EXPONENTIAL EQUATION (y = aebx) FITTING :-
ALGORITHM, FLOWCHART AND NUMERICAL
Presented by,
VAIBHAV SANJAY
PATIL
TEME – A – 13
(CIA – 3)
Guided by,
Prof. D.R. Satpute
Flowchart &
Algorithm
(A)
(B)
* Fill the exponential curve y = aebx to the following data
• Step 1 : - Convert the equation into y = ax + b form
To find out the values of a & b we require here to prepare a table as below. The equation being
y = aebx
Taking loge both sides i.e.
In y = In a + bx In (e) since In (e) = 1
We get In y = In a + bx
Y = a’x + b’
Where Y = In y
a’ = b
b ’ = In a
X 2 4 6 8
Y 25 38 56 84
• Step 2 : - Prepare the table indicating the value of ΣX, ΣY, ΣX2, ΣXY
• Step 3 : - Write the desired equation for straight line
Σy = A Σx + nB
15.312631 = 20(A) + 4(B) ……………………………………………………(1)
Σxy = A Σx2 + B Σx
80.58673 = 120(A) + 20(B) ………………………….………………………….(2)
x y Y xy x2
2
4
6
8
25
38
56
84
3.218876
3.637586
4.025352
4.430817
3.43775
14.55034
24.15211
35.44653
4
16
36
64
Σx = 20 ΣY=15.312631 Σxy=80.58673 Σx2 = 120
• Step 4 : - Find the value of Δ
Δ = |20 4 |
|120 20|
Δ = 80
• Step 5 : - Find the value of Δ a’
Δ a’ = | 15.312631 4 |
| 80.58673 20|
Δ a’ = 16.0943
• Step 6 : - Find the value of Δ b’
Δ b’ = | 20 15.312631|
| 120 80.58673 |
Δ b’ = 225.78112
• Step 7 : - Then the bet curve is given by the equation
 ∴ a‘ = Δ a’ ÷ Δ
= 16.0743 ÷ 80
a’ = 0.201178
b’ = Δ b’ ÷ Δ
= 225.78112 ÷ 80
b’ = 2.822264
∴ Δ b’ = Δ a’ = 0.201178
a = Antilog b’ = 16.81488
∴ The equation becomes
y = 16.81488 e0.201178 x
THANK YOU…..

Numerical methods and programming

  • 1.
    EXPONENTIAL EQUATION (y= aebx) FITTING :- ALGORITHM, FLOWCHART AND NUMERICAL Presented by, VAIBHAV SANJAY PATIL TEME – A – 13 (CIA – 3) Guided by, Prof. D.R. Satpute
  • 2.
  • 3.
  • 4.
    * Fill theexponential curve y = aebx to the following data • Step 1 : - Convert the equation into y = ax + b form To find out the values of a & b we require here to prepare a table as below. The equation being y = aebx Taking loge both sides i.e. In y = In a + bx In (e) since In (e) = 1 We get In y = In a + bx Y = a’x + b’ Where Y = In y a’ = b b ’ = In a X 2 4 6 8 Y 25 38 56 84
  • 5.
    • Step 2: - Prepare the table indicating the value of ΣX, ΣY, ΣX2, ΣXY • Step 3 : - Write the desired equation for straight line Σy = A Σx + nB 15.312631 = 20(A) + 4(B) ……………………………………………………(1) Σxy = A Σx2 + B Σx 80.58673 = 120(A) + 20(B) ………………………….………………………….(2) x y Y xy x2 2 4 6 8 25 38 56 84 3.218876 3.637586 4.025352 4.430817 3.43775 14.55034 24.15211 35.44653 4 16 36 64 Σx = 20 ΣY=15.312631 Σxy=80.58673 Σx2 = 120
  • 6.
    • Step 4: - Find the value of Δ Δ = |20 4 | |120 20| Δ = 80 • Step 5 : - Find the value of Δ a’ Δ a’ = | 15.312631 4 | | 80.58673 20| Δ a’ = 16.0943 • Step 6 : - Find the value of Δ b’ Δ b’ = | 20 15.312631| | 120 80.58673 | Δ b’ = 225.78112
  • 7.
    • Step 7: - Then the bet curve is given by the equation  ∴ a‘ = Δ a’ ÷ Δ = 16.0743 ÷ 80 a’ = 0.201178 b’ = Δ b’ ÷ Δ = 225.78112 ÷ 80 b’ = 2.822264 ∴ Δ b’ = Δ a’ = 0.201178 a = Antilog b’ = 16.81488 ∴ The equation becomes y = 16.81488 e0.201178 x
  • 8.