Maclaurin and Taylor Series.

Maclaurin and Taylor Series

Introduction A wide range of functions of x can be expressed as an infinite series (a bit like the Binomial Expansion) These are sometimes called a ‘power series’ These are used very often as they can allow us to find values of things such as Sinx or Cosx to whatever degree of accuracy we want Remember that many values of Sin and Cos are irrational numbers, hence cannot always be calculated exactly The series sum idea is possibly what your calculator does to calculate values such as Sin20 etc…

Teachings for Exercise 6a

Maclaurin and Taylor Series
You need to be able to find and use higher derivatives of functions You will be familiar with the notation for differentiation You need to be able to find more and more differentials Fortunately, the notation is pretty much what you’d expect! 𝑦 𝑓(𝑥) 1st Differential 𝑑𝑦 𝑑𝑥 𝑓′(𝑥) 2nd Differential 𝑑 2 𝑦 𝑑 𝑥 2 𝑓′′(𝑥) 3rd Differential 𝑑 3 𝑦 𝑑 𝑥 3 𝑓′′′(𝑥) 4th Differential 𝑑 4 𝑦 𝑑 𝑥 4 𝑓′′′′(𝑥) nth Differential 𝑑 𝑛 𝑦 𝑑 𝑥 𝑛 𝑓 𝑛 (𝑥) 6A

You need to be able to find and use higher derivatives of functions You also need to understand this next piece of notation: 𝑑 4 𝑦 𝑑𝑥 𝑑 𝑛 𝑦 𝑑𝑥 𝑛 𝑎 So this means find the value of the 4th differential when x = 6 The nth differential Sub in x = a 𝑓 𝑛 (𝑎) This means the same as the above 6A

Chain rule Differentiate 𝑓′(𝑥) 𝑓(𝑥) 𝑓(𝑥) 𝑛 𝑛 𝑓(𝑥) 𝑛−1 𝑓′(𝑥) ln⁡𝑓(𝑥) Maclaurin and Taylor Series You need to be able to find and use higher derivatives of functions Given that: Find the value of:  So we need to find the value of the 3rd differential when x = 1/2 is substituted into it 𝑦=𝑙𝑛 1−𝑥 Differentiate using the rule for ln (above) 𝑑𝑦 𝑑𝑥 = −1 1−𝑥 Differentiate using the chain rule (rewrite as a bracket to a power first) 𝑦=𝑙𝑛 1−𝑥 𝑑 2 𝑦 𝑑𝑥 2 = −1 1−𝑥 2 Differentiate using the chain rule again (rewrite as a bracket to a power first) 𝑑 3 𝑦 𝑑𝑥 𝑑 3 𝑦 𝑑𝑥 3 = −2 1−𝑥 3 If you prefer to leave the differentials in the ‘negative power’ form we use, that is ok. But be aware that sometimes you can simplify these and you might miss this if they aren’t written as above! − 1−𝑥 −2 − 1−𝑥 −1 Differentiate Differentiate =2 1−𝑥 −3 (−1) = 1−𝑥 −2 (−1) Simplify Simplify =−2 1−𝑥 −3 =− 1−𝑥 −2 6A

Chain rule Differentiate 𝑓′(𝑥) 𝑓(𝑥) 𝑓(𝑥) 𝑛 𝑛 𝑓(𝑥) 𝑛−1 𝑓′(𝑥) ln⁡𝑓(𝑥) Maclaurin and Taylor Series You need to be able to find and use higher derivatives of functions Given that: Find the value of:  So we need to find the value of the 3rd differential when x = 1/2 is substituted into it 𝑑 3 𝑦 𝑑𝑥 3 = −2 1−𝑥 3 Sub in x = 1/2 𝑑 3 𝑦 𝑑𝑥 = −2 1− 𝑦=𝑙𝑛 1−𝑥 Calculate 𝑑 3 𝑦 𝑑𝑥 = −16 6A

𝑒 𝑓(𝑥) Differentiate 𝑓 ′ (𝑥)𝑒 𝑓(𝑥) Maclaurin and Taylor Series You need to be able to find and use higher derivatives of functions Given that: Show that: 𝑓 𝑥 = 𝑒 𝑥 2 Differentiate using the rule for ‘e’ (above) 𝑓′ 𝑥 = 2𝑥𝑒 𝑥 2 Use the equation from the start to replace the exponential 𝑓 𝑥 = 𝑒 𝑥 2 𝑓′ 𝑥 =2𝑥𝑓(𝑥) 𝑓′ 𝑥 =2𝑥𝑓(𝑥) 6A

You need to be able to find and use higher derivatives of functions Given that: Show that: By differentiating the result twice more with respect to x, find f’’(x) and f’’’(x)  In this chapter you will need to use the product rule A LOT! 𝑓′ 𝑥 =2𝑥𝑓(𝑥) Differentiate using the product rule 𝑓′′ 𝑥 = 2𝑓(𝑥) + 2𝑥𝑓′(𝑥) Differentiate (using the product rule for the second term) 𝑓 𝑥 = 𝑒 𝑥 2 𝑓′′′ 𝑥 = 2𝑓′(𝑥) + 2𝑥𝑓′′(𝑥) + 2𝑓′(𝑥) Simplify by grouping like terms 𝑓′ 𝑥 =2𝑥𝑓(𝑥) 𝑓′′′ 𝑥 = 2𝑥𝑓′′(𝑥) + 4𝑓′(𝑥) Note that you could also get to f’’’(x) using the dy/dx notation… 𝑑 3 𝑦 𝑑 3 𝑥 =2𝑥 𝑑 2 𝑦 𝑑𝑥 𝑑𝑦 𝑑𝑥 Product Rule for 2xf(x) Product Rule for 2xf’(x) 2𝑥 𝑓(𝑥) 2𝑥 𝑓′(𝑥) 2 𝑓′(𝑥) 2 𝑓′′(𝑥) 6A

You need to be able to find and use higher derivatives of functions Given that: Show that: By differentiating the result twice more with respect to x, find f’’(x) and f’’’(x) c) Deduce the values of f(0), f’(0), f’’(0) and f’’’(0) Start by finding f(0), and then using the result to work out f’(0) 𝑓 𝑥 = 𝑒 𝑥 2 𝑓′ 𝑥 =2𝑥𝑓(𝑥) Sub in 0 Sub in 0 𝑓 0 = 𝑒 (0) 2 𝑓′ 0 =2(0)𝑓(0) Calculate, using f(0) Calculate 𝑓 0 =1 𝑓′ 0 =0 𝑓 𝑥 = 𝑒 𝑥 2 𝑓′′ 𝑥 =2𝑓(𝑥)+ 2𝑥𝑓′(𝑥) 𝑓′ 𝑥 =2𝑥𝑓(𝑥) Sub in 0 𝑓′′ 0 =2𝑓(0)+ 2(0)𝑓′(0) Calculate, using f(0) and f’(0) 𝑓′′ 0 =2 𝑓 ′′′ (𝑥)=2𝑥 𝑓 ′′ 𝑥 +4𝑓′(𝑥) Sub in 0 𝑓 ′′′ (𝑥)=2(0) 𝑓 ′′ 0 +4𝑓′(0) Calculate, using f(0), f’(0) and f’’(0) 𝑓 ′′′ (𝑥)=0 6A

Teachings for Exercise 6b

1+𝑥 = 1+𝑥 1 2 You can express many functions of x as an infinite series in ascending powers of x, by using Maclaurin’s expansion In C4 differentiation, you have seen that functions such as: and Can be expressed as a series sum, when x is within certain limits. x = x In C4 we started by considering the general form of the binomial expansion 1+𝑛𝑥+ 𝑛(𝑛−1) 2! 𝑥 2 + 𝑛(𝑛−1)(𝑛−2) 3! 𝑥 3 … Sub in n = 1/2 and x = x 1+𝑥 1−2𝑥 −1 𝑥 − 𝑥 − − 𝑥 3 … Simplify = 𝑥− 1 8 𝑥 𝑥 3 … This series can be used to calculate an approximate value for the original expression, for particular values of x For the sequence to converge, x needs to be between -1 and 1. If not, the terms which are squared/cubed will increase exponentially and cause the sequence to diverge If we want a more accurate approximation, we can just calculate more terms in the series and include these as well! 6B

1−2𝑥 −1 You can express many functions of x as an infinite series in ascending powers of x, by using Maclaurin’s expansion In C4 differentiation, you have seen that functions such as: and Can be expressed as a series sum, when x is within certain limits. x = -2x In C4 we started by considering the general form of the binomial expansion 1+𝑛𝑥+ 𝑛(𝑛−1) 2! 𝑥 2 + 𝑛(𝑛−1)(𝑛−2) 3! 𝑥 3 … Sub in n = -1 and x = -2x 1+𝑥 1−2𝑥 −1 1+ −1 (−2𝑥)+ −1 − (−2𝑥) 2 + −1 −2 −3 6 (−2𝑥) 3 … Simplify =1+2𝑥+4 𝑥 2 +8 𝑥 3 … For this series to converge, we need to look at the term which is raised to a power  Again, the value of this expression must be between -1 and 1 for it to converge. As the term itself is ‘-2x’… −1<−2𝑥<1 −1<2𝑥<1 − 1 2 <𝑥< 1 2 6B

You can express many functions of x as an infinite series in ascending powers of x, by using Maclaurin’s expansion More functions can be expressed this way, as series sums This is an extremely valuable tool as it allows us to calculate functions to any degree of accuracy It is also essentially the process a calculator goes through when calculating such values! 6B

You can express many functions of x as an infinite series in ascending powers of x, by using Maclaurin’s expansion Given that f(x) = ex can be written in the form: And that it is valid to differentiate an infinite series term by term, show that:  Effectively, we need to find the values of the ‘a’ terms 𝒇(𝒙) 𝑒 𝑥 = 𝑎 0 + 𝑎 1 𝑥+ 𝑎 2 𝑥 2 + 𝑎 3 𝑥 3 + 𝑎 4 𝑥 4 + 𝑎 5 𝑥 𝑎 𝑟 𝑥 𝑟 Differentiate (non-x terms will disappear) 𝒇′(𝒙) 𝑒 𝑥 = 𝑎 1 + 2𝑎 2 𝑥 + 3𝑎 3 𝑥 2 + 4𝑎 4 𝑥 3 + 5𝑎 5 𝑥 4 + …+ 𝑟𝑎 𝑟 𝑥 𝑟−1 And again 𝒇′′(𝒙) 𝑒 𝑥 = 2𝑎 2 + 6𝑎 3 𝑥 + 12𝑎 4 𝑥 2 + 20𝑎 5 𝑥 3 + …+ 𝑟(𝑟−1)𝑎 𝑟 𝑥 𝑟−2 And again 𝒇′′′(𝒙) 𝑒 𝑥 = 6𝑎 3 + 24𝑎 4 𝑥 + 60𝑎 5 𝑥 3 + …+ 𝑟(𝑟−1)(𝑟−2)𝑎 𝑟 𝑥 𝑟−3 𝑒 𝑥 = 𝑎 0 + 𝑎 1 𝑥+ 𝑎 2 𝑥 2 + 𝑎 3 𝑥 𝑎 𝑟 𝑥 𝑟 Now we can start finding some of the ‘a’ terms by substituting x = 0 into each differential 𝑒 𝑥 =1+𝑥+ 𝑥 2 2! + 𝑥 3 3! + … + 𝑥 𝑟 𝑟! 𝒇(𝒙) 𝑒 𝑥 = 𝑎 0 + 𝑎 1 𝑥+ 𝑎 2 𝑥 2 + 𝑎 3 𝑥 3 + 𝑎 4 𝑥 4 + 𝑎 5 𝑥 𝑎 𝑟 𝑥 𝑟 Sub in x = 0 𝒇(𝟎) 𝑒 0 = 𝑎 0 e0 = 1  All the terms with x in will be cancelled out! 1= 𝑎 0 𝑎 0 =1 6B

You can express many functions of x as an infinite series in ascending powers of x, by using Maclaurin’s expansion Given that f(x) = ex can be written in the form: And that it is valid to differentiate an infinite series term by term, show that:  Effectively, we need to find the values of the ‘a’ terms 𝒇(𝒙) 𝑒 𝑥 = 𝑎 0 + 𝑎 1 𝑥+ 𝑎 2 𝑥 2 + 𝑎 3 𝑥 3 + 𝑎 4 𝑥 4 + 𝑎 5 𝑥 𝑎 𝑟 𝑥 𝑟 Differentiate (non-x terms will disappear) 𝒇′(𝒙) 𝑒 𝑥 = 𝑎 1 + 2𝑎 2 𝑥 + 3𝑎 3 𝑥 2 + 4𝑎 4 𝑥 3 + 5𝑎 5 𝑥 4 + …+ 𝑟𝑎 𝑟 𝑥 𝑟−1 And again 𝒇′′(𝒙) 𝑒 𝑥 = 2𝑎 2 + 6𝑎 3 𝑥 + 12𝑎 4 𝑥 2 + 20𝑎 5 𝑥 3 + …+ 𝑟(𝑟−1)𝑎 𝑟 𝑥 𝑟−2 And again 𝒇′′′(𝒙) 𝑒 𝑥 = 6𝑎 3 + 24𝑎 4 𝑥 + 60𝑎 5 𝑥 3 + …+ 𝑟(𝑟−1)(𝑟−2)𝑎 𝑟 𝑥 𝑟−3 𝑒 𝑥 = 𝑎 0 + 𝑎 1 𝑥+ 𝑎 2 𝑥 2 + 𝑎 3 𝑥 𝑎 𝑟 𝑥 𝑟 Now we can start finding some of the ‘a’ terms by substituting x = 0 into each differential 𝑒 𝑥 =1+𝑥+ 𝑥 2 2! + 𝑥 3 3! + … + 𝑥 𝑟 𝑟! 𝒇′(𝒙) 𝑒 𝑥 = 𝑎 1 + 2𝑎 2 𝑥 + 3𝑎 3 𝑥 2 + 4𝑎 4 𝑥 3 + 5𝑎 5 𝑥 4 + …+ 𝑟𝑎 𝑟 𝑥 𝑟−1 Sub in x = 0 𝒇′(𝟎) 𝑒 0 = 𝑎 1 e0 = 1  All the terms with x in will be cancelled out! 1= 𝑎 1 𝑎 0 =1 𝑎 1 =1 6B

You can express many functions of x as an infinite series in ascending powers of x, by using Maclaurin’s expansion Given that f(x) = ex can be written in the form: And that it is valid to differentiate an infinite series term by term, show that:  Effectively, we need to find the values of the ‘a’ terms 𝒇(𝒙) 𝑒 𝑥 = 𝑎 0 + 𝑎 1 𝑥+ 𝑎 2 𝑥 2 + 𝑎 3 𝑥 3 + 𝑎 4 𝑥 4 + 𝑎 5 𝑥 𝑎 𝑟 𝑥 𝑟 Differentiate (non-x terms will disappear) 𝒇′(𝒙) 𝑒 𝑥 = 𝑎 1 + 2𝑎 2 𝑥 + 3𝑎 3 𝑥 2 + 4𝑎 4 𝑥 3 + 5𝑎 5 𝑥 4 + …+ 𝑟𝑎 𝑟 𝑥 𝑟−1 And again 𝒇′′(𝒙) 𝑒 𝑥 = 2𝑎 2 + 6𝑎 3 𝑥 + 12𝑎 4 𝑥 2 + 20𝑎 5 𝑥 3 + …+ 𝑟(𝑟−1)𝑎 𝑟 𝑥 𝑟−2 And again 𝒇′′′(𝒙) 𝑒 𝑥 = 6𝑎 3 + 24𝑎 4 𝑥 + 60𝑎 5 𝑥 3 + …+ 𝑟(𝑟−1)(𝑟−2)𝑎 𝑟 𝑥 𝑟−3 𝑒 𝑥 = 𝑎 0 + 𝑎 1 𝑥+ 𝑎 2 𝑥 2 + 𝑎 3 𝑥 𝑎 𝑟 𝑥 𝑟 Now we can start finding some of the ‘a’ terms by substituting x = 0 into each differential 𝑒 𝑥 =1+𝑥+ 𝑥 2 2! + 𝑥 3 3! + … + 𝑥 𝑟 𝑟! 𝒇′′(𝒙) 𝑒 𝑥 = 2𝑎 2 + 6𝑎 3 𝑥 + 12𝑎 4 𝑥 2 + 20𝑎 5 𝑥 3 + …+ 𝑟(𝑟−1)𝑎 𝑟 𝑥 𝑟−2 Sub in x = 0 𝒇′′(𝟎) 𝑒 0 = 2𝑎 2 e0 = 1 1= 2𝑎 2 𝑎 0 =1 𝑎 1 =1 Divide by 2 𝑎 2 = 1 2 1 2 = 𝑎 2 6B

You can express many functions of x as an infinite series in ascending powers of x, by using Maclaurin’s expansion Given that f(x) = ex can be written in the form: And that it is valid to differentiate an infinite series term by term, show that:  Effectively, we need to find the values of the ‘a’ terms 𝒇(𝒙) 𝑒 𝑥 = 𝑎 0 + 𝑎 1 𝑥+ 𝑎 2 𝑥 2 + 𝑎 3 𝑥 3 + 𝑎 4 𝑥 4 + 𝑎 5 𝑥 𝑎 𝑟 𝑥 𝑟 Differentiate (non-x terms will disappear) 𝒇′(𝒙) 𝑒 𝑥 = 𝑎 1 + 2𝑎 2 𝑥 + 3𝑎 3 𝑥 2 + 4𝑎 4 𝑥 3 + 5𝑎 5 𝑥 4 + …+ 𝑟𝑎 𝑟 𝑥 𝑟−1 And again 𝒇′′(𝒙) 𝑒 𝑥 = 2𝑎 2 + 6𝑎 3 𝑥 + 12𝑎 4 𝑥 2 + 20𝑎 5 𝑥 3 + …+ 𝑟(𝑟−1)𝑎 𝑟 𝑥 𝑟−2 And again 𝒇′′′(𝒙) 𝑒 𝑥 = 6𝑎 3 + 24𝑎 4 𝑥 + 60𝑎 5 𝑥 3 + …+ 𝑟(𝑟−1)(𝑟−2)𝑎 𝑟 𝑥 𝑟−3 𝑒 𝑥 = 𝑎 0 + 𝑎 1 𝑥+ 𝑎 2 𝑥 2 + 𝑎 3 𝑥 𝑎 𝑟 𝑥 𝑟 Now we can start finding some of the ‘a’ terms by substituting x = 0 into each differential 𝑒 𝑥 =1+𝑥+ 𝑥 2 2! + 𝑥 3 3! + … + 𝑥 𝑟 𝑟! 𝒇′′′(𝒙) 𝑒 𝑥 = 6𝑎 3 + 24𝑎 4 𝑥 + 60𝑎 5 𝑥 3 + …+ 𝑟(𝑟−1)(𝑟−2)𝑎 𝑟 𝑥 𝑟−3 Sub in x = 0 𝒇′′′(𝟎) 𝑒 0 = 6𝑎 3 e0 = 1 1= 6𝑎 3 𝑎 0 =1 𝑎 1 =1 Divide by 6 𝑎 2 = 1 2 𝑎 3 = 1 6 1 6 = 𝑎 3 6B

𝑒 𝑥 = 𝑎 0 + 𝑎 1 𝑥+ 𝑎 2 𝑥 2 + 𝑎 3 𝑥 You can express many functions of x as an infinite series in ascending powers of x, by using Maclaurin’s expansion Given that f(x) = ex can be written in the form: And that it is valid to differentiate an infinite series term by term, show that: We can now replace the ‘a’ terms with the values we calculated… 𝑒 𝑥 =1+𝑥 𝑥 𝑥 However, 1/2 = 1/2!, and 1/6 = 1/3! etc… 𝑒 𝑥 = 𝑎 0 + 𝑎 1 𝑥+ 𝑎 2 𝑥 2 + 𝑎 3 𝑥 𝑎 𝑟 𝑥 𝑟 𝑒 𝑥 =1+𝑥+ 1 2! 𝑥 ! 𝑥 𝑟! 𝑥 𝑟 Note that for the first term, r is defined as being ‘0’ 𝑒 𝑥 =1+𝑥+ 𝑥 2 2! + 𝑥 3 3! + … + 𝑥 𝑟 𝑟! It is possible to go through similar processes to find series summations of other functions  However in the early 18th century a Scottish mathematician called Colin Maclaurin created a generalised solution that can be used in many expansions… 𝑎 0 =1 𝑎 1 =1 𝑎 2 = 1 2 𝑎 3 = 1 6 6B

𝑒 𝑥 =1+𝑥+ 𝑥 2 2! + 𝑥 3 3! + … A quick graphical comparison of the summation against the exact value of ex 𝒚= 𝒆 𝒙 𝒚= 𝒆 𝒙 𝒚= 𝒆 𝒙 𝒚=𝟏 𝒚=𝟏+𝒙 𝒚=𝟏+𝒙+ 𝒙 𝟐 𝟐! 𝒚= 𝒆 𝒙 𝒚= 𝒆 𝒙 𝒚= 𝒆 𝒙 𝒚=𝟏+𝒙+ 𝒙 𝟐 𝟐! + 𝒙 𝟑 𝟑! 𝒚=𝟏+𝒙+ 𝒙 𝟐 𝟐! + 𝒙 𝟑 𝟑! + 𝒙 𝟒 𝟒! 𝒚=𝟏+𝒙+ 𝒙 𝟐 𝟐! + 𝒙 𝟑 𝟑! + 𝒙 𝟒 𝟒! + 𝒙 𝟓 𝟓! As you can see from these graphs, the more terms we have in the sum, the closer the values are to the actual value of ex  Note that as we move away from x = 0, we need more and more terms to make the summation accurate 6B

𝑓 𝑥 =𝑓 0 + 𝑓 ′ 0 𝑥+ 𝑓′′(0) 2! 𝑥 2 + 𝑓′′′(0) 3! 𝑥 3 … Maclaurin and Taylor Series You can express many functions of x as an infinite series in ascending powers of x, by using Maclaurin’s expansion For the continuous function: Provided that f(0), f’(0), f’’(0), …, fr(0) all have finite values, then: Some functions are only valid for certain values of x (similar to the binomial expansion)  The ranges are given where appropriate in FP2 𝑓 𝑥 , 𝑥∈𝑅 You get given this result in the formula booklet! 𝑓 𝑥 =𝑓 0 + 𝑓 ′ 0 𝑥+ 𝑓′′(0) 2! 𝑥 2 + 𝑓′′′(0) 3! 𝑥 3 … 6B

𝑓 𝑥 =𝑓 0 + 𝑓 ′ 0 𝑥+ 𝑓′′(0) 2! 𝑥 2 + 𝑓′′′(0) 3! 𝑥 3 … Differentiate 𝑓′(𝑥) 𝑓(𝑥) ln⁡𝑓(𝑥) Maclaurin and Taylor Series 𝑓 𝑥 =ln⁡(1+𝑥) You can express many functions of x as an infinite series in ascending powers of x, by using Maclaurin’s expansion Express ln(1 + x) as an infinite series in ascending powers of x, up to and including the term in x3 Using the form above, you will need to find f(x) (which you’re given), f’(x), f’’(x) and f’’’(x) Once you have these, you can substitute x = 0 into all of them to find the values which need to be used in the expansion Differentiate using the rule for the natural logarithm 𝑓′ 𝑥 = 1 1+𝑥 Differentiate by writing as a power and using the chain rule 𝑓 ′′ (𝑥)=− 𝑥 2 Differentiate by writing as a power and using the chain rule again 𝑓 ′′′ (𝑥)= 𝑥 3 𝑓′ 𝑥 = 1+𝑥 −1 𝑓 ′′ 𝑥 =− 1+𝑥 −2 Use the chain rule Use the chain rule 𝑓′′ 𝑥 = − 1+𝑥 −2 (1) 𝑓′′′ 𝑥 =2 1+𝑥 −3 (1) 6B

𝑓 𝑥 =𝑓 0 + 𝑓 ′ 0 𝑥+ 𝑓′′(0) 2! 𝑥 2 + 𝑓′′′(0) 3! 𝑥 3 … Differentiate 𝑓′(𝑥) 𝑓(𝑥) ln⁡𝑓(𝑥) Maclaurin and Taylor Series 𝑓 𝑥 =ln⁡(1+𝑥) You can express many functions of x as an infinite series in ascending powers of x, by using Maclaurin’s expansion Express ln(1 + x) as an infinite series in ascending powers of x, up to and including the term in x3 Using the form above, you will need to find f(x) (which you’re given), f’(x), f’’(x) and f’’’(x) Once you have these, you can substitute x = 0 into all of them to find the values which need to be used in the expansion Differentiate using the rule for the natural logarithm 𝑓′ 𝑥 = 1 1+𝑥 Differentiate by writing as a power and using the chain rule 𝑓 ′′ (𝑥)=− 𝑥 2 Differentiate by writing as a power and using the chain rule again 𝑓 ′′′ (𝑥)= 𝑥 3 Substitute x = 0 into each of these to find the coefficients for the form above 𝑓 0 =ln⁡(1+0) 𝑓 0 =0 𝑓′ 0 = 1 1+0 𝑓′ 0 =1 𝑓 ′′ (0)=− 𝑓 0 =0 𝑓′ 0 =1 𝑓 ′′ (0)=−1 𝑓 ′′ (0)=−1 𝑓 ′′′ (0)=2 𝑓 ′′′ (0)= 𝑓 ′′′ (0)=2 6B

𝑓 𝑥 =𝑓 0 + 𝑓 ′ 0 𝑥+ 𝑓′′(0) 2! 𝑥 2 + 𝑓′′′(0) 3! 𝑥 3 … Maclaurin and Taylor Series You can express many functions of x as an infinite series in ascending powers of x, by using Maclaurin’s expansion Express ln(1 + x) as an infinite series in ascending powers of x, up to and including the term in x3 Using the form above, you will need to find f(x) (which you’re given), f’(x), f’’(x) and f’’’(x) Once you have these, you can substitute x = 0 into all of them to find the values which need to be used in the expansion Put these values into Maclaurin’s form above 𝑓 𝑥 =𝑓 0 + 𝑓 ′ 0 𝑥+ 𝑓′′(0) 2! 𝑥 2 + 𝑓′′′(0) 3! 𝑥 3 Sub in the values we just calculated 𝑓 𝑥 =(0)+(1)𝑥+ (−1) 2! 𝑥 2 + (2) 3! 𝑥 3 Simplify, calculating the factorials where needed (ie – 2/3! = 2/6 = 1/3) 𝑓 𝑥 =𝑥− 1 2 𝑥 𝑥 3 𝑓 0 =0 𝑓′ 0 =1 𝑓 ′′ (0)=−1 𝑓 ′′′ (0)=2 6B

𝑓 𝑥 =𝑓 0 + 𝑓 ′ 0 𝑥+ 𝑓′′(0) 2! 𝑥 2 + 𝑓′′′(0) 3! 𝑥 3 … Maclaurin and Taylor Series ln⁡(1+𝑥)=𝑥− 1 2 𝑥 𝑥 3 You can express many functions of x as an infinite series in ascending powers of x, by using Maclaurin’s expansion Express ln(1 + x) as an infinite series in ascending powers of x, up to and including the term in x3 Using this series, find approximate values for: ln(1.05) ln(1.25) ln(1.8) Sub in x = 0.05 ln⁡(1.05)=(0.05)− 1 2 (0.05) (0.05) 3 Calculate ln⁡(1.05)= This is accurate to 5 dp ln⁡(1+𝑥)=𝑥− 1 2 𝑥 𝑥 3 Sub in x = 0.25 𝑓 𝑥 =𝑥− 1 2 𝑥 𝑥 3 ln⁡(1.25)=(0.25)− 1 2 (0.25) (0.25) 3 Calculate ln⁡(1.25)= This is accurate to 2 dp ln⁡(1+𝑥)=𝑥− 1 2 𝑥 𝑥 3 Sub in x = 0.8 ln⁡(1.8)=(0.8)− 1 2 (0.8) (0.8) 3 The Maclaurin’s expansion requires more terms to be accurate, as the value of x moves away from 0… Calculate ln⁡(1.8)= This is not even accurate to 1 dp! 6B

𝑓 𝑥 =𝑓 0 + 𝑓 ′ 0 𝑥+ 𝑓′′(0) 2! 𝑥 2 + 𝑓′′′(0) 3! 𝑥 3 … Maclaurin and Taylor Series Find successive differentials of f(x), and substitute x = 0 into each You can express many functions of x as an infinite series in ascending powers of x, by using Maclaurin’s expansion Find the Maclaurin expansion for sinx, up to the term in x5. Then use your expansion to find an approximation for sin10˚. As on the previous example, you need to find the differentials of f(x), up to f’’’’’(x). Then you need to substitute x = 0 into these to find the values to use in the expansion above. 𝑓 𝑥 =𝑠𝑖𝑛𝑥 𝑓 0 =0 𝑓′ 𝑥 =𝑐𝑜𝑠𝑥 𝑓′ 0 =1 𝑓 ′′ (𝑥)=−𝑠𝑖𝑛𝑥 𝑓 ′′ (0)=0 𝑓 ′′′ (𝑥)=−𝑐𝑜𝑠𝑥 𝑓 ′′′ 0 =−1 𝑓 ′′′′ (𝑥)=𝑠𝑖𝑛𝑥 𝑓 ′′′′ (0)=0 𝑓 ′′′′′ (𝑥)=𝑐𝑜𝑠𝑥 𝑓 ′′′′′ (0)=1 Now sub these into Maclaurin’s expansion 𝑓 𝑥 =𝑓 0 + 𝑓 ′ 0 𝑥+ 𝑓′′(0) 2! 𝑥 2 + 𝑓′′′(0) 3! 𝑥 3 + 𝑓′′′′(0) 4! 𝑥 4 + 𝑓′′′′(0) 5! 𝑥 5 𝑓 𝑥 = 𝑥 ! 𝑥 2 + −1 3! 𝑥 3 + (0) 4! 𝑥 4 + (1) 5! 𝑥 5 𝑓 𝑥 =𝑥− 1 3! 𝑥 ! 𝑥 5 𝑓 𝑥 =𝑥− 1 3! 𝑥 ! 𝑥 5 6B

𝑓 𝑥 =𝑓 0 + 𝑓 ′ 0 𝑥+ 𝑓′′(0) 2! 𝑥 2 + 𝑓′′′(0) 3! 𝑥 3 … Maclaurin and Taylor Series The angle MUST be in radians for this process to work… 10˚ = π/18 radians (since 180˚ = π) You can express many functions of x as an infinite series in ascending powers of x, by using Maclaurin’s expansion Find the Maclaurin expansion for sinx, up to the term in x5. Then use your expansion to find an approximation for sin10˚. As on the previous example, you need to find the differentials of f(x), up to f’’’’’(x). Then you need to substitute x = 0 into these to find the values to use in the expansion above. 𝑠𝑖𝑛 𝑥 =𝑥− 1 3! 𝑥 ! 𝑥 5 Sub in x = π/18 𝑠𝑖𝑛 𝜋 18 = 𝜋 18 − 1 3! 𝜋 ! 𝜋 Calculate = … 𝑓 𝑥 =𝑥− 1 3! 𝑥 ! 𝑥 5 6B

Teachings for Exercise 6c

You can find the series expansions of composite functions using known Maclaurin’s expansions You can use some known expansions to find more complicated ones  These expansions are given to you in the formula booklet, so you just need to be able to recognise when to use them! 𝑒 𝑥 =1+𝑥+ 𝑥 2 2! + 𝑥 3 3! + … + 𝑥 𝑟 𝑟! Valid for all values of x Valid for -1 < x ≤ 1 ln 1+𝑥 =𝑥− 𝑥 𝑥 3 3 − 𝑥 … + (−1) 𝑟−1 𝑥 𝑟 𝑟 𝑠𝑖𝑛𝑥=𝑥− 𝑥 3 3! + 𝑥 5 5! − 𝑥 7 7! + … + −1 𝑟 𝑥 2𝑟+1 2𝑟+1 ! Valid for all values of x 𝑐𝑜𝑠𝑥=1− 𝑥 2 2! + 𝑥 4 4! − 𝑥 6 6! + … + −1 𝑟 𝑥 2𝑟 2𝑟 ! Valid for all values of x Be aware that this formula is actually in the C2 section of the booklet, rather than FP2! (1+𝑥) 𝑛 =1+𝑛𝑥+ 𝑛(𝑛−1) 2! 𝑥 2 + 𝑛(𝑛−1)(𝑛−2) 3! 𝑥 3 … Valid for -1 < x < 1 You can use any of these with the x terms replaced with 2x, -5x, x/2, x2 etc… 6C

𝑐𝑜𝑠𝑥=1− 𝑥 2 2! + 𝑥 4 4! − 𝑥 6 6! + … + −1 𝑟 𝑥 2𝑟 2𝑟 ! Maclaurin and Taylor Series 𝑐𝑜𝑠𝑥=1− 𝑥 2 2! + 𝑥 4 4! − 𝑥 6 6! You can find the series expansions of composite functions using known Maclaurin’s expansions Write down the first 4 non-zero terms in the series expansion of cos(2x2) (in ascending powers of x) Start by writing out the expansion of cosx Then just replace all the ‘x’ terms with ‘2x2’ terms Make sure you use brackets!! Replace ‘x’ terms with ‘2x2’ 𝑐𝑜𝑠(2 𝑥 2 )=1− (2 𝑥 2 ) 2 2! + (2 𝑥 2 ) 4 4! − (2 𝑥 2 ) 6 6! Work out the brackets and factorials 𝑐𝑜𝑠(2 𝑥 2 )=1− 4𝑥 𝑥 − 64𝑥 Simplify where possible 𝑐𝑜𝑠(2 𝑥 2 )=1−2 𝑥 𝑥 8 3 − 4𝑥 6C

ln 1+𝑥 =𝑥− 𝑥 𝑥 3 3 − 𝑥 … + (−1) 𝑟−1 𝑥 𝑟 𝑟 Maclaurin and Taylor Series 𝑙𝑛 𝑥 1−3𝑥 You can find the series expansions of composite functions using known Maclaurin’s expansions Find the first 4 non-zero terms in the series expansion of: This example does not follow the pattern we have, but we can manipulate it so it does… Rewrite as a subtraction (log laws are from C2) = 𝑙𝑛 1+2𝑥 −𝑙𝑛 1−3𝑥 Write the first one using a power = 𝑙𝑛 1+2𝑥 −𝑙𝑛 1−3𝑥 𝑙𝑛 𝑥 1−3𝑥 Use the power law = 1 2 𝑙𝑛 1+2𝑥 −𝑙𝑛 1−3𝑥 This is equivalent to the original logarithm, but now both are in the form that we have an expansion for! 1 2 𝑙𝑛 1+2𝑥 −𝑙𝑛 1−3𝑥 6C

ln 1+𝑥 =𝑥− 𝑥 𝑥 3 3 − 𝑥 … + (−1) 𝑟−1 𝑥 𝑟 𝑟 Maclaurin and Taylor Series = 2𝑥 − (2𝑥) (2𝑥) 3 3 − 2𝑥 4 4 You can find the series expansions of composite functions using known Maclaurin’s expansions Find the first 4 non-zero terms in the series expansion of: This example does not follow the pattern we have, but we can manipulate it so it does… Find the expansion for both of these for the first 4 non-zero terms, using the expansion of ln(1 + x) 𝑙𝑛 1+2𝑥 Multiply out brackets and simplify =2𝑥− 4𝑥 𝑥 3 −4 𝑥 4 𝑙𝑛 1+2𝑥 Divide by 2 as we want 1/2ln(1 + 2x) 1 2 𝑙𝑛 1+2𝑥 =𝑥− 2𝑥 𝑥 3 −2 𝑥 4 𝑙𝑛 𝑥 1−3𝑥 = −3𝑥 − (−3𝑥) (−3𝑥) 3 3 − −3𝑥 4 4 Multiply out brackets and simplify (be careful with negatives!!!) 𝑙𝑛 1−3𝑥 =−3𝑥− 9𝑥 2 2 − 9𝑥 3 − 81𝑥 4 4 𝑙𝑛 1−3𝑥 1 2 𝑙𝑛 1+2𝑥 −𝑙𝑛 1−3𝑥 1 2 𝑙𝑛 1+2𝑥 −𝑙𝑛 1−3𝑥 Replace each part with its expansion above = 𝑥− 2𝑥 𝑥 3 −2 𝑥 4 − −3𝑥− 9𝑥 2 2 − 9𝑥 3 − 81𝑥 4 4 Group like terms (careful again with negatives!) =4𝑥 𝑥 𝑥 𝑥 4 6C

𝑠𝑖𝑛𝑥=𝑥− 𝑥 3 3! + 𝑥 5 5! − 𝑥 7 7! … 𝑒 𝑥 =1+𝑥+ 𝑥 2 2! + 𝑥 3 3! + 𝑥 4 4! … Maclaurin and Taylor Series You can find the series expansions of composite functions using known Maclaurin’s expansions Given that terms in xn, n > 4 can be ignored, show, using the series expansions of ex and sinx, that:  In this example you can put one expansion into another… 𝑒 𝑠𝑖𝑛𝑥 Replace sinx with the expansion as far as x3 (as we are not using powers of greater than 4) =𝑒 𝑥− 𝑥 3 3! This can be written as a multiplication (as when you multiply, you add the powers) =𝑒 𝑥 × 𝑒 − 𝑥 3 3! Work out the factorial 𝑒 𝑠𝑖𝑛𝑥 ≈1+𝑥+ 𝑥 2 2 − 𝑥 4 8 =𝑒 𝑥 × 𝑒 − 𝑥 3 6 So we will multiply together these two expansions of e 𝑒 𝑥 × 𝑒 − 𝑥 3 6 6C

𝑠𝑖𝑛𝑥=𝑥− 𝑥 3 3! + 𝑥 5 5! − 𝑥 7 7! … 𝑒 𝑥 =1+𝑥+ 𝑥 2 2! + 𝑥 3 3! + 𝑥 4 4! … Maclaurin and Taylor Series 𝑒 𝑥 =1+𝑥+ 𝑥 𝑥 𝑥 4 24 You can find the series expansions of composite functions using known Maclaurin’s expansions Given that terms in xn, n > 4 can be ignored, show, using the series expansions of ex and sinx, that:  In this example you can put one expansion into another… Use this expansion to work out the second power of e (you only need the first two terms as the next would have a power higher than 4) 𝑒 − 𝑥 =1+ − 𝑥 3 6 Simplify 𝑒 − 𝑥 =1− 𝑥 3 6 𝑒 𝑠𝑖𝑛𝑥 ≈1+𝑥+ 𝑥 2 2 − 𝑥 4 8 Now we have to multiply these expansions together. Again we can ignore any situations where the power would be greater than 4. 1+𝑥+ 𝑥 𝑥 𝑥 − 𝑥 3 6 Multiply out (ignore any terms that would have a power more than 4) 𝑒 𝑥 × 𝑒 − 𝑥 3 6 + 𝑥 2 2 + 𝑥 3 6 + 𝑥 4 24 − 𝑥 3 6 − 𝑥 4 6 =1 + 𝑥 Simplify =1+𝑥+ 𝑥 2 2 − 𝑥 4 8 6C

Teachings for Exercise 6d

You can find an approximation to a function of x close to x = a, where a ≠ 0, using Taylor’s expansion of the function The Maclaurin expansion is very effective and relatively straightforward to use, but it has some limitations Some functions cannot be expanded like this In addition to this, you have seen that the expansion focuses of values of x that are equal to or very close to 0 We can use Taylor’s expansion in a wider range of cases The idea of the Taylor Series came from the Scottish mathematician James Gregory, but was formally introduced later by the English mathematician Brook Taylor 6D

𝑓 𝑥 =𝑓 0 + 𝑓 ′ 0 𝑥+ 𝑓′′(0) 2! 𝑥 2 + 𝑓′′′(0) 3! 𝑥 3 … Maclaurin and Taylor Series 𝑔 𝑥+𝑎 =𝑓(𝑥) You can find an approximation to a function of x close to x = a, where a ≠ 0, using Taylor’s expansion of the function Imagine we have two functions, f and g, linked as shown to the right. Note that we are using g here, but Taylor’s expansion will be in terms of f, same as Maclaurin’s If these are equal, the differentials will be as well 𝑔 𝑟 𝑥+𝑎 = 𝑓 𝑟 (𝑥) 𝑟=1,2,3 𝑒𝑡𝑐 Let x = 0 𝑔 𝑟 𝑎 = 𝑓 𝑟 (0) What this means is: If we write our function f(x) as g(x + a) instead We have to change all the fr(0) terms to gr(a) terms (so all the differentials change as r can take any integer value) This is one form of Taylor’s expansion 𝑓 𝑥 =𝑓 0 + 𝑓 ′ 0 𝑥+ 𝑓′′(0) 2! 𝑥 2 + 𝑓′′′(0) 3! 𝑥 3 … Replace ‘x’ with ‘x + a’, and all fr(0) with gr(a) 𝑔 𝑥+𝑎 =𝑔 𝑎 + 𝑔 ′ 𝑎 𝑥+ 𝑔′′(𝑎) 2! 𝑥 2 + 𝑔′′′(𝑎) 3! 𝑥 3 … 6D

𝑓 𝑥+𝑎 =𝑓 𝑎 + 𝑓 ′ 𝑎 𝑥+ 𝑓′′(𝑎) 2! 𝑥 2 + 𝑓′′′(𝑎) 3! 𝑥 3 … You can find an approximation to a function of x close to x = a, where a ≠ 0, using Taylor’s expansion of the function You can also change the Taylor expansion into a slightly different form… Replace all ‘x’ terms with ‘x – a’ terms 𝑓 𝑥 =𝑓 𝑎 + 𝑓 ′ 𝑎 (𝑥−𝑎)+ 𝑓′′(𝑎) 2! (𝑥−𝑎) 2 + 𝑓′′′(𝑎) 3! (𝑥−𝑎) 3 … Both these forms can be used under different circumstances as Taylor expansions  You will need to decide which to use based on the question!  They are also in the formula booklet!! 6D

𝑓 𝑥 =𝑓 𝑎 + 𝑓 ′ 𝑎 (𝑥−𝑎)+ 𝑓′′(𝑎) 2! (𝑥−𝑎) 2 + 𝑓′′′(𝑎) 3! (𝑥−𝑎) 3 … 𝑓 𝑥+𝑎 =𝑓 𝑎 + 𝑓 ′ 𝑎 𝑥+ 𝑓′′(𝑎) 2! 𝑥 2 + 𝑓′′′(𝑎) 3! 𝑥 3 … Maclaurin and Taylor Series You can find an approximation to a function of x close to x = a, where a ≠ 0, using Taylor’s expansion of the function Find the Taylor expansion of e-x in powers of (x + 4) up to and including the terms in (x + 4)3 First you need to decide which form to use… As you are asked for the expansion using powers of a bracket (x + 4), you should use the form to the left at the top of the screen From here you effectively work through in a similar way as for the Maclaurin expansion You need to find successive differentials up to f’’’(x) Then substitute ‘a’ into these (instead of 0 as you would have done with the Maclaurin expansion) The value ‘a’ will be -4 (compare the form of the expansion with the bracket!) 𝑓 𝑥 = 𝑒 −𝑥 𝑓 −4 = 𝑒 4 𝑓′ 𝑥 = −𝑒 −𝑥 𝑓′ −4 = −𝑒 4 𝑓′′ 𝑥 = 𝑒 −𝑥 𝑓′′ −4 = 𝑒 4 𝑓′′′ 𝑥 = −𝑒 −𝑥 𝑓′′′ −4 = −𝑒 4 Now we can substitute these values into the Taylor expansion form… 6D

𝑓 𝑥 =𝑓 𝑎 + 𝑓 ′ 𝑎 (𝑥−𝑎)+ 𝑓′′(𝑎) 2! (𝑥−𝑎) 2 + 𝑓′′′(𝑎) 3! (𝑥−𝑎) 3 … 𝑓 𝑥+𝑎 =𝑓 𝑎 + 𝑓 ′ 𝑎 𝑥+ 𝑓′′(𝑎) 2! 𝑥 2 + 𝑓′′′(𝑎) 3! 𝑥 3 … Maclaurin and Taylor Series You can find an approximation to a function of x close to x = a, where a ≠ 0, using Taylor’s expansion of the function Find the Taylor expansion of e-x in powers of (x + 4) up to and including the terms in (x + 4)3 First you need to decide which form to use… As you are asked for the expansion using powers of a bracket (x + 4), you should use the form to the left at the top of the screen From here you effectively work through in a similar way as for the Maclaurin expansion 𝑓 𝑥 = 𝑒 −𝑥 𝑓 −4 = 𝑒 4 𝑓′ 𝑥 = −𝑒 −𝑥 𝑓′ −4 = −𝑒 4 𝑓′′ 𝑥 = 𝑒 −𝑥 𝑓′′ −4 = 𝑒 4 𝑓′′′ 𝑥 = −𝑒 −𝑥 𝑓′′′ −4 = −𝑒 4 Now we can substitute these values into the Taylor expansion form… 𝑓 𝑥 =𝑓 𝑎 + 𝑓 ′ 𝑎 (𝑥−𝑎)+ 𝑓′′(𝑎) 2! (𝑥−𝑎) 2 + 𝑓′′′(𝑎) 3! (𝑥−𝑎) 3 Sub in the values for a, as well as the differentials 𝑒 −𝑥 = 𝑒 4 − 𝑒 4 𝑥+4 + 𝑒 𝑥+4 2 − 𝑒 (𝑥+4) 3 You can factorise this expansion 𝑒 −𝑥 = 𝑒 4 1− 𝑥 𝑥 (𝑥+4) 3 6D

𝑓 𝑥 =𝑓 𝑎 + 𝑓 ′ 𝑎 (𝑥−𝑎)+ 𝑓′′(𝑎) 2! (𝑥−𝑎) 2 + 𝑓′′′(𝑎) 3! (𝑥−𝑎) 3 … 𝑓 𝑥+𝑎 =𝑓 𝑎 + 𝑓 ′ 𝑎 𝑥+ 𝑓′′(𝑎) 2! 𝑥 2 + 𝑓′′′(𝑎) 3! 𝑥 3 … Maclaurin and Taylor Series For f(x) you can just use tanx  The Taylor series takes into account the transformation to tan (x + π/4) You can find an approximation to a function of x close to x = a, where a ≠ 0, using Taylor’s expansion of the function Express: As a series in ascending powers of x, up to the term in x3 For this question, the function you want to express as a series has an additional part added to it So in this case it makes sense to use the form to the right, at the top of the screen… 𝑓 𝑥 =𝑡𝑎𝑛𝑥 Differentiate 𝑓′ 𝑥 = 𝑠𝑒𝑐 2 𝑥 Differentiate using the chain rule 𝑡𝑎𝑛 𝑥+ 𝜋 4 𝑓′′ 𝑥 = 2𝑠𝑒 𝑐 2 𝑥𝑡𝑎𝑛𝑥 Differentiate using the chain rule and product rule 𝑓′′′ 𝑥 = 4 𝑠𝑒𝑐 2 𝑥𝑡𝑎 𝑛 2 𝑥 + 2 𝑠𝑒𝑐 4 𝑥 𝑠𝑒𝑐𝑥 2 Product rule for 2sec2xtanx Chain rule =2(𝑠𝑒𝑐𝑥)(𝑠𝑒𝑐𝑥𝑡𝑎𝑛𝑥) 2 𝑠𝑒𝑐𝑥 2 𝑡𝑎𝑛𝑥 Simplify 4(𝑠𝑒𝑐𝑥)(𝑠𝑒𝑐𝑥𝑡𝑎𝑛𝑥) 4 𝑠𝑒𝑐 2 𝑥𝑡𝑎𝑛𝑥 𝑠𝑒𝑐 2 𝑥 =2 𝑠𝑒𝑐 2 𝑥𝑡𝑎𝑛𝑥 6D

𝑓 𝑥 =𝑓 𝑎 + 𝑓 ′ 𝑎 (𝑥−𝑎)+ 𝑓′′(𝑎) 2! (𝑥−𝑎) 2 + 𝑓′′′(𝑎) 3! (𝑥−𝑎) 3 … 𝑓 𝑥+𝑎 =𝑓 𝑎 + 𝑓 ′ 𝑎 𝑥+ 𝑓′′(𝑎) 2! 𝑥 2 + 𝑓′′′(𝑎) 3! 𝑥 3 … Maclaurin and Taylor Series 𝑓 𝜋 4 =1 You can find an approximation to a function of x close to x = a, where a ≠ 0, using Taylor’s expansion of the function Express: As a series in ascending powers of x, up to the term in x3 For this question, the function you want to express as a series has an additional part added to it So in this case it makes sense to use the form to the right, at the top of the screen… 𝑓 𝑥 =𝑡𝑎𝑛𝑥 𝑓 ′ 𝜋 4 =2 𝑓′ 𝑥 = 𝑠𝑒𝑐 2 𝑥 𝑓 ′′ 𝜋 4 =4 𝑓′′ 𝑥 = 2𝑠𝑒 𝑐 2 𝑥𝑡𝑎𝑛𝑥 𝑡𝑎𝑛 𝑥+ 𝜋 4 𝑓 ′′′ 𝜋 4 =16 𝑓′′′ 𝑥 = 4 𝑠𝑒𝑐 2 𝑥𝑡𝑎 𝑛 2 𝑥 + 2 𝑠𝑒𝑐 4 𝑥 𝑓 𝑥+𝑎 =𝑓 𝑎 + 𝑓 ′ 𝑎 𝑥+ 𝑓′′(𝑎) 2! 𝑥 2 + 𝑓′′′(𝑎) 3! 𝑥 3 Replace the differentials from above 𝑡𝑎𝑛 𝑥+ 𝜋 4 =(1)+(2)𝑥+ (4) 2 𝑥 𝑥 3 Simplify 𝑡𝑎𝑛 𝑥+ 𝜋 4 =1+2𝑥+2 𝑥 𝑥 3 6D

𝑓 𝑥 =𝑓 𝑎 + 𝑓 ′ 𝑎 (𝑥−𝑎)+ 𝑓′′(𝑎) 2! (𝑥−𝑎) 2 + 𝑓′′′(𝑎) 3! (𝑥−𝑎) 3 … 𝑓 𝑥+𝑎 =𝑓 𝑎 + 𝑓 ′ 𝑎 𝑥+ 𝑓′′(𝑎) 2! 𝑥 2 + 𝑓′′′(𝑎) 3! 𝑥 3 … Maclaurin and Taylor Series 𝑓 𝜋 6 = 1 2 You can find an approximation to a function of x close to x = a, where a ≠ 0, using Taylor’s expansion of the function Show that the Taylor expansion of sinx in ascending powers of (x – π/6) up to the term in (x – π/6)2 is: Then use the series to find an approximation for sin40, in terms of π. Based on the question, we should use the form to the left at the top… Find the differentials we need and substitute π/6 into them 𝑓 𝑥 =𝑠𝑖𝑛𝑥 𝑓′ 𝜋 6 = 𝑓′ 𝑥 =𝑐𝑜𝑠𝑥 𝑓 ′′ 𝜋 6 =− 1 2 𝑓 ′′ (𝑥)=−𝑠𝑖𝑛𝑥 𝑓 𝑥 =𝑓 𝑎 + 𝑓 ′ 𝑎 (𝑥−𝑎)+ 𝑓′′(𝑎) 2! (𝑥−𝑎) 2 𝑠𝑖𝑛𝑥= 𝑥− 𝜋 6 − 𝑥− 𝜋 6 2 Sub in the values we have calculated 𝑠𝑖𝑛𝑥= 𝑥− 𝜋 6 + − 𝑥− 𝜋 6 2 Simplify where possible 𝑠𝑖𝑛𝑥= 𝑥− 𝜋 6 − 𝑥− 𝜋 6 2 6D

𝑓 𝑥 =𝑓 𝑎 + 𝑓 ′ 𝑎 (𝑥−𝑎)+ 𝑓′′(𝑎) 2! (𝑥−𝑎) 2 + 𝑓′′′(𝑎) 3! (𝑥−𝑎) 3 … 𝑓 𝑥+𝑎 =𝑓 𝑎 + 𝑓 ′ 𝑎 𝑥+ 𝑓′′(𝑎) 2! 𝑥 2 + 𝑓′′′(𝑎) 3! 𝑥 3 … Maclaurin and Taylor Series The angle MUST be in radians for this process to work… 40º = 2π/9 radians (since 40º is 2/9 of 180º) You can find an approximation to a function of x close to x = a, where a ≠ 0, using Taylor’s expansion of the function Show that the Taylor expansion of sinx in ascending powers of (x – π/6) up to the term in (x – π/6)2 is: Then use the series to find an approximation for sin40, in terms of π. Based on the question, we should use the form to the left at the top… Find the differentials we need and substitute π/6 into them 𝑠𝑖𝑛𝑥= 𝑥− 𝜋 6 − 𝑥− 𝜋 6 2 Sub in x = 2π/9 𝑠𝑖𝑛𝑥= 𝑥− 𝜋 6 − 𝑥− 𝜋 6 2 𝑠𝑖𝑛𝑥= 𝜋 9 − 𝜋 6 − 𝜋 9 − 𝜋 6 2 Calculate the bracket parts 𝑠𝑖𝑛𝑥= 1 2 + 𝜋 18 − 𝜋 Simplify 𝑠𝑖𝑛𝑥= 𝜋 − 𝜋 6D

Teachings for Exercise 6e

You can find the solution, in the form of a series, to a differential equation using the Taylor series method Some differential equations cannot be solved using the methods from chapters 4 and 5 However, for these it is often possible to find a series solution instead Suppose you have a first order differential equation of the form: 𝑑𝑦 𝑑𝑥 =𝑓(𝑥,𝑦) and you know that initially, 𝑥= 𝑥 0 and 𝑦= 𝑦 0 So the differential is a function of both x and y And we know two initial values Based on knowing these two initial values, we can then calculate 𝑑𝑦 𝑑𝑥 If we then differentiate the whole equation, we will obtain an equation with a term in 𝑑 2 𝑦 𝑑𝑥 2 We can use the initial values alongside the 𝑑𝑦 𝑑𝑥 term we calculated, to find the value of 𝑑 2 𝑦 𝑑𝑥 2 We can continue this to find as many differentials as we need! 6E

You can find the solution, in the form of a series, to a differential equation using the Taylor series method Once you have found the differentials you need, based on initial values x0 and y0, you substitute them into the expansion to the right 𝑦= 𝑦 0 + 𝑥− 𝑥 𝑑𝑦 𝑑𝑥 𝑥 𝑥− 𝑥 ! 𝑑 2 𝑦 𝑑𝑥 𝑥 𝑥− 𝑥 ! 𝑑 3 𝑦 𝑑𝑥 𝑥 0 … Note that this is effectively just the Taylor series expansion 𝑓 𝑥 =𝑓 𝑎 + 𝑓 ′ 𝑎 (𝑥−𝑎)+ 𝑓′′(𝑎) 2! (𝑥−𝑎) 2 + 𝑓′′′(𝑎) 3! (𝑥−𝑎) 3 … Also note that if x0 is 0, the expansion at the top effectively becomes the Maclaurin expansion Be aware that you do not get given the equation above – so remember it! 6E

𝑦= 𝑦 0 + 𝑥− 𝑥 𝑑𝑦 𝑑𝑥 𝑥 𝑥− 𝑥 ! 𝑑 2 𝑦 𝑑𝑥 𝑥 𝑥− 𝑥 ! 𝑑 3 𝑦 𝑑𝑥 𝑥 0 … Maclaurin and Taylor Series 𝑑𝑦 𝑑𝑥 =2 𝑑 2 𝑦 𝑑𝑥 2 =1 You can find the solution, in the form of a series, to a differential equation using the Taylor series method Use the Taylor method to find a series solution, in ascending powers of x up to and including the term in x3, of: Given that when x = 0, y = 1 and 𝑑𝑦 𝑑𝑥 = 2 So we will need to calculate differentials up to 𝑑 3 𝑦 𝑑𝑥 3 Start by using the known values to find 𝑑 2 𝑦 𝑑𝑥 2 𝑥 0 =0 𝑦 0 =1 𝑑 2 𝑦 𝑑𝑥 2 =𝑦−𝑠𝑖𝑛𝑥 Sub in x and y (these are labelled above as x0 and y0 as they are the initial values) 𝑑 2 𝑦 𝑑𝑥 2 =1−𝑠𝑖𝑛0 Calculate 𝑑 2 𝑦 𝑑𝑥 2 =𝑦−𝑠𝑖𝑛𝑥 𝑑 2 𝑦 𝑑𝑥 2 =1 6E

𝑦= 𝑦 0 + 𝑥− 𝑥 𝑑𝑦 𝑑𝑥 𝑥 𝑥− 𝑥 ! 𝑑 2 𝑦 𝑑𝑥 𝑥 𝑥− 𝑥 ! 𝑑 3 𝑦 𝑑𝑥 𝑥 0 … Maclaurin and Taylor Series 𝑑𝑦 𝑑𝑥 =2 𝑑 2 𝑦 𝑑𝑥 2 =1 𝑑 3 𝑦 𝑑𝑥 3 =1 You can find the solution, in the form of a series, to a differential equation using the Taylor series method Use the Taylor method to find a series solution, in ascending powers of x up to and including the term in x3, of: Given that when x = 0, y = 1 and 𝑑𝑦 𝑑𝑥 = 2 Now differentiate the equation term by term And then use the values we have to find 𝑑 3 𝑦 𝑑𝑥 3 𝑥 0 =0 𝑦 0 =1 𝑑 2 𝑦 𝑑𝑥 2 =𝑦−𝑠𝑖𝑛𝑥 Differentiate each term with respect to x 𝑑 3 𝑦 𝑑𝑥 3 = 𝑑𝑦 𝑑𝑥 − 𝑐𝑜𝑠𝑥 Use the values of dy/dx and x from above 𝑑 2 𝑦 𝑑𝑥 2 =𝑦−𝑠𝑖𝑛𝑥 𝑑 3 𝑦 𝑑𝑥 3 = 2−𝑐𝑜𝑠0 Calculate 𝑑 3 𝑦 𝑑𝑥 3 =1 6E

𝑦= 𝑦 0 + 𝑥− 𝑥 𝑑𝑦 𝑑𝑥 𝑥 𝑥− 𝑥 ! 𝑑 2 𝑦 𝑑𝑥 𝑥 𝑥− 𝑥 ! 𝑑 3 𝑦 𝑑𝑥 𝑥 0 … Maclaurin and Taylor Series 𝑑𝑦 𝑑𝑥 =2 𝑑 2 𝑦 𝑑𝑥 2 =1 𝑑 3 𝑦 𝑑𝑥 3 =1 You can find the solution, in the form of a series, to a differential equation using the Taylor series method Use the Taylor method to find a series solution, in ascending powers of x up to and including the term in x3, of: Given that when x = 0, y = 1 and 𝑑𝑦 𝑑𝑥 = 2 Now we know the differentials we need, we can substitute them into the expansion above… 𝑥 0 =0 𝑦 0 =1 𝑦= 𝑦 0 + 𝑥− 𝑥 𝑑𝑦 𝑑𝑥 𝑥 𝑥− 𝑥 ! 𝑑 2 𝑦 𝑑𝑥 𝑥 𝑥− 𝑥 ! 𝑑 3 𝑦 𝑑𝑥 𝑥 0 Sub in values from above 𝑦=1+ 𝑥− 𝑥− 𝑥− Simplify 𝑑 2 𝑦 𝑑𝑥 2 =𝑦−𝑠𝑖𝑛𝑥 𝑦=1+2𝑥 𝑥 𝑥 3 6E

𝑦= 𝑦 0 + 𝑥− 𝑥 𝑑𝑦 𝑑𝑥 𝑥 𝑥− 𝑥 ! 𝑑 2 𝑦 𝑑𝑥 𝑥 𝑥− 𝑥 ! 𝑑 3 𝑦 𝑑𝑥 𝑥 0 … Maclaurin and Taylor Series 𝑑𝑦 𝑑𝑥 =2 𝑑 2 𝑦 𝑑𝑥 2 =−3 You can find the solution, in the form of a series, to a differential equation using the Taylor series method Given that: And that y = 1 and dy/dx = 2 when x = 1, express y as a series in ascending powers of (x – 1), up to and including the term in (x – 1)4 As before, you need to start by finding the differentials we need (in this case up to 𝑑 4 𝑦 𝑑𝑥 4 ) As you work through, use the known conditions as a starting point to calculate values 𝑥 0 =1 𝑦 0 =1 𝑑 2 𝑦 𝑑𝑥 𝑑𝑦 𝑑𝑥 =𝑥𝑦 Sub in values from above 𝑑 2 𝑦 𝑑𝑥 𝑑𝑦 𝑑𝑥 =𝑥𝑦 𝑑 2 𝑦 𝑑𝑥 2 +2(2)=(1)(1) Simplify 𝑑 2 𝑦 𝑑𝑥 2 +4=1 Calculate 𝑑 2 𝑦 𝑑𝑥 2 =−3 6E

𝑦= 𝑦 0 + 𝑥− 𝑥 𝑑𝑦 𝑑𝑥 𝑥 𝑥− 𝑥 ! 𝑑 2 𝑦 𝑑𝑥 𝑥 𝑥− 𝑥 ! 𝑑 3 𝑦 𝑑𝑥 𝑥 0 … Maclaurin and Taylor Series 𝑑𝑦 𝑑𝑥 =2 𝑑 2 𝑦 𝑑𝑥 2 =−3 𝑑 3 𝑦 𝑑𝑥 3 =9 You can find the solution, in the form of a series, to a differential equation using the Taylor series method Given that: And that y = 1 and dy/dx = 2 when x = 1, express y as a series in ascending powers of (x – 1), up to and including the term in (x – 1)4 Now we need to differentiate the equation to obtain a term in 𝑑 3 𝑦 𝑑𝑥 3 You will need to use the product rule (and you will use it lots in this section!) 𝑥 0 =1 𝑦 0 =1 𝑑 2 𝑦 𝑑𝑥 𝑑𝑦 𝑑𝑥 =𝑥𝑦 Differentiate, using the product rule where needed 𝑑 2 𝑦 𝑑𝑥 𝑑𝑦 𝑑𝑥 =𝑥𝑦 𝑑 3 𝑦 𝑑𝑥 3 + 2 𝑑 2 𝑦 𝑑𝑥 2 = 𝑦+𝑥 𝑑𝑦 𝑑𝑥 Sub in the values we have calculated up to this point Product Rule for xy 𝑥 𝑦 𝑑 3 𝑦 𝑑𝑥 −3 = 1 +(1)(2) 𝑑𝑦 𝑑𝑥 1 Simplify 𝑑 3 𝑦 𝑑𝑥 3 −6=3 Calculate 𝑑 3 𝑦 𝑑𝑥 3 =9 6E

𝑦= 𝑦 0 + 𝑥− 𝑥 𝑑𝑦 𝑑𝑥 𝑥 𝑥− 𝑥 ! 𝑑 2 𝑦 𝑑𝑥 𝑥 𝑥− 𝑥 ! 𝑑 3 𝑦 𝑑𝑥 𝑥 0 … Maclaurin and Taylor Series 𝑑𝑦 𝑑𝑥 =2 𝑑 2 𝑦 𝑑𝑥 2 =−3 𝑑 3 𝑦 𝑑𝑥 3 =9 𝑑 4 𝑦 𝑑𝑥 4 =−17 You can find the solution, in the form of a series, to a differential equation using the Taylor series method Given that: And that y = 1 and dy/dx = 2 when x = 1, express y as a series in ascending powers of (x – 1), up to and including the term in (x – 1)4 Now we need to differentiate one last time to obtain a term in 𝑑 4 𝑦 𝑑𝑥 4 𝑥 0 =1 𝑦 0 =1 𝑑 3 𝑦 𝑑𝑥 𝑑 2 𝑦 𝑑𝑥 2 =𝑦+𝑥 𝑑𝑦 𝑑𝑥 Differentiate each term (again using the product rule where appropriate) 𝑑 2 𝑦 𝑑𝑥 𝑑𝑦 𝑑𝑥 =𝑥𝑦 𝑑 4 𝑦 𝑑𝑥 4 + 2 𝑑 3 𝑦 𝑑𝑥 3 = 𝑑𝑦 𝑑𝑥 + 𝑑𝑦 𝑑𝑥 +𝑥 𝑑 2 𝑦 𝑑𝑥 2 Sub in the values we have so far (or group terms first – your choice!) Product Rule for xdy/dx 𝑑𝑦 𝑑𝑥 𝑑 4 𝑦 𝑑𝑥 = 2 +(2)+(1)(−3) 𝑥 Simplify 𝑑 2 𝑦 𝑑 𝑥 2 𝑑 4 𝑦 𝑑𝑥 =1 1 Calculate 𝑑 4 𝑦 𝑑𝑥 4 =−17 6E

𝑦= 𝑦 0 + 𝑥− 𝑥 𝑑𝑦 𝑑𝑥 𝑥 𝑥− 𝑥 ! 𝑑 2 𝑦 𝑑𝑥 𝑥 𝑥− 𝑥 ! 𝑑 3 𝑦 𝑑𝑥 𝑥 0 … Maclaurin and Taylor Series 𝑑𝑦 𝑑𝑥 =2 𝑑 2 𝑦 𝑑𝑥 2 =−3 𝑑 3 𝑦 𝑑𝑥 3 =9 𝑑 4 𝑦 𝑑𝑥 4 =−17 You can find the solution, in the form of a series, to a differential equation using the Taylor series method Given that: And that y = 1 and dy/dx = 2 when x = 1, express y as a series in ascending powers of (x – 1), up to and including the term in (x – 1)4 And you can now use the values you have calculated in the expansion above… 𝑥 0 =1 𝑦 0 =1 𝑦= 𝑦 0 + 𝑥− 𝑥 𝑑𝑦 𝑑𝑥 𝑥 𝑥− 𝑥 ! 𝑑 2 𝑦 𝑑𝑥 𝑥 𝑥− 𝑥 ! 𝑑 3 𝑦 𝑑𝑥 𝑥 𝑥− 𝑥 ! 𝑑 4 𝑦 𝑑𝑥 𝑥 0 𝑑 2 𝑦 𝑑𝑥 𝑑𝑦 𝑑𝑥 =𝑥𝑦 Substitute in all the values we calculated… 𝑦=(1)+ 𝑥−1 (2)+ 𝑥− (−3)+ 𝑥− (9)+ 𝑥− (−17) Simplify 𝑦=1+2 𝑥−1 − 𝑥− 𝑥−1 3 − 𝑥−1 4 6E

𝑦= 𝑦 0 + 𝑥− 𝑥 𝑑𝑦 𝑑𝑥 𝑥 𝑥− 𝑥 ! 𝑑 2 𝑦 𝑑𝑥 𝑥 𝑥− 𝑥 ! 𝑑 3 𝑦 𝑑𝑥 𝑥 0 … Maclaurin and Taylor Series 𝑑𝑦 𝑑𝑥 =1 You can find the solution, in the form of a series, to a differential equation using the Taylor series method Given that y satisfies the differential equation: And that y = 1 and x = 0, find a series expansion for y in ascending powers of x, up to and including the term in x3  As before, start by finding 𝑑𝑦 𝑑𝑥 , then differentiating and repeating the process to find the other values you need, in this case up to 𝑑 3 𝑦 𝑑𝑥 3 𝑥 0 =0 𝑦 0 =1 𝑑𝑦 𝑑𝑥 = 𝑦 2 −𝑥 Sub in values 𝑑𝑦 𝑑𝑥 = (1) 2 −(0) 𝑑𝑦 𝑑𝑥 = 𝑦 2 −𝑥 Calculate 𝑑𝑦 𝑑𝑥 =1 6E

𝑦= 𝑦 0 + 𝑥− 𝑥 𝑑𝑦 𝑑𝑥 𝑥 𝑥− 𝑥 ! 𝑑 2 𝑦 𝑑𝑥 𝑥 𝑥− 𝑥 ! 𝑑 3 𝑦 𝑑𝑥 𝑥 0 … Maclaurin and Taylor Series 𝑑𝑦 𝑑𝑥 =1 𝑑 2 𝑦 𝑑𝑥 2 =1 You can find the solution, in the form of a series, to a differential equation using the Taylor series method Given that y satisfies the differential equation: And that y = 1 and x = 0, find a series expansion for y in ascending powers of x, up to and including the term in x3  As before, start by finding 𝑑𝑦 𝑑𝑥 , then differentiating and repeating the process to find the other values you need, in this case up to 𝑑 3 𝑦 𝑑𝑥 3 𝑥 0 =0 𝑦 0 =1 𝑑𝑦 𝑑𝑥 = 𝑦 2 −𝑥 Differentiate each term 𝑑 2 𝑦 𝑑𝑥 2 = 2𝑦 𝑑𝑦 𝑑𝑥 − 1 𝑑𝑦 𝑑𝑥 = 𝑦 2 −𝑥 Sub in values 𝑑 2 𝑦 𝑑𝑥 2 = −(1) Calculate 𝑑 2 𝑦 𝑑𝑥 2 =1 6E

𝑦= 𝑦 0 + 𝑥− 𝑥 𝑑𝑦 𝑑𝑥 𝑥 𝑥− 𝑥 ! 𝑑 2 𝑦 𝑑𝑥 𝑥 𝑥− 𝑥 ! 𝑑 3 𝑦 𝑑𝑥 𝑥 0 … Maclaurin and Taylor Series 𝑑𝑦 𝑑𝑥 =1 𝑑 2 𝑦 𝑑𝑥 2 =1 𝑑 3 𝑦 𝑑𝑥 3 =4 You can find the solution, in the form of a series, to a differential equation using the Taylor series method Given that y satisfies the differential equation: And that y = 1 and x = 0, find a series expansion for y in ascending powers of x, up to and including the term in x3  As before, start by finding 𝑑𝑦 𝑑𝑥 , then differentiating and repeating the process to find the other values you need, in this case up to 𝑑 3 𝑦 𝑑𝑥 3 𝑥 0 =0 𝑦 0 =1 𝑑 2 𝑦 𝑑𝑥 2 =2𝑦 𝑑𝑦 𝑑𝑥 −1 Differentiate each term, using the product rule where needed…  Remember that 𝑑𝑦 𝑑𝑥 x 𝑑𝑦 𝑑𝑥 is not 𝑑 2 𝑦 𝑑𝑥 2 𝑑 3 𝑦 𝑑𝑥 3 = 2𝑦 𝑑 2 𝑦 𝑑𝑥 2 + 2 𝑑𝑦 𝑑𝑥 2 𝑑𝑦 𝑑𝑥 = 𝑦 2 −𝑥 Product rule for 2ydy/dx Sub in values 𝑑 3 𝑦 𝑑𝑥 3 = 𝑑𝑦 𝑑𝑥 2𝑦 Calculate 2 𝑑𝑦 𝑑𝑥 𝑑 2 𝑦 𝑑𝑥 2 𝑑 3 𝑦 𝑑𝑥 3 =4 6E

𝑦= 𝑦 0 + 𝑥− 𝑥 𝑑𝑦 𝑑𝑥 𝑥 𝑥− 𝑥 ! 𝑑 2 𝑦 𝑑𝑥 𝑥 𝑥− 𝑥 ! 𝑑 3 𝑦 𝑑𝑥 𝑥 0 … Maclaurin and Taylor Series 𝑑𝑦 𝑑𝑥 =1 𝑑 2 𝑦 𝑑𝑥 2 =1 𝑑 3 𝑦 𝑑𝑥 3 =4 You can find the solution, in the form of a series, to a differential equation using the Taylor series method Given that y satisfies the differential equation: And that y = 1 and x = 0, find a series expansion for y in ascending powers of x, up to and including the term in x3  As before, start by finding 𝑑𝑦 𝑑𝑥 , then differentiating and repeating the process to find the other values you need, in this case up to 𝑑 3 𝑦 𝑑𝑥 3 𝑥 0 =0 𝑦 0 =1 Now you can substitute these into the form above… 𝑦= 𝑦 0 + 𝑥− 𝑥 𝑑𝑦 𝑑𝑥 𝑥 𝑥− 𝑥 ! 𝑑 2 𝑦 𝑑𝑥 𝑥 𝑥− 𝑥 ! 𝑑 3 𝑦 𝑑𝑥 𝑥 0 Sub in values 𝑑𝑦 𝑑𝑥 = 𝑦 2 −𝑥 𝑦=(1)+ 𝑥−0 (1)+ 𝑥− (1)+ 𝑥− (4) Simplify 𝑦=1+𝑥 𝑥 𝑥 3 6E

Summary You have learnt the forms of the Maclaurin and Taylor series
You have seen how they create an infinite sequence that can converge to a function You have seen how to use these in differential equations

Maclaurin and Taylor Series.

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Maclaurin and Taylor Series.

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Presentation on theme: "Maclaurin and Taylor Series."— Presentation transcript:

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