www.bookspar.com | VTU NOTES | QUESTION PAPERS | NEWS | RESULTS | FORUM www.bookspar.com | VTU Notes LAPLACE TRANSFORMS INTRODUCTION Laplace transform is an integral transform employed in solving physical problems. Many physical problems when analysed assumes the form of a differential equation subjected to a set of initial conditions or boundary conditions. By initial conditions we mean that the conditions on the dependent variable are specified at a single value of the independent variable. If the conditions of the dependent variable are specified at two different values of the independent variable, the conditions are called boundary conditions. The problem with initial conditions is referred to as the Initial value problem. The problem with boundary conditions is referred to as the Boundary value problem. Example 1 : The problem of solving the equation x y dx dy dx y d = + + 2 2 with conditions y(0) = y ′ (0) = 1 is an initial value problem Example 2 : The problem of solving the equation x y dx dy dx y d cos 2 3 2 2 = + + with y(1)=1, y(2)=3 is called Boundary value problem. Laplace transform is essentially employed to solve initial value problems. This technique is of great utility in applications dealing with mechanical systems and electric circuits. Besides the technique may also be employed to find certain integral values also. The transform is named after the French Mathematician P.S. de’ Laplace (1749 – 1827). The subject is divided into the following sub topics. LAPLACE TRANSFORMS Definition and Properties Transforms of some functions Convolution theorem Inverse transforms Solution of differential equations 1 of 133
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LAPLACE TRANSFORMS - BookSpar · LAPLACE TRANSFORMS . INTRODUCTION Laplace transform is an integral transform employed in solving physical problems. Many physical problems when analysed
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Laplace transform is an integral transform employed in solving physical problems.
Many physical problems when analysed assumes the form of a differential equation subjected to a set of initial conditions or boundary conditions.
By initial conditions we mean that the conditions on the dependent variable are specified at a single value of the independent variable.
If the conditions of the dependent variable are specified at two different values of the independent variable, the conditions are called boundary conditions.
The problem with initial conditions is referred to as the Initial value problem.
The problem with boundary conditions is referred to as the Boundary value problem.
Example 1 : The problem of solving the equation xydxdy
dxyd
=++2
2
with conditions y(0) =
y′ (0) = 1 is an initial value problem
Example 2 : The problem of solving the equation xydxdy
dxyd cos23 2
2
=++ with y(1)=1,
y(2)=3 is called Boundary value problem.
Laplace transform is essentially employed to solve initial value problems. This technique is of great utility in applications dealing with mechanical systems and electric circuits. Besides the technique may also be employed to find certain integral values also. The transform is named after the French Mathematician P.S. de’ Laplace (1749 – 1827).
The subject is divided into the following sub topics.
-------------------------------------------- 28.04.05 -------------------------- Transforms of the derivatives of f(t) Consider
L )(tf ′ = ∫∞
− ′0
)( dttfe st
= [ ] ∫∞
−∞− −−0
0 )()()( dttfestfe stst , by using integration by parts
= [ ] )()0()(( tsLfftfeLt st
t+−−
∞→
= 0 - f (0) + s Lf(t) Thus L )(tf ′ = s Lf(t) – f(0) Similarly, L )(tf ′′ = s2 L f(t) – s f(0) - )0(f ′ In general, we have )0(.......)0()0()()( 121 −−− −−′−−= nnnnn ffsfstLfstLf
Transform of a periodic function A function f(t) is said to be a periodic function of period T > 0 if f(t) = f(t + nT) where n=1,2,3,….. The graph of the periodic function repeats itself in equal intervals. For example, sint, cost are periodic functions of period 2π since sin(t + 2nπ) = sin t, cos(t + 2nπ) = cos t.
The graph of f(t) = sin t is shown below :
Note that the graph of the function between 0 and 2π is the same as that between 2π and 4π and so on.
Note that the graph of the function between 0 and 2π is the same as that between 2π and 4π and so on. Formula : Let f(t) be a periodic function of period T. Then
1 , identifying the above series as a geometric series.
Thus
L f(t) = ∫ −−
−
Tst
sT dttfee 0
)(1
1
This is the desired result. Examples:- 1. For the periodic function f(t) of period 4, defined by f(t) = 3t, 0 < t < 2 6 , 2 < t < 4 find L f(t) Here, period of f(t) = T = 4 We have,
L f(t) = ∫ −−
−
Tst
sT dttfee 0
)(1
1 = ∫ −−
−
4
04 )(
11 dttfee
sts
=
+
− ∫∫ −−−
4
2
2
04 63
11 dtedttee
ststs
=
−
+
−−
−−
−−−
− ∫4
2
2
0
2
04 6.13
11
sedt
se
set
e
ststst
s
= ( )
−−−
−−
− 2
42
4
2131
1s
seee
ss
s
Thus,
Lf(t) = )1(
)21(342
42
s
ss
essee
−
−−
−−−
2. A periodic function of period ωπ2 is defined by
----------------------------------------------- 29.04.05 -------------------------------------- Step Function : In many Engineering applications, we deal with an important discontinuous function H(t-a) defined as follows :
0, t ≤ a H (t-a) =
1, t > a
where a is a non-negative constant.
This function is known as the unit step function or the Heaviside function. The function is named after the British electrical engineer Oliver Heaviside. The function is also denoted by u(t-a). The graph of the function is shown below:
H(t-a)
1 0 a t Note that the value of the function suddenly jumps from value zero to the value 1 as at → from the left and retains the value 1 for all t>a. Hence the function H(t-a) is called the unit step function. In particular, when a=0, the function H(t-a) become H(t), where 0 , t ≤ 0 H(t) = 1 , t > 0 Transform of step function By definition, we have
Heaviside shift theorem Statement :- L [f(t-a) H(t-a)] = e-as Lf(t) Proof :- We have
L [f(t-a) H(t-a)] = ∫∞
−−−0
)()( dteatHatf st
= ∫∞
− −a
st dtatfe )(
Setting t-a = u, we get
L[f(t-a) H(t-a)] = duufe uas )(0
)(∫∞
+−
= e-as L f(t) This is the desired shift theorem. Also, L-1 [e-as L f(t)] = f(t-a) H(t-a) Examples : 1. Find L[et-2 + sin(t-2)] H(t-2) Let f(t-2) = [et-2 + sin(t-2)] Then f(t) = [et + sint] so that
L f(t) = 1
11
12 +
+− ss
By Heaviside shift theorem, we have L[f(t-2) H(t-2)] = e-2s Lf(t) Thus,
3. Find Le-t H(t-2) Let f(t-2) = e-t , so that, f(t) = e-(t+2) Thus,
L f(t) = 1
2
+
−
se
By shift theorem, we have
1
)()]2()2([)1(2
2
+==−−
+−−
setLfetHtfL
ss
Thus
[ ]1
)2()1(2
+=−
+−−
setHeL
st
4. Let f(t) = f1 (t) , t ≤ a f2 (t), t > a Verify that f(t) = f1(t) + [f2(t) – f1(t)]H(t-a) Consider f1(t) + [f2(t) – f1(t)]H(t-a) = f1(t) + f2 (t) – f1(t), t > a 0 , t ≤ a = f2 (t), t > a f1(t), t ≤ a = f(t), given Thus the required result is verified. 5. Express the following functions in terms of unit step function and hence find their Laplace transforms. 1. t2 , 1 < t ≤ 2 f(t) = 4t , t > 2 2. cost, 0 < t < π
The convolution of two functions f(t) and g(t) denoted by f(t) ∗ g(t) is defined as
f(t) ∗ g(t) = ∫ −t
duugutf0
)()(
Property : f(t) ∗ g(t) = g(t) ∗ f(t) Proof :- By definition, we have
f(t) ∗ g(t) = ∫ −t
duugutf0
)()(
Setting t-u = x, we get
f(t) ∗ g(t) = ∫ −−0
))(()(t
dxxtgxf
= ∫ ∗=−t
tftgdxxfxtg0
)()()()(
This is the desired property. Note that the operation ∗ is commutative. Convolution theorem :- L[f(t) ∗ g(t)] = L f(t) . L g(t) Proof :- Let us denote
f(t) ∗ g(t) = φ(t) = ∫ −t
duugutf0
)()(
Consider
∫ ∫∞
− −=0 0
)]()([)]([t
st dtugutfetL φ
= ∫ ∫∞
− −0 0
)()(t
st duugutfe (1)
We note that the region for this double integral is the entire area lying between the lines u =0 and u = t. On changing the order of integration, we find that t varies from u to ∞ and u varies from 0 to ∞. u u=t t=u t = ∞ 0 u=0 t Hence (1) becomes
= L g(t) . L f(t) Thus L f(t) . L g(t) = L[f(t) ∗ g(t)] This is desired property. Examples : 1. Verify Convolution theorem for the functions f(t) and g(t) in the following cases : (i) f(t) = t, g(t) = sint (ii) f(t) =t, g(t) = et (i) Here,
f ∗ g = ∫ −t
duutguf0
)()( = ∫ −t
duutu0
)sin(
Employing integration by parts, we get f ∗ g = t – sint so that
L [f ∗ g] = )1(
11
112222 +
=+
−ssss
(1)
Next consider
L f(t) . L g(t) = )1(
11
112222 +
=+
⋅ssss
(2)
From (1) and (2), we find that L [f ∗ g] = L f(t) . L g(t) Thus convolution theorem is verified. (ii) Here
f ∗ g = ∫ −t
ut duue0
Employing integration by parts, we get f ∗ g = et – t – 1 so that
ASSIGNMENT 1. By using the Laplace transform of coshat, find the Laplace transform of sinhat.
2. Find (i) ∫t
dtt
tL0
sin (ii) ∫ −t
t tdteL0
cos (iii) ∫ −t
t tdteL0
cos
(iv) ∫
t
t dtet
tL0
sin (v) ∫t
atdttL0
2 sin
3. If f(t) = t2, 0 < t < 2 and f(t+2) = f(t) for t > 2, find L f(t) 4. Find L f(t) given f(t) = t , 0 ≤ t ≤ a f(2a+t) = f(t) 2a – t, a < t ≤ 2a
5. Find L f(t) given f(t) = 1, 0 < t < 2a f(a+t) = f(t)
-1, 2a < t < a
6. Find the Laplace transform of the following functions : (i) et-1 H(t-2) (ii) t2 H(t-2) (iii) (t2 + t + 1) H(t +2) (iv) (e-t sint) H(t - π) 7. Express the following functions in terms of unit step function and hence find their Laplace transforms : (i) 2t, 0 < t ≤ π (ii) t2 , 2 < t ≤ 3 f(t) = f(t) = 1 , t > π 3t , t > 3 (iii) sin2t, 0 < t ≤ π (iv) sint, 0 < t ≤ π/2 f(t) = f(t) = 0 , t > π cost, t > π/2 8. Let f1 (t) , t ≤ a f(t) = f2 (t) , a<t ≤ b f3 (t) , t > b Verify that f(t) = f1(t) + [f2(t) – f1(t)]H(t-a) + [f3(t) - f2(t)] H(t-b) 9. Express the following function in terms of unit step function and hence find its Laplace transform. Sint, 0 < t ≤ π f(t) = Sin2t, π < t ≤ 2π Sin3t, t > 2π
10. Verify convolution theorem for the following pair of functions: (i) f(t) = cosat, g(t) = cosbt (ii) f(t) = t, g(t) = t e-t (iii) f(t) = et g(t) = sint 11. Using the convolution theorem, prove the following:
INVERSE LAPLACE TRANSFORMS Let L f(t) = F(s). Then f(t) is defined as the inverse Laplace transform of F(s) and is denoted by L-1 F(s). Thus L-1 F(s) = f(t).
Linearity Property Let L-1 F(s) = f(t) and L-1 G(s) = g(t) and a and b be any two constants. Then
L-1 [a F(s) + b G(s)] = a L-1 F(s) + b L-1 G(s)
Table of Inverse Laplace Transforms
F(s) )()( 1 sFLtf −=
0,1>s
s
1
asas
>−
,1 ate
0,22 >+
sas
s
Cos at
0,122 >
+s
as aatSin
asas
>−
,122
aath Sin
as
ass
>−
,22
ath Cos
0,11 >+ s
sn
n = 0, 1, 2, 3, . . . !n
tn
0,11 >+ s
sn
n > -1 ( )1+Γ ntn
Examples 1. Find the inverse Laplace transforms of the following:
2s2+5s-4 = A(s+2) (s-1) + Bs (s-1) + Cs (s+2) For s = 0, we get A = 2, for s = 1, we get C = 1 and for s = -2, we get B = -1. Using these values in (1), we get
For s = -1, we get A = 1, for s = -2, we get C = -3 Comparing the coefficients of s2, we get B + C = 0, so that B = 3. Using these values in (1), we get
( ) ( ) ( ) ( ) 23
13
11
2154
22 +−
++
+=
++++
ssssss
Hence
( ) ( ) sLe
sLe
sLe
sssL ttt 13131
2154 121
21
21 −−−−−−− −+=
++++
ttt eete 233 −−− −+=
44
31: Evaluate 5.
assL−
−
Let
)1(2244
3
asDCs
asB
asA
ass
++
++
+−
=−
Hence
s3 = A(s + a) (s2 + a2) + B (s-a)(s2+a2)+(Cs + D) (s2 – a2)
For s = a, we get A = ¼; for s = -a, we get B = ¼; comparing the constant terms, we get D = a(A-B) = 0; comparing the coefficients of s3, we get 1 = A + B + C and so C = ½. Using these values in (1), we get
LAPLACE TRANSFORM METHOD FOR DIFFERENTIAL EQUATIONS
As noted earlier, Laplace transform technique is employed to solve initial-value problems. The solution of such a problem is obtained by using the Laplace Transform of the derivatives of function and then the inverse Laplace Transform. The following are the expressions for the derivatives derived earlier.
(o)f - (o)f s - f(o) s - f(t) L s (t)f[ L
(o)f - f(o) s - f(t) L s (t)fL[
f(o) - f(t) L s (t)]fL[
23
2
′′′=′′′
′=′′
=′
Examples
1) Solve by using Laplace transform method 2 y(o) t y y , ==+′ −te
s. transformLaplace using t any timeat particle theofnt displaceme thefind 10, is speed initial theand 20at x is particle theofposition initial theIf t.at time particle theofnt displaceme thedenotesx
where025dtdx
6dt
xdequation the,satisfyingpath a along moving is particleA 4)(
that theShow R. resistance and L inductance ofcircuit a to0at t applied is Ee A voltage (5) -at
The circuit is an LR circuit. The differential equation with respect to the circuit is
)(tERidtdiL =+
Here L denotes the inductance, i denotes current at any time t and E(t) denotes the E.M.F. It is given that E(t) = E e-at. With this, we have Thus, we have
t of terms in y andx for equations aldifferenti ussimultaneo following the Solve
LAPLACE TRANSFORMS
INTRODUCTION
Laplace transform is an integral transform employed in solving physical problems.
Many physical problems when analysed assumes the form of a differential equation subjected to a set of initial conditions or boundary conditions.
By initial conditions we mean that the conditions on the dependent variable are specified at a single value of the independent variable.
If the conditions of the dependent variable are specified at two different values of the independent variable, the conditions are called boundary conditions.
The problem with initial conditions is referred to as the Initial value problem.
The problem with boundary conditions is referred to as the Boundary value problem.
Example 1 : The problem of solving the equation xydxdy
Example 2 : The problem of solving the equation xydxdy
dxyd cos23 2
2
=++ with y(1)=1,
y(2)=3 is called Boundary value problem.
Laplace transform is essentially employed to solve initial value problems. This technique is of great utility in applications dealing with mechanical systems and electric circuits. Besides the technique may also be employed to find certain integral values also. The transform is named after the French Mathematician P.S. de’ Laplace (1749 – 1827).
The subject is divided into the following sub topics.
Definition :
Let f(t) be a real-valued function defined for all t ≥ 0 and s be a parameter, real or complex.
Suppose the integral ∫∞
−
0
)( dttfe st exists (converges). Then this integral is called the Laplace
transform of f(t) and is denoted by Lf(t).
Thus,
Lf(t) = ∫∞
−
0
)( dttfe st (1)
We note that the value of the integral on the right hand side of (1) depends on s. Hence Lf(t) is a function of s denoted by F(s) or )(sf .
Recapitulation Closed interval: An interval of the form a x b≤ ≤ , that includes every point between a and b and also the end points, is called a closed interval and is denoted by [ ]ba, . Open Interval: An interval of the form a x b< < , that includes every point between a and b but not the end points, is called an open interval and is denoted by ( )ba, Continuity: A real valued function )(xf is said to be continuous at a point 0x if
0
0lim ( ) ( )x x
f x f x→
=
The function )(xf is said to be continuous in an interval if it is continuous at every point in the interval. Roughly speaking, if we can draw a curve without lifting the pen, then it is a continuous curve otherwise it is discontinuous, having discontinuities at those points at which the curve will have breaks or jumps. We note that all elementary functions such as algebraic, exponential, trigonometric, logarithmic, hyperbolic functions are continuous functions. Also the sum, difference, product of continuous functions is continuous. The quotient of continuous functions is continuous at all those points at which the denominator does not become zero. Differentiability: A real valued function )(xf is said to be differentiable at point 0x if
A real valued function f(x) is said to be differentiable in an interval if it is differentiable at
every in the interval or if 0
( ) ( )limh
f x h f xh→
+ − exists uniquely. This is denoted by )(' xf .
We say that either )(' xf exists or f(x) is differentiable. Geometrically, it means that the curve is a smooth curve. In other words a curve is said to be smooth if there exists a unique tangent to the curve at every point on it. For example a circle is a smooth curve. Triangle, rectangle, square etc are not smooth, since we can draw more number of tangents at every corner point. We note that if a function is differentiable in an interval then it is necessarily continuous in that interval. The converse of this need not be true. That means a function is continuous need not imply that it is differentiable. Rolle’s Theorem: (French Mathematician Michelle Rolle 1652-1679)
Suppose a function )(xf satisfies the following three conditions: (i) )(xf is continuous in a closed interval [ ]ba, (ii) )(xf is differentiable in the open interval ( )ba, (iii) )()( bfaf =
Then there exists at least one point c in the open interval ( )ba, such that 0)(' =cf Geometrical Meaning of Rolle’s Theorem: Consider a curve )(xf that satisfies the conditions of the Rolle’s Theorem as shown in figure:
As we see the curve )(xf is continuous in the closed interval [ ]ba, , the curve is smooth i.e. there can be a unique tangent to the curve at any point in the open interval ( )ba, and also )()( bfaf = . Hence by Rolle’s Theorem there exist at least one point c belonging to ( )ba, such that 0)(' =cf . In other words there exists at least one point at which the tangent drawn to the curve will have its slope zero or lies parallel to x-axis. Notation: Often we use the statement: there exists a point c belonging to ( )ba, such that ……. We present the same in mathematical symbols as: ∃ c ( )ba,∈ : 0)(' =cf From now onwards let us start using the symbolic notation.
Example 1: Verify Rolle’s Theorem for 2)( xxf = in [ ]1,1− First we check whether the conditions of Rolle’s theorem hold good for the given function:
(i) 2)( xxf = is an elementary algebraic function, hence it is continuous every where and so also in [ ]1,1− .
(ii) xxf 2)(' = exists in the interval (-1, 1) i.e. the function is differentiable in (-1, 1).
(iii) Also we see that 1)1()1( 2 =−=−f and 11)1( 2 ==f i.e., )1()1( ff =− Hence the three conditions of the Rolle’s Theorem hold good. ∴By Rolle’s Theorem ∃ c ( )1,1−∈ : 0)(' =cf that means 2c = 0 ⇒ c= 0 ( )1,1−∈
Hence Rolle’s Theorem is verified.
Example 2: Verify Rolle’s Theorem for 2/)3()( xexxxf −+= in [-3, 0]
)(xf is a product of elementary algebraic and exponential functions which are continuous and hence it is continuous in [ 3,0]−
2/22/ )3(21)32()( xx exxexxf −− +−+=′
2/2 )]3(21)32[( xexxx −+−+=
2/2/2 )2)(3(21]6[
21 xx exxexx −− +−−=−−−= exists in (-3, 0)
0)3( =−f and also 0)0( =f )0()3( ff =−∴ That is, the three conditions of the Rolle’s Theorem hold good. ∴ ∃ c ( )0,3−∈ : 0)(' =cf
∞−=⇒=+−−∴ − ,2,30)2)(3(21 2/ cecc c
Out of these values of c, since 2 ( 3,0)− ∈ − , the Rolle’s Theorem is verified.
Lagrange’s Mean Value Theorem (LMVT): (Also known as the First Mean Value Theorem) (Italian-French Mathematician J. L. Lagrange 1736-1813)
Suppose a function )(xf satisfies the following two conditions: (i) )(xf is continuous in a closed interval [ ]ba,
(ii) )(xf is differentiable in the open interval ( )ba,
Then there exists at least one point c in the open interval ( )ba, such that ab
afbfcf−−
=)()()('
Proof: Consider the function )(xφ defined by kxxfx −= )()(φ where k is a constant to be found such that ( ) ( )a bϕ φ= . Since )(xf is continuous in the closed interval [ ]ba, , )(xφ which is a sum of continuous functions is also continuous in the closed interval [ ]ba,
' '( ) ( ) (1)x f x kxφ = − → exists in the interval ( )ba, as )(xf is differentiable in ( )ba, . We have ( ) ( )a f a kaφ = − and ( ) ( )b f b kbφ = −
( ) ( )( ) ( ) ( ) ( ) (2)f b f aa b f a ka f b kb kb a
φ φ −∴ = ⇒ − = − ⇒ = →
−
This means that when k is chosen as in (2) we will have ( ) ( )a bφ φ= Hence the conditions of Rolle’s Theorem hold good for )(xφ in [ ]ba, ∴By Rolle’s Theorem ∃ c ( ),a b∈ : '( ) 0cφ =
''( ) 0 ( ) 0 ( ) (3)c f c k k f cφ ′= ⇒ − = ⇒ = →
From (2) and (3), we infer that ∃ c ( ),a b∈ : ab
afbfcf−−
=)()()('
Thus the Lagrange’s Mean Value Theorem (LMVT) is proved.
Geometrical Meaning of Lagrange’s Mean Value Theorem: Consider a curve )(xf that satisfies the conditions of the LMVT as shown in figure:
From the figure, we observe that the curve )(xf is continuous in the closed interval [ ]ba, ; the curve is smooth i.e. there can be a unique tangent to the curve at any point in the open interval ( )ba, . Hence by LMVT there exist at least one point c belonging to ( )ba, such
that ab
afbfcf−−
=)()()(' . In other words there exists at least one point at which the tangent
drawn to the curve lies parallel to the chord joining the points [ ], ( )a f a and[ ], ( )b f b . Other form of LMVT: The Lagrange’s Mean Value Theorem can also be stated as follows: Suppose )(xf is continuous in the closed interval [ ],a a h+ and is differentiable in the open
interval ( ), )a a h+ then (0,1) : ( ) ( ) ( )f a h f a hf a hθ θ′∃ ∈ + = + + When (0,1)θ ∈ we see that ( , )a h a a hθ+ ∈ + i.e., here c a hθ= +
Using the earlier form of LMVT we may write that ( ) ( )( )( )
f a h f af a ha h a
θ + −′ + =+ −
On simplification this becomes ( ) ( ) ( )f a h f a hf a hθ′+ = + + .
Example 5: Verify LMVT for ( ) ( 1)( 2)( 3)f x x x x= − − − in [0, 4]
3 2( ) ( 1)( 2)( 3) 6 11 6f x x x x x x x= − − − = − + − is an algebraic function hence it is continuous in [0, 4]
2( ) 3 12 11f x x x′ = − + exists in (0, 4) i.e., ( )f x is differentiable in (0, 4) . i.e., both the conditions of LMVT hold good for ( )f x in[0,4] .
Hence (4) (0)(0,4) : ( )4 0
f fc f c −′∃ ∈ =−
i.e., 2 (3)(2)(1) ( 1)( 2)( 3)3 1 114 0
c c − − − −− + =
−
i.e, 2 2 23 12 11 3 3 12 8 0 2 (0,4)3
c c c c c− + = ⇒ − + = ⇒ = ± ∈
Hence the LMVT is verified.
Example 6: Verify LMVT for ( ) logef x x= in [1, ]e
( ) logef x x= is an elementary logarithmic function hence continuous in [1, ]e .
1( )f xx
′ = exists in ( )1,e or ( )f x is differentiable in ( )1,e .
i.e., both the conditions of LMVT hold good for ( )f x in [1, ]e
Example 7: Verify LMVT for 1( ) sinf x x−= in [0,1]
We find 2
1( )1
f xx
′ =−
exists in ( )0,1 , and hence ( )f x is continuous in [0,1] .
i.e., both the conditions of LMVT hold good for ( )f x in [0,1]
Hence 1 1
2(1) (0) 1 sin 1 sin 0(0,1) : ( )
1 0 11
f fc f cc
− −− −′∃ ∈ = ⇒ =− −
2
2 222
1 2 4121
c cc
π ππ π
−⇒ = ⇒ − = ⇒ =
−
2 4 0.7712 (0,1)c ππ
−∴ = = ∈
Hence the LMVT is verified.
Example 8: Find θ of LMVT for 2( )f x ax bx c= + + or
Verify LMVT by finding θ for 2( )f x ax bx c= + + 2( )f x ax bx c= + + is an algebraic function hence continuous in [ , ]a a h+ .
( ) 2f x ax b′ = + exists in ( ),a a h+
i.e., both the conditions of LMVT hold good for ( )f x in [ , ]a a h+ Then the second form of LMVT (0,1) : ( ) ( ) ( ) (1)f a h f a hf a hθ θ′∃ ∈ + = + + →
Here ; 2 3 2 2( ) ( ) ( ) 2 (2)f a h a a h b a h c a ah a h ab bh c+ = + + + + = + + + + + → 3( ) (3)f a a ab c= + + → ; 2( ) 2 ( ) 2 2 (4)f a h a a h b a a h bθ θ θ′ + = + + = + + →
Using (2), (3) and (4) in (1) we have:
3 2 2 3 2 12 (2 2 ) (0,1)2
a ah a h ab bh c a ab c h a a h bθ θ+ + + + + = + + + + + ⇒ = ∈
Mean Value Theorems Cauchy’s Mean Value Theorem (CMVT): (French Mathematician A. L. Cauchy 1789-1857)
Suppose two functions )(xf and ( )g x satisfies the following conditions:
(i) )(xf and ( )g x are continuous in a closed interval [ ]ba, (ii) )(xf and ( )g x are differentiable in the open interval ( )ba,
(iii) ( ) 0 .[ ]g x for all x for all can be denoted by the symbol′ ≠ ∀
Then there exists at least one point c in the open interval ( )ba,
such that '( ) ( ) ( )'( ) ( ) ( )
f c f b f ag c g b g a
−=
−
Proof: Consider the function )(xφ defined by ( ) ( ) ( )x f x k g xφ = − where k is a constant to be found such that ( ) ( )a bϕ φ= . Since )(xf and ( )g x are continuous in the closed interval [ ]ba, , )(xφ which is a sum of continuous functions is also continuous in the closed interval [ ]ba,
' '( ) ( ) ( ) (1)x f x k g xφ ′= − → exists in the interval ( )ba, as )(xf and ( )g x are differentiable in ( )ba, . We have ( ) ( ) ( )a f a kg aφ = − and ( ) ( ) ( )b f b k g bφ = −
( ) ( )( ) ( ) ( ) ( ) ( ) ( ) (2)( ) ( )
f b f aa b f a kg a f b k g b kg b g a
φ φ −∴ = ⇒ − = − ⇒ = →
−
This means that when k is chosen as in (2) we will have ( ) ( )a bφ φ= Hence the conditions of Rolle’s Theorem hold good for )(xφ in [ ]ba, ∴By Rolle’s Theorem ∃ c ( ),a b∈ : '( ) 0cφ =
' ( )'( ) 0 ( ) ( ) 0 (3)( )
f cc f c k g c kg c
φ′
′= ⇒ − = ⇒ = →′
From (2) and (3), we infer that ∃ c ( ),a b∈ : '( ) ( ) ( )'( ) ( ) ( )
f c f b f ag c g b g a
−=
−
Thus the Cauchy’s Mean Value Theorem (CMVT) is proved.
Example 12: Verify the Cauchy’s MVT for 2( )f x x= and 4( )g x x= in [ , ]a b
2( )f x x= and 4( )g x x= are algebraic polynomials hence continuous in [ , ]a b
( ) 2f x x′ = and 3( ) 4g x x′ = exist in ( ),a b also we see that ( ) 0 ( , )g x for all x a b′ ≠ ∈ since 0 a b< < i.e., the conditions of CMVT hold good for ( )f x and ( )g x in[ , ]a b .
Hence ∃ c ( ),a b∈ : '( ) ( ) ( )'( ) ( ) ( )
f c f b f ag c g b g a
−=
−
i.e., 2 2 2 2
3 4 4 2 2 22 1 1 ( , )
24 2c b a b ac a bc b a c b a
− += ⇒ = ⇒ = ∈
− +
Hence the CMVT is verified.
Example 13: Verify the Cauchy’s MVT for ( ) logf x x= and 1( )g xx
= in [1, ]e
( ) logf x x= and 1( )g xx
= are elementary logarithmic and rational algebraic functions that
are continuous in [1, ]e
1( )f xx
′ = and 21( )g x
x−′ = exist in ( )1,e
also we see that ( ) 0 (1, )g x for all x e′ ≠ ∈ i.e., the conditions of CMVT hold good for ( )f x and ( )g x in ( )1,e .
Hence ( )1,c e∃ ∈ : '( ) ( ) (1)'( ) ( ) (1)
f c f e fg c g e g
−=
−
i.e., 2
1 log log1 1 (1, )1 1 1 11 1
e ec c c ee
c e e
−= ⇒ − = ⇒ = ∈
− −− −
Hence the CMVT is verified. Example 14: Verify the Cauchy’s MVT for ( ) xf x e= and ( ) xg x e−= in [ , ]a b
( ) xf x e= and ( ) xg x e−= are elementary exponential functions that are continuous in [ , ]a b
( ) xf x e′ = and ( ) xg x e−′ = − exist in ( ),a b also we see that ( ) 0 ( , )g x for all x a b′ ≠ ∈ , since 0 a b< < . i.e., the conditions of CMVT hold good for ( )f x and ( )g x in ( ),a b .
Hence ( ),c a b∃ ∈ : '( ) ( ) ( )'( ) ( ) ( )
f c f b f ag c g b g a
−=
−
i.e., 2 21 1
c b a b a b ac c
c b a a bb a a b
e e e e e e ee ee e e e e
e e e
− − −
+
− − −= ⇒ − = ⇒ − =
− − −−
2 ( , )
2c a b a be e c a b+ +
= ⇒ = ∈
Hence the CMVT is verified.
Example 15: Verify the Cauchy’s MVT for ( )f x Sin x= and ( )g x Cos x= in 0,2π
( )f x Sin x= and ( )g x Cos x= are elementary trignometric functions
that are continuous in 0,2π
.
( )f x Cos x′ = and ( )g x Sin x′ = − exist in 0,2π
also we see that ( ) 0 0,2
g x for all x π ′ ≠ ∈
.
i.e., the conditions of CMVT hold good for ( )f x and ( )g x in 0,2π
When ,n → ∞ we can show that 0nR → , thus we can write the Taylor’s series as
2 1( 1)
( )
1
( ) ( )( ) ( ) ( ) ( ) ( ) ... ( ) ....2 1
( )( ) ( ) (4)
nn
nn
n
x a x af x f a x a f a f a f an
x af a f an
−−
∞
=
− −′ ′′= + − + + + +−
−= + →∑
Using (4) we can write a Taylor’s series expansion for the given function f(x) in powers of (x-a) or about the point ‘a’. Maclaurin’s series: (Scottish Mathematician Colin Maclaurin 1698-1746) When a=0, expression (4) reduces to a Maclaurin’s expansion given by
2 1( 1)
( )
1
( ) (0) (0) (0) ... (0) ....2 1
(0) (0) (5)
nn
nn
n
x xf x f xf f fn
xf fn
−−
∞
=
′ ′′= + + + + +−
= + →∑
Example 16: Obtain a Taylor’s expansion for ( )f x Sin x= in the ascending powers of
2 3 4 5x x x xf x f x f f f f f′ ′′ ′′′= + + + + + →
Here ( ) (0) 0 0f x Sin x f Sin= ⇒ = = ( ) (0) 0 1f x Cos x f Cos′ ′= ⇒ = =
( ) (0) 0 0f x Sin x f Sin′′ ′′= − ⇒ = − = ( ) (0) 0 1f x Cos x f Cos′′′ ′′′= − ⇒ = − = −
(4) (4)( ) (0) 0 0f x Sin x f Sin= ⇒ = = (5) (5)( ) (0) 0 1f x Cos x f Cos= ⇒ = = Substituting these values in (1), we get the Maclaurin’s series for ( )f x Sin x= as
2 3 4 5 3 5( ) 0 (1) (0) ( 1) (0) (1) .... ....
2 3 4 5 3 5x x x x x xf x Sin x x Sin x x= = + + + − + + ⇒ = − +
Note: As done in the above example, we can find the Maclaurin’s series for various functions, for ex:
2 4 6( ) 1 ......
2 4 6x x xi Cosx = − + −
2 3 4 5 6( ) 1 ......
2 3 4 5 6x x x x x xii e x= + + + + + +
2 3 4( 1) ( 1)( 2) ( 1)( 2)( 3)( ) (1 ) 1 ......
2 3 4m m m m m m m m m miii x mx x x x− − − − − −
+ = + + + +
Taking 1 ( )m in iii= − we can get 1 2 3 4 5 6(1 ) 1 ......x x x x x x x−+ = + + + + + +
Replacing x by (-x) in this we get 1 2 3 4 5 6(1 ) 1 ......x x x x x x x−− = − + − + − +
We use these expansions in the study of various topics.
Indeterminate Forms: While evaluating certain limits, we come across expressions of the form
0 00 , , 0 , , 0 , 10
and ∞∞×∞ ∞ − ∞ ∞
∞which do not represent any value. Such expressions are
called Indeterminate Forms. We can evaluate such limits that lead to indeterminate forms by using L’Hospital’s Rule (French Mathematician 1661-1704). L’Hospital’s Rule: If f(x) and g(x) are two functions such that (i) lim ( ) 0 lim ( ) 0
x a x af x and g x
→ →= =
(ii) ( ) ( ) ( ) 0f x andg x exist and g a′ ′ ′ ≠
Then ( ) ( )lim lim( ) ( )x a x a
f x f xg x g x→ →
′=
′
The above rule can be extended, i.e, if ( ) ( ) ( )( ) 0 ( ) 0 lim lim lim .....( ) ( ) ( )x ax a x a
f x f x f xf a and g a theng x g x g x→→ →
′ ′′′ ′= = = = =
′ ′′
Note:
1. We apply L’Hospital’s Rule only to evaluate the limits that in 0 ,
0∞∞
forms. Here we
differentiate the numerator and denominator separately to write ( )( )
f xg x
′′
and apply the
limit to see whether it is a finite value. If it is still in 00
or ∞∞
form we continue to
differentiate the numerator and denominator and write further ( )( )
f xg x
′′′′
and apply the
limit to see whether it is a finite value. We can continue the above procedure till we get a definite value of the limit.
2. To evaluate the indeterminate forms of the form 0 , ,×∞ ∞ − ∞ we rewrite the
functions involved or take L.C.M. to arrange the expression in either 00
or ∞∞
and then
apply L’Hospital’s Rule.
3. To evaluate the limits of the form 0 00 , 1and ∞∞ i.e, where function to the power of function exists, call such an expression as some constant, then take logarithm on both
Example 39: Find the values of ‘a’ and ‘b’ such that 20
cosh coslim 1x
a x b xx→
−= .
20
cosh coslim0x
a x b x a bLet A finitex→
− −= = ≠
We can continue to apply the L’Hospital’s Rule, if a-b=0, since the denominator=0 .
0For a b− = ,
20 0 0
cosh cos 0 sinh sin 0 cosh coslim lim lim0 2 0 2 2x x x
a x b x a x b x a x b x a bAx x→ → →
− + + + = = = =
1. 2But thisis given as a b∴ + =
0 2 1 1.Solving the equations a b and a b we obtain a and b− = + = = = Limits of the form 0 00 , 1and ∞∞ : To evaluate such limits, where function to the power of function exists, we call such an expression as some constant, then take logarithm on both