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Round-Off Errors; and Iterative Methods - MATLAB Video Lecture | MATLAB Programming for Numerical Computation - Software Development
FAQs on Round-Off Errors; and Iterative Methods - MATLAB Video Lecture - MATLAB Programming for Numerical Computation - Software Development
| 1. What are round-off errors in MATLAB? |  |
Ans. Round-off errors in MATLAB refer to the errors that occur due to the limited precision of floating-point numbers. These errors occur when a number is stored in a computer's memory using a finite number of bits, leading to the loss of precision in calculations.
| 2. How can round-off errors affect the accuracy of iterative methods in MATLAB? | |
Ans. Round-off errors can accumulate during the iterations of iterative methods in MATLAB, leading to a loss of accuracy. As the iterations progress, the errors can magnify and cause the calculated result to deviate from the true solution. Therefore, it is important to consider the impact of round-off errors when using iterative methods.
| 3. What are some common iterative methods used in MATLAB to solve equations? | |
Ans. Some common iterative methods used in MATLAB to solve equations include the Jacobi method, Gauss-Seidel method, and successive over-relaxation (SOR) method. These methods involve iteratively updating the solution until a desired level of accuracy is achieved.
| 4. How can I mitigate round-off errors in MATLAB when using iterative methods? | |
Ans. To mitigate round-off errors in MATLAB when using iterative methods, you can use techniques such as double-precision arithmetic, increasing the precision of calculations, or employing specialized numerical libraries that handle round-off errors more effectively. Additionally, using convergence criteria and adaptive methods can help control the impact of round-off errors.
| 5. Are there any built-in functions in MATLAB to handle round-off errors in iterative methods? | |
Ans. MATLAB provides various built-in functions and tools to handle round-off errors in iterative methods. The 'vpa' function allows for variable-precision arithmetic, enabling higher precision calculations. The 'eps' function helps determine the machine epsilon, which gives an estimate of the round-off error. Additionally, the Symbolic Math Toolbox provides advanced numerical techniques to handle round-off errors.
Text Transcript from Video
hello and welcome to this week to of math
matlab programming for numerical computation
course aah in this module we are going to
cover errors and approximations ah this is
lecture lecture 2 point 1 and in this lecture
we are going to look at where do errors originate
in numerical methods
ah one of the things that happens always when
we talk about numerical methods when you use
computers for ah u using numerical techniques
ah you will always encounter errors of ad
various forms and these errors are going to
have 2 basic aah origins 1 is what is known
as truncation errors and the other is known
as the round of errors ah where do this error
come from and how to they affect the overall
numerical methods is something that we will
introduce to ourselves in this particular
lecture
ok so let us look at the 1st term type of
an error that is called the truncation error
and what are listed over here is the maclaurin
series for calculating e to the power x the
maclaurin series of e to the power x upto
nth order term is return as 1 e to the power
a i am sorry is return as 1 plus a plus a
square by 2 factorial plus a cube by 3 factorial
and so on
so what we are going to do is we are going
to compute e to the power 0 point 1 using
multiple number of terms in this maclaurin
series and we will compare it with the actual
value ah the error we have we will define
as the difference between the true value and
the approximate value we do not care whether
this difference is positive or negative the
error is therefore just the absolute value
of the difference so it does not matter whether
the approximate value is greater than or lesser
than the true value as long as it s different
than the true value we want to capture how
much that difference is and that is how we
define the absolute error ok
and what we will show using matlab is that
if we introduce more number of terms in this
maclaurin series we are going to get lower
error ok so let us head on over to matlab
in order to compute this using
aah the maclaurin series approximation so
let us open ah script file ah let us call
it maclaurinExp ok will create the file maclaurin
series ffr for exp of 0 point 1 ok so let
us look at maclaurin series with different
number of terms so let us say we want to look
at terms upto n equal to 5 so we will say
n equal to 5 our value of a was 0 point 1
ok and vah the result that is that is called
as as expVal equal to 1 point 0 why 1 pont
0 is
the 1st term over here is 1 the first order
term that is a to the power 1 term is a a
to the power 2 term is a square bha divided
by 2 factorial so on and so forth
so if you look at the 1st term and compare
it with the 2nd term it is the 1st term is
just multiplied aah by a divided by 2 if you
compare these 2 terms is just multiplied by
a divided by 3 so on and so forth
so that is what we are going to do we will
call currentTerm as 1 point 0 and we will
use a for loop for i equal to 1 to n ok and
currentTerm equal to currentTerm multiplied
by a divided by i ok so let us see how exactly
this works
so we have e to the power a ah that we need
to compute approximately when i is going to
be equal to 1 at that time ah our value of
currentTerm is going to be 1 multiplied by
a divided by 1 that is a in the next loop
it is going to be a multiplied by a by 2 that
is a square by 2 factorial the next term will
be a multiplied ah a by a square by 2 multiplied
by a by 3 that would be a cube by 3 factorial
so on and so forth ok
and our expVal equal to expVal plus currentTerm
ok and i will end this loop over here so that
is going to be the value of expVal let us
say trueval equal to exp 0 point 1 and error
equal to abs trueval minus expVal ok so we
save this and will run let us go to matlab
and run it and we get this result so we actually
see that the error is ah 1 ah 1 point 4 multiplied
by e to the power 10 minus 9 that is the error
expVal is 1 point 1052 and trueval is also
1 point 1052 because the error is of the order
of n to the power minus 9 ok now what let
us do this is let us change the particular
code so that we store all the various values
of expVal so let us say expVal i equal to
expVal i minus 1 of let us put it this way
expVal i plus 1 equal to expVal i plus currentTerm
so what does the first expVal contain it is
it is the 0 ordered approximation
that is just 1 so that is not really an approximation
expVal 2 is going to have 1 plus a expVal
3 is going to have 1 plus a plus a square
by 2 factorial so on and so forth ff when
we calculate this error trueval is a scalar
and expVal is a vector so we can subtract
a scalar from a vector or a vector from a
scalar and we will get a result in vector
so this should not be any problem for us the
error will actually ah end of the of 6 dimensional
vector a 1 by 6 vector
ok so let us go ahead and clear ok and let
us now run maclaurin ex exp ok so now we have
the error and as you can see the error is
10 to the power minus 3 10 to the power minus
4 and as we go to higher order terms the error
keeps decreasing quite significantly so what
we see over here is that as we introducing
more and more terms our error keeps decreasing
quite significantly ok let u s look at error
4 which is basically when n equal to 3 our
error is 10 to the power minus 6 when n equal
to 4 our error is going to be 10 to the power
minus 8 ok and we have seen already that when
we introduce all 5 terms of the maclaurin
series our error drops aah even further ok
so this is how we have computed
ah aah the e to the power a approximation
using maclaurin series so what we have seen
over here is that as we truncate the overall
series to nth order terms we realized that
the error keeps on decreasing so more the
number of terms of maclaurin series we introduce
aah ther th the smaller is going to be the
error ok why does the error keep changing
or why is the er an error in the first place
the reason why there is an error in the first
place is because we are truncating this maclaurin
series so this series is being truncated at
the 1st term or the 2nd term or the 3rd term
and so on and so forth
if we truncated just the 1st order term we
are going get a greater amount of error the
error was of the order of tu to 10 to the
power of minus 3 when we truncated it to 2nd
ordered term we got error of 10 to the power
minus 4 when we truncated it to the 3rd ordered
term we got the error of the order of 10 to
the power minus 6 so on and so forth so what
we saw over here is that once we introduce
greater number of terms we are going to reduce
the error this is going to be something that
is going to we are going to see consistently
in this particular course so if ah ve wha
i can summarize these results what this results
are is when you have a infinite series the
greater number of terms that you retain in
the infinite series the greater is the accuracy
or conversely smaller is the error ok ah in
the next lecture we will expand this particular
idea to ah taylor series expansion tylor series
expansion is indeed where we get this maclaurin
series from so that particular idea of introducing
ah more number of terms in order to reduce
the overall truncation error is going to be
a common motive that we are going to follow
in the rest of this particular lecture series
ok so this was the first type of an error
ah which is the truncation error the second
type of an error truncs out because of what
is known as machine precision we like to think
of aah vah a computer as having what is known
as an infinite precision however that is not
really true the machine aah the computer also
has a machine precision which in some ways
is like a least count so let us consider a
ruler that i have shown over here aah th the
least count of this ruler is 1 millimeter
or 0 point wha 1 centimeter so that s kind
of the minimum resolution of that ruler you
cannot measure length that is less than 1
millimeter ok if you want to measure then
with the precision greater than 1 millimeter
or a precision value of sub millimeter
we can for example use a vernier caliper a
this particular example of vernier calipers
has the least count of 0 point 01 centimeter
so we can say that this vernier calipers has
a better precision than a ruler ok
so precision is basically a term which we
can approximately we can use it in of the
term least count ok so it is kind of like
least count now in the same manner in which
these 2 ah devices have a least count our
computer
is also has a least count ok ah just like
this devices the real numbers in a computer
representation have a least count so between
a number a and a plusthat does not exist any
number and that is what the machine precision
is about it is about the least count of that
particular computer ok and that least count
is dependent on the number of bytes that we
used in order to store a real number so for
example ah aah ah a standard real number is
ah uses ah 4 bytes which is basically 32 bits
so that particular real number has a certain
precision
a double precision real number which uses
8 bytes or 64 bits has a better machine precision
ok
aah and that rep representation is based on
what is known as the floating point representation
of a number ah for more information on this
you can go to the computational techniques
course the link for which is given over here
and look at the machine precision and floating
point representation so let us look at floating
point representation of a decimal number
let us say we have 5 boxes in order to store
a number and 1 box in order to store an exponent
ok
the number is always going to be represented
as 0 point x x x x x multiplied by 10 to the
power n ok
so this is where you we put in those 5 digits
of our decimal number so let us see
how we represent a number of 23 point 217
the number 23 point 217 becomes 0 point 23217
multiplied by 10 to the power 2 so what is
the next number that comes in the next number
is when the lowest digit over here is incremented
so that becomes the next number becomes 23218
multiplied by 10 to the power 2 that is the
next number ok
so what if we had a number of the form
23.2172 the problem is the twen the number
23.2172 cannot be represented with 5 digits
of what is known as mantissa and 1 digit exponent
the reason is that number will be represented
as 0.23217 multiplied by 10 to the power 2
there is no space to store this trailing digit
ok as a result this number and this particular
number have the exact same representation
the reason why we are not able to represent
a number with a 6 significant digit is because
we do not have enough precision in our decimal
machine ok
so the least count of this particular decimal
machine turns out to
be 10 to the power minus 5 multiplied by the
absolute value of x so that is the 10 to the
power minus 5 is what is known as the machine
precision
in case of matlab the machine precision is
given by a keyword eps and the machine precision
in matlab is 2 multiplied by 10 to the power
minus 16 ok this machine precision is 2 to
the power minus 52 the reason is because
just the way we had 5 digit representation
of this mantissa or in this decimal number
likewise the double precision real number
that matlab uses ah the which is the ibm standard
uses a 52 bit mantissa so the mantissa has
52 bits in it so the precision is 2 to the
power minus 52
note over here that the precision was 10 to
the power minus 5 ok so let us the case where
we have let us call
a equal to 1 plus 2 to the power minus 52
so ok
that is the value of our a and if you subtract
a minus 1 we are going to extract this particular
value ok
so what is happening over here is when we
use the value of this this precision aah that
particular number a 1 plus 2 into 10 to the
power minus 6 is indeed represented as 1 point
000 15 times followed by a 2 however what
if we write a equal to 1 minus 2 to the power
minus 53
this number is going to be indistinguishable
from 1 if we say y a minus 1 we are going
to get a result of 0 ok so what does this
mean is this means that our matlab has a least
count of 2 to the power minus 16 and that
is what is known as the machine precision
ok so with this we come to the end of this
particular lecture in this lecture we cove
covered 2 important context ah ah concepts
the 2 concepts that we covered in this lecture
were is that the truncation error because
we are going to truncate an infinite series
to a finite number of terms there is going
to be an inherent error that is associated
with that truncation the second thing that
we realized was that there is a machine precision
we cannot represent any real number in any
form that we want there is going to be a least
count res with respect to representing those
real numbers so what does
this imply when it comes to numerical techniques
something we are going to cover in the next
lecture thank you and see in the next lecture
matlab programming for numerical computation
course aah in this module we are going to
cover errors and approximations ah this is
lecture lecture 2 point 1 and in this lecture
we are going to look at where do errors originate
in numerical methods
ah one of the things that happens always when
we talk about numerical methods when you use
computers for ah u using numerical techniques
ah you will always encounter errors of ad
various forms and these errors are going to
have 2 basic aah origins 1 is what is known
as truncation errors and the other is known
as the round of errors ah where do this error
come from and how to they affect the overall
numerical methods is something that we will
introduce to ourselves in this particular
lecture
ok so let us look at the 1st term type of
an error that is called the truncation error
and what are listed over here is the maclaurin
series for calculating e to the power x the
maclaurin series of e to the power x upto
nth order term is return as 1 e to the power
a i am sorry is return as 1 plus a plus a
square by 2 factorial plus a cube by 3 factorial
and so on
so what we are going to do is we are going
to compute e to the power 0 point 1 using
multiple number of terms in this maclaurin
series and we will compare it with the actual
value ah the error we have we will define
as the difference between the true value and
the approximate value we do not care whether
this difference is positive or negative the
error is therefore just the absolute value
of the difference so it does not matter whether
the approximate value is greater than or lesser
than the true value as long as it s different
than the true value we want to capture how
much that difference is and that is how we
define the absolute error ok
and what we will show using matlab is that
if we introduce more number of terms in this
maclaurin series we are going to get lower
error ok so let us head on over to matlab
in order to compute this using
aah the maclaurin series approximation so
let us open ah script file ah let us call
it maclaurinExp ok will create the file maclaurin
series ffr for exp of 0 point 1 ok so let
us look at maclaurin series with different
number of terms so let us say we want to look
at terms upto n equal to 5 so we will say
n equal to 5 our value of a was 0 point 1
ok and vah the result that is that is called
as as expVal equal to 1 point 0 why 1 pont
0 is
the 1st term over here is 1 the first order
term that is a to the power 1 term is a a
to the power 2 term is a square bha divided
by 2 factorial so on and so forth
so if you look at the 1st term and compare
it with the 2nd term it is the 1st term is
just multiplied aah by a divided by 2 if you
compare these 2 terms is just multiplied by
a divided by 3 so on and so forth
so that is what we are going to do we will
call currentTerm as 1 point 0 and we will
use a for loop for i equal to 1 to n ok and
currentTerm equal to currentTerm multiplied
by a divided by i ok so let us see how exactly
this works
so we have e to the power a ah that we need
to compute approximately when i is going to
be equal to 1 at that time ah our value of
currentTerm is going to be 1 multiplied by
a divided by 1 that is a in the next loop
it is going to be a multiplied by a by 2 that
is a square by 2 factorial the next term will
be a multiplied ah a by a square by 2 multiplied
by a by 3 that would be a cube by 3 factorial
so on and so forth ok
and our expVal equal to expVal plus currentTerm
ok and i will end this loop over here so that
is going to be the value of expVal let us
say trueval equal to exp 0 point 1 and error
equal to abs trueval minus expVal ok so we
save this and will run let us go to matlab
and run it and we get this result so we actually
see that the error is ah 1 ah 1 point 4 multiplied
by e to the power 10 minus 9 that is the error
expVal is 1 point 1052 and trueval is also
1 point 1052 because the error is of the order
of n to the power minus 9 ok now what let
us do this is let us change the particular
code so that we store all the various values
of expVal so let us say expVal i equal to
expVal i minus 1 of let us put it this way
expVal i plus 1 equal to expVal i plus currentTerm
so what does the first expVal contain it is
it is the 0 ordered approximation
that is just 1 so that is not really an approximation
expVal 2 is going to have 1 plus a expVal
3 is going to have 1 plus a plus a square
by 2 factorial so on and so forth ff when
we calculate this error trueval is a scalar
and expVal is a vector so we can subtract
a scalar from a vector or a vector from a
scalar and we will get a result in vector
so this should not be any problem for us the
error will actually ah end of the of 6 dimensional
vector a 1 by 6 vector
ok so let us go ahead and clear ok and let
us now run maclaurin ex exp ok so now we have
the error and as you can see the error is
10 to the power minus 3 10 to the power minus
4 and as we go to higher order terms the error
keeps decreasing quite significantly so what
we see over here is that as we introducing
more and more terms our error keeps decreasing
quite significantly ok let u s look at error
4 which is basically when n equal to 3 our
error is 10 to the power minus 6 when n equal
to 4 our error is going to be 10 to the power
minus 8 ok and we have seen already that when
we introduce all 5 terms of the maclaurin
series our error drops aah even further ok
so this is how we have computed
ah aah the e to the power a approximation
using maclaurin series so what we have seen
over here is that as we truncate the overall
series to nth order terms we realized that
the error keeps on decreasing so more the
number of terms of maclaurin series we introduce
aah ther th the smaller is going to be the
error ok why does the error keep changing
or why is the er an error in the first place
the reason why there is an error in the first
place is because we are truncating this maclaurin
series so this series is being truncated at
the 1st term or the 2nd term or the 3rd term
and so on and so forth
if we truncated just the 1st order term we
are going get a greater amount of error the
error was of the order of tu to 10 to the
power of minus 3 when we truncated it to 2nd
ordered term we got error of 10 to the power
minus 4 when we truncated it to the 3rd ordered
term we got the error of the order of 10 to
the power minus 6 so on and so forth so what
we saw over here is that once we introduce
greater number of terms we are going to reduce
the error this is going to be something that
is going to we are going to see consistently
in this particular course so if ah ve wha
i can summarize these results what this results
are is when you have a infinite series the
greater number of terms that you retain in
the infinite series the greater is the accuracy
or conversely smaller is the error ok ah in
the next lecture we will expand this particular
idea to ah taylor series expansion tylor series
expansion is indeed where we get this maclaurin
series from so that particular idea of introducing
ah more number of terms in order to reduce
the overall truncation error is going to be
a common motive that we are going to follow
in the rest of this particular lecture series
ok so this was the first type of an error
ah which is the truncation error the second
type of an error truncs out because of what
is known as machine precision we like to think
of aah vah a computer as having what is known
as an infinite precision however that is not
really true the machine aah the computer also
has a machine precision which in some ways
is like a least count so let us consider a
ruler that i have shown over here aah th the
least count of this ruler is 1 millimeter
or 0 point wha 1 centimeter so that s kind
of the minimum resolution of that ruler you
cannot measure length that is less than 1
millimeter ok if you want to measure then
with the precision greater than 1 millimeter
or a precision value of sub millimeter
we can for example use a vernier caliper a
this particular example of vernier calipers
has the least count of 0 point 01 centimeter
so we can say that this vernier calipers has
a better precision than a ruler ok
so precision is basically a term which we
can approximately we can use it in of the
term least count ok so it is kind of like
least count now in the same manner in which
these 2 ah devices have a least count our
computer
is also has a least count ok ah just like
this devices the real numbers in a computer
representation have a least count so between
a number a and a plusthat does not exist any
number and that is what the machine precision
is about it is about the least count of that
particular computer ok and that least count
is dependent on the number of bytes that we
used in order to store a real number so for
example ah aah ah a standard real number is
ah uses ah 4 bytes which is basically 32 bits
so that particular real number has a certain
precision
a double precision real number which uses
8 bytes or 64 bits has a better machine precision
ok
aah and that rep representation is based on
what is known as the floating point representation
of a number ah for more information on this
you can go to the computational techniques
course the link for which is given over here
and look at the machine precision and floating
point representation so let us look at floating
point representation of a decimal number
let us say we have 5 boxes in order to store
a number and 1 box in order to store an exponent
ok
the number is always going to be represented
as 0 point x x x x x multiplied by 10 to the
power n ok
so this is where you we put in those 5 digits
of our decimal number so let us see
how we represent a number of 23 point 217
the number 23 point 217 becomes 0 point 23217
multiplied by 10 to the power 2 so what is
the next number that comes in the next number
is when the lowest digit over here is incremented
so that becomes the next number becomes 23218
multiplied by 10 to the power 2 that is the
next number ok
so what if we had a number of the form
23.2172 the problem is the twen the number
23.2172 cannot be represented with 5 digits
of what is known as mantissa and 1 digit exponent
the reason is that number will be represented
as 0.23217 multiplied by 10 to the power 2
there is no space to store this trailing digit
ok as a result this number and this particular
number have the exact same representation
the reason why we are not able to represent
a number with a 6 significant digit is because
we do not have enough precision in our decimal
machine ok
so the least count of this particular decimal
machine turns out to
be 10 to the power minus 5 multiplied by the
absolute value of x so that is the 10 to the
power minus 5 is what is known as the machine
precision
in case of matlab the machine precision is
given by a keyword eps and the machine precision
in matlab is 2 multiplied by 10 to the power
minus 16 ok this machine precision is 2 to
the power minus 52 the reason is because
just the way we had 5 digit representation
of this mantissa or in this decimal number
likewise the double precision real number
that matlab uses ah the which is the ibm standard
uses a 52 bit mantissa so the mantissa has
52 bits in it so the precision is 2 to the
power minus 52
note over here that the precision was 10 to
the power minus 5 ok so let us the case where
we have let us call
a equal to 1 plus 2 to the power minus 52
so ok
that is the value of our a and if you subtract
a minus 1 we are going to extract this particular
value ok
so what is happening over here is when we
use the value of this this precision aah that
particular number a 1 plus 2 into 10 to the
power minus 6 is indeed represented as 1 point
000 15 times followed by a 2 however what
if we write a equal to 1 minus 2 to the power
minus 53
this number is going to be indistinguishable
from 1 if we say y a minus 1 we are going
to get a result of 0 ok so what does this
mean is this means that our matlab has a least
count of 2 to the power minus 16 and that
is what is known as the machine precision
ok so with this we come to the end of this
particular lecture in this lecture we cove
covered 2 important context ah ah concepts
the 2 concepts that we covered in this lecture
were is that the truncation error because
we are going to truncate an infinite series
to a finite number of terms there is going
to be an inherent error that is associated
with that truncation the second thing that
we realized was that there is a machine precision
we cannot represent any real number in any
form that we want there is going to be a least
count res with respect to representing those
real numbers so what does
this imply when it comes to numerical techniques
something we are going to cover in the next
lecture thank you and see in the next lecture
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