I continued for a few more degrees, and the pattern continued as well. I thought maybe it is somehow possible that I made a mistake that only appears for odd degrees, so I checked if the answer held up with the original problem. I asked Maple for a random polynomial of degree ≤ 3, and integrated it against f ( x ) = − 15 x 2 + 15 x − 3 2 . The answer was the random polynomial at x = 1 2 . Perhaps no mistake was made. But why is this happening? To answer this we will state the question again, but in more general terms. Scott Cameron A Linear Algebra Problem Related to Legendre Polynomials
• Generalizing the Problem Find f n ( x ) such that � 1 g ( c ) = f n ( x ) g ( x ) dx 0 where g ( x ) and f n ( x ) are polynomials of degree ≤ n . Scott Cameron A Linear Algebra Problem Related to Legendre Polynomials
• Generalizing the Problem Find f n ( x ) such that � 1 g ( c ) = f n ( x ) g ( x ) dx 0 where g ( x ) and f n ( x ) are polynomials of degree ≤ n . Now let us write our question in terms of the following proposition. Scott Cameron A Linear Algebra Problem Related to Legendre Polynomials
Proposition Let f n ( x ) be as previously defined. Then when c = 1 2 we have f 2 m + 1 ( x ) = f 2 m ( x ) for m ∈ N . Scott Cameron A Linear Algebra Problem Related to Legendre Polynomials
Proposition Let f n ( x ) be as previously defined. Then when c = 1 2 we have f 2 m + 1 ( x ) = f 2 m ( x ) for m ∈ N . So then let n � a k x k f n ( x ) = k = 0 Scott Cameron A Linear Algebra Problem Related to Legendre Polynomials
Proposition Let f n ( x ) be as previously defined. Then when c = 1 2 we have f 2 m + 1 ( x ) = f 2 m ( x ) for m ∈ N . So then let n � a k x k f n ( x ) = k = 0 and n � b k x k g ( x ) = k = 0 Scott Cameron A Linear Algebra Problem Related to Legendre Polynomials
Proposition Let f n ( x ) be as previously defined. Then when c = 1 2 we have f 2 m + 1 ( x ) = f 2 m ( x ) for m ∈ N . So then let n � a k x k f n ( x ) = k = 0 and n � b k x k g ( x ) = k = 0 Solving in the same manner leads to Scott Cameron A Linear Algebra Problem Related to Legendre Polynomials
1 1 1 1 . . . a 0 1 2 3 n + 1 1 1 1 1 . . . a 1 c 2 3 4 n + 2 1 1 1 1 c 2 a 2 . . . = 3 4 5 n + 3 . . . . . . ... . . . . . . . . . . . . c n 1 1 1 1 a n . . . n + 1 n + 2 n + 3 2 n + 1 Scott Cameron A Linear Algebra Problem Related to Legendre Polynomials
1 1 1 1 . . . a 0 1 2 3 n + 1 1 1 1 1 . . . a 1 c 2 3 4 n + 2 1 1 1 1 c 2 a 2 . . . = 3 4 5 n + 3 . . . . . . ... . . . . . . . . . . . . c n 1 1 1 1 a n . . . n + 1 n + 2 n + 3 2 n + 1 Now write this as Ha = c . Scott Cameron A Linear Algebra Problem Related to Legendre Polynomials
1 1 1 1 . . . a 0 1 2 3 n + 1 1 1 1 1 . . . a 1 c 2 3 4 n + 2 1 1 1 1 c 2 a 2 . . . = 3 4 5 n + 3 . . . . . . ... . . . . . . . . . . . . c n 1 1 1 1 a n . . . n + 1 n + 2 n + 3 2 n + 1 Now write this as Ha = c . 1 The choice of H is because matrices of this form ( H i , j = i + j − 1 ) are known as Hilbert matrices. Scott Cameron A Linear Algebra Problem Related to Legendre Polynomials
To continue we need to know the inverse of the matrix H. Scott Cameron A Linear Algebra Problem Related to Legendre Polynomials
To continue we need to know the inverse of the matrix H. Luckily Hilbert matrices have a known formula for the entries of their inverse. Scott Cameron A Linear Algebra Problem Related to Legendre Polynomials
To continue we need to know the inverse of the matrix H. Luckily Hilbert matrices have a known formula for the entries of their inverse. � n + i �� n + j � 2 �� i + j − 2 ( H i , j ) − 1 = ( − 1 ) i + j − 1 ( i + j − 1 ) . i + j − 1 i + j − 1 i − 1 Scott Cameron A Linear Algebra Problem Related to Legendre Polynomials
To continue we need to know the inverse of the matrix H. Luckily Hilbert matrices have a known formula for the entries of their inverse. � n + i �� n + j � 2 �� i + j − 2 ( H i , j ) − 1 = ( − 1 ) i + j − 1 ( i + j − 1 ) . i + j − 1 i + j − 1 i − 1 Before continuing we need a definition. Scott Cameron A Linear Algebra Problem Related to Legendre Polynomials
Definition Let h n , i ( c ) be the polynomial created by taking the dot product of the i th row of H − 1 and c . Scott Cameron A Linear Algebra Problem Related to Legendre Polynomials
Definition Let h n , i ( c ) be the polynomial created by taking the dot product of the i th row of H − 1 and c . These polynomials determine the coefficients of f n ( x ) . That is, n + 1 � h n , k ( c ) x k − 1 h n , i ( c ) = a i − 1 , or f n ( x ) = k = 1 Scott Cameron A Linear Algebra Problem Related to Legendre Polynomials
Definition Let h n , i ( c ) be the polynomial created by taking the dot product of the i th row of H − 1 and c . These polynomials determine the coefficients of f n ( x ) . That is, n + 1 � h n , k ( c ) x k − 1 h n , i ( c ) = a i − 1 , or f n ( x ) = k = 1 Let’s do an example to clarify. Scott Cameron A Linear Algebra Problem Related to Legendre Polynomials
Example: If n = 2, then 1 1 1 2 3 1 1 1 . H = 2 3 4 1 1 1 3 4 5 Scott Cameron A Linear Algebra Problem Related to Legendre Polynomials
Example: If n = 2, then 1 1 1 2 3 1 1 1 . H = 2 3 4 1 1 1 3 4 5 Using our formula, 9 − 36 30 H − 1 = . − 36 192 − 180 30 − 180 180 Scott Cameron A Linear Algebra Problem Related to Legendre Polynomials
Example: If n = 2, then 1 1 1 2 3 1 1 1 . H = 2 3 4 1 1 1 3 4 5 Using our formula, 9 − 36 30 H − 1 = . − 36 192 − 180 30 − 180 180 So we have n = 2 and can see that 1 ≤ i ≤ 3. Thus h 2 , 1 ( c ) = 9 − 36 c + 30 c 2 , h 2 , 2 ( c ) = − 36 + 192 c − 180 c 2 , h 2 , 3 ( c ) = 30 − 180 c + 180 c 2 , f n ( x ) = h 2 , 1 ( c ) + h 2 , 2 ( c ) x + h 2 , 3 ( c ) x 2 Scott Cameron A Linear Algebra Problem Related to Legendre Polynomials
Now back to our problem. Scott Cameron A Linear Algebra Problem Related to Legendre Polynomials
Now back to our problem. If n is odd, and f n ( x ) = f n − 1 ( x ) when c = 1 2 , then the degree of f n ( x ) is n − 1. This means that a n = 0 in f n ( x ) . Scott Cameron A Linear Algebra Problem Related to Legendre Polynomials
Now back to our problem. If n is odd, and f n ( x ) = f n − 1 ( x ) when c = 1 2 , then the degree of f n ( x ) is n − 1. This means that a n = 0 in f n ( x ) . This would correspond to the polynomial formed by the bottom row of H − 1 having a root at c = 1 2 . Using our definition, this can be written as h n , n + 1 ( c ) having a root at c = 1 2 . Scott Cameron A Linear Algebra Problem Related to Legendre Polynomials
Now back to our problem. If n is odd, and f n ( x ) = f n − 1 ( x ) when c = 1 2 , then the degree of f n ( x ) is n − 1. This means that a n = 0 in f n ( x ) . This would correspond to the polynomial formed by the bottom row of H − 1 having a root at c = 1 2 . Using our definition, this can be written as h n , n + 1 ( c ) having a root at c = 1 2 . Using the formula for H − 1 we can write h n , n + 1 ( c ) as h n , n + 1 ( c ) = n + 1 � 2 � 2 n + 1 �� n + j �� n + j − 1 � ( − 1 ) n + j + 1 ( n + j ) c j − 1 n + j n + j n j = 1 Scott Cameron A Linear Algebra Problem Related to Legendre Polynomials
Rearranging, simplifying, and shifting indicies gives us Scott Cameron A Linear Algebra Problem Related to Legendre Polynomials
Rearranging, simplifying, and shifting indicies gives us n � 2 n + 1 � � n �� n + k � h n , n + 1 ( c ) = ( − 1 ) n ( 2 n + 1 ) � ( − 1 ) k c k . n k k k = 0 Scott Cameron A Linear Algebra Problem Related to Legendre Polynomials
Rearranging, simplifying, and shifting indicies gives us n � 2 n + 1 � � n �� n + k � h n , n + 1 ( c ) = ( − 1 ) n ( 2 n + 1 ) � ( − 1 ) k c k . n k k k = 0 This form is exactly what we need. The sum is our polynomial, and then we have a scaling factor outside. Scott Cameron A Linear Algebra Problem Related to Legendre Polynomials
Rearranging, simplifying, and shifting indicies gives us n � 2 n + 1 � � n �� n + k � h n , n + 1 ( c ) = ( − 1 ) n ( 2 n + 1 ) � ( − 1 ) k c k . n k k k = 0 This form is exactly what we need. The sum is our polynomial, and then we have a scaling factor outside. All we need now is a definition to solve our problem. Scott Cameron A Linear Algebra Problem Related to Legendre Polynomials
Definition The shifted Legendre polynomials, denoted ˜ P n ( x ) , are given by n � n �� n + k � P n ( x ) = ( − 1 ) n ˜ � ( − 1 ) k x k . k k k = 0 Scott Cameron A Linear Algebra Problem Related to Legendre Polynomials
Definition The shifted Legendre polynomials, denoted ˜ P n ( x ) , are given by n � n �� n + k � P n ( x ) = ( − 1 ) n ˜ � ( − 1 ) k x k . k k k = 0 Looking back at our expresion for h n , n + 1 ( c ) , n � 2 n + 1 � � n �� n + k � h n , n + 1 ( c ) = ( − 1 ) n ( 2 n + 1 ) � ( − 1 ) k c k , n k k k = 0 Scott Cameron A Linear Algebra Problem Related to Legendre Polynomials
Definition The shifted Legendre polynomials, denoted ˜ P n ( x ) , are given by n � n �� n + k � P n ( x ) = ( − 1 ) n ˜ � ( − 1 ) k x k . k k k = 0 Looking back at our expresion for h n , n + 1 ( c ) , n � 2 n + 1 � � n �� n + k � h n , n + 1 ( c ) = ( − 1 ) n ( 2 n + 1 ) � ( − 1 ) k c k , n k k k = 0 we can now see that h n , n + 1 ( c ) is just a multiple of ˜ P n ( c ) . Scott Cameron A Linear Algebra Problem Related to Legendre Polynomials
The shifted Legendre polynomials are so named because they are, unsurprisingly, Legendre polynomials which have been shifted. Scott Cameron A Linear Algebra Problem Related to Legendre Polynomials
The shifted Legendre polynomials are so named because they are, unsurprisingly, Legendre polynomials which have been shifted. The shift is given by sending x to 2 x − 1. That is, if we denote the Legendre polynomials by P n ( x ) , then P n ( 2 x − 1 ) = ˜ P n ( x ) . Scott Cameron A Linear Algebra Problem Related to Legendre Polynomials
The shifted Legendre polynomials are so named because they are, unsurprisingly, Legendre polynomials which have been shifted. The shift is given by sending x to 2 x − 1. That is, if we denote the Legendre polynomials by P n ( x ) , then P n ( 2 x − 1 ) = ˜ P n ( x ) . The Legendre polynomials are known to have x = 0 as a root when their degree is odd. Therefore, the shifted Legendre polynomials must have a root at x = 1 2 when their degree is odd. Scott Cameron A Linear Algebra Problem Related to Legendre Polynomials
So, h n , n + 1 ( c ) = a n and these are just multiples of the shifted Legendre polynomials. Scott Cameron A Linear Algebra Problem Related to Legendre Polynomials
So, h n , n + 1 ( c ) = a n and these are just multiples of the shifted Legendre polynomials. The shifted Legendre polynomials have a root at 1 2 when their degree is odd. Scott Cameron A Linear Algebra Problem Related to Legendre Polynomials
So, h n , n + 1 ( c ) = a n and these are just multiples of the shifted Legendre polynomials. The shifted Legendre polynomials have a root at 1 2 when their degree is odd. Therefore, if n is odd and c = 1 2 , then a n = 0. This ends up forcing f n ( x ) = f n − 1 ( x ) , thus answering our question. Scott Cameron A Linear Algebra Problem Related to Legendre Polynomials
The appearance of Legendre polynomials is unsurprising since the original problem involves an inner product of polynomials. Scott Cameron A Linear Algebra Problem Related to Legendre Polynomials
The appearance of Legendre polynomials is unsurprising since the original problem involves an inner product of polynomials. However, when I found this solution, It gave me another question. Scott Cameron A Linear Algebra Problem Related to Legendre Polynomials
The appearance of Legendre polynomials is unsurprising since the original problem involves an inner product of polynomials. However, when I found this solution, It gave me another question. If h n , n + 1 ( c ) is always just a multiple of a shifted Legendre polynomial, what do other rows correspond to? Scott Cameron A Linear Algebra Problem Related to Legendre Polynomials
The appearance of Legendre polynomials is unsurprising since the original problem involves an inner product of polynomials. However, when I found this solution, It gave me another question. If h n , n + 1 ( c ) is always just a multiple of a shifted Legendre polynomial, what do other rows correspond to? That is, how does h n , 1 ( c ) change as we change n ? h n , 2 ( c ) ? etc. Scott Cameron A Linear Algebra Problem Related to Legendre Polynomials
• Another Approach to the Problem and the Other Rows of H − 1 Until now, I have been using just basic calculus and matrix operations to answer these questions. There is however a better way. Scott Cameron A Linear Algebra Problem Related to Legendre Polynomials
• Another Approach to the Problem and the Other Rows of H − 1 Until now, I have been using just basic calculus and matrix operations to answer these questions. There is however a better way. Theorem (Riesz Represention Theorem) If we have some finite dimensional vector space, V, and some linear functional φ on V, then there is a unique vector u ∈ V such that φ ( v ) = � v , u � for all v ∈ V. Scott Cameron A Linear Algebra Problem Related to Legendre Polynomials
• Another Approach to the Problem and the Other Rows of H − 1 Until now, I have been using just basic calculus and matrix operations to answer these questions. There is however a better way. Theorem (Riesz Represention Theorem) If we have some finite dimensional vector space, V, and some linear functional φ on V, then there is a unique vector u ∈ V such that φ ( v ) = � v , u � for all v ∈ V. We can interpret our problem in terms of this theorem. The integral is an inner product, g ( x ) corresponds to v , f n ( x ) corresponds to u , and evaluation at c is a linear functional. Scott Cameron A Linear Algebra Problem Related to Legendre Polynomials
This theorem has the consequence of allowing us to write n P k ( c )˜ ˜ P k ( x ) � f n ( x ) = � 1 ˜ P k ( x ) 2 dx k = 0 0 n � ( 2 k + 1 )˜ P k ( c )˜ = P k ( x ) k = 0 Scott Cameron A Linear Algebra Problem Related to Legendre Polynomials
This theorem has the consequence of allowing us to write n P k ( c )˜ ˜ P k ( x ) � f n ( x ) = � 1 ˜ P k ( x ) 2 dx k = 0 0 n � ( 2 k + 1 )˜ P k ( c )˜ = P k ( x ) k = 0 It should be noted that this expression for f n ( x ) shows us that it is actually a familier concept in the study of orthogonal polynomials. Scott Cameron A Linear Algebra Problem Related to Legendre Polynomials
This theorem has the consequence of allowing us to write n P k ( c )˜ ˜ P k ( x ) � f n ( x ) = � 1 ˜ P k ( x ) 2 dx k = 0 0 n � ( 2 k + 1 )˜ P k ( c )˜ = P k ( x ) k = 0 It should be noted that this expression for f n ( x ) shows us that it is actually a familier concept in the study of orthogonal polynomials. In this form f n ( x ) would be called the kernel of the shifted Legendre polynomials. Therefore what I am studying can be interpreted as looking at how the coefficients of this kernel change with n , and with c . Scott Cameron A Linear Algebra Problem Related to Legendre Polynomials
Moving back to h n , i ( c ) , we can use the previous expression of f n ( x ) and the fact that n + 1 � h n , k ( c ) x k − 1 f n ( x ) = k = 1 Scott Cameron A Linear Algebra Problem Related to Legendre Polynomials
Moving back to h n , i ( c ) , we can use the previous expression of f n ( x ) and the fact that n + 1 � h n , k ( c ) x k − 1 f n ( x ) = k = 1 To find that � k n �� k + i − 1 � ( 2 k + 1 )˜ � ( − 1 ) k + i − 1 h n , i ( c ) = P k ( c ) i − 1 i − 1 k = i − 1 Scott Cameron A Linear Algebra Problem Related to Legendre Polynomials
Moving back to h n , i ( c ) , we can use the previous expression of f n ( x ) and the fact that n + 1 � h n , k ( c ) x k − 1 f n ( x ) = k = 1 To find that � k n �� k + i − 1 � ( 2 k + 1 )˜ � ( − 1 ) k + i − 1 h n , i ( c ) = P k ( c ) i − 1 i − 1 k = i − 1 Using this equation, I wanted to find a generating function for these polynomials. Scott Cameron A Linear Algebra Problem Related to Legendre Polynomials
First let us focus on i = 1. Scott Cameron A Linear Algebra Problem Related to Legendre Polynomials
First let us focus on i = 1. If i = 1 we have n � ( − 1 ) k ( 2 k + 1 )˜ h n , 1 ( c ) = P k ( c ) . k = 0 Scott Cameron A Linear Algebra Problem Related to Legendre Polynomials
First let us focus on i = 1. If i = 1 we have n � ( − 1 ) k ( 2 k + 1 )˜ h n , 1 ( c ) = P k ( c ) . k = 0 Now we need to make use of a relationship between the shifted Legendre polynomials. ( n + 1 )˜ P n + 1 ( x ) = ( 2 n + 1 )( 2 x − 1 )˜ P n ( x ) − n ˜ P n − 1 ( x ) Scott Cameron A Linear Algebra Problem Related to Legendre Polynomials
First let us focus on i = 1. If i = 1 we have n � ( − 1 ) k ( 2 k + 1 )˜ h n , 1 ( c ) = P k ( c ) . k = 0 Now we need to make use of a relationship between the shifted Legendre polynomials. ( n + 1 )˜ P n + 1 ( x ) = ( 2 n + 1 )( 2 x − 1 )˜ P n ( x ) − n ˜ P n − 1 ( x ) Combining the two expressions, it can be shown that h n , 1 ( c ) = ( − 1 ) n ( n + 1 ) � � P n ( c ) + ˜ ˜ P n + 1 ( c ) 2 c Scott Cameron A Linear Algebra Problem Related to Legendre Polynomials
First let us focus on i = 1. If i = 1 we have n � ( − 1 ) k ( 2 k + 1 )˜ h n , 1 ( c ) = P k ( c ) . k = 0 Now we need to make use of a relationship between the shifted Legendre polynomials. ( n + 1 )˜ P n + 1 ( x ) = ( 2 n + 1 )( 2 x − 1 )˜ P n ( x ) − n ˜ P n − 1 ( x ) Combining the two expressions, it can be shown that h n , 1 ( c ) = ( − 1 ) n ( n + 1 ) � � P n ( c ) + ˜ ˜ P n + 1 ( c ) 2 c Using this we can find a generating function for h n , 1 ( c ) . Scott Cameron A Linear Algebra Problem Related to Legendre Polynomials
Rearranging, multiplying by x n + 1 , and summing over n yields h n , 1 ( c ) x n + 1 ∞ ∞ 1 � P n ( c )( − x ) n + 1 + ˜ P n + 1 ( c )( − x ) n + 1 � � � ˜ = − n + 1 2 c n = 0 n = 0 � � ∞ ∞ = − 1 P n ( c )( − x ) n + P n ( c )( − x ) n − ˜ ˜ ˜ � � − x P 0 ( c ) 2 c n = 0 n = 0 � � ∞ = − 1 P n ( c ) x n − 1 ˜ � ( 1 − x ) 2 c n = 0 � � = − 1 1 − x 1 + 2 ( 2 c − 1 ) x + x 2 − 1 . � 2 c Scott Cameron A Linear Algebra Problem Related to Legendre Polynomials
Rearranging, multiplying by x n + 1 , and summing over n yields h n , 1 ( c ) x n + 1 ∞ ∞ 1 � P n ( c )( − x ) n + 1 + ˜ P n + 1 ( c )( − x ) n + 1 � � � ˜ = − n + 1 2 c n = 0 n = 0 � � ∞ ∞ = − 1 P n ( c )( − x ) n + P n ( c )( − x ) n − ˜ ˜ ˜ � � − x P 0 ( c ) 2 c n = 0 n = 0 � � ∞ = − 1 P n ( c ) x n − 1 ˜ � ( 1 − x ) 2 c n = 0 � � = − 1 1 − x 1 + 2 ( 2 c − 1 ) x + x 2 − 1 . � 2 c Now we take the derivative with respect to x of both sides which gives us the generating function. Scott Cameron A Linear Algebra Problem Related to Legendre Polynomials
Let H 1 ( x ) be the generating function for h n , 1 ( c ) . Scott Cameron A Linear Algebra Problem Related to Legendre Polynomials
Let H 1 ( x ) be the generating function for h n , 1 ( c ) . Then we have from the previous slide � � H 1 ( x ) = − 1 d 1 − x 1 + 2 ( 2 c − 1 ) x + x 2 − 1 � 2 c dx 1 + x = 3 ( 1 + 2 ( 2 c − 1 ) x + x 2 ) 2 Scott Cameron A Linear Algebra Problem Related to Legendre Polynomials
Now using our expression for H 1 ( x ) , along with � k n �� k + i − 1 � ( 2 k + 1 )˜ � ( − 1 ) k + i − 1 h n , i ( c ) = P k ( c ) i − 1 i − 1 k = i − 1 we can find an expression for the generating function of any h n , i ( c ) , denoted H i ( x ) . Scott Cameron A Linear Algebra Problem Related to Legendre Polynomials
Sparing the details of the calculation, as it is more complicated but similar to the derivation of H 1 ( x ) , the final final result is given by � � ( − x ) i − 1 d 2 i − 2 x i − 1 ( 1 − x )( 1 + x ) H i ( x ) = ( 1 − x )(( i − 1 )!) 2 dx 2 i − 2 3 ( 1 + 2 ( 2 c − 1 ) x + x 2 ) 2 Scott Cameron A Linear Algebra Problem Related to Legendre Polynomials
Sparing the details of the calculation, as it is more complicated but similar to the derivation of H 1 ( x ) , the final final result is given by � � ( − x ) i − 1 d 2 i − 2 x i − 1 ( 1 − x )( 1 + x ) H i ( x ) = ( 1 − x )(( i − 1 )!) 2 dx 2 i − 2 3 ( 1 + 2 ( 2 c − 1 ) x + x 2 ) 2 If we let j = i − 1 then this takes on a nicer form of � � d 2 j ( − x ) j x j ( 1 − x )( 1 + x ) H j + 1 ( x ) = ( 1 − x )( j !) 2 dx 2 j 3 ( 1 + 2 ( 2 c − 1 ) x + x 2 ) 2 Scott Cameron A Linear Algebra Problem Related to Legendre Polynomials
Sparing the details of the calculation, as it is more complicated but similar to the derivation of H 1 ( x ) , the final final result is given by � � ( − x ) i − 1 d 2 i − 2 x i − 1 ( 1 − x )( 1 + x ) H i ( x ) = ( 1 − x )(( i − 1 )!) 2 dx 2 i − 2 3 ( 1 + 2 ( 2 c − 1 ) x + x 2 ) 2 If we let j = i − 1 then this takes on a nicer form of � � d 2 j ( − x ) j x j ( 1 − x )( 1 + x ) H j + 1 ( x ) = ( 1 − x )( j !) 2 dx 2 j 3 ( 1 + 2 ( 2 c − 1 ) x + x 2 ) 2 So we have accomplished our goal of finding the generating function for the polynomials h n , i ( c ) . Scott Cameron A Linear Algebra Problem Related to Legendre Polynomials
Here are the first couple generating functions. Scott Cameron A Linear Algebra Problem Related to Legendre Polynomials
Here are the first couple generating functions. 1 + x H 1 ( x ) = 3 ( 1 + 2 ( 2 c − 1 ) x + x 2 ) 2 Scott Cameron A Linear Algebra Problem Related to Legendre Polynomials
Here are the first couple generating functions. 1 + x H 1 ( x ) = 3 ( 1 + 2 ( 2 c − 1 ) x + x 2 ) 2 2 ) x 2 − ( c 2 − c − 1 ) x + c − 1 ( c − 1 � � H 2 ( x ) = 12 x ( 1 + x ) 2 7 ( 1 + 2 ( 2 c − 1 ) x + x 2 ) 2 Scott Cameron A Linear Algebra Problem Related to Legendre Polynomials
Here are the first couple generating functions. 1 + x H 1 ( x ) = 3 ( 1 + 2 ( 2 c − 1 ) x + x 2 ) 2 2 ) x 2 − ( c 2 − c − 1 ) x + c − 1 ( c − 1 � � H 2 ( x ) = 12 x ( 1 + x ) 2 7 ( 1 + 2 ( 2 c − 1 ) x + x 2 ) 2 H 3 ( x ) is a bit long so I will break it up a bit. Scott Cameron A Linear Algebra Problem Related to Legendre Polynomials
Here are the first couple generating functions. 1 + x H 1 ( x ) = 3 ( 1 + 2 ( 2 c − 1 ) x + x 2 ) 2 2 ) x 2 − ( c 2 − c − 1 ) x + c − 1 ( c − 1 � � H 2 ( x ) = 12 x ( 1 + x ) 2 7 ( 1 + 2 ( 2 c − 1 ) x + x 2 ) 2 H 3 ( x ) is a bit long so I will break it up a bit. The denominator is the same as the others but with exponent 11 2 . Scott Cameron A Linear Algebra Problem Related to Legendre Polynomials
Here are the first couple generating functions. 1 + x H 1 ( x ) = 3 ( 1 + 2 ( 2 c − 1 ) x + x 2 ) 2 2 ) x 2 − ( c 2 − c − 1 ) x + c − 1 ( c − 1 � � H 2 ( x ) = 12 x ( 1 + x ) 2 7 ( 1 + 2 ( 2 c − 1 ) x + x 2 ) 2 H 3 ( x ) is a bit long so I will break it up a bit. The denominator is the same as the others but with exponent 11 2 . There is also the term 180 x 2 ( 1 + x ) which is also similar to the others. Scott Cameron A Linear Algebra Problem Related to Legendre Polynomials
Here are the first couple generating functions. 1 + x H 1 ( x ) = 3 ( 1 + 2 ( 2 c − 1 ) x + x 2 ) 2 2 ) x 2 − ( c 2 − c − 1 ) x + c − 1 ( c − 1 � � H 2 ( x ) = 12 x ( 1 + x ) 2 7 ( 1 + 2 ( 2 c − 1 ) x + x 2 ) 2 H 3 ( x ) is a bit long so I will break it up a bit. The denominator is the same as the others but with exponent 11 2 . There is also the term 180 x 2 ( 1 + x ) which is also similar to the others. The part I want to show however, is the polynomial in the numerator. Scott Cameron A Linear Algebra Problem Related to Legendre Polynomials
For H 1 ( x ) the polynomial would just be 1. Scott Cameron A Linear Algebra Problem Related to Legendre Polynomials
Recommend
More recommend
