Seventh degree polynomialApostol Calculus Vol.1 Exercise 9 , Chapter 1.5 (Prove property of polynomial function)How to find polynomials $a(x)$ and $b(x)$ such that $c(x) = a(x) / b(x)$?Prove there exists a unique $n$-th degree polynomial that passes through $n+1$ points in the planeInductive proof of the degree of a polynomialProving fundamental properties of polynomialsPolynomial whose roots are also the coefficentsHow to solve $c_1x+c_2x^2+…+c_nx^n = K $ type equation (internal rate of return)Degree of interpolation polynomialParameterize a polynomial with no real rootsA polynomial coefficient
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Seventh degree polynomial
Apostol Calculus Vol.1 Exercise 9 , Chapter 1.5 (Prove property of polynomial function)How to find polynomials $a(x)$ and $b(x)$ such that $c(x) = a(x) / b(x)$?Prove there exists a unique $n$-th degree polynomial that passes through $n+1$ points in the planeInductive proof of the degree of a polynomialProving fundamental properties of polynomialsPolynomial whose roots are also the coefficentsHow to solve $c_1x+c_2x^2+…+c_nx^n = K $ type equation (internal rate of return)Degree of interpolation polynomialParameterize a polynomial with no real rootsA polynomial coefficient
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There is a unique polynomial $P(x)$ of the form $$ P(x) = x^7 + c_1x^6 + c_2x^5 + cdots + c_6x + c_7 $$ such that $P(1) = 1, P(2) = 2, ldots$, and $P(7) = 7$. Find $P(8)$.
How do I start this? I'm confused.
polynomials
asked 8 hours ago
AllenAllen
446 bronze badges
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add a comment
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$begingroup$
There is a unique polynomial $P(x)$ of the form $$ P(x) = x^7 + c_1x^6 + c_2x^5 + cdots + c_6x + c_7 $$ such that $P(1) = 1, P(2) = 2, ldots$, and $P(7) = 7$. Find $P(8)$.
How do I start this? I'm confused.
polynomials
asked 8 hours ago
AllenAllen
446 bronze badges
$endgroup$
add a comment
|
$begingroup$
There is a unique polynomial $P(x)$ of the form $$ P(x) = x^7 + c_1x^6 + c_2x^5 + cdots + c_6x + c_7 $$ such that $P(1) = 1, P(2) = 2, ldots$, and $P(7) = 7$. Find $P(8)$.
How do I start this? I'm confused.
polynomials
asked 8 hours ago
AllenAllen
446 bronze badges
$endgroup$
There is a unique polynomial $P(x)$ of the form $$ P(x) = x^7 + c_1x^6 + c_2x^5 + cdots + c_6x + c_7 $$ such that $P(1) = 1, P(2) = 2, ldots$, and $P(7) = 7$. Find $P(8)$.
How do I start this? I'm confused.
polynomials
polynomials
asked 8 hours ago
AllenAllen
446 bronze badges
asked 8 hours ago
AllenAllen
446 bronze badges
asked 8 hours ago
AllenAllen
446 bronze badges
asked 8 hours ago
AllenAllen
446 bronze badges
asked 8 hours ago
AllenAllen
446 bronze badges
446 bronze badges
add a comment
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2 Answers
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$P(x)-x$ is a monic, seventh degree polynomial with roots at $1,2,3,4,5,6,7$. It follows that
$$P(x)-x=(x-1)(x-2)cdotldotscdot(x-7),$$
so
$$ P(8) = 8+7! = colorblue5048. $$
answered 8 hours ago
Jack D'AurizioJack D'Aurizio
300k33 gold badges295 silver badges692 bronze badges
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$begingroup$
Extreme nit-picking here. Red-green color blindness is commonplace. I would have colored the answer blue.
$endgroup$
– Oscar Lanzi
7 hours ago
add a comment
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$begingroup$
You have seven unknowns ($c_1,dots,c_7$) and seven equations. In fact, these equations are linear. For instance, knowing $P(2)=2$ tells you:
$$2 = 2^7+2^6 c_1 + 2^5 c_2 + 2^4 c_3 + 2^3c_4+2^2 c_5+2c_6 + c_7.$$
Now you can just use linear algebra to solve this system for $c_1,dots, c_7$. Finally, use these values to compute $P(8)$.
answered 8 hours ago
kccukccu
14.4k1 gold badge14 silver badges32 bronze badges
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2
$begingroup$
There is no need to actually perform the inversion of a $7times 7$ Vandermonde matrix...
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– Jack D'Aurizio
8 hours ago
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
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$begingroup$
$P(x)-x$ is a monic, seventh degree polynomial with roots at $1,2,3,4,5,6,7$. It follows that
$$P(x)-x=(x-1)(x-2)cdotldotscdot(x-7),$$
so
$$ P(8) = 8+7! = colorblue5048. $$
answered 8 hours ago
Jack D'AurizioJack D'Aurizio
300k33 gold badges295 silver badges692 bronze badges
$endgroup$
$begingroup$
Extreme nit-picking here. Red-green color blindness is commonplace. I would have colored the answer blue.
$endgroup$
– Oscar Lanzi
7 hours ago
add a comment
|
$begingroup$
$P(x)-x$ is a monic, seventh degree polynomial with roots at $1,2,3,4,5,6,7$. It follows that
$$P(x)-x=(x-1)(x-2)cdotldotscdot(x-7),$$
so
$$ P(8) = 8+7! = colorblue5048. $$
answered 8 hours ago
Jack D'AurizioJack D'Aurizio
300k33 gold badges295 silver badges692 bronze badges
$endgroup$
$begingroup$
Extreme nit-picking here. Red-green color blindness is commonplace. I would have colored the answer blue.
$endgroup$
– Oscar Lanzi
7 hours ago
add a comment
|
$begingroup$
$P(x)-x$ is a monic, seventh degree polynomial with roots at $1,2,3,4,5,6,7$. It follows that
$$P(x)-x=(x-1)(x-2)cdotldotscdot(x-7),$$
so
$$ P(8) = 8+7! = colorblue5048. $$
answered 8 hours ago
Jack D'AurizioJack D'Aurizio
300k33 gold badges295 silver badges692 bronze badges
$endgroup$
$P(x)-x$ is a monic, seventh degree polynomial with roots at $1,2,3,4,5,6,7$. It follows that
$$P(x)-x=(x-1)(x-2)cdotldotscdot(x-7),$$
so
$$ P(8) = 8+7! = colorblue5048. $$
answered 8 hours ago
Jack D'AurizioJack D'Aurizio
300k33 gold badges295 silver badges692 bronze badges
edited 7 hours ago
answered 8 hours ago
Jack D'AurizioJack D'Aurizio
300k33 gold badges295 silver badges692 bronze badges
answered 8 hours ago
Jack D'AurizioJack D'Aurizio
300k33 gold badges295 silver badges692 bronze badges
answered 8 hours ago
Jack D'AurizioJack D'Aurizio
300k33 gold badges295 silver badges692 bronze badges
300k33 gold badges295 silver badges692 bronze badges
$begingroup$
Extreme nit-picking here. Red-green color blindness is commonplace. I would have colored the answer blue.
$endgroup$
– Oscar Lanzi
7 hours ago
add a comment
|
$begingroup$
Extreme nit-picking here. Red-green color blindness is commonplace. I would have colored the answer blue.
$endgroup$
– Oscar Lanzi
7 hours ago
$begingroup$
Extreme nit-picking here. Red-green color blindness is commonplace. I would have colored the answer blue.
$endgroup$
– Oscar Lanzi
7 hours ago
$begingroup$
Extreme nit-picking here. Red-green color blindness is commonplace. I would have colored the answer blue.
$endgroup$
– Oscar Lanzi
7 hours ago
add a comment
|
$begingroup$
You have seven unknowns ($c_1,dots,c_7$) and seven equations. In fact, these equations are linear. For instance, knowing $P(2)=2$ tells you:
$$2 = 2^7+2^6 c_1 + 2^5 c_2 + 2^4 c_3 + 2^3c_4+2^2 c_5+2c_6 + c_7.$$
Now you can just use linear algebra to solve this system for $c_1,dots, c_7$. Finally, use these values to compute $P(8)$.
answered 8 hours ago
kccukccu
14.4k1 gold badge14 silver badges32 bronze badges
$endgroup$
2
$begingroup$
There is no need to actually perform the inversion of a $7times 7$ Vandermonde matrix...
$endgroup$
– Jack D'Aurizio
8 hours ago
add a comment
|
$begingroup$
You have seven unknowns ($c_1,dots,c_7$) and seven equations. In fact, these equations are linear. For instance, knowing $P(2)=2$ tells you:
$$2 = 2^7+2^6 c_1 + 2^5 c_2 + 2^4 c_3 + 2^3c_4+2^2 c_5+2c_6 + c_7.$$
Now you can just use linear algebra to solve this system for $c_1,dots, c_7$. Finally, use these values to compute $P(8)$.
answered 8 hours ago
kccukccu
14.4k1 gold badge14 silver badges32 bronze badges
$endgroup$
2
$begingroup$
There is no need to actually perform the inversion of a $7times 7$ Vandermonde matrix...
$endgroup$
– Jack D'Aurizio
8 hours ago
add a comment
|
$begingroup$
You have seven unknowns ($c_1,dots,c_7$) and seven equations. In fact, these equations are linear. For instance, knowing $P(2)=2$ tells you:
$$2 = 2^7+2^6 c_1 + 2^5 c_2 + 2^4 c_3 + 2^3c_4+2^2 c_5+2c_6 + c_7.$$
Now you can just use linear algebra to solve this system for $c_1,dots, c_7$. Finally, use these values to compute $P(8)$.
answered 8 hours ago
kccukccu
14.4k1 gold badge14 silver badges32 bronze badges
$endgroup$
You have seven unknowns ($c_1,dots,c_7$) and seven equations. In fact, these equations are linear. For instance, knowing $P(2)=2$ tells you:
$$2 = 2^7+2^6 c_1 + 2^5 c_2 + 2^4 c_3 + 2^3c_4+2^2 c_5+2c_6 + c_7.$$
Now you can just use linear algebra to solve this system for $c_1,dots, c_7$. Finally, use these values to compute $P(8)$.
answered 8 hours ago
kccukccu
14.4k1 gold badge14 silver badges32 bronze badges
answered 8 hours ago
kccukccu
14.4k1 gold badge14 silver badges32 bronze badges
answered 8 hours ago
kccukccu
14.4k1 gold badge14 silver badges32 bronze badges
answered 8 hours ago
kccukccu
14.4k1 gold badge14 silver badges32 bronze badges
14.4k1 gold badge14 silver badges32 bronze badges
2
$begingroup$
There is no need to actually perform the inversion of a $7times 7$ Vandermonde matrix...
$endgroup$
– Jack D'Aurizio
8 hours ago
add a comment
|
2
$begingroup$
There is no need to actually perform the inversion of a $7times 7$ Vandermonde matrix...
$endgroup$
– Jack D'Aurizio
8 hours ago
2
2
$begingroup$
There is no need to actually perform the inversion of a $7times 7$ Vandermonde matrix...
$endgroup$
– Jack D'Aurizio
8 hours ago
$begingroup$
There is no need to actually perform the inversion of a $7times 7$ Vandermonde matrix...
$endgroup$
– Jack D'Aurizio
8 hours ago
add a comment
|
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