Degree of exponential polynomial
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What is the degree of the exponential polynomial :$$ e^x * (x^2 + x + 1)$$?
I am getting confused because of the exponential term there...
polynomials exponential-function
edited Aug 28 at 17:22
Deepesh Meena
3,0982824
asked Aug 28 at 17:19
Harshada Kelkar
265
add a comment |Â
up vote
0
down vote
favorite
What is the degree of the exponential polynomial :$$ e^x * (x^2 + x + 1)$$?
I am getting confused because of the exponential term there...
polynomials exponential-function
edited Aug 28 at 17:22
Deepesh Meena
3,0982824
asked Aug 28 at 17:19
Harshada Kelkar
265
4
never heard of the word degree used in this situation.
– Will Jagy
Aug 28 at 17:21
3
It's a polynomial of degree $2$ times an exponential. In what context are you trying to assign it a degree?
– Robert Israel
Aug 28 at 17:24
Actually, I am trying to solve integration of x^2*e^x using method of undetermined coefficients with A0, A1,A2....... previously, we had used this method for polynomials with degree <= 2.... was trying to apply the same logic here.... Analytically integrating x^2*e^x gives e^x*(x^2-2x+2) ..... So thought if this is a polynomial of degree <= 2 or not.... So posted this question in that context..
– Harshada Kelkar
Aug 28 at 17:41
If $P$ is a polynomial then any anti-derivative of $P(x)e^x$ is $K+Q(x)e^x$ where $Q$ is a polynomial with $deg (Q)=deg (P)$. Use integration by parts to prove this by induction on $deg (P). $ That is, $int e^xP(x)dx=$ $int P(x)de^x=$ $e^xP(x)-int e^xdP(x)=$ $e^xP(x)-int e^xP'(x)dx.$
– DanielWainfleet
Aug 28 at 18:33
Thanks, Daniel and Christian, you both could perfectly figure out my point of confusion :-)
– Harshada Kelkar
Aug 29 at 2:43
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
What is the degree of the exponential polynomial :$$ e^x * (x^2 + x + 1)$$?
I am getting confused because of the exponential term there...
polynomials exponential-function
edited Aug 28 at 17:22
Deepesh Meena
3,0982824
asked Aug 28 at 17:19
Harshada Kelkar
265
What is the degree of the exponential polynomial :$$ e^x * (x^2 + x + 1)$$?
I am getting confused because of the exponential term there...
polynomials exponential-function
edited Aug 28 at 17:22
Deepesh Meena
3,0982824
asked Aug 28 at 17:19
Harshada Kelkar
265
edited Aug 28 at 17:22
Deepesh Meena
3,0982824
edited Aug 28 at 17:22
Deepesh Meena
3,0982824
edited Aug 28 at 17:22
Deepesh Meena
3,0982824
3,0982824
asked Aug 28 at 17:19
Harshada Kelkar
265
asked Aug 28 at 17:19
Harshada Kelkar
265
asked Aug 28 at 17:19
Harshada Kelkar
265
265
4
never heard of the word degree used in this situation.
– Will Jagy
Aug 28 at 17:21
3
It's a polynomial of degree $2$ times an exponential. In what context are you trying to assign it a degree?
– Robert Israel
Aug 28 at 17:24
Actually, I am trying to solve integration of x^2*e^x using method of undetermined coefficients with A0, A1,A2....... previously, we had used this method for polynomials with degree <= 2.... was trying to apply the same logic here.... Analytically integrating x^2*e^x gives e^x*(x^2-2x+2) ..... So thought if this is a polynomial of degree <= 2 or not.... So posted this question in that context..
– Harshada Kelkar
Aug 28 at 17:41
If $P$ is a polynomial then any anti-derivative of $P(x)e^x$ is $K+Q(x)e^x$ where $Q$ is a polynomial with $deg (Q)=deg (P)$. Use integration by parts to prove this by induction on $deg (P). $ That is, $int e^xP(x)dx=$ $int P(x)de^x=$ $e^xP(x)-int e^xdP(x)=$ $e^xP(x)-int e^xP'(x)dx.$
– DanielWainfleet
Aug 28 at 18:33
Thanks, Daniel and Christian, you both could perfectly figure out my point of confusion :-)
– Harshada Kelkar
Aug 29 at 2:43
add a comment |Â
4
never heard of the word degree used in this situation.
– Will Jagy
Aug 28 at 17:21
3
It's a polynomial of degree $2$ times an exponential. In what context are you trying to assign it a degree?
– Robert Israel
Aug 28 at 17:24
Actually, I am trying to solve integration of x^2*e^x using method of undetermined coefficients with A0, A1,A2....... previously, we had used this method for polynomials with degree <= 2.... was trying to apply the same logic here.... Analytically integrating x^2*e^x gives e^x*(x^2-2x+2) ..... So thought if this is a polynomial of degree <= 2 or not.... So posted this question in that context..
– Harshada Kelkar
Aug 28 at 17:41
If $P$ is a polynomial then any anti-derivative of $P(x)e^x$ is $K+Q(x)e^x$ where $Q$ is a polynomial with $deg (Q)=deg (P)$. Use integration by parts to prove this by induction on $deg (P). $ That is, $int e^xP(x)dx=$ $int P(x)de^x=$ $e^xP(x)-int e^xdP(x)=$ $e^xP(x)-int e^xP'(x)dx.$
– DanielWainfleet
Aug 28 at 18:33
Thanks, Daniel and Christian, you both could perfectly figure out my point of confusion :-)
– Harshada Kelkar
Aug 29 at 2:43
4
4
never heard of the word degree used in this situation.
– Will Jagy
Aug 28 at 17:21
never heard of the word degree used in this situation.
– Will Jagy
Aug 28 at 17:21
3
3
It's a polynomial of degree $2$ times an exponential. In what context are you trying to assign it a degree?
– Robert Israel
Aug 28 at 17:24
It's a polynomial of degree $2$ times an exponential. In what context are you trying to assign it a degree?
– Robert Israel
Aug 28 at 17:24
Actually, I am trying to solve integration of x^2*e^x using method of undetermined coefficients with A0, A1,A2....... previously, we had used this method for polynomials with degree <= 2.... was trying to apply the same logic here.... Analytically integrating x^2*e^x gives e^x*(x^2-2x+2) ..... So thought if this is a polynomial of degree <= 2 or not.... So posted this question in that context..
– Harshada Kelkar
Aug 28 at 17:41
Actually, I am trying to solve integration of x^2*e^x using method of undetermined coefficients with A0, A1,A2....... previously, we had used this method for polynomials with degree <= 2.... was trying to apply the same logic here.... Analytically integrating x^2*e^x gives e^x*(x^2-2x+2) ..... So thought if this is a polynomial of degree <= 2 or not.... So posted this question in that context..
– Harshada Kelkar
Aug 28 at 17:41
If $P$ is a polynomial then any anti-derivative of $P(x)e^x$ is $K+Q(x)e^x$ where $Q$ is a polynomial with $deg (Q)=deg (P)$. Use integration by parts to prove this by induction on $deg (P). $ That is, $int e^xP(x)dx=$ $int P(x)de^x=$ $e^xP(x)-int e^xdP(x)=$ $e^xP(x)-int e^xP'(x)dx.$
– DanielWainfleet
Aug 28 at 18:33
If $P$ is a polynomial then any anti-derivative of $P(x)e^x$ is $K+Q(x)e^x$ where $Q$ is a polynomial with $deg (Q)=deg (P)$. Use integration by parts to prove this by induction on $deg (P). $ That is, $int e^xP(x)dx=$ $int P(x)de^x=$ $e^xP(x)-int e^xdP(x)=$ $e^xP(x)-int e^xP'(x)dx.$
– DanielWainfleet
Aug 28 at 18:33
Thanks, Daniel and Christian, you both could perfectly figure out my point of confusion :-)
– Harshada Kelkar
Aug 29 at 2:43
Thanks, Daniel and Christian, you both could perfectly figure out my point of confusion :-)
– Harshada Kelkar
Aug 29 at 2:43
add a comment |Â
1 Answer
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The idea is that one considers the vector space $V$ of all functions $f$ of the form $f(x)=e^x,p(x)$, where $xmapsto p(x)$ is a polynomial of degree $leq 2$. This $V$ has dimension $3$. The operator $D:>fmapsto f'$ is linear, maps $V$ to itself, and has rank $3$, since $D(e^x,x^k)=e^x q(x)$ is again in $V$, and $q$ is of degree $k$. It follows that $D:>Vto V$ is nonsingular, in particular: surjective. This implies that any given function $f(x)=e^x(a_0+a_1x+a_2 x^2)in V$ is the derivative of some function $F(x)=e^x(b_0+b_1x+b_2 x^2)in V$. The $b_k$ can be determined by comparing coefficients of $F'$ and the given $f$.
answered Aug 28 at 18:34
Christian Blatter
165k7109311
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
The idea is that one considers the vector space $V$ of all functions $f$ of the form $f(x)=e^x,p(x)$, where $xmapsto p(x)$ is a polynomial of degree $leq 2$. This $V$ has dimension $3$. The operator $D:>fmapsto f'$ is linear, maps $V$ to itself, and has rank $3$, since $D(e^x,x^k)=e^x q(x)$ is again in $V$, and $q$ is of degree $k$. It follows that $D:>Vto V$ is nonsingular, in particular: surjective. This implies that any given function $f(x)=e^x(a_0+a_1x+a_2 x^2)in V$ is the derivative of some function $F(x)=e^x(b_0+b_1x+b_2 x^2)in V$. The $b_k$ can be determined by comparing coefficients of $F'$ and the given $f$.
answered Aug 28 at 18:34
Christian Blatter
165k7109311
add a comment |Â
up vote
0
down vote
accepted
The idea is that one considers the vector space $V$ of all functions $f$ of the form $f(x)=e^x,p(x)$, where $xmapsto p(x)$ is a polynomial of degree $leq 2$. This $V$ has dimension $3$. The operator $D:>fmapsto f'$ is linear, maps $V$ to itself, and has rank $3$, since $D(e^x,x^k)=e^x q(x)$ is again in $V$, and $q$ is of degree $k$. It follows that $D:>Vto V$ is nonsingular, in particular: surjective. This implies that any given function $f(x)=e^x(a_0+a_1x+a_2 x^2)in V$ is the derivative of some function $F(x)=e^x(b_0+b_1x+b_2 x^2)in V$. The $b_k$ can be determined by comparing coefficients of $F'$ and the given $f$.
answered Aug 28 at 18:34
Christian Blatter
165k7109311
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
The idea is that one considers the vector space $V$ of all functions $f$ of the form $f(x)=e^x,p(x)$, where $xmapsto p(x)$ is a polynomial of degree $leq 2$. This $V$ has dimension $3$. The operator $D:>fmapsto f'$ is linear, maps $V$ to itself, and has rank $3$, since $D(e^x,x^k)=e^x q(x)$ is again in $V$, and $q$ is of degree $k$. It follows that $D:>Vto V$ is nonsingular, in particular: surjective. This implies that any given function $f(x)=e^x(a_0+a_1x+a_2 x^2)in V$ is the derivative of some function $F(x)=e^x(b_0+b_1x+b_2 x^2)in V$. The $b_k$ can be determined by comparing coefficients of $F'$ and the given $f$.
answered Aug 28 at 18:34
Christian Blatter
165k7109311
The idea is that one considers the vector space $V$ of all functions $f$ of the form $f(x)=e^x,p(x)$, where $xmapsto p(x)$ is a polynomial of degree $leq 2$. This $V$ has dimension $3$. The operator $D:>fmapsto f'$ is linear, maps $V$ to itself, and has rank $3$, since $D(e^x,x^k)=e^x q(x)$ is again in $V$, and $q$ is of degree $k$. It follows that $D:>Vto V$ is nonsingular, in particular: surjective. This implies that any given function $f(x)=e^x(a_0+a_1x+a_2 x^2)in V$ is the derivative of some function $F(x)=e^x(b_0+b_1x+b_2 x^2)in V$. The $b_k$ can be determined by comparing coefficients of $F'$ and the given $f$.
answered Aug 28 at 18:34
Christian Blatter
165k7109311
answered Aug 28 at 18:34
Christian Blatter
165k7109311
answered Aug 28 at 18:34
Christian Blatter
165k7109311
answered Aug 28 at 18:34
Christian Blatter
165k7109311
165k7109311
add a comment |Â
add a comment |Â
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4
never heard of the word degree used in this situation.
– Will Jagy
Aug 28 at 17:21
3
It's a polynomial of degree $2$ times an exponential. In what context are you trying to assign it a degree?
– Robert Israel
Aug 28 at 17:24
Actually, I am trying to solve integration of x^2*e^x using method of undetermined coefficients with A0, A1,A2....... previously, we had used this method for polynomials with degree <= 2.... was trying to apply the same logic here.... Analytically integrating x^2*e^x gives e^x*(x^2-2x+2) ..... So thought if this is a polynomial of degree <= 2 or not.... So posted this question in that context..
– Harshada Kelkar
Aug 28 at 17:41
If $P$ is a polynomial then any anti-derivative of $P(x)e^x$ is $K+Q(x)e^x$ where $Q$ is a polynomial with $deg (Q)=deg (P)$. Use integration by parts to prove this by induction on $deg (P). $ That is, $int e^xP(x)dx=$ $int P(x)de^x=$ $e^xP(x)-int e^xdP(x)=$ $e^xP(x)-int e^xP'(x)dx.$
– DanielWainfleet
Aug 28 at 18:33
Thanks, Daniel and Christian, you both could perfectly figure out my point of confusion :-)
– Harshada Kelkar
Aug 29 at 2:43