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Euler’s Theorem
(For B.Sc./B.A. Part-I, Hons. And Subsidiary Courses of Mathematics)
Poonam Kumari Department of Mathematics, Magadh Mahila College Patna University
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Contents 1. Homogeneous Function 2. Euler’s Theorem on Homogeneous Function of Two Variables 3. Euler’s Theorem on Homogeneous Function of Three Variables
1. Homogeneous Function A function f of two independent variables x, y is said to be a homogeneous function of degree n if it can be put in either of the following two forms : y y f (x, y) = x n , where denotes a function of x x or f (t x, t y) = t n f (x, y) , where t is any positive real number.
Similarly, a function of three independent variables x, y, z is said to be a homogeneous function of degree if it can be put in either of the following two forms : y z z f (x, y, z) = x n , , where denotes a function of and x x x or f (t x, t y,t z) = t n f (x, y, z) , where is any positive real number. The above definition can be extended to a function of any number of variables. Example : The function x 4 + y 4 f (x, y) = x − y is a homogeneous function of degree 3, since 4 4 y y x 4 1+ 1+ 4 4 x x x + y 3 3 y f (x, y) = = = x = x x − y y y x x 1− 1− x x or alternatively, 4 4 4 4 4 4 4 (t x) + (t y ) t (x + y ) 3 (x + y ) 3 f (t x,t y) = = = t = t f (x, y) . t x − t y t (x − y) (x − y)
Similarly, the function x 3 + y 3 + z 3 f (x, y, z) = x + y + z is a homogeneous function of degree 2, since 3 3 3 3 y z y z x3 1+ + 1+ + 3 3 3 x x x x x + y + z 2 2 y z f (x, y, z) = = = x = x , x + y + z y z y z x x x 1+ + 1+ + x x x x or alternatively,
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3 3 3 3 3 3 3 3 3 3 (t x) + (t y ) + (t z) t (x + y + z ) 2 (x + y + z ) 2 f (t x,t y,t z) = = = t = t f (x, y, z) . t x + t y + t z t (x + y + z) (x + y + z)
Note: A polynomial function is a homogeneous function of degree n if all of its terms are of the same degree n.
Proof : Let f be a polynomial function in two independent variables x, y, i.e.,
n n−1 n−2 2 n−1 n f (x, y) = a0 x + a1x y + a2 x y + ...... + an−1xy + an y .
y y 2 y n−1 y n Then n f (x, y) = x a0 + a1 + a2 2 + ...... + an−1 n−1 + an n x x x x
2 n−1 n n y y y y = x a0 + a1 + a2 + ...... + an−1 + an x x x x
y y = x n , where is a function of . x x Example : The function
f (x, y) = x5 + 6x 4 y + 7x3 y 2 + 2y 5 is a homogeneous function of degree 5, since
f (x, y) = x5 + 6x 4 y + 7x3 y 2 + 2y 5
2 5 y y y = x5 1+ 6 + 7 + 2 x x x
y y = x5 ,where is a function of . x x
Similarly, the function
f (x, y, z) = x 4 + 3x 2 y 2 + 4xyz 2 + 5yz 3 + 6y 4 + 7z 4 is a homogeneous function of degree 4, since
f (x, y, z) = x 4 + 3x 2 y 2 + 4xyz 2 + 5yz 3 + 6y 4 + 7z 4
2 2 3 4 4 y y z y z y z = x 4 1+ 3 + 4 + 5 + 6 + 7 x x x x x x x
y z y z = x 4 , , where is a function of and . x x x x
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2. Euler’s Theorem on Homogeneous Function of Two Variables Statement : If u be a homogeneous function of degree n in two independent variables x, y, then u u x + y = n u. x y Proof : Let
1 1 2 2 3 3 n n u = A1x y + A2 x y + A3 x y + ...... + An x y ...... (1)
where 1 + 1 = 2 + 2 = 3 + 3 = ...... = n + n = n
Differentiating both sides of equation (1) partially w. r. t. x, we get
u 1−1 1 2 −1 2 3−1 3 n −1 n = A1 (1 x ) y + A2 (2 x ) y + A3 (3 x ) y + ...... + An ( n x ) y x
u 1 1 2 2 3 3 n n This x = A x y + A x y + A x y +...... + A x y ...... (2) x 1 1 2 2 3 3 n n
Now, differentiating both sides of equation (1) partially w. r. t. y, we get
u 1 1−1 2 2−1 3 3−1 n n −1 = A1x ( 1 y )+ A2 x ( 2 y )+ A3 x ( 3 y ) +...... + An x ( n y ) y
u 1 1 2 2 3 3 n n This y = A x y + A x y + A x y +...... + A x y ...... (3) y 1 1 2 2 3 3 n n
Adding equations (2) and (3), we get
u u 1 1 2 2 3 3 x + y = (1 + 1 )A1x y + ( 2 + 2 )A2 x y + (3 + 3 )A3 x y x y n n + ...... + ( n + n )An x y
1 1 2 2 3 3 n n = n A1x y + n A2 x y + n A3 x y +...... + n An x y
( 1 + 1 = 2 + 2 = 3 + 3 = ...... = n + n = n )
1 1 2 2 3 3 n n = n (A1x y + A2 x y + A3 x y +...... + An x y )
= n u (using equation (1))
i.e., u u x + y = n u x y
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Corollary : If u be a homogeneous function of degree n in two independent variables x, y, then
2u 2u u (i) x + y = (n −1) x2 x y x 2u 2u u (ii) x + y = (n −1) x y y 2 y 2u 2u 2u (iii) x 2 + 2xy + y 2 = n(n −1) u. x 2 x y y 2 Proof : (i) Since is a homogeneous function of degree n in two independent variables x, y, therefore, by Euler’s Theorem u u x + y = n u ...... (1) x y Differentiating both sides of equation (1) partially w. r. t. x, we get
2 u u 2 u u x + (1) + y = n x 2 x x y x
2 u 2 u u This x + y = (n −1) ...... (2) x 2 x y x Hence (i) is proved. (ii) Differentiating both sides of equation (1) partially w. r. t. y, we get
2 u 2 u u u x + y + (1) = n y x y 2 y y
2 u 2 u u u 2 u 2 u This x + y + = n = 2 x y y y y y x x y
2 u 2 u u x + y = (n −1) ...... (3) x y y 2 y
Hence (ii) is proved. (iii) Multiplying equations (2) and (3) by x and y respectively and adding, we get
2u 2u 2u u u x 2 + 2xy + y 2 = (n −1) x + y 2 2 x x y y x y = (n −1) (nu) (using equation (1)) 2u 2u 2u i.e., x 2 + 2xy + y 2 = n (n −1) u. x 2 x y y 2 This proves (iii).
3. Euler’s Theorem on Homogeneous Function of Three Variables Statement : If be a homogeneous function of degree n in three independent variables x, y, z, then u u u x + y + z = n u. x y z
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Proof : Let 1 1 1 2 2 2 3 3 3 n n n u = A1x y z + A2 x y z + A3 x y z + ...... + An x y z ...... (1) where 1 + 1 + 1 = 2 + 2 + 2 = 3 + 3 + 3 = ...... = n + n + n = n Differentiating both sides of equation (1) partially w. r. t. x, we get
u 1− 1 1 1 2 − 1 2 2 3− 1 3 3 = A1 (1 x ) y z + A2 (2 x ) y z + A3 (3 x ) y z x
n − 1 n n + ...... + An ( n x ) y z
u 1 1 1 2 2 2 3 3 3 This x = A11x y z + A2 2 x y z + A33 x y z x ...... (2) n n n +...... + An n x y z Now, differentiating both sides of equation (1) partially w. r. t. y, we get
u 1 1−1 1 2 2 −1 2 3 3−1 3 = A1x ( 1 y ) z + A2 x ( 2 y ) z + A3 x ( 3 y ) z y n n −1 n + ...... + An x ( n y ) z
u This y = A x 1 y 1 z 1 + A x 2 y 2 z 2 + A x 3 y 3 z 3 y 1 1 2 2 3 3 ...... (3)
n n n +...... + An n x y z Similarly, differentiating both sides of equation (1) partially w. r. t. z, we get
u 1 1 1−1 2 2 2 −1 3 3 3−1 = A1x y ( 1 z )+ A2 x y ( 2 z )+ A3 x y ( 3 z ) z
n n n −1 + ...... + An x y ( n z )
u This z = A x 1 y 1 z 1 + A x 2 y 2 z 2 + A x 3 y 3 z 3 z 1 1 2 2 3 3 ...... (4)
n n n +...... + An n x y z Adding equations (2), (3) and (4), we get
u u u 1 1 1 2 2 2 x + y + z = (1 + 1 + 1 )A1x y z + ( 2 + 2 + 2 )A2 x y z + x y z 3 3 3 n n n (3 + 3 + 3 )A3 x y z + ...... + ( n + n + n )An x y z
1 1 1 2 2 2 3 3 3 = n A1x y z + n A2 x y z + n A3 x y z + ...... +
n n n n An x y z
(1 + 1 + 1 = 2 + 2 + 2 = ...... = n + n + n = n) 1 1 1 2 2 2 3 3 3 n n n = n (A1x y z + A2 x y z + A3 x y z +.....+ An x y z ) = n u (using equation (1))
u u u i.e., x + y + z = n u x y z
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x (x3 − y3 ) Example 1 : Verify Euler’s Theorem when u = . x3 + y3 Solution : According to Euler’s Theorem, if u be a homogeneous function of degree n in two independent variables x, y, then u u x + y = n u. x y Given that x (x3 − y3 ) u = .....(1) x3 + y3 3 3 y y x 4 1− 1− x x y y u = = x = x , i.e., 3 3 where is a function of . y y x x x3 1+ 1+ x x This The given function u is a homogeneous function of degree 1 in two independent variables x, y.Therefore Euler’s Theorem will be verified if we can prove that u u x + y = u. x y Taking logarithm of both sides of equation (1), we get
logu = log x +log (x3 − y3 ) −log (x3 + y3 ) ...... (2) Now, differentiating both sides of equation (2) partially w. r. t. x, we get 1 u 1 = + 1 (3x 2 ) − 1 (3x 2 ) u x x x3 − y 3 x3 + y 3
1 u 3x 3 3x 3 This x = 1+ − ...... (3) u x x 3 − y 3 x 3 + y 3 Similarly, differentiating both sides of equation (2) partially w. r. t. y, we get 1 u = 0 + 1 (−3y 2 ) − 1 (3y 2 ) u y x3 − y 3 x3 + y 3
1 u 3y 3 3y 3 This y = − − 3 3 3 3 ...... (4) u y x − y x + y Adding equations (3) and (4), we get 1 u u 3 (x3 − y 3 ) 3 (x3 + y 3 ) x + y =1+ − 3 3 3 3 u x y x − y x + y =1+ 3−3 = 1 u u This x + y = u x y
Euler’s Theorem is verified for the given function.
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1 1 x 4 + y 4 Example 2 : Verify Euler’s Theorem when u = 1 1 . x 5 + y 5 Solution : According to Euler’s Theorem, if u be a homogeneous function of degree n in two independent variables x, y, then u u x + y = n u. x y Given that 1 1 x 4 + y 4 u = 1 1 .....(1) x 5 + y 5 1 1 1 y 4 y 4 x 4 1+ 1+ x 1 x 1 y y u = = x 20 = x 20 , i.e., 1 1 where is a function of . 1 x x y 5 y 5 x 5 1+ 1+ x x 1 This The given function u is a homogeneous function of degree in two independent variables 20 x, y.Therefore Euler’s Theorem will be verified if we can prove that u u 1 x + y = u. x y 20 Taking logarithm of both sides of equation (1), we get 1 1 1 1 4 4 5 5 logu = log x + y −log x + y ...... (2) Now, differentiating both sides of equation (2) partially w. r. t. x, we get
3 4 1 u 1 − 1 − = 1 x 4 − 1 x 5 u x 1 1 4 1 1 5 x 4 + y 4 x 5 + y 5 1 1 1 u x 4 x 5 This x = − ...... (3) u x 1 1 1 1 4 4 5 5 4 x + y 5 x + y Similarly, differentiating both sides of equation (2) partially w. r. t. y, we get
3 4 1 u 1 − 1 − = 1 y 4 − 1 y 5 u y 1 1 4 1 1 5 x 4 + y 4 x 5 + y 5 1 1 1 u y 4 y 5 This y = − ...... (4) 1 1 1 1 u y 4 4 5 5 4 x + y 5 x + y Adding equations (3) and (4), we get
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1 1 1 1 1 u u x 4 + y 4 x 5 + y 5 x + y = − 1 1 1 1 u x y 4 4 5 5 4 x + y 5 x + y 1 1 = − 4 5 1 = 20 u u 1 This x + y = u x y 20 Euler’s Theorem is verified for the given function.
x 2 + y 2 u u 1 Example 3 : If u = tan −1 , prove that x + y = sin 2u. x + y x y 2 Solution : Given that x 2 + y 2 u = tan −1 . x + y 2 2 y y x 2 1+ + 1+ + x 2 + y 2 x x y y This tan u = = = x = x , where is a function of . x + y y y x x x 1+ + 1+ + x x tan u is a homogeneous function of degree 1 in two independent variables x, y.
Let v = tan u ...... (1) Then v is a homogeneous function of degree 1 in two independent variables x, y. Therefore, by Euler’s Theorem, v v x + y = v. x y
2 u 2 u This x sec u + y sec u = tan u x y
v 2 u v 2 u = sec u , = sec u and v = tan u, by equation (1) x x y y tan u x u + y u = x y sec2 u sin u = (cos2 u) cosu = sinucosu 1 = (2sin u cosu) 2 1 = sin 2u. 2
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−1 x + y u u 1 Example 4 : If u = cos , prove that x + y + cotu = 0. x + y x y 2 Solution : Given that
−1 x + y u = cos . x + y y y x 1+ + 1 1+ + 1 x + y x x y y This cosu = = = x 2 = x 2 , where is a function of . x + y y y x x x 1+ + 1+ + x x 1 cosu is a homogeneous function of degree in two independent variables x, y. 2 Let v = cosu ...... (1) Then v is a homogeneous function of degree in two independent variables x, y.Therefore, by
Euler’s Theorem, v v 1 x + y = v. x y 2
u u 1 This x − sin u + y − sin u = cosu x y 2 v u v u = −sin u , = −sin u and v = cosu, by equation (1) x x y y 1 cosu x u + y u = − x y 2 sin u 1 = − cotu 2 1 x u + y u + cotu = 0. x y 2
−1 x −1 y u u Example 5 : If u = sin + tan , prove that x + y = 0. y x x y Solution : Given that
−1 x −1 y u = sin + tan ...... (1) y x
−1 x Let sin = . y x Then sin = . y x This tan = y 2 − x 2
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x = tan −1 y 2 − x 2
−1 x −1 x sin = tan .....(2) y y 2 − x 2
−1 x Substituting the value of sin from equation (2) in equation (1), we get y x y u = tan −1 + tan−1 y 2 − x 2 x
x y + y 2 − x 2 x = tan −1 x y 1− 2 2 x y − x
x 2 + y y 2 − x 2 = tan −1 . x y 2 − x 2 − xy
2 2 y y y y x 2 1+ −1 1+ −1 2 2 2 x x x x x + y y − x 0 y This tan u = = = = x , 2 2 2 2 x x y − x − xy y y y y x 2 −1 − −1 − x x x x y where is a function of . x tan u is a homogeneous function of degree zero in two independent variables x, y.
Let v = tan u ...... (3)
Then v is a homogeneous function of degree zero in two independent variables x, y.Therefore, by Euler’s Theorem,
v v x + y = 0. x y
2 u 2 u This x sec u + y sec u = 0 x y
v 2 u v 2 u = sec u , = sec u and v = tan u, by equation (3) x x y y
0 x u + y u = x y sec2 u
= 0.
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u u 1 Example 6 : If u = sin ( x + y ) , prove that x + y = ( x + y ) cos ( x + y ) . x y 2 Solution : Given that u = sin ( x + y ) ...... (1) 1 −1 y y y This sin u = x + y = x 1+ + = x 2 , where is a function of . x x x 1 sin −1 u is a homogeneous function of degree in two independent variables x, y. 2 Let v = sin −1 u ...... (2) Then v is a homogeneous function of degree in two independent variables x, y.Therefore, by
Euler’s Theorem, v v 1 x + y = v. x y 2
1 1 1 This x u + y u = sin −1 u 2 2 1 − u x 1 − u y 2 1 1 v = u , v = u and v = sin −1 u, by equation (2) 2 2 x 1 − u x y 1 − u y 1 x u + y u = (sin −1 u) ( 1− u 2 ) x y 2 1 = ( x + y ) 1− sin 2 ( x + y ) (using equation (1) ) 2 1 = ( x + y ) cos ( x + y ) . 2
Exercises y 1. Verify Euler’s Theorem when u = x3 log . x
x − y u u 2. If u = sin , prove that Euler’s formula x + y = nu holds good. x + y x y
x − y u y u 3. u = sin −1 , prove that = − . If x + y x x y
x3 + y 3 u u 4. If sin u = , prove that x + y = 2 tan u. x + y x y