Fitting of an Exponential Curve of the Type 𝒚=𝒂𝒆(^𝒃𝒙)
Summary
TLDRIn 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
📈 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.
🔍 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.
📚 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.
🔢 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.
📉 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.
Mindmap
Keywords
💡Exponential Curve
💡Least Squares Method
💡Logarithm
💡Linear Form
💡Substitution
💡Error
💡Normal Equation
💡Base of Logarithm
💡Data Points
💡Fitting
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
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