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Taylor Series (Edexcel A Level Further Maths: Further Pure 1): Flashcards

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Define a Taylor series.

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Define a Taylor series.
A Taylor series is a representation of a function as an infinite sum of ascending powers of $x - a$ , built from the value of the function and of all its derivatives at $x = a$ .
Because the powers are of $x - a$ rather than of $x$ , the expansion is built around the point $x = a$ rather than around the origin.
A Taylor series can be built about any point $x = a$ . What decides which value of $a$ to use?
The series only represents $f (x)$ well for values of $x$ close to $a$ , so $a$ is taken near the values you are interested in.
Expanding $x \sin x$ about $x = \frac{π}{2}$ gives a good approximation near $\frac{π}{2}$ , where an expansion about the origin would need a great many terms to do the same job.
What has to be true of $f$ at $x = a$ for it to have a Taylor series about that point?
Every one of its derivatives must exist at $x = a$ , that is $f^{(n)} (a)$ must exist for all $n$ .
The coefficients are built from those derivatives, so if any one of them fails to exist at that point the series breaks down there.
True or False?
Every Maclaurin series is also a Taylor series.
True.
A Maclaurin series is what a Taylor series becomes when $a = 0$ , so the powers of $x - a$ become plain powers of $x$ and every derivative is evaluated at the origin.
The two are one formula rather than two, and the Maclaurin form is simply the case that comes up most often.
When finding a Taylor expansion of $f (x)$ about $x = a$ , at what point in the working do you substitute $x = a$ ?
Only after each derivative has been found in full as a function of $x$ .
Substituting first would turn $f (x)$ into a single number, and every derivative of a number is zero, so nothing would be left to build the series from.
Expanding $f (x) = x \sin x$ about $x = \frac{π}{2}$ gives $f (\frac{π}{2}) = \frac{π}{2}$ , $f' (\frac{π}{2}) = 1$ , $f'' (\frac{π}{2}) = - \frac{π}{2}$ and $f''' (\frac{π}{2}) = - 3$ . Complete the two missing coefficients:
$x \sin x = \frac{π}{2} + (x - \frac{π}{2}) -______{(x - \frac{π}{2})}^{2} -______{(x - \frac{π}{2})}^{3} + \dots$
The completed expansion is:
$x \sin x = \frac{π}{2} + (x - \frac{π}{2}) - \frac{π}{4} {(x - \frac{π}{2})}^{2} - \frac{1}{2} {(x - \frac{π}{2})}^{3} + \dots$
Each derivative value is divided by the factorial of the power it belongs to, so $- \frac{π}{2}$ becomes $- \frac{π}{2} \div 2! = - \frac{π}{4}$ and $- 3$ becomes $- 3 \div 3! = - \frac{1}{2}$ .
Why does every derivative of $x \sin x$ need the product rule?
Because it is a product of $x$ with a trigonometric function, and differentiating such a product always leaves another one of the same kind.
The first derivative is $\sin x + x \cos x$ , whose second term is again $x$ times a trigonometric function, so the rule is needed once more at the next stage.
In the series method a function is replaced by its Taylor series about $x = a$ before the limit is taken. Complete the reason this works:
Every term after the first contains a power of $______$ as a factor, and each of those terms tends to $______$ as $x$ approaches $a$ , so only the constant term survives.
The completed reason is:
Every term after the first contains a power of $x - a$ as a factor, and each of those terms tends to $0$ as $x$ approaches $a$ , so only the constant term survives.
That constant term is $f (a)$ , which is why expanding about the right point hands you the limit directly.
When a limit as $x \to a$ is found by expanding a quotient as two series, what has to be done before $x$ is allowed to approach $a$ ?
The common power of $x - a$ has to be cancelled from the numerator and the denominator first.
Until it is, letting $x \to a$ sends the denominator to zero and nothing can be read off; once it is cancelled, what is left has an ordinary constant term.
The Taylor expansion of $\ln x$ about $x = 1$ is $(x - 1) - \frac{1}{2} {(x - 1)}^{2} + \frac{1}{3} {(x - 1)}^{3} - \dots$ Use it to find $\lim_{x \to 1} (\frac{\ln x - (x - 1) + \frac{1}{2} {(x - 1)}^{2}}{{(x - 1)}^{3}})$
The limit is $\frac{1}{3}$ .
Substituting the expansion cancels the $(x - 1)$ and $\frac{1}{2} {(x - 1)}^{2}$ terms in the numerator, leaving $\frac{1}{3} {(x - 1)}^{3} - \frac{1}{4} {(x - 1)}^{4} + \dots$
Dividing through by ${(x - 1)}^{3}$ leaves $\frac{1}{3} - \frac{1}{4} (x - 1) + \dots$ , whose value at the limit is the constant term.
True or False?
A limit found by expanding as a series is only an approximation, because a truncated series is only an approximation.
False.
The limit obtained is exact.
Every term that is discarded carries a positive power of $x - a$ , so each of them contributes exactly zero once $x \to a$ and dropping them changes nothing at all.
Why is $\lim_{x \to 0} (\frac{\sin (x^{3})}{x^{3}})$ equal to $1$ ?
The expansion of $\sin (x^{3})$ begins $x^{3} - \frac{x^{9}}{3!} + \frac{x^{15}}{5!} - \dots$ , so $x^{3}$ is a factor of every term.
Cancelling it against the denominator leaves $1 - \frac{x^{6}}{3!} + \frac{x^{12}}{5!} - \dots$ , and only the $1$ survives as $x \to 0$ .
While working towards a limit by expanding series, what must be written on every line until the last?
The limit sign itself, $\lim_{x \to a}$ , has to stay in front of the whole expression on every line.
Nothing becomes a number until $x$ is allowed to approach $a$ , so all the expanding, cancelling and simplifying is done to the expression inside the limit.
Define a series solution to a differential equation.
A series solution is an approximation to $y$ written as a Taylor series in ascending powers of $x - a$ , where the question gives $y = b$ when $x = a$ .
It is used for differential equations that cannot be solved exactly, and its coefficients come from the values of $y$ and of its successive derivatives at $x = a$ .
In a series solution about $x = a$ , where does the first term $y (a)$ come from?
Straight from the condition given in the question: if $y = b$ when $x = a$ then $y (a) = b$ , with no work needed at all.
Every later coefficient has to be found by differentiating, so this is the one term that is handed to you.
A differential equation is $y^{3} \frac{d y}{d x} = \ln y + x^{2}$ , with $y = 1$ when $x = 2$ . Find $y' (2)$ .
The value is $y' (2) = 4$ .
Substituting $x = 2$ and $y = 1$ into the equation gives $1^{3} \frac{d y}{d x} = \ln 1 + 2^{2}$ , and since $\ln 1 = 0$ that leaves $\frac{d y}{d x} = 4$ .
Higher coefficients of a series solution bring in derivatives of derivatives. Complete both results:
$\frac{d}{d x} (\frac{d y}{d x}) =______and \frac{d}{d x} (\frac{d^{2} y}{d x^{2}}) =______$
The completed results are:
$\frac{d}{d x} (\frac{d y}{d x}) = \frac{d^{2} y}{d x^{2}} and \frac{d}{d x} (\frac{d^{2} y}{d x^{2}}) = \frac{d^{3} y}{d x^{3}}$
Differentiating a derivative with respect to $x$ raises its order by one, which is what lets the working carry on for as many terms as the question asks for.
The differential equation gives you $\frac{d y}{d x}$ . Where does $\frac{d^{2} y}{d x^{2}}$ come from?
Differentiate both sides of the differential equation itself with respect to $x$ , which calls for implicit differentiation on every term written in $y$ .
Then substitute the known values of $x$ , $y$ and $\frac{d y}{d x}$ into the result and rearrange for $\frac{d^{2} y}{d x^{2}}$ .
True or False?
${(\frac{d y}{d x})}^{2}$ and $\frac{d^{2} y}{d x^{2}}$ are two ways of writing the same thing.
False.
The first is the first derivative squared, while the second is the second derivative of $y$ with respect to $x$ .
Both turn up on the same line of working when a series solution is built, so the two have to be kept carefully apart.

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