, I want to solve it myself.
calculus
edited Jul 16 at 8:00
Martin Sleziak
43.5k6113259
asked Jul 15 at 22:34
Abcd
2,3831624
add a comment |Â
up vote
8
down vote
favorite
up vote
8
down vote
favorite
Let $p(x)$ be fifth degree polynomial such that $p(x)+1$ is divisible by $(x-1)^3$ and $p(x)-1 $is divisible by $(x+1)^3 $. Then find the value of the definite integral $$int _-10^10p(x)dx$$
Attempt:
$p(x)-1 = (x+1)^3 Q(x)$
$p(x)+1 = (x-1)^3 H(x)$
Where $Q(x)$ and $H(x)$ are unknown quadratics.
But there's not sufficient information to find $Q$ and $H$ thats why I am unable to proceed.
Please provide only a guiding hint, I want to solve it myself.
calculus
edited Jul 16 at 8:00
Martin Sleziak
43.5k6113259
asked Jul 15 at 22:34
Abcd
2,3831624
Let $p(x)$ be fifth degree polynomial such that $p(x)+1$ is divisible by $(x-1)^3$ and $p(x)-1 $is divisible by $(x+1)^3 $. Then find the value of the definite integral $$int _-10^10p(x)dx$$
Attempt:
$p(x)-1 = (x+1)^3 Q(x)$
$p(x)+1 = (x-1)^3 H(x)$
Where $Q(x)$ and $H(x)$ are unknown quadratics.
But there's not sufficient information to find $Q$ and $H$ thats why I am unable to proceed.
Please provide only a guiding hint, I want to solve it myself.
calculus
edited Jul 16 at 8:00
Martin Sleziak
43.5k6113259
asked Jul 15 at 22:34
Abcd
2,3831624
edited Jul 16 at 8:00
Martin Sleziak
43.5k6113259
edited Jul 16 at 8:00
Martin Sleziak
43.5k6113259
edited Jul 16 at 8:00
Martin Sleziak
43.5k6113259
43.5k6113259
asked Jul 15 at 22:34
Abcd
2,3831624
asked Jul 15 at 22:34
Abcd
2,3831624
asked Jul 15 at 22:34
Abcd
2,3831624
2,3831624
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
6
down vote
accepted
Note that if $(x-1)^3$ divides $p(x)+1$, then $(x-1)^2$ divides $p^prime(x)$
Similarly,
if $(x+1)^3$ divides $p(x)-1$, then $(x-1)^2$ divides $p^prime(x)$
Since given that the degree is $5$ then when we differentiate it becomes $4$.
Therefore, $p^prime(x)=t(x+1)^2(x-1)^2=t(x^4-2x^2+1)$ where $tinmathbbR$
Then $$p(x)=dfrac t5x^5-dfrac2t3x^3+tx+bmbox $$ for $binmathbbR$$$$$
Since $(x+1)^3$ divides, we have $$p(-1)=-frac a5+dfrac2a3-a+b=1......(1)$$
Since $(x-1)^3$ divides, we have $$p(1)=dfrac a5-dfrac2a3+a+b=-1.......(2)$$
Now add $(1)+(2)$ gives $b=1$ and $a=-dfrac158$
Therefore, $p(x)=-dfrac38x^5+dfrac54x^3-dfrac158x$
Now $$int_-10^10p(x)dx=int_-10^10left(-dfrac38x^5+dfrac54x^3-dfrac158xright)dx=0$$
answered Jul 15 at 22:50
Key Flex
4,416525
add a comment |Â
up vote
4
down vote
Thanks to @KeyFlex answer, if both $(x-1)^2$ and $(x+1)^2$ divide $p^prime(x)$, then since $p'$ is degree $4$ so
$$p'(x)=k(x^2-1)^2$$
for a real $k$, hence $p'$ is even and then $p$ is odd which concludes $$int_-10^10p(x)dx=0$$
answered Jul 15 at 23:20
Nosrati
19.9k41644
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
accepted
Note that if $(x-1)^3$ divides $p(x)+1$, then $(x-1)^2$ divides $p^prime(x)$
Similarly,
if $(x+1)^3$ divides $p(x)-1$, then $(x-1)^2$ divides $p^prime(x)$
Since given that the degree is $5$ then when we differentiate it becomes $4$.
Therefore, $p^prime(x)=t(x+1)^2(x-1)^2=t(x^4-2x^2+1)$ where $tinmathbbR$
Then $$p(x)=dfrac t5x^5-dfrac2t3x^3+tx+bmbox $$ for $binmathbbR$$$$$
Since $(x+1)^3$ divides, we have $$p(-1)=-frac a5+dfrac2a3-a+b=1......(1)$$
Since $(x-1)^3$ divides, we have $$p(1)=dfrac a5-dfrac2a3+a+b=-1.......(2)$$
Now add $(1)+(2)$ gives $b=1$ and $a=-dfrac158$
Therefore, $p(x)=-dfrac38x^5+dfrac54x^3-dfrac158x$
Now $$int_-10^10p(x)dx=int_-10^10left(-dfrac38x^5+dfrac54x^3-dfrac158xright)dx=0$$
answered Jul 15 at 22:50
Key Flex
4,416525
add a comment |Â
up vote
6
down vote
accepted
Note that if $(x-1)^3$ divides $p(x)+1$, then $(x-1)^2$ divides $p^prime(x)$
Similarly,
if $(x+1)^3$ divides $p(x)-1$, then $(x-1)^2$ divides $p^prime(x)$
Since given that the degree is $5$ then when we differentiate it becomes $4$.
Therefore, $p^prime(x)=t(x+1)^2(x-1)^2=t(x^4-2x^2+1)$ where $tinmathbbR$
Then $$p(x)=dfrac t5x^5-dfrac2t3x^3+tx+bmbox $$ for $binmathbbR$$$$$
Since $(x+1)^3$ divides, we have $$p(-1)=-frac a5+dfrac2a3-a+b=1......(1)$$
Since $(x-1)^3$ divides, we have $$p(1)=dfrac a5-dfrac2a3+a+b=-1.......(2)$$
Now add $(1)+(2)$ gives $b=1$ and $a=-dfrac158$
Therefore, $p(x)=-dfrac38x^5+dfrac54x^3-dfrac158x$
Now $$int_-10^10p(x)dx=int_-10^10left(-dfrac38x^5+dfrac54x^3-dfrac158xright)dx=0$$
answered Jul 15 at 22:50
Key Flex
4,416525
add a comment |Â
up vote
6
down vote
accepted
up vote
6
down vote
accepted
Note that if $(x-1)^3$ divides $p(x)+1$, then $(x-1)^2$ divides $p^prime(x)$
Similarly,
if $(x+1)^3$ divides $p(x)-1$, then $(x-1)^2$ divides $p^prime(x)$
Since given that the degree is $5$ then when we differentiate it becomes $4$.
Therefore, $p^prime(x)=t(x+1)^2(x-1)^2=t(x^4-2x^2+1)$ where $tinmathbbR$
Then $$p(x)=dfrac t5x^5-dfrac2t3x^3+tx+bmbox $$ for $binmathbbR$$$$$
Since $(x+1)^3$ divides, we have $$p(-1)=-frac a5+dfrac2a3-a+b=1......(1)$$
Since $(x-1)^3$ divides, we have $$p(1)=dfrac a5-dfrac2a3+a+b=-1.......(2)$$
Now add $(1)+(2)$ gives $b=1$ and $a=-dfrac158$
Therefore, $p(x)=-dfrac38x^5+dfrac54x^3-dfrac158x$
Now $$int_-10^10p(x)dx=int_-10^10left(-dfrac38x^5+dfrac54x^3-dfrac158xright)dx=0$$
answered Jul 15 at 22:50
Key Flex
4,416525
Note that if $(x-1)^3$ divides $p(x)+1$, then $(x-1)^2$ divides $p^prime(x)$
Similarly,
if $(x+1)^3$ divides $p(x)-1$, then $(x-1)^2$ divides $p^prime(x)$
Since given that the degree is $5$ then when we differentiate it becomes $4$.
Therefore, $p^prime(x)=t(x+1)^2(x-1)^2=t(x^4-2x^2+1)$ where $tinmathbbR$
Then $$p(x)=dfrac t5x^5-dfrac2t3x^3+tx+bmbox $$ for $binmathbbR$$$$$
Since $(x+1)^3$ divides, we have $$p(-1)=-frac a5+dfrac2a3-a+b=1......(1)$$
Since $(x-1)^3$ divides, we have $$p(1)=dfrac a5-dfrac2a3+a+b=-1.......(2)$$
Now add $(1)+(2)$ gives $b=1$ and $a=-dfrac158$
Therefore, $p(x)=-dfrac38x^5+dfrac54x^3-dfrac158x$
Now $$int_-10^10p(x)dx=int_-10^10left(-dfrac38x^5+dfrac54x^3-dfrac158xright)dx=0$$
answered Jul 15 at 22:50
Key Flex
4,416525
edited Jul 15 at 22:56
answered Jul 15 at 22:50
Key Flex
4,416525
answered Jul 15 at 22:50
Key Flex
4,416525
answered Jul 15 at 22:50
Key Flex
4,416525
4,416525
add a comment |Â
add a comment |Â
up vote
4
down vote
Thanks to @KeyFlex answer, if both $(x-1)^2$ and $(x+1)^2$ divide $p^prime(x)$, then since $p'$ is degree $4$ so
$$p'(x)=k(x^2-1)^2$$
for a real $k$, hence $p'$ is even and then $p$ is odd which concludes $$int_-10^10p(x)dx=0$$
answered Jul 15 at 23:20
Nosrati
19.9k41644
add a comment |Â
up vote
4
down vote
Thanks to @KeyFlex answer, if both $(x-1)^2$ and $(x+1)^2$ divide $p^prime(x)$, then since $p'$ is degree $4$ so
$$p'(x)=k(x^2-1)^2$$
for a real $k$, hence $p'$ is even and then $p$ is odd which concludes $$int_-10^10p(x)dx=0$$
answered Jul 15 at 23:20
Nosrati
19.9k41644
add a comment |Â
up vote
4
down vote
up vote
4
down vote
Thanks to @KeyFlex answer, if both $(x-1)^2$ and $(x+1)^2$ divide $p^prime(x)$, then since $p'$ is degree $4$ so
$$p'(x)=k(x^2-1)^2$$
for a real $k$, hence $p'$ is even and then $p$ is odd which concludes $$int_-10^10p(x)dx=0$$
answered Jul 15 at 23:20
Nosrati
19.9k41644
Thanks to @KeyFlex answer, if both $(x-1)^2$ and $(x+1)^2$ divide $p^prime(x)$, then since $p'$ is degree $4$ so
$$p'(x)=k(x^2-1)^2$$
for a real $k$, hence $p'$ is even and then $p$ is odd which concludes $$int_-10^10p(x)dx=0$$
answered Jul 15 at 23:20
Nosrati
19.9k41644
answered Jul 15 at 23:20
Nosrati
19.9k41644
answered Jul 15 at 23:20
Nosrati
19.9k41644
answered Jul 15 at 23:20
Nosrati
19.9k41644
19.9k41644
add a comment |Â
add a comment |Â
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