Introduction This chapter gives you several methods which can be used to solve complicated equations to given levels of accuracy These are similar to methods which computers and calculators will use and hence can be used in computer programming ID: 605907
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Slide1
Numerical solutions of equationsSlide2
IntroductionThis chapter gives you several methods which can be used to solve complicated equations to given levels of accuracyThese are similar to methods which computers and calculators will use, and hence can be used in computer programmingSlide3
Teachings for Exercise 2ASlide4
Numerical solutions of equationsYou can solve equations of the form f(x) = 0 using interval bisection
Interval bisection is a variation on Trial and Improvement which you will have seen a
t GCSE level
Interval Bisection is an iterative process which allows us to find a root to whatever degree of accuracy we wish (usually 1-2 decimal places!)
An iterative process is one which is a short set of instructions which are then repeated as many times as needed
As a result such processes can be used in computers and calculators so they can solve equations
2ASlide5
Numerical solutions of equationsYou can solve equations of the form f(x) = 0 using interval bisection
Use Interval Bisection to find √11 to 1 decimal place
Set this up as an equation:
2A
Square both sides
Subtract 11
Sub in integers until we find a change of sign
So an solution lies between 3 and 4Slide6
Numerical solutions of equationsYou can solve equations of the form f(x) = 0 using interval bisection
Use Interval Bisection to find √11 to 1 decimal place
So a solution lies between 3 and 4.
Now we set up a table, subbing these 2 values into f(x), as well as the midpoint of these
When you have found the midpoint and substituted it in, choose the positive and negative answers closest to 0
The answer will be between these. Now repeat the process for these 2 numbers
2A
Our answer must be between 3.3125 and 3.34375
To one decimal place, the answer therefore must be 3.3!Slide7
Numerical solutions of equationsYou can solve equations of the form f(x) = 0 using interval bisection
Show that a root of the equation:
l
ies between 0 and 1
Use interval bisection 4 times to find an approximation for this root
2A
Sub in 0 and 1 to show the sign of the answer changes
As the sign has changed, a solution must lie between 0 and 1…Slide8
Numerical solutions of equationsYou can solve equations of the form f(x) = 0 using interval bisection
Show that a root of the equation:
l
ies between 0 and 1
Use interval bisection 4 times to find an approximation for this root
2A
Our approximation is the final bisection
0.6875 (or round if necessary)Slide9
Teachings for Exercise 2BSlide10
Numerical solutions of equationsYou can solve equations of the form f(x) = 0 using linear interpolation
In linear interpolation, you first draw a sketch of the function between 2 intervals
Then, you draw a straight line between the interval coordinates (this will be a rough approximation to the curve
You can then use similar triangles to find the place the straight line crosses the x-axis (the ‘root’ as it were)
You then update the interval and repeat the process…
2BSlide11
Numerical solutions of equationsYou can solve equations of the form f(x) = 0 using linear interpolation
A solution of the equation:
lies in the interval [1,2]. Use linear interpolation to find this root, correct to one decimal place.
2B
Sub in 1 and 2 to show the sign of the answer changes
As the sign has changed, a solution must lie between 1 and 2…Slide12
Numerical solutions of equations
You can solve equations of the form f(x) = 0 using linear interpolation
A solution of the equation:
lies in the interval [1,2]. Use linear interpolation to find this root, correct to one decimal place.
2B
Now sketch the graph between x = 1 and x = 2 (the limits you were given) It does not have to be really accurate!
(1,-4)
(2,7)
After sketching the graph between the limits, draw a straight line between them
The place this crosses the x-axis is an approximation for the root
You can call it x and then use similar triangles to find its value
x
x
ySlide13
Numerical solutions of equations
You can solve equations of the form f(x) = 0 using linear interpolation
A solution of the equation:
lies in the interval [1,2]. Use linear interpolation to find this root, correct to one decimal place.
2B
(1,-4)
(2,7)
x
x
y
4
7
x-1
2-x
Imagine creating triangles using the x-axis and the coordinates marked
Label the sides, using x as the place the straight line crosses the x-axis
These two triangles are similar –
ie
) They have the same angles (both have a right angle and two other pairs that are the same – you can see this from the vertically opposite angles at the centre and the ‘alternate’ angles ta the top and bottom!)
In similar triangles, a long side divided by a shorter side will give the same answer (provided that equivalent sides are used!) Slide14
Numerical solutions of equations
You can solve equations of the form f(x) = 0 using linear interpolation
A solution of the equation:
lies in the interval [1,2]. Use linear interpolation to find this root, correct to one decimal place.
2B
(1,-4)
(2,7)
x
x
y
4
7
x-1
2-x
Short side ÷ longer side in each triangle gives the same answer…
Multiply the whole left by 7 and the whole right by 4
Multiply by 28 to leave the numerators
Add 4x, Add 7
Divide by 11
So the value of x is
15
/
11
or 1.3636…Slide15
Numerical solutions of equations
You can solve equations of the form f(x) = 0 using linear interpolation
A solution of the equation:
lies in the interval [1,2]. Use linear interpolation to find this root, correct to one decimal place.
2B
(1,-4)
(2,7)
x
x
y
4
7
x-1
2-x
So the value of x is
15
/
11
or 1.3636…
15
/
11
Sub in
15
/
11
to find the value on the curve at this point
Calculate
(
15
/
11
,-1.009….)
We now repeat the process from this new point…Slide16
Numerical solutions of equationsYou can solve equations of the form f(x) = 0 using linear interpolation
A solution of the equation:
lies in the interval [1,2]. Use linear interpolation to find this root, correct to one decimal place.
2B
(2,7)
x
y
(
15
/
11
,-1.009….)
As you can see, the new estimate for the root is closer than the first approximation
Repeat the process using these new values (strictly speaking the original estimate was x
1
, and this one is x2 – use this notation when you solve these problems!)
xSlide17
Numerical solutions of equationsYou can solve equations of the form f(x) = 0 using linear interpolation
A solution of the equation:
lies in the interval [1,2]. Use linear interpolation to find this root, correct to one decimal place.
2B
(2,7)
x
y
(
15
/
11
,-1.009….)
x
7
1.009…
x-
15
/
11
2-x
Cross-multiply
Expand brackets
Rearrange
Solve
Repeat this process several times until your answer is accurate to the requested degree
Try to be as accurate as possible at each stage, avoiding rounding too much
Draw a new sketch at each stage
Use x
1
, x
2
, x
3
to represent each approximation!Slide18
Teachings for Exercise 2CSlide19
Numerical solutions of equationsYou can solve equations of the form f(x) = 0 using the Newton-Raphson
process
The Newton-Raphson formula is as follows:
2C
Our current approximation for the root
Our next approximation for the root
The function we are solving, with our current approximation substituted in
The derivative of the function we are solving, with our current approximation substituted in
Slide20
Numerical solutions of equationsYou can solve equations of the form f(x) = 0 using the Newton-Raphson
process
Use the Newton-Raphson process to find the root of the equation:
Use x
0
= 3 and give your answer to 2 decimal places.
Find the function and its derivative first…
2C
Rearrange
Differentiate
Slide21
Numerical solutions of equationsYou can solve equations of the form f(x) = 0 using the Newton-Raphson
process
Use the Newton-Raphson process to find the root of the equation:
Use x
0
= 3 and give your answer to 2 decimal places.
2C
Our current approximation is x
0
, replace the fraction with equivalent expressions
Sub in x
0
= 3
CalculateSlide22
Numerical solutions of equationsYou can solve equations of the form f(x) = 0 using the Newton-Raphson
process
Use the Newton-Raphson process to find the root of the equation:
Use x
0
= 3 and give your answer to 2 decimal places.
2C
Sub in x
1
= 2.912
Calculate
Repeat the process, but now we use the value of x
1
to find x
2
As both x
1
and x
2
round to 2.91, then this is the solution to 2 decimal places!Slide23
Numerical solutions of equationsYou can solve equations of the form f(x) = 0 using the Newton-Raphson
process
Use the Newton-Raphson process twice to find approximate a root of the equation:
Use x
0
= 2 as your first approximation and give your answer to 3 decimal places.
2C
Differentiate
Slide24
Numerical solutions of equationsYou can solve equations of the form f(x) = 0 using the Newton-Raphson
process
Use the Newton-Raphson process twice to find approximate a root of the equation:
Use x
0
= 2 as your first approximation and give your answer to 3 decimal places.
2C
Sub in x
0
as our first approximation and replace the parts of the fraction
Sub in x
0
= 2
CalculateSlide25
Numerical solutions of equationsYou can solve equations of the form f(x) = 0 using the Newton-Raphson
process
Use the Newton-Raphson process twice to find approximate a root of the equation:
Use x
0
= 2 as your first approximation and give your answer to 3 decimal places.
2C
Sub in x
1
= 1.866
Calculate
It is important to note that the Newton-
Raphson
method will not always work, sometimes tending away from a root rather than towards it
Usually, choosing a different first approximation will correct this!Slide26
SummaryYou have learnt several iterative methods for solving equationsAlthough these may seem cumbersome, the ideas involved in these are used in computers and calculators
They will also work when other methods fail to find answers!
2C