Math 263 Equation Sheet
Math 263 Equation Sheet 1 ⎛ ∂M ∂N ⎞ − ⎜ ⎟ = g(x) N ⎝ ∂y ∂x ⎠ 1 ⎛ ∂M ∂N ⎞ − ⎜ ⎟ = h(y) M ⎝ ∂y ∂x ⎠
MATH 263 EQUATION SHEET First Order ODE:
y'= f (x, y)
standard form
e ∫ g(x )dx e− ∫ h(y )dy 1 xM − yN
M = yf (xy) and N = xg (xy )
If M ( x, y ) y ' = f ( x, y ) = − ⇒ N ( x, y ) dy = − M ( x, y ) dx ⇒ M ( x, y )dx + N ( x, y ) dy = 0 N ( x, y )
Linear: y '+ p ( x ) y = q ( x ) Separable:
1.
y′ =
Complex Numbers
p(x )dx p(x )dx y =e ∫ ∫ ⎢⎣⎡e ∫ q(x) + C⎥⎦⎤dx −
A(x) B(y)
Write as B ( y ) dy = A( x) dx (get all the
y' s
equations, and the
Integrate both sides of the equation:
3.
After integrating, isolate for
Homogeneous equations:
•
argument. Let z = x + iy
x' s to one side of the
e =e z
to the other side.)
2.
•
i = −1 , so i ⋅ i = −1. (x1 + iy1 ) + (x 2 + iy 2 ) = (x1 + x 2 ) + i(y1 + y 2 ) x + iy = r (cos α + i sin α ) , r is the modulus, and
• •
x + iy
= r(cosα + isinα ) . Then, = e e = e x (cos y + i sin y) x iy
Second Order ODE
∫ B(y)dy = ∫ A(x)dx .
y if possible.
Linear:
y'= f (x, y) and f ( xt , yt ) = f ( x, y ) .
a(x) y ′′ + b(x) y ′ + c(x)y = f (x)
y = y h + y p . y h is the homogeneous solution,
Solution of the form
the solution to the homogeneous equation a( x ) y ′′ + b( x ) y ′ + c ( x ) y = 0 . The particular solution
y = vx
1.
Let
2.
Taking the derivative of
3.
When we substitute
y = vx
we get: y'= v + x dv
y and y ' back into the original equation,
Constant coefficients:
the homogeneous D.E. becomes separable, and we can solve it using the technique outlined above
polynomial: aλ Roots
λ1 = λ2
Substituting into the Bernoulli equation, we get a linear equation
dx
for
v : dv + (1− n) p(x)v = (1− n)q(x) .
Solve for
v , then y = v
1 1−n
1.
Most O.D.E’s having the form M(x, y)dx + N(x, y)dy = 0 are not exact. but can be made exact by multiplying by all terms by an integrating factor. Condition Integrating Factor
C1e λ1 x + C2 xe λ1 x
1
{eαx cos(βx ),eαx sin(βx)}
f (x) , assume y p yp
Based on
f (x)
an x n + an−1 x n−1 + L + a1 x + a0 e
αx
(a x n
n
+ an−1 x
n−1
+ L + a1 x + a0 )
eαx cos(βx)(an x n + an−1 x n−1 + L + a1 x + a0 ) +e sin(βx)(bm x + bm−1 x αx
Integrating Factors
C1e λ1 x + C2e λ2 x
C1eα cos(βx ) + C 2eαx sin(βx ) x
Undetermined Coefficients
dx
dx
4.
complex
v = y1− n , so dv = (1− n )y −n dy .
3.
Homogeneous Solution
λ2 x
1
λ = α ± iβ
y '+ p ( x) y = q ( x) y n −n dy −n Multiply both sides by y : y + p(x)y1−n = q(x) dx
+ bλ + c = 0 .
{e ,e } {e λ x , xe λ x }
distinct
Bernoulli Equations:
2
λ1 x
equal
Let
ay ′′ + by ′ + cy = f ( x ), a, b, c constant
Fundamental Solutions
λ1 ≠ λ2
]
2.
is found
The homogeneous solution is determined by the roots of the characteristic
Exact: M ( x, y)dx + N ( x, y)dy = 0 is exact if ∂M ( x, y) = ∂N ( x, y) . An ∂y ∂x exact solution has an implicit solution g( x, y) = c , c constant. ⎛ ⎞ d M (x, y)dx ⎟dy g(x, y) = ∫ M(x, y)dx + ∫ ⎜ N(x, y) − ∫ dy ⎝ ⎠
[
yp
by the method of undetermined coefficients or by variation of parameters.
dx
1.
α is the
m
m−1
2.
Substitute
3.
Solve for
+ L + b1 x + b0 )
takes the form:
An x n + An−1 x n−1 + L + A1 x + A0 eαx (An x n + An−1 x n−1 + L + A1 x + A0 )
eαx cos(βx)(An x n + An−1 x n−1 + L + A1 x + A0 )
+eαx sin(βx)(Bm x m + Bm−1 x m−1 + L + B1 x + B0 )
ay ′′ + by ′ + cy = f (x) . A0, A1,K, An and B0,B1,K,Bm .
yp
into
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Cauchy-Euler Equation:
x 2 y''+axy'+by = f (x), a,b constant
The homogeneous solution is determined by the roots of the auxiliary equation: λ2 + (a −1)λ + b = 0 .
x
∫ 0
ω
f ( x)
Roots
Fundamental Solutions
Homogeneous Solution
distinct
{x λ , x λ }
C1x λ1 + C2 x λ2
equal
{x λ , x λ log x}
C1 x λ + C2 x λ log x
x
x [C1 cos(β log x ) + C2 sin(β log x )]
e− ax
λ1 ≠ λ2
1
λ1 = λ2
λ = α ± iβ
{x
complex
α
2
cos(β log x ), x sin(β log x )} α
α
F(s) s
f (t)dt
periodic with period
ω
1.
Find the general solution y h (x) = c1 y1 (x) + L + c n y n (x) to the homogeneous equation
dn y d n−1 y dy an (x) n + an−1 (x) n−1 + L + a1 (x) + a0 (x) = 0 . dx dx dx 2.
Look for a particular solution
3.
Plug
y p (x) into the original equation and solve for
v1(x),K,v n (x) .
v1(x),K,v n (x) into
Plug
The final answer is y(x) = y h (x) + y p (x) .
Wronskian: W [y1 (x), y 2 (x)] = •
1 s+ a
s s2 + a 2 a s2 + a 2
2. 3. 4.
f (x) * g(x) =
x
∫
e−csF ( s) 1
f (t)g(x − t)dt
L{ f ( x )}⋅ L{g( x )}
Solving ODE by Laplace Transforms
3.
• •
an (x)
dn y d n−1 y dy + an−1 (x) n−1 + L + a1 (x) + a0 (x) = F(x) , n dx dx dx
Apply Laplace transform to both sides of ODE, using initial conditions to determine y(0), y ′(0),..., y ( n −1) (0) . Solve for Y ( s) = L{y( x )} . Find a function tables.
y ( x ) with Laplace transform Y ( s) using
F(s) = L{ f (x)} =
∞
∫e
− sx
Eigenvalue of a matrix A: any scalar λ such that Ax = λx for some nonzero vector x Eigenvector of a matrix A: any nonzero vector x such that Ax = λx for some scalar λ Characteristic polynomial of a matrix A:
p(λ ) = det (λI − A) = (λ − λ1 )
•
(λ − λ2 ) L (λ − λk ) Characteristic equation of a matrix A: det (λI − A) = 0
•
Algebraic multiplicity: in the characteristic polynomial,
m1
multiplicity of
f (x)dx
0
L{c1 f1 (x)} + L{c 2 f 2 (x)}
F(s − a)
dn [F(s)] ds n
x f (x)
(−1) n
dn f dx n
s n F(s) − s n−1 f (0) − s n−2 f '(0) − ...− f ( n−1) (0)
n
uc ( x ) f ( x − c ) δ( x )
•
The Laplace Transform
e ax f (x)
e−cs s
Eigenvalues
If y ' ' = f ( x, y ' ) , let v = y ' , so v ' = y ' ' . dv dy dv . If y ' ' = f ( y , y ' ) , let v = y ' , y''= ⋅ = ⋅v dy dt dy The equation for v is linear. Solve for v . Integrate v = y ' to find y .
c1 f1 (x) + c 2 f 2 (x)
⎧0 if x < c uc (x) = ⎨ ⎩1 if c < x
2.
y1 (x)y 2 '(x) − y 2 (x)y1 '(x)
y1(x), y 2 (x) are independent if and only if W [y1(x), y 2 (x)] ≠ 0 .
f ( x)
sin(ax )
1.
Reduction of Order 1.
1 − e−ωs
s2
Given
y p (x) .
4.
f (x)dx
0
y p (x) = v1 (x)y1 (x) + L + v n (x)y n (x) .
5.
−sx
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Variation of Parameters To solve
0
1 s
1
cos( ax ) dn y d n−1 y dy an (x) n + an−1 (x) n−1 + L + a1 (x) + a0 (x) = F(x) : dx dx dx
∫e
λi
m2
mk
mi is the
λi : solutions x satisfying the
•
Eigenspace corresponding to system (λ i I − A)x = 0
•
Eigenspace of a matrix: combined space of all eigenspaces corresponding to all its eigenvalues
Computing Eigenvalues, Eigenvectors, and Eigenspaces 1.
Find characteristic polynomial p(λ ) = det (λI − A).
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2. 3. 4.
The eigenvalues are the roots
λ1 ,L , λk
of characteristic
equation det (λI − A) = 0 . For each eigenvalue, solve (λi I − A)x = 0 - find a basis for the set of solutions. The set of all the basis vectors in step 3 form a basis for the eigenspace of A.
Gram-Schmidt algorithm: Given a basis
B'= (w1,w 2 ,w 3 ,..., w n ) be the set of vectors
w1 = v1 w2 = v 2 −
v 2,w1 w1 w1,w1
w3 = v 3 −
v 3,w1 v ,w w1 − 3 2 w 2 w1,w1 w 2,w 2
Diagonalization
•
Diagonal matrix A: all the elements not on the main diagonal are zero; aij = 0 if i ≠ j
•
Similar matrices A and B: there exists an invertible matrix P such that B = P −1 AP o similar matrices have the same determinant, rank, nullity, and eigenvalues Diagonalizable matrix A: there exists an invertible matrix P such that D = P −1 AP is diagonal o an n x n matrix is diagonalizable if and only if the eigenspace of this matrix has n basis vectors
•
B = (v1,v 2 ,v 3 ,...,v n ) for V, let
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... wn = v n − •
v n ,w1 v ,w v ,w w1 − n 2 w 2 − ...− n n−1 w n−1 w1,w1 w 2,w 2 w n−1,w n−1
B' is an orthogonal basis for V. The vectors B''= (u1,u2 ,..., un ) , u = w n n
, are an orthonormal
wn
basis for V. Diagonalizing a Matrix 1. 2.
3.
Find the eigenvalues λ1,K, λn of A. Find n linearly independent eigenvectors
x1,x 2 ,K,x n of A by
solving ( λi I − A)x i = 0 for each eigenvalue λ . For A to be diagonalizable there must be n of them. The matrix D = P −1 AP is diagonal, where ⎡λ1 0 0 ⎤ and P = [x1 x 2 L x n ] ⎥ ⎢ 0 D=⎢ ⎢0 ⎢ ⎣0
λ2 0 0
O
0⎥ 0⎥ ⎥ λn ⎦
Solving a System of Linear ODEs by Diagonalization Given a linear system of ODEs:
y1'(x) = a11 y1 (x) + L + a1n y n (x)
y 2'(x) = a21 y1 (x) + L + a2n y n (x) M
y n '(x) = an1 y1 (x) + L + ann y n (x)
Let
A = [aij ]. If A is diagonalizable with eigenvalues λ1,K, λn and x1,x 2 ,K,x n , then y(x) = x1e λ x + x 2e λ x + L + x n e λ x
eigenvectors
1
2
n
Gram-Schmidt
•
n
If
v = (a1,K,an ) and w = (b1,K,bn ), v,w = ∑ aibi and i=1 n
v = v,v = ∑ ai2 . i= 1
•
Orthogonal basis: a set of basis vectors B = (w1,w 2 ,w 3 ,...,w n ) in which any two vectors are orthogonal ( w i ,w j = 0, i ≠ j )
•
Orthonormal basis: orthogonal basis in which each vector has length 1.
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MATH 263 EQUATION SHEET First Order ODE:
y'= f (x, y)
standard form
e ∫ g(x )dx e− ∫ h(y )dy 1 xM − yN
M = yf (xy) and N = xg (xy )
If M ( x, y ) y ' = f ( x, y ) = − ⇒ N ( x, y ) dy = − M ( x, y ) dx ⇒ M ( x, y )dx + N ( x, y ) dy = 0 N ( x, y )
Linear: y '+ p ( x ) y = q ( x ) Separable:
1.
y′ =
Complex Numbers
p(x )dx p(x )dx y =e ∫ ∫ ⎢⎣⎡e ∫ q(x) + C⎥⎦⎤dx −
A(x) B(y)
Write as B ( y ) dy = A( x) dx (get all the
y' s
equations, and the
Integrate both sides of the equation:
3.
After integrating, isolate for
Homogeneous equations:
•
argument. Let z = x + iy
x' s to one side of the
e =e z
to the other side.)
2.
•
i = −1 , so i ⋅ i = −1. (x1 + iy1 ) + (x 2 + iy 2 ) = (x1 + x 2 ) + i(y1 + y 2 ) x + iy = r (cos α + i sin α ) , r is the modulus, and
• •
x + iy
= r(cosα + isinα ) . Then, = e e = e x (cos y + i sin y) x iy
Second Order ODE
∫ B(y)dy = ∫ A(x)dx .
y if possible.
Linear:
y'= f (x, y) and f ( xt , yt ) = f ( x, y ) .
a(x) y ′′ + b(x) y ′ + c(x)y = f (x)
y = y h + y p . y h is the homogeneous solution,
Solution of the form
the solution to the homogeneous equation a( x ) y ′′ + b( x ) y ′ + c ( x ) y = 0 . The particular solution
y = vx
1.
Let
2.
Taking the derivative of
3.
When we substitute
y = vx
we get: y'= v + x dv
y and y ' back into the original equation,
Constant coefficients:
the homogeneous D.E. becomes separable, and we can solve it using the technique outlined above
polynomial: aλ Roots
λ1 = λ2
Substituting into the Bernoulli equation, we get a linear equation
dx
for
v : dv + (1− n) p(x)v = (1− n)q(x) .
Solve for
v , then y = v
1 1−n
1.
Most O.D.E’s having the form M(x, y)dx + N(x, y)dy = 0 are not exact. but can be made exact by multiplying by all terms by an integrating factor. Condition Integrating Factor
C1e λ1 x + C2 xe λ1 x
1
{eαx cos(βx ),eαx sin(βx)}
f (x) , assume y p yp
Based on
f (x)
an x n + an−1 x n−1 + L + a1 x + a0 e
αx
(a x n
n
+ an−1 x
n−1
+ L + a1 x + a0 )
eαx cos(βx)(an x n + an−1 x n−1 + L + a1 x + a0 ) +e sin(βx)(bm x + bm−1 x αx
Integrating Factors
C1e λ1 x + C2e λ2 x
C1eα cos(βx ) + C 2eαx sin(βx ) x
Undetermined Coefficients
dx
dx
4.
complex
v = y1− n , so dv = (1− n )y −n dy .
3.
Homogeneous Solution
λ2 x
1
λ = α ± iβ
y '+ p ( x) y = q ( x) y n −n dy −n Multiply both sides by y : y + p(x)y1−n = q(x) dx
+ bλ + c = 0 .
{e ,e } {e λ x , xe λ x }
distinct
Bernoulli Equations:
2
λ1 x
equal
Let
ay ′′ + by ′ + cy = f ( x ), a, b, c constant
Fundamental Solutions
λ1 ≠ λ2
]
2.
is found
The homogeneous solution is determined by the roots of the characteristic
Exact: M ( x, y)dx + N ( x, y)dy = 0 is exact if ∂M ( x, y) = ∂N ( x, y) . An ∂y ∂x exact solution has an implicit solution g( x, y) = c , c constant. ⎛ ⎞ d M (x, y)dx ⎟dy g(x, y) = ∫ M(x, y)dx + ∫ ⎜ N(x, y) − ∫ dy ⎝ ⎠
[
yp
by the method of undetermined coefficients or by variation of parameters.
dx
1.
α is the
m
m−1
2.
Substitute
3.
Solve for
+ L + b1 x + b0 )
takes the form:
An x n + An−1 x n−1 + L + A1 x + A0 eαx (An x n + An−1 x n−1 + L + A1 x + A0 )
eαx cos(βx)(An x n + An−1 x n−1 + L + A1 x + A0 )
+eαx sin(βx)(Bm x m + Bm−1 x m−1 + L + B1 x + B0 )
ay ′′ + by ′ + cy = f (x) . A0, A1,K, An and B0,B1,K,Bm .
yp
into
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Need help for exams?
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Cauchy-Euler Equation:
x 2 y''+axy'+by = f (x), a,b constant
The homogeneous solution is determined by the roots of the auxiliary equation: λ2 + (a −1)λ + b = 0 .
x
∫ 0
ω
f ( x)
Roots
Fundamental Solutions
Homogeneous Solution
distinct
{x λ , x λ }
C1x λ1 + C2 x λ2
equal
{x λ , x λ log x}
C1 x λ + C2 x λ log x
x
x [C1 cos(β log x ) + C2 sin(β log x )]
e− ax
λ1 ≠ λ2
1
λ1 = λ2
λ = α ± iβ
{x
complex
α
2
cos(β log x ), x sin(β log x )} α
α
F(s) s
f (t)dt
periodic with period
ω
1.
Find the general solution y h (x) = c1 y1 (x) + L + c n y n (x) to the homogeneous equation
dn y d n−1 y dy an (x) n + an−1 (x) n−1 + L + a1 (x) + a0 (x) = 0 . dx dx dx 2.
Look for a particular solution
3.
Plug
y p (x) into the original equation and solve for
v1(x),K,v n (x) .
v1(x),K,v n (x) into
Plug
The final answer is y(x) = y h (x) + y p (x) .
Wronskian: W [y1 (x), y 2 (x)] = •
1 s+ a
s s2 + a 2 a s2 + a 2
2. 3. 4.
f (x) * g(x) =
x
∫
e−csF ( s) 1
f (t)g(x − t)dt
L{ f ( x )}⋅ L{g( x )}
Solving ODE by Laplace Transforms
3.
• •
an (x)
dn y d n−1 y dy + an−1 (x) n−1 + L + a1 (x) + a0 (x) = F(x) , n dx dx dx
Apply Laplace transform to both sides of ODE, using initial conditions to determine y(0), y ′(0),..., y ( n −1) (0) . Solve for Y ( s) = L{y( x )} . Find a function tables.
y ( x ) with Laplace transform Y ( s) using
F(s) = L{ f (x)} =
∞
∫e
− sx
Eigenvalue of a matrix A: any scalar λ such that Ax = λx for some nonzero vector x Eigenvector of a matrix A: any nonzero vector x such that Ax = λx for some scalar λ Characteristic polynomial of a matrix A:
p(λ ) = det (λI − A) = (λ − λ1 )
•
(λ − λ2 ) L (λ − λk ) Characteristic equation of a matrix A: det (λI − A) = 0
•
Algebraic multiplicity: in the characteristic polynomial,
m1
multiplicity of
f (x)dx
0
L{c1 f1 (x)} + L{c 2 f 2 (x)}
F(s − a)
dn [F(s)] ds n
x f (x)
(−1) n
dn f dx n
s n F(s) − s n−1 f (0) − s n−2 f '(0) − ...− f ( n−1) (0)
n
uc ( x ) f ( x − c ) δ( x )
•
The Laplace Transform
e ax f (x)
e−cs s
Eigenvalues
If y ' ' = f ( x, y ' ) , let v = y ' , so v ' = y ' ' . dv dy dv . If y ' ' = f ( y , y ' ) , let v = y ' , y''= ⋅ = ⋅v dy dt dy The equation for v is linear. Solve for v . Integrate v = y ' to find y .
c1 f1 (x) + c 2 f 2 (x)
⎧0 if x < c uc (x) = ⎨ ⎩1 if c < x
2.
y1 (x)y 2 '(x) − y 2 (x)y1 '(x)
y1(x), y 2 (x) are independent if and only if W [y1(x), y 2 (x)] ≠ 0 .
f ( x)
sin(ax )
1.
Reduction of Order 1.
1 − e−ωs
s2
Given
y p (x) .
4.
f (x)dx
0
y p (x) = v1 (x)y1 (x) + L + v n (x)y n (x) .
5.
−sx
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Variation of Parameters To solve
0
1 s
1
cos( ax ) dn y d n−1 y dy an (x) n + an−1 (x) n−1 + L + a1 (x) + a0 (x) = F(x) : dx dx dx
∫e
λi
m2
mk
mi is the
λi : solutions x satisfying the
•
Eigenspace corresponding to system (λ i I − A)x = 0
•
Eigenspace of a matrix: combined space of all eigenspaces corresponding to all its eigenvalues
Computing Eigenvalues, Eigenvectors, and Eigenspaces 1.
Find characteristic polynomial p(λ ) = det (λI − A).
Our Course Booklets - free at prep sessions - are the “Perfect Study Guides.”
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2. 3. 4.
The eigenvalues are the roots
λ1 ,L , λk
of characteristic
equation det (λI − A) = 0 . For each eigenvalue, solve (λi I − A)x = 0 - find a basis for the set of solutions. The set of all the basis vectors in step 3 form a basis for the eigenspace of A.
Gram-Schmidt algorithm: Given a basis
B'= (w1,w 2 ,w 3 ,..., w n ) be the set of vectors
w1 = v1 w2 = v 2 −
v 2,w1 w1 w1,w1
w3 = v 3 −
v 3,w1 v ,w w1 − 3 2 w 2 w1,w1 w 2,w 2
Diagonalization
•
Diagonal matrix A: all the elements not on the main diagonal are zero; aij = 0 if i ≠ j
•
Similar matrices A and B: there exists an invertible matrix P such that B = P −1 AP o similar matrices have the same determinant, rank, nullity, and eigenvalues Diagonalizable matrix A: there exists an invertible matrix P such that D = P −1 AP is diagonal o an n x n matrix is diagonalizable if and only if the eigenspace of this matrix has n basis vectors
•
B = (v1,v 2 ,v 3 ,...,v n ) for V, let
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... wn = v n − •
v n ,w1 v ,w v ,w w1 − n 2 w 2 − ...− n n−1 w n−1 w1,w1 w 2,w 2 w n−1,w n−1
B' is an orthogonal basis for V. The vectors B''= (u1,u2 ,..., un ) , u = w n n
, are an orthonormal
wn
basis for V. Diagonalizing a Matrix 1. 2.
3.
Find the eigenvalues λ1,K, λn of A. Find n linearly independent eigenvectors
x1,x 2 ,K,x n of A by
solving ( λi I − A)x i = 0 for each eigenvalue λ . For A to be diagonalizable there must be n of them. The matrix D = P −1 AP is diagonal, where ⎡λ1 0 0 ⎤ and P = [x1 x 2 L x n ] ⎥ ⎢ 0 D=⎢ ⎢0 ⎢ ⎣0
λ2 0 0
O
0⎥ 0⎥ ⎥ λn ⎦
Solving a System of Linear ODEs by Diagonalization Given a linear system of ODEs:
y1'(x) = a11 y1 (x) + L + a1n y n (x)
y 2'(x) = a21 y1 (x) + L + a2n y n (x) M
y n '(x) = an1 y1 (x) + L + ann y n (x)
Let
A = [aij ]. If A is diagonalizable with eigenvalues λ1,K, λn and x1,x 2 ,K,x n , then y(x) = x1e λ x + x 2e λ x + L + x n e λ x
eigenvectors
1
2
n
Gram-Schmidt
•
n
If
v = (a1,K,an ) and w = (b1,K,bn ), v,w = ∑ aibi and i=1 n
v = v,v = ∑ ai2 . i= 1
•
Orthogonal basis: a set of basis vectors B = (w1,w 2 ,w 3 ,...,w n ) in which any two vectors are orthogonal ( w i ,w j = 0, i ≠ j )
•
Orthonormal basis: orthogonal basis in which each vector has length 1.
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