Fitting of an Exponential Curve of the Type 𝒚=𝒂𝒆(^𝒃𝒙)

Summary

In this video lecture, the speaker, SN Patil, explains how to fit an exponential curve of the form y = ae^(bx) using the least squares method. The process involves transforming the exponential equation into a linear form by taking logarithms and then applying the least squares method to find the best-fit parameters 'a' and 'b'. The lecture covers both natural (base e) and common (base 10) logarithms, providing examples for each. The speaker also discusses the calculation of errors and the derivation of normal equations to solve for 'a' and 'b'.

Takeaways

• 📈 The lecture discusses the process of fitting an exponential curve of the form y = ae^{bx} using the method of least squares.
• 🔍 To linearize the exponential equation, logarithms are taken on both sides, resulting in log y = log a + bx log e.
• 🔄 The base of the logarithm can be natural (e) or common (10), affecting how coefficients are calculated.
• 📐 The linearized form allows the use of least squares method to find the best-fit line, represented as Y = a + bx where Y is log y.
• 📉 The method involves calculating the sum of squares of the errors to find the coefficients that minimize this sum.
• ⚖️ The normal equations derived from the least squares method are used to solve for the coefficients a and b.
• 🔢 The values of a and b are found by differentiating the sum of squared errors with respect to a and b, then setting the derivatives to zero.
• 🧮 Once a and b are calculated, they are converted back from their logarithmic forms to their original exponential forms.
• 🌐 Examples are provided to demonstrate fitting an exponential curve to a set of data points using both natural and common logarithms.
• 📝 The final fitted exponential equation is given by y = ae^{bx} or y = a · 10^{bx} depending on the logarithm base used.

Q & A

What is the main topic of the video lecture?

The main topic of the video lecture is discussing the fitting of an exponential curve of the type y = a * e^(bx) using the method of least squares.

How do you convert the exponential equation into a linear form for fitting?

You convert the exponential equation into a linear form by taking the logarithm of both sides, resulting in log(y) = log(a) + bx * log(e).

What is the significance of the least squares method in curve fitting?

The least squares method is used to find the best fit line by minimizing the sum of the squares of the vertical distances (residuals) between the observed values and the values predicted by the model.

What are the two types of bases discussed for the exponential curve fitting?

The two types of bases discussed for the exponential curve fitting are the natural base (e) and base 10.

How do you define the error in the context of the least squares method?

The error is defined as the difference between the observed values (y) and the values predicted by the model (f(x)), i.e., error = y - (a + bx).

What are the normal equations derived from the least squares method?

The normal equations derived from the least squares method are: Σy = n*a + b*Σx and Σxy = a*Σx + b*Σx^2.

How do you find the values of 'a' and 'b' in the exponential curve fitting?

You find the values of 'a' and 'b' by solving the normal equations simultaneously after substituting the calculated sums and averages.

What is the difference between 'capital a' and 'small a' in the context of the video?

'Capital a' refers to the logarithmic form of 'a' used in the linearized equation, while 'small a' is the actual parameter in the exponential equation, calculated as e^(capital a).

What is the final form of the exponential function after fitting using the least squares method?

The final form of the exponential function after fitting is y = a * e^(bx), where 'a' and 'b' are the parameters estimated from the data.

How do you handle the base of the logarithm when converting the exponential equation to linear form?

When converting the exponential equation to linear form, if the base of the logarithm is e, then log(a) becomes 'capital a' and b * log(e) becomes 'capital b'. If the base is 10, log(a) becomes 'capital a' and b * log(e) is 'capital b' divided by log(e) base 10.

Can you provide an example of how to apply the least squares method to fit an exponential curve?

Yes, the video provides an example where data points are used to calculate the sums and averages required for the normal equations. Then, 'a' and 'b' are solved for, and finally, the exponential curve is written in the form y = a * e^(bx) using the estimated parameters.

Outlines

00:00

📈 Introduction to Fitting Exponential Curves

The speaker, SN Patil, introduces the topic of fitting exponential curves of the form y = a * e^(bx) using the least squares method. The video builds upon previous discussions on fitting straight lines and quadratic equations. The process involves transforming the exponential equation into a linear form by taking the logarithm of both sides, resulting in log(y) = log(a) + bx * log(e). The speaker emphasizes the importance of using logarithmic properties to simplify the equation and prepare it for the least squares method. The method requires defining an error function and finding the values of 'a' and 'b' that minimize the sum of the squares of the errors.

05:02

🔍 Detailed Explanation of the Least Squares Method

The video delves deeper into the least squares method, explaining how to define the error function and set up the normal equations necessary for finding the best-fit parameters 'a' and 'b'. The speaker demonstrates how to calculate the sum of the squares of the errors and how to differentiate this function with respect to 'a' and 'b' to find the values that minimize the error. The process involves substituting values and solving simultaneous equations to determine 'a' and 'b'. The speaker also discusses the implications of using different bases for the logarithm, such as natural (e) and base 10.

10:02

📚 Solving an Exponential Fitting Example Using Base e

The speaker provides a step-by-step guide to fitting an exponential curve to a given dataset using the natural base (e). The process begins with transforming the exponential equation into a linear form by taking the logarithm of both sides. The speaker then demonstrates how to use substitution to simplify the equation further and apply the least squares method to derive the normal equations. The video includes a detailed walkthrough of calculating the necessary sums and products of the dataset, as well as solving the normal equations to find the values of 'a' and 'b'. The final step involves back-substituting these values to obtain the exponential function that best fits the data.

15:03

🔢 Calculation and Rounding of Exponential Coefficients

The speaker continues the example from the previous paragraph, focusing on the calculation and rounding of the coefficients 'a' and 'b' for the exponential function. After obtaining the values of 'a' and 'b' from the normal equations, the speaker explains how to convert these values back to their original form, taking into account the base of the logarithm used. The video demonstrates the calculations for both natural base (e) and base 10, showing how to round the coefficients to a reasonable number of significant figures. The speaker also discusses the implications of these values on the shape of the fitted exponential curve.

20:04

📉 Fitting Exponential Curves Using Base 10

In this segment, the speaker shifts the focus to fitting exponential curves using base 10 logarithms. The process is similar to the one using base e, but with adjustments for the base 10 logarithm. The speaker explains how to take the logarithm of both sides of the exponential equation to linearize it, then uses substitution to simplify the equation. The video includes a detailed example, demonstrating how to prepare a chart with the necessary sums and products, and how to solve the normal equations to find the coefficients 'a' and 'b'. The speaker concludes by showing how to substitute these coefficients back into the original exponential equation to obtain the best-fit curve for the given data.

Keywords

💡Exponential Curve

An exponential curve is a mathematical function that describes the growth or decay of a quantity that increases or decreases at a rate proportional to its current value. In the video, the exponential curve is represented by the equation y = a * e^(bx), where 'a' and 'b' are constants, and 'e' is the base of the natural logarithm. The video discusses how to fit such a curve to data points using the method of least squares.

💡Least Squares Method

The least squares method is a standard approach in statistics to determine the best fit of a model to a set of data points by minimizing the sum of the squares of the vertical distances (residuals) between the observed values and the values predicted by the model. In the context of the video, this method is used to fit an exponential curve to a set of data points by finding the optimal values of 'a' and 'b'.

💡Logarithm

A logarithm is the inverse operation to exponentiation. It is the power to which a given number, the base, must be raised to produce a given number. In the script, logarithms are used to transform the exponential equation into a linear form, which simplifies the process of fitting the curve to the data using the least squares method.

💡Linear Form

A linear form refers to an equation where the variables are to the first power. In the video, the exponential equation y = a * e^(bx) is transformed into a linear form by taking the logarithm of both sides, resulting in log(y) = log(a) + bx * log(e). This transformation allows the use of linear regression techniques to fit an exponential curve.

💡Substitution

Substitution is a method used in algebra to solve equations by replacing a variable with an expression or a value. In the video, substitution is used to simplify the linear form of the exponential equation, where 'log(y)' is substituted with 'Y', 'log(a)' with 'A', and 'b * log(e)' with 'B', leading to the linear equation Y = A + Bx.

💡Error

In the context of the video, error refers to the difference between the actual observed values (y) and the values predicted by the model (f(x)). The least squares method minimizes the sum of the squares of these errors to find the best fit line. The script mentions that the error is calculated as the difference between the observed y values and the values predicted by the linearized model (A + Bx).

💡Normal Equation

The normal equations are a set of equations derived from the least squares method that are used to find the coefficients of the best fit line in linear regression. In the video, the normal equations are derived from the conditions that the partial derivatives of the sum of squared errors with respect to 'A' and 'B' are set to zero, leading to two equations that can be solved to find the optimal values of 'A' and 'B'.

💡Base of Logarithm

The base of a logarithm is the number to which the logarithm is applied. In the video, two bases are discussed: the natural base 'e' and base 10. The choice of base affects how the exponential curve is fitted and the values of 'a' and 'b' that are calculated. The video explains how to adjust the calculations based on the chosen base.

💡Data Points

Data points are the individual sets of values in a data set. In the video, data points are used to fit an exponential curve. The script describes how to prepare a chart with columns for the x values, the y values (or their logarithms), and other necessary calculations to apply the least squares method.

💡Fitting

Fitting, in the context of the video, refers to the process of finding the best curve that matches a set of data points. The video specifically discusses fitting an exponential curve to data points using the least squares method, which involves finding the coefficients 'a' and 'b' that minimize the sum of the squared differences between the observed and predicted values.

Highlights

Introduction to the lecture on fitting an exponential curve of the type y = ae^(bx).

Previous lecture discussed fitting of straight lines and quadratic equations.

Method of least squares is used to fit the exponential curve.

Equation y = ae^(bx) is transformed into a linear form by taking logarithms.

Logarithmic properties are used to simplify the equation.

Substitution is used to further simplify the equation into a linear form.

The method of least squares involves defining an error function.

The error is minimized by taking the sum of squares of the error.

The normal equations are derived from the method of least squares.

The values of 'a' and 'b' are found by solving the normal equations.

The base of the logarithm can be either natural (e) or base 10.

The values of 'a' and 'b' are simplified based on the base of the logarithm.

An example is solved using the least squares method for a natural base.

The process involves reducing the exponential equation to a linear form and using substitution.

The normal equations are set up using the method of least squares.

The values of 'a' and 'b' are calculated using the normal equations.

The final exponential curve is derived using the calculated values of 'a' and 'b'.

A second example is solved using the least squares method for base 10.

The process is repeated with adjustments for the base 10 logarithm.

The final exponential curve is derived for base 10.

Conclusion and thanks for watching the lecture.

Transcripts

hello everyone my name is sn patil in

this video lecture we will discuss

spitting of an exponential curve of the

type y is equal to a e raised to

b x in previous

to video lecture we already discussed

the fitting

of straight line and fitting of

non-linear equation in which we

discussed the two degree polynomial

means quadratic equation

okay so here we find the fitting of an

exponential curve

using the method of least square

now we have the equation y is equal to

a e raised to

b x so first we reduce this equation to

the linear form

for that we take the log both sides so

taking log both sides what we get

if we take the log then log of y is

equal to log of

log of a e raised to b x now here apply

the logarithmic property log of a plus

log log of a into b is log of a plus

log of b and log of

log of a raise to b is b into log of a

using this property

we get the equation

as a log of y is equal to log of a plus

b x log of e basis m so here we consider

two type of base natural and base a

okay now uh for uh

for our simplicity means we convert this

equation into a linear form

for that we use the substitution

put

log y is equal to capital y log a is

equal to capital a b into log e is equal

to capital

b then equation to become

is linear that is y is equal to a plus

b

x now here we use method of least square

those we have already discussed in a

previous two video lecture fitting of

straight line and fitting of quadratic

equation okay and we

find out

using that method we find out the

normal equation okay let me remind the

what we discuss

in method of least square first we

define the error suppose we have the

function y is equal to f of x and f of x

is i'm considering this equation

uh a plus b x

okay that is the error is equal to

capital y minus a

minus b x means difference between these

two y minus a profits

y minus f of x okay

now here the error is minimum when we

squaring this

okay

that is

y minus a

minus bx bracket square

okay

the

sum of

sum of

square of the error should be a minimum

here

that is uh we write this equation as a

y uh

as a sum of the square means what

the left hand side we consider

as a yes yes

in bracket a comma b means the right

hand side is a function of capital a and

b and the summation

summation

y a

minus

y minus a minus b

x

here we consider the n points x 1 x 2

and their corresponding value y 1 y 2

okay

and it squared

so

when

when we get the

when we get the capital a and capital b

so here we use the necessary condition

for this function is minimum

okay for that we use we differentiate

this function partially with respect to

capital a

and equal to the zero and partially

differentiate with s with respect to

capital b is equal to zero we

differentiate this equation

with respect to a and b and equal to the

zero and solid then we get the two

normal equation

that is the first normal equation

and

second normal equation when we get when

we differentiate with respect to a and

with respect to b

okay and solve this we get this to

normal equation

okay

now solving this equation 4 and 5 we get

the value of a and b but what is a

and what is b

simplify this okay

that is capital a is log

a

base is m and b is b into log of m base

is e

okay and here simplify we get value of a

and b and put this a and b in equation

number in equation number one we get the

required equation of the exponential

function

okay now here discuss the

regarding the base so first we consider

the natural base that is m is equal to e

then log a

base e is equal to capital a then small

a we get e raised to

a capital a and b

when base is e

then here we get 1 that is b is equal to

capital b and when we consider base as a

10

when m is equal to 10

then

then log of a base 10 is equal to

capital a then small a is equal to 10

raised to a

and b into log e base 10 is equal to

capital b then b is equal to

the log

e base 10 is divided here that is

capital b divided by the

the value of the log e base 10 that is

0.434294

okay

let us solve the example first

we solve the example base e and then we

solve the example we will solve on the

base 10

okay

now fit an exponential curve

line by least square method to the

following data

okay

that is the

exponential curve

the first we write the exponential curve

y is equal to a e to the power

bx

so we

first

we reduce this equation to the linear

form for that we take the log both side

then we get

we get log y is equal to log a plus b x

log of e and throughout base is e

now use the substitution put

log y is equal to capital y log a is

equal to capital a b into log e that is

it become 1 simply that is b is equal to

capital

b

and put this equation

put this value sorry put this value in

equation number 2

then we get the linear equation that is

capital y is equal to a plus bx

okay once we get the linear equation

using the least square method we get the

two normal equation

capital y is equal to n a plus b capital

b summation x and the second is

summation x y is equal to

a capital a summation x plus b summation

x square

okay now here the n means what number of

points here number of points one two

three four add five okay then n become

here five

n is equal to five

now here

now we prepare the chart as per as the

requirement of the four and five is ten

set

the we want the

summation x for that

for that we must have the column x now

summation y summation y is log of y so

prepare the column for log y then we get

the capital one

okay now uh

here then x into y then we have the

column of this capital y and column of x

take their multiplication okay we

as now here the x square

prepare the column for x square we get

the summation x square now see the table

now here this is the summation

this is the x so here we get summation

x

okay this is the y we want the log of y

so through kelsey calculate the value of

log of y

okay log 5.0600

substitute here we get 1.6214

similarly calculate the all remaining

value this is

this is capital y

this is capital y this is summation x

squared

and this is summation

x into capital

y now put all these value in normal

equation see this is the

normal equation

four and five substitute the value

summation value in equation 4 and 5

then we get

we get

put

in

normal equation normal

equation that is four and five

so equation four become

equation four equation for the

the left hand side

summation capital one

okay

now put this value here

that is

some summation capital y is seventeen

point

three eight

four one

okay is equal to

is equal to

5 times ena means 5 times a

now i'm writing the

equation here so that it will be helpful

for us summation capital y is equal to

n a

plus

plus

capital b summation x

now summation x into capital y is equal

to

a

into summation x plus b

summation x square

okay this is the 4 and this is the 5

okay

now substitute here then here 5a now

plus

plus summation x summation x is just 25

into b

into b now this equation 5. equation 5

summation x into y see this is the x

into

y

we get

105.9291

okay this value is what summation of

this all point

okay

is equal to is equal to 25 a 25 means

what summation

x

now plus

plus summation x square means

165

summation x1 into

b

okay say equation number five six

and seven now solve six and seven by

simultaneous method yeah any known

method yeah through kelsey

solve

solve

equation

six

and seven

okay

then we get a is equal to now i am

calculating this value through kelsey a

is equal to

1.0

1.1

0

7

4 5

and capital b is equal to zero point

four seven

five

two

okay now we write this value as a one

point one zero 0 we round up this number

that is 5

okay

and again round off this

that is the

8

when we round up this it become 5 and

when we round up up to this digit then

it become 8

ok

otherwise keep as it is no problem

now we get the capital a and capital b

but the we require small a and small b

but

log of a base e

is equal to a

so that is

log of

e

a

the a is

1.1008

that is a is equal to a is equal to

here

if we calculate here

a is equal to e to the power

a

okay that is e to the power one point

one zero zero

eight

just

calculating through kelsey

eight to the power

one point one zero zero

it

we get three point

a is equal to three point

zero zero

six five

six five seven so we round up this that

is a is equal to 3.0066

now b is equal to

b is equal to

now here this is the base is natural so

that b is equal to capital b okay

that is small b is equal to zero point

four seven five two

okay we discuss

if base is next natural then capital b

is same as small b

okay put this value in equation 1

put in

equation 1

equation 1 is

y is equal to a e raised to b x that is

y is equal to the value of a

is 3.0066

e to the power b is 0.47

5 to

x

which is required fitting of exponential

curve from the

given data

okay this is the problem for

natural base now next problem we will

discuss the basis 10.

okay

fit an exponential curve line by least

square method to the following data

now the first process is as usual those

we have discussed in a previous problem

first we consider the exponential

function y is equal to a e to the power

b x taking log both sides why we take

the log both side we reduce this

equation to the linear

equation

okay

then we get the equation in terms of log

then we use substitution

that is log y is equal to

capital y and now right now we consider

the basis n

log a is equal to capital a b log e is

equal to capital b substitute in

equation 2 we get the linear equation

y is equal to capital a plus bx now we

have the linear equation then it's

normal equation

for normal equation just multiply the

summation here that is summation y

summation operate on the constant then

it run number of points are there if

there are n points then n a if there are

three right now three point then three a

so we get the n is three

n is

three

okay now for second equation first

multiplied throughout x

first multiplied throughout x and then

multiplied summation sign

so we get the second normal equation

okay now as per the requirement of the 4

and 5 we prepare the chart see we

prepare the chart for x we want the

summation x now x

so for column of x we get the summation

x now some

summation y now it require

summation y means what capital y it

requires the log one so it require the

log y so first we must have the y and

then we prepare the log y column okay

log y means capital

y

then x uh x square

just squaring this

okay and then

x into capital y this is the x and this

is the capital y x into capital y in

this way prepare the chart

okay now c

so this is the chart x y

capital y x square x into y so we

calculate this value log of y base 10

through c

okay

now first we write the normal equation

in previous slide

there is a little bit confusion so that

first we write the normal equation

normal equation

capita summation 1

see i'm writing the same equation on the

next slide

okay

that is summation y n a n means 3

a

plus

capital b summation x this is the

equation number four

then

we say for second normal equation just

multiply x

summation x into capital y

okay a

summation x

plus

b

x into x x square c

five now put this value

put the summation x this is summation

capital y

this is summation

x square

this is summation x into capital y

substitute here

then we get equation 4

as a

capital y means three point two

three point two

zero zero

okay

is equal to three a

plus summation x is six

b

now 5 become

summation x into y that is 7 point

7.9998

is equal to

summation x is 6

6 a

plus

summation x square that is 20

b

now solve this equation say equation

number 6

say equation number seven

solve six and seven

and seven we get a is equal to

now i am solving this

using the kelsey

that is three

is equal to six

seven point nine nine nine eight

okay

so we get the first value is

zero point

we get first value a is equal to

0.66671

and b is equal to

b is equal to

one zero point

zero point

one nine nine

one nine nine

nine seven five okay

now these are the capital a and b and we

require the small a and b

okay

but

but

capital a is equal to

log of

a

a base 10

so here a is equal to

a is equal to

10 raised to e

we get

10 raised to

0.66671

we get a is equal to

4 point through calc four point six

four two

one

now b is equal to

b is equal to

capital b is equal to

log small b log

base state

and e

okay

that is b is equal to

b is equal to capital b

divided by the log e base 10

capital b is what zero point one

nine nine nine seven five

and log e base ten its value is zero

point

four three

four two

nine four

we get b is equal to

zero point four six

0 4 5 so this is the small b

put this value

in equation 1

put in

equation 1 then equation 1 become what

is equation 1 y a

y is equal to a e to the power b x

y is equal to a a is 4.6421

e

raised to

0.4605

into x which is required fitting of

exponential

function

thank you thanks for watching