Iteration - GCSE Higher Maths
Summary
TLDRThis educational video introduces the concept of iteration, a method for solving cubic equations that cannot be addressed with the quadratic formula. It demonstrates how to show a solution exists between two values by observing a change of sign in the equation's output. The video then guides through the process of rearranging the equation for iteration, using an iterative formula to approximate the solution progressively closer. It concludes with a practical example, calculating successive terms to find a solution accurate to five decimal places, and verifying the solution's accuracy by substituting it back into the original equation.
Takeaways
- 🔍 The video introduces the concept of iteration, a method for solving equations, particularly cubic equations that cannot be solved using the quadratic formula.
- 📌 Iteration begins by identifying a range where a solution exists, such as between x=1 and x=2, by observing a change of sign when substituting these values into the equation.
- 📈 A change of sign in the equation's output when substituting values indicates a solution exists between those values, which can be visualized on a graph where the curve crosses the x-axis.
- 📝 To apply iteration, the equation must be rearranged so that x is isolated on one side, setting the stage for the iterative formula.
- ✂️ The iterative formula involves taking the cube root of a term involving x, which is derived from rearranging the original equation.
- 🔑 The subscript n+1 denotes the next term in the sequence, while subscript n refers to the current term, allowing for the calculation of successive approximations of the solution.
- 🔄 The iterative process involves substituting the current term into the formula to find the next term, which is then used to find the subsequent term, and so on.
- 📊 The initial term, x0, is given or chosen, and successive terms (x1, x2, x3, etc.) are calculated using the iterative formula.
- 🔢 Iteration can be performed using a calculator, making it easy to compute multiple iterations quickly and accurately.
- 🎯 The goal of iteration is to approximate the solution to a high degree of accuracy, with each iteration bringing the approximation closer to the exact solution.
- 📋 To verify the accuracy of the solution, the calculated value can be substituted back into the original equation, with a result close to zero indicating a precise approximation.
Q & A
What is the main topic of the video?
-The main topic of the video is the concept of iteration, specifically how to use it to solve a cubic equation.
Why can't the quadratic formula be used for the given equation?
-The quadratic formula cannot be used because the given equation is a cubic equation, with the highest power of x being 3, not 2 as in quadratic equations.
What is the first step to show that a solution to the equation lies between x equals one and x equals two?
-The first step is to substitute x with 1 and 2 in the equation and observe the change in sign from negative to positive, which indicates a solution between these two values.
What does a change of sign in the values of the equation when substituting x with 1 and 2 indicate?
-A change of sign indicates that there is at least one solution to the equation between the two substituted values, in this case, between x = 1 and x = 2.
How can the iterative formula be derived from the given cubic equation?
-The iterative formula is derived by rearranging the equation to make x the subject, then taking the cube root of both sides, and finally writing x with subscripts to represent the iterative process.
What does the subscript 'n+1' represent in the context of the iterative formula?
-In the context of the iterative formula, the subscript 'n+1' represents the next term in the sequence of approximations to the solution.
What is the purpose of the iterative formula in solving the cubic equation?
-The purpose of the iterative formula is to provide a method to calculate successive approximations to the solution of the equation, with each iteration getting closer to the actual solution.
How is the initial term 'x0' used in the iterative process?
-The initial term 'x0' is the starting point for the iterative process. It is used to calculate the first term 'x1', and subsequent terms are calculated using the iterative formula.
What is the significance of calculating x1, x2, and x3 in the iterative process?
-Calculating x1, x2, and x3 is part of the iterative process that gradually refines the approximation of the solution to the cubic equation. Each term is closer to the actual solution than the previous one.
How can you determine the accuracy of the solution obtained through iteration?
-The accuracy of the solution can be determined by substituting the iterative solution back into the original equation and observing how close the result is to zero. A very small non-zero result indicates a high level of accuracy.
What is the final step in solving the cubic equation using iteration as described in the video?
-The final step is to continue the iterative process until the solution stabilizes, and then round the result to the desired number of decimal places, as specified in the question.
Outlines
📚 Introduction to Iteration for Solving Cubic Equations
This paragraph introduces the concept of iteration as a method for solving cubic equations, which cannot be addressed with the quadratic formula. It demonstrates how to show that a solution exists between x=1 and x=2 by substituting these values into the equation and observing a change of sign from negative to positive. The explanation includes the graphical representation of the equation, where the graph crosses the x-axis, indicating a solution. The paragraph also explains the process of rearranging the equation to isolate x and the introduction of an iterative formula to approximate the solution through successive iterations.
🔍 Iterative Formula Application and Calculation Process
The second paragraph delves into the application of the iterative formula to calculate the successive terms of the sequence, starting with an initial term x0=2. It explains the iterative process of using the formula to find x1, x2, and x3 by substituting the current term into the formula to find the next. The importance of using the entire number from the calculator to maintain accuracy is emphasized. The paragraph also outlines how to continue the iterative process to approximate the solution to the equation more accurately, eventually reaching a value that, when rounded to five decimal places, gives the solution 1.89329.
Mindmap
Keywords
💡Iteration
💡Cubic Equation
💡Change of Sign
💡Graphical Representation
💡Rearranging Equation
💡Cube Root
💡Iterative Formula
💡Initial Term
💡Approximation
💡Rounding
💡Accuracy
Highlights
Introduction to the topic of iteration for solving cubic equations.
Explanation of why the quadratic formula cannot be used for cubic equations.
Demonstration of how to show a solution lies between x=1 and x=2 using iteration.
The concept of 'change of sign' indicating the presence of a solution between two values.
Graphical representation of the change of sign on a graph to visualize the solution.
Rearranging the equation to make x the subject for the iteration process.
Use of the cube root to simplify the equation for iteration.
Introduction of the iterative formula and its role in finding successive approximations.
How to derive the iterative formula from the original equation.
Calculating the first three terms of the sequence using the iterative formula.
The iterative process explained as a sequence of calculations to approximate the solution.
Using an initial term x0 and calculating subsequent terms x1, x2, and x3.
The importance of using the entire number from a calculator for accurate iteration.
Solving the cubic equation to five decimal places using iteration.
Technique for using a calculator to perform multiple iterations quickly.
The final solution rounded to five decimal places and its significance.
Verification of the solution's accuracy by substituting it back into the original equation.
Conclusion on the accuracy of the solution and its closeness to zero.
Encouragement for viewers to watch more educational content and subscribe for updates.
Transcripts
in this video we're going to learn about
a topic called iteration
imagine we were trying to solve this
equation here
this equation is known as a cubic
equation because it has an x to the
power 3 as the highest power so it's not
a quadratic equation so we can't use the
quadratic formula for instance
fortunately iteration is a method for
solving equations that will help us
a typical start to an iteration question
might look something like this
part a show that a solution to the
equation lies between x equals one and x
equals two
so the questions actually told us that
one of the solutions is between one and
two
we're asked to show that this is the
case though
to do this all you do is take the
equation and substitute in values x
equals one and x equals two so if we
substitute in one we'd get one cubed
minus two times one minus three which
comes out as negative four
if we substitute in two we get two cubed
minus two times two minus three which
gives us one
at this point we can observe what we
call a change of sign it's a change of
sign because we've gone from negative to
positive it would also be a change of
sign if it went from positive to
negative the reason this shows a
solution is easier to see on a graph
imagine we were sketching this graph and
we were plotting the points for one and
two
when we plotted one it was at negative
four
when we plotted two it was at positive
one so if we were to draw the graph of
this we'd somehow connect up these two
points now it probably wouldn't be a
straight line but somewhere between
these two points it must cross over the
x-axis
when it crosses the x-axis we know that
would be a solution to the equation so a
solution must be between x equals one
and x equals two
so what we can say here is that the
change of sign shows there must be a
solution between the values x equals one
and x equals two
you should note that this method only
works when the equation is equal to zero
if the equation isn't already equal to
zero just rearrange it so that you get
zero on the right hand side then if you
notice a change of sign you know there's
a solution between those two x values
now that we know a solution lies between
1 and 2 we can use iteration to find the
solution really accurately
the first thing we're going to do is
imagine we were asked to make this x the
subject so we're going to rearrange it
so that x is on the left-hand side
to begin we're going to add 2x and 3 to
both sides
if we add this to the left hand side it
cancels out the negative 2x and negative
3 so we're just left with x cubed on the
right hand side 0 add 2x add 3 is 2x at
3.
the next thing to try and do is remove
the cubed
the inverse of cubed is cube root so
we'll cube root both sides
if we cube root the left we just get x
and if we cube root the right we get the
cube root of 2x plus
this might seem like quite an unusual
approach but it will help us out
the next thing to do is take the x on
the left hand side and write it with a
subscript n plus one
and any x's on the right hand side
become x with a subscript n this is now
known as an iterative formula
often the iterative formula is given to
you in the question but you may be asked
to show that it does work like i have
done in this question
the follow-up question could look
something like this
the iterative formula can be used to
solve the equation
part b given that x0 equals 2 calculate
the values of x1 x2 and x3
now to understand how to do this you
need to understand what an iterative
formula is
x m plus 1 here refers to the next term
of a sequence of terms whereas x n
refers to the current term we're on
if for example we started with our
current term as x 0 and we substituted
in that number
and then worked it out it would tell you
what x 1 is
you could then take that term and
substitute it in
work it out and you'd get x2
you could then take that term substitute
it in work it out and you'd get x3 and
you can carry on this iterative process
x3 gives you x4 and so on so all an
iterative formula tells you is the
formula you need to use to work out the
next term and that's what we're going to
do now
in the question we were told that x0
equals 2 this is known as our initial
term
we need to calculate the next three
terms
so we need to calculate x1 first to
calculate x1 we're going to write out
the formula
cube root of 2 times
x0
plus 3 and we're using x zero because
that's the term we currently have
we already know what x zero is though x
zero is two so if we substitute x zero
for two we can calculate this
you can now type this into your
calculator and you'll get this number
here
in an iteration question you're probably
going to need to write your whole
calculator display down
we can now use this value for x1 to get
the next value for x2 so imagine x1 is
our current term
x2 is our next term so we're going to do
the cube root of 2 times our current
term which is x1 and then add 3.
so we substitute x1 for this long number
here
and then calculate it
it's important at this point that you
don't round the long number on your
calculator you want to type in the whole
number or even better use the answer
button on your calculator
so x2 would come out as this number
we can then imagine that x2 is our
current term and then calculate x3
so x3 would be the cube root of 2 times
our current term which is x2 add 3.
we can then substitute the x2 for our
most recent number
which is this one and then type this
into a calculator again being sure to
use the whole number
so x3 would come out as this
we've now completed the question because
we found the values x1 x2 and x3
part c asks us to solve the equation
giving our answer to five decimal places
if we think back this was our original
plan with the question we wanted to
solve the equation
so far we were given the x0 value of 2
and we calculated x1 x2 and x3
each of these successive iterations is
an approximation to the solution the
further you go the better the answer so
to find the solution we need to keep
this going we can find x4 x5 and so on
fortunately this is made a lot easier
using a calculator
we can calculate iterations really
quickly using a calculator to do this
i'm going to press 2
and then equals because 2 was the
initial value of x0 that was given to us
in the question the effect of this is
this number 2 here is stored in the
calculator's memory it was the most
recent number it calculated
now we'll go ahead and do the iterations
so we need to do the cube root because
it started with a cube root that's on
shift so shift cube root that it was two
times the previous answer so i know the
previous answer x0 that was 2 so i could
just press another 2 at this point or i
could press the answer button because
that's the number the calculator most
recently calculated
then it was plus 3 and then press equals
and you can see the number on screen
matches the one that we calculated
now we need to get x2 and you could do
it by repeating all of those steps but
the calculation you need is already on
the screen if we simply press the equals
button now the calculator will take the
most recent answer the one displayed on
screen and put it into this formula here
so if we press equals you'll see we get
the next answer which is x2 and then we
press equals
x3
equals
x4
x5 x6 x7 x8 x9 and so on
in fact you can go even further than
this if you continue to press equals
you'll see the number gets closer and
closer to the exact answer and
eventually when you get to this point
here no matter what you do if you keep
pressing equals the number has settled
so this is the number we need to round
to get our final answer the question
said five decimal places so one two
three four five
so the fifth decimal point here is an
eight but it's got a 9 after it so we'll
need to round up so the final answer
rounded to 5 decimal places is 1.89329
so we've managed to find one of the
solutions to this equation
a common follow-up question is to ask
you just how accurate your solution is
it might read something like this
part d by substituting your answer from
part c into x cubed minus 2x minus 3
comment on the accuracy of your solution
to the equation x cubed minus 2x minus 3
equals 0.
all we need to do here is write out the
equation again but instead of x we're
going to write our solution 1.89329
1.89329 cubed minus two lots of one
point eight nine three two nine minus
three
now if you type this into a calculator
you'll get zero point zero zero zero
zero zero seven zero three five two five
so a very very small number
so we could say that this is very close
to zero so our solution is a good
estimate for the real solution
thank you for watching this video i hope
you found it useful check out what i
think you should watch next and also
subscribe so you don't miss out on
future uploads
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