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<!DOCTYPE html>
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<head>
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<meta name="viewport" content="width=device-width, initial-scale=1.0">
<title>Further Pure 3</title>
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<body><div class="container"><h1 id="series-and-limits">Series and limits</h1>


<h2 id="limits">Limits</h2>

<p>If <script type="math/tex" id="MathJax-Element-1">x</script> is a real number, which varies such that it approaches a particular value <script type="math/tex" id="MathJax-Element-2">a</script>, but is never equal to <script type="math/tex" id="MathJax-Element-3">a</script>, this is signified by writing <script type="math/tex" id="MathJax-Element-4">x \to a</script>. <br>
The variation of <script type="math/tex" id="MathJax-Element-5">x</script> must be such that for any positive number, <script type="math/tex" id="MathJax-Element-6">\delta</script>, <script type="math/tex" id="MathJax-Element-7">x</script> can be chosen so that <script type="math/tex" id="MathJax-Element-8">0 < |x-a| < \delta</script>, no matter how small <script type="math/tex" id="MathJax-Element-9">\delta</script> may be.</p>

<p>When a function <script type="math/tex" id="MathJax-Element-10">f(x)</script> is such that <script type="math/tex" id="MathJax-Element-11">f(x) \to l</script> when <script type="math/tex" id="MathJax-Element-12">x \to a</script>, where <script type="math/tex" id="MathJax-Element-13">l</script> is finite, the number <script type="math/tex" id="MathJax-Element-14">l</script> is called the limit or limiting value of <script type="math/tex" id="MathJax-Element-15">f(x)</script> as <script type="math/tex" id="MathJax-Element-16">x \to a</script>.</p>

<p>This is expressed as <script type="math/tex; mode=display" id="MathJax-Element-17">\lim_{x \to a} f(x) = l</script></p>

<p>The statements <script type="math/tex" id="MathJax-Element-18">x \to \infty</script> and <script type="math/tex" id="MathJax-Element-19">f(x) \to \infty</script> mean that the values of <script type="math/tex" id="MathJax-Element-20">x</script> and <script type="math/tex" id="MathJax-Element-21">f(x)</script> increase indefinitely. However, it should never be written that <script type="math/tex" id="MathJax-Element-22">x = \infty</script> as <script type="math/tex" id="MathJax-Element-23">\infty</script> is not a number.</p>


<h3 id="direction-of-approach">Direction of approach</h3>

<p>Unless stated otherwise, <script type="math/tex" id="MathJax-Element-24">x \to a</script> means that <script type="math/tex" id="MathJax-Element-25">x</script> can approach <script type="math/tex" id="MathJax-Element-26">a</script> from either direction on the real axis.</p>

<p>The notation <script type="math/tex" id="MathJax-Element-27">x \to a+</script> is used to signify that <script type="math/tex" id="MathJax-Element-28">a</script> approaches from the right, while <script type="math/tex" id="MathJax-Element-29">x \to a-</script> is used to signify an approach from the left.</p>

<p>This distinction can be important. <br>
consider the behaviour of <script type="math/tex" id="MathJax-Element-30">\frac 1 x</script> as <script type="math/tex" id="MathJax-Element-31">x \to 0</script>. When <script type="math/tex" id="MathJax-Element-32">x \to 0+, \frac 1 x \to \infty</script>, but when <script type="math/tex" id="MathJax-Element-33">x \to 0-, \frac 1 x \to -\infty</script>.</p>


<h3 id="finding-simple-limits">Finding simple limits</h3>

<p>Some cases are simple</p>


<p><script type="math/tex; mode=display" id="MathJax-Element-34">\lim_{x \to 0} \frac{1+x}{2-x} = \frac 1 2 </script> <br>
because <script type="math/tex" id="MathJax-Element-35">1+x \to 1</script> and <script type="math/tex" id="MathJax-Element-36">2-x \to 2</script> as <script type="math/tex" id="MathJax-Element-37">x \to 0</script>.</p>


<p><script type="math/tex; mode=display" id="MathJax-Element-38">\lim_{x \to \frac \pi 2} \frac{\sin(x)}{1-\cos(x)} = 1</script></p>

<p>because <script type="math/tex" id="MathJax-Element-39">\sin(x) \to 1</script> and <script type="math/tex" id="MathJax-Element-40">\cos(x) \to 0</script> as <script type="math/tex" id="MathJax-Element-41">x \to \frac \pi 2</script>.</p>

<p>Finally</p>

<p>as <script type="math/tex" id="MathJax-Element-42">x \to 1+, \ \frac{1+x}{1-x} \to -\infty</script></p>

<p>because <script type="math/tex" id="MathJax-Element-43">1+x \to 2+</script> and <script type="math/tex" id="MathJax-Element-44">1-x \to 0-</script> as <script type="math/tex" id="MathJax-Element-45">x \to 1+</script>.</p>

<p>However, it would be incorrect to state</p>


<p><script type="math/tex; mode=display" id="MathJax-Element-46">\lim_{x \to 1+} \frac{1+x}{1-x} = -\infty</script></p>

<p>because <script type="math/tex" id="MathJax-Element-47">\infty</script> is not a finite value.</p>


<h2 id="maclaurin-series-expansions">Maclaurin Series Expansions</h2>

<p>The Maclaurin series for a function <script type="math/tex" id="MathJax-Element-48">f(x)</script> is defined as</p>

<p><script type="math/tex" id="MathJax-Element-49">f(x) = f(0) + f'(0)x + \frac{f''(0}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \ ... + \frac{f^{(r)} (0)}{r!}x^r + \ ...</script></p>

<p>where <script type="math/tex" id="MathJax-Element-50">f'</script>, <script type="math/tex" id="MathJax-Element-51">f''</script>, <script type="math/tex" id="MathJax-Element-52">f'''</script>  denote the first, second, and third derivatives of <script type="math/tex" id="MathJax-Element-53">f</script> respectively, and <script type="math/tex" id="MathJax-Element-54">f^{(r)}</script> denotes the derivative of order <script type="math/tex" id="MathJax-Element-55">r</script>.</p>


<h3 id="derivation">Derivation</h3>

<p>To derive the Maclaurin series, the following assumptions are made</p>

<ul>
<li>The function <script type="math/tex" id="MathJax-Element-56">f(x)</script> can be expressed as a series of the form <script type="math/tex" id="MathJax-Element-57">f(x) = a_0 + a_1x + a_2x^2 + a_3x^3 + \ ... + a_rx^r + \ ... </script>, where <script type="math/tex" id="MathJax-Element-58">a_1,\ a_2,\ a_3, \ ...,\ a_r, \ ...</script> are constants.</li>
<li>The series can be differentiated term by term</li>
<li>The function <script type="math/tex" id="MathJax-Element-59">f(x)</script> and all its derivatives exist at <script type="math/tex" id="MathJax-Element-60">x=0</script></li>
</ul>

<p>Successive differentiations of each side of the polynomial representation of <script type="math/tex" id="MathJax-Element-61">f(x)</script> give</p>


<p><script type="math/tex; mode=display" id="MathJax-Element-62">\begin{align} \\
&f'(x) = a_1 + 2a_2x + 3a_3x^2 + \ ...\\
&f''(x) = 2a_2 + 2\times 3a_3x + 3\times 4a_4x^2 + \ ... \\
&f'''(x) = 2\times 3a_3 + 2\times3\times4a_4x + \ ...
\end{align}</script></p>

<p>and in general</p>

<p><script type="math/tex" id="MathJax-Element-63">f^{(r)} = 2\times3\times \ ... \times(r-1)\times ra_r </script> terms in <script type="math/tex" id="MathJax-Element-64">x, x^2, x^3, \ ...</script></p>

<p>Putting <script type="math/tex" id="MathJax-Element-65">x=0</script> in the expressions for <script type="math/tex" id="MathJax-Element-66">f(x)</script> and its derivatives, we see that</p>


<p><script type="math/tex; mode=display" id="MathJax-Element-67">\begin{align}\\
&a_0 = f(0)\\
&a_1 = f'(0)\\
&a_2 = \frac{f''(0)}{2} = \frac{f''(0)}{2!} \\
&a_3 = \frac{f'''(0)}{2\times 3} = \frac{f'''(0)}{3!}
 \end{align}</script></p>

<p>and, in general</p>

<p><script type="math/tex" id="MathJax-Element-68">a_r = \frac{f^{(r)}(0)}{1\times 2\times 3 \times \ ... \times (r-1)r} = \frac{f^{(r)}(0)}{r!}</script></p>

<p>Substituting these values into the polynomial series gives the Maclaurin series of f(x).</p>


<h3 id="example">Example</h3>

<p>Find the Maclaurin series of <script type="math/tex" id="MathJax-Element-69">\sin(x)</script></p>


<p><script type="math/tex; mode=display" id="MathJax-Element-70">\begin{alignat*}{3} \\
&f(x) &&= \sin(x) &&&\Rightarrow f(0) &&&&= 0 \\
&f'(x) &&= \cos(x) &&&\Rightarrow f'(0) &&&&= 1 \\
&f''(x) &&= -\sin(x) &&&\Rightarrow f''(0) &&&&= 0 \\
&f'''(x) &&= -\cos(x) &&&\Rightarrow f'''(0) &&&&= -1 \\
&f^{(4)}(x) &&= \sin(x) &&&\Rightarrow f^{(4)}(0) &&&& = 0 \\
&f^{(5)}(x) &&= \cos(x) &&&\Rightarrow f^{(5)}(x) &&&&= 1
\end{alignat*}</script></p>

<p>Substituting these values into the general form of the Maclaurin series gives</p>


<p><script type="math/tex; mode=display" id="MathJax-Element-71">\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} + \ ...</script></p>

<p>this can be generalised to </p>


<p><script type="math/tex; mode=display" id="MathJax-Element-72">(-1)^r \frac{x^{2r +1}}{(2r+1)!}</script></p>

<p>Where <script type="math/tex" id="MathJax-Element-73">r=0</script> gives the first term <script type="math/tex" id="MathJax-Element-74">r=1</script> the second and so forth, the permissible values of <script type="math/tex" id="MathJax-Element-75">r</script> being <script type="math/tex" id="MathJax-Element-76">r \in \mathbb{Z^*}</script>.</p>


<h3 id="range-of-validity-of-a-series-expansion">Range of validity of a series expansion</h3>

<p>The Maclaurin series may not be valid for all values of <script type="math/tex" id="MathJax-Element-77">x</script>. <br>
for example, the expansion of <script type="math/tex" id="MathJax-Element-78">\frac{1}{1-x}</script> is <script type="math/tex" id="MathJax-Element-79">1 + x + x^2 +\ ... + x^n + \ ...</script>.</p>

<p>For <script type="math/tex" id="MathJax-Element-80">x=2</script> the original equation has the value <script type="math/tex" id="MathJax-Element-81">-1</script>. However, for <script type="math/tex" id="MathJax-Element-82">x=2</script> the expansion takes the value <script type="math/tex" id="MathJax-Element-83">1 + 2 + 2^2 + 2^3 + \ ...</script> which is not equal to <script type="math/tex" id="MathJax-Element-84">-1</script>. The expansion is not valid for <script type="math/tex" id="MathJax-Element-85">x=2</script>.</p>

<p>The range of validity for this expansion can be determined as follows.</p>

<p>The expansion is an infinite geometric series with the first term <script type="math/tex" id="MathJax-Element-86">1</script> and a common ratio <script type="math/tex" id="MathJax-Element-87">x</script>. The sum <script type="math/tex" id="MathJax-Element-88">S_n(x)</script> for the first <script type="math/tex" id="MathJax-Element-89">n</script> terms is given by</p>

<p><script type="math/tex" id="MathJax-Element-90">S_n(x) = \frac{1 - x^{n+1}}{1-x} = \frac{1}{1-x} - \frac{x^{n+1}}{1-x}</script></p>

<p>Hence <script type="math/tex" id="MathJax-Element-91">\frac{1}{1-x} S_n(x) + \frac{x^{n+1}}{1-x}</script> </p>

<p>For the expansion to be valid, S<script type="math/tex" id="MathJax-Element-92">_n(x)</script> must have the same value as <script type="math/tex" id="MathJax-Element-93">\frac{1}{1-x}</script> in the limit when <script type="math/tex" id="MathJax-Element-94">n \to \infty</script>. For this to happen <script type="math/tex" id="MathJax-Element-95">\frac{x^{n+1}}{1-x} \to 0</script> as <script type="math/tex" id="MathJax-Element-96">n \to \infty</script>, which occurs only when <script type="math/tex" id="MathJax-Element-97">|x| < 1</script>.</p>

<p>We can therefore conclude that the Maclaurin series expansion of <script type="math/tex" id="MathJax-Element-98">\frac{1}{1-x}</script> is valid provided <script type="math/tex" id="MathJax-Element-99">|x| < 1.</script></p>

<p>In general, suppose the Maclaurin series of <script type="math/tex" id="MathJax-Element-100">f(x)</script> is given by <script type="math/tex" id="MathJax-Element-101">f(x) = S_n(x) + R_n(x)</script> where <script type="math/tex" id="MathJax-Element-102">S_n(x)</script> is the sum of the first <script type="math/tex" id="MathJax-Element-103">n</script> terms of the series and <script type="math/tex" id="MathJax-Element-104">R_n(x)</script> is the sum of all the remaining terms. <br>
For a particular value of <script type="math/tex" id="MathJax-Element-105">x</script>, the series expansion will be provided that <script type="math/tex" id="MathJax-Element-106">R_n(x) \to 0</script> as <script type="math/tex" id="MathJax-Element-107">n \to \infty</script>.</p>


<h3 id="the-basic-series-expansions">The basic series expansions</h3>

<p>The following expansions are given in the formula booklet, and are often used to derive other expansions</p>


<p><script type="math/tex; mode=display" id="MathJax-Element-108">\begin{align}[1]\ &e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \ ... + \frac{x^r}{r!} + \ ... \quad &r = (0, 1, 2, \ ...)\\
[2]\ &\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!}-\ ... + (-1)^r \frac{x^{2r+1}}{(2r+1)!}+\ ... \quad &r = (0, 1, 2, \ ...)\\
[3]\ &\cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!}- \ ... + (-1)^r \frac{x^{2r}}{(2r)!}+\ ... \quad &r = (0, 1, 2, \ ...)\\
[4]\ &\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3}- \ ... + (-1)^{r+1}\frac{x^{2r}}{(2r)!}+\ ... \quad &r=(1, 2, 3, \ ...) \\
[5]\ &(1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \ ... + \begin{pmatrix}n \\ r\end{pmatrix}x^r + \ ... \quad &r=(0, 1, 2, \ ...)\end{align}</script></p>

<p>Expansions 1 to 3 are valid for all real values of <script type="math/tex" id="MathJax-Element-109">x</script>.</p>

<p>The expansion of <script type="math/tex" id="MathJax-Element-110">\ln(1+x)</script> is valid for <script type="math/tex" id="MathJax-Element-111">-1 < x \leq 1</script>.</p>

<p>The expansion of <script type="math/tex" id="MathJax-Element-112">(1+x)^n</script> is the binomial series and is valid for any real value of <script type="math/tex" id="MathJax-Element-113">n</script> when <script type="math/tex" id="MathJax-Element-114">-1 < x < 1</script>.  <br>
When <script type="math/tex" id="MathJax-Element-115">n</script> is a positive integer, the series is a finite polynomial of degree n. This makes it value for all real values of <script type="math/tex" id="MathJax-Element-116">x</script> when <script type="math/tex" id="MathJax-Element-117">n</script> is an integer. <br>
One other case is that when <script type="math/tex" id="MathJax-Element-118">x=1</script> the expansion is still valid if <script type="math/tex" id="MathJax-Element-119">n \geq - \frac 1 2</script>.</p>

<p>These expansions can then be used to determine other more complicated expansions.</p>

<blockquote>
  <p>Obtain the first three non-zero terms in the series expansions of <script type="math/tex" id="MathJax-Element-120">x\sin(x^2)</script> and <script type="math/tex" id="MathJax-Element-121">\frac{1}{(1+2x)^{\frac 1 3}}</script> </p>

  <p>Firstly, for <script type="math/tex" id="MathJax-Element-122">x\sin(x^2)</script> we use the expansion for <script type="math/tex" id="MathJax-Element-123">\sin(x)</script> substituting <script type="math/tex" id="MathJax-Element-124">x^2</script> in place of <script type="math/tex" id="MathJax-Element-125">x</script>. <br>
   This gives  <br>
   <script type="math/tex" id="MathJax-Element-126">x\sin(x^2) = x\left(x^2 - \frac{(x^2)^3}{3!} + \frac{(x^2)^5}{5!}\right)</script> <br>
   The expansion is valid for all values of <script type="math/tex" id="MathJax-Element-127">x^2</script> and therefore for all values of <script type="math/tex" id="MathJax-Element-128">x</script> </p>

  <p>Secondly for <script type="math/tex" id="MathJax-Element-129">\frac{1}{(1+2x)^{\frac 1 3}}</script>  we have  <br>
   <script type="math/tex; mode=display" id="MathJax-Element-130">\begin{align}\frac{1}{(1+2x)^{\frac 1 3}} &= (1+2x)^{-\frac 2 3}\\
   &= 1 -\frac 1 3 (2x) + \frac{-\frac 1 3 \times - \frac 4 3}{1\times 2}(2x)^2 + \ ... \\
   &= 1 - \frac 2 3 x + \frac 8 9 x^2 + \ ...\end{align}</script> <br>
  The expansion is valid for <script type="math/tex" id="MathJax-Element-131">-1 < 2x < 1 \Rightarrow -\frac 1 2 < x < \frac 1 2</script> </p>
</blockquote>


<h3 id="use-of-series-expansions-to-find-limits">Use of series expansions to find limits</h3>

<p>Consider the function <script type="math/tex" id="MathJax-Element-132">f(x) = \frac{x}{1- \sqrt{1-x}}</script> as <script type="math/tex" id="MathJax-Element-133">x \to 0</script>.</p>

<p>Using the binomial expansion </p>


<p><script type="math/tex; mode=display" id="MathJax-Element-134">\begin{align}\sqrt{1-x} &= (1-x)^{\frac 1 2} \\
&=  1 + \frac1 2 (-x) \frac{\frac 1 2 \times \frac 1 2}{2!}(-x)^2 + \frac{\frac 1 2 \times -\frac 1 2 \times -\frac 3 2}{3!}(-x)^3 + \ ... \\
&= 1 - \frac 1 2 x - \frac 1 8 x^2 - \frac{1}{16} x^3 + \ ...\end{align}</script></p>

<p>Hence </p>


<p><script type="math/tex; mode=display" id="MathJax-Element-135">\begin{align}f(x) &= \frac{x}{1- \left(1 - \frac 1 2 x - \frac 1 8 x^2 - \frac{1}{16}x^3 + \ ... \right)} \\
&= \frac{x}{\frac 1 2 x + \frac 1 8 x^2 + \frac{1}{16}x^3 + \ ...} \\
&= \frac{1}{\frac 1 2 + \frac 1 8 x + \frac{1}{16} x^3} \end{align}</script></p>

<p>It can now be seen that when <script type="math/tex" id="MathJax-Element-136">x\to 0</script>, all the terms in the denominator, except the first, tend to zero. <br>
Hence,  <br>
<script type="math/tex; mode=display" id="MathJax-Element-137">\lim_{x\to 0} f(x) = \frac{1}{\frac 1 2} = 2</script></p>


<h3 id="important-limits">Important limits</h3>


<h4 id="the-limit-of-xke-x-x-to-infty">The limit of <script type="math/tex" id="MathJax-Element-138">x^ke^{-x}, x \to \infty</script></h4>

<p>First consider the function <script type="math/tex" id="MathJax-Element-139">x^2e^{-x}</script>. When x becomes very large, <script type="math/tex" id="MathJax-Element-140">x^2</script> becomes very large, and <script type="math/tex" id="MathJax-Element-141">e^{-x}</script> becomes very small. <br>
As <script type="math/tex" id="MathJax-Element-142">e^{-x}</script> gets smaller at a greater rate than <script type="math/tex" id="MathJax-Element-143">x^2</script> increases, the result tends to 0. <br>
when <script type="math/tex" id="MathJax-Element-144">x \to \infty</script>, <script type="math/tex" id="MathJax-Element-145">x^ke^{-k} \to 0, \ k \in \mathbb{R}</script></p>


<h5 id="proof">Proof</h5>

<blockquote>
  <p>First note that when <script type="math/tex" id="MathJax-Element-146">k = 0</script>, and <script type="math/tex" id="MathJax-Element-147">x \neq 0 </script> the expression <script type="math/tex" id="MathJax-Element-148">x^k e^{-x}</script> becomes <script type="math/tex" id="MathJax-Element-149">e^{-x}</script> which tends to zero as <script type="math/tex" id="MathJax-Element-150">x \to \infty</script>. <br>
  The result therefore holds in this case. <br>
  Further, when <script type="math/tex" id="MathJax-Element-151">k < 0</script>, both <script type="math/tex" id="MathJax-Element-152">x^k</script> and <script type="math/tex" id="MathJax-Element-153">e^{-x}</script> tend to zero as <script type="math/tex" id="MathJax-Element-154">x \to \infty</script> so the product of the two must also tend to 0. Therefore the result holds for this case as well.</p>

  <p>Finally, suppose that <script type="math/tex" id="MathJax-Element-155">k > 0</script>. Let <script type="math/tex" id="MathJax-Element-156">n</script> be an integer such that <script type="math/tex" id="MathJax-Element-157">n > k</script>. <br>
  Using the series expansion for <script type="math/tex" id="MathJax-Element-158">e^x</script> we have <br>
  <script type="math/tex; mode=display" id="MathJax-Element-159">\begin{align}x^ke^{-x} &= \dfrac{x^k}{e^x} \\
 &= \dfrac{x^k}{1 + x + \dfrac{x^2}{2!} + {\ ...} + \dfrac{x^n}{n!} + \dfrac{x^{n+1}}{(n+1)!} + \dfrac{x^{n+2}}{(n+2)!}+ {\ ...}} \\
 &= \dfrac{x^{k-n}}{x^{-n} + x^{1-n} + \dfrac{x^{2-n}}{2!} +{ \ ...} + \dfrac{1}{n!} + \dfrac{x}{(n+1)!} + \dfrac{x^2}{(n+2)!}+ {\ ...}} \end{align}</script> <br>
  As <script type="math/tex" id="MathJax-Element-160">n>k</script>, <script type="math/tex" id="MathJax-Element-161">k-n</script> is negative. Hence, when <script type="math/tex" id="MathJax-Element-162">x \to \infty</script>, <script type="math/tex" id="MathJax-Element-163">x^{k-n} \to 0</script>. <br>
  In the denominator, all of the terms are positive, and all of those after <script type="math/tex" id="MathJax-Element-164">\frac{1}{n!}</script> tend to infinity as <script type="math/tex" id="MathJax-Element-165">x \to \infty</script>. <br>
  Hence, when <script type="math/tex" id="MathJax-Element-166">x \to \infty</script> the denominator of the expression tends to infinity. <br>
  It follows that <br>
  <script type="math/tex; mode=display" id="MathJax-Element-167">x^ke^{-k} \to \frac 0 \infty = 0</script> <br>
  No matter how large the value of <script type="math/tex" id="MathJax-Element-168">k</script> may be, <script type="math/tex" id="MathJax-Element-169">x^ke^{-k} \to 0</script> as <script type="math/tex" id="MathJax-Element-170">x \to \infty</script> still holds. Therefore, for large values of <script type="math/tex" id="MathJax-Element-171">x</script>, the influence of <script type="math/tex" id="MathJax-Element-172">e^{-x}</script> is stronger than any power of <script type="math/tex" id="MathJax-Element-173">x</script>.</p>
</blockquote>


<h4 id="the-limit-of-xk-lnx-as-x-to-0">The limit of <script type="math/tex" id="MathJax-Element-174">x^k \ln(x)</script> as <script type="math/tex" id="MathJax-Element-175">x \to 0</script></h4>

<p>Consider the function <script type="math/tex" id="MathJax-Element-176">x^k \ln(x)</script> where <script type="math/tex" id="MathJax-Element-177">k >0</script>. <script type="math/tex" id="MathJax-Element-178">x</script> must be restricted to <script type="math/tex" id="MathJax-Element-179">\mathbb{R^+}</script> as <script type="math/tex" id="MathJax-Element-180">\ln(x)</script> is only real for this set.</p>

<p>When <script type="math/tex" id="MathJax-Element-181">x \to 0</script>, <script type="math/tex" id="MathJax-Element-182">x^k \to 0</script> and <script type="math/tex" id="MathJax-Element-183">\ln(x) \to -\infty</script>. It is not obvious what happens to the product as <script type="math/tex" id="MathJax-Element-184">x \to 0</script>.</p>

<p>As <script type="math/tex" id="MathJax-Element-185">x \to 0</script>, <script type="math/tex" id="MathJax-Element-186">x^k\ln(x) \to 0</script> for all <script type="math/tex" id="MathJax-Element-187">k > 0</script></p>


<h5 id="proof-1">Proof</h5>

<blockquote>
  <p>Let <script type="math/tex" id="MathJax-Element-188"> x = e^{-\frac y k}</script>. Then when <script type="math/tex" id="MathJax-Element-189">y \to \infty</script>, <script type="math/tex" id="MathJax-Element-190">x \to 0</script>. Also <script type="math/tex" id="MathJax-Element-191">x^k = e^{-y}</script> and <script type="math/tex" id="MathJax-Element-192">\ln(x)  = -\frac y k</script>. Hence, <br>
  <script type="math/tex; mode=display" id="MathJax-Element-193">\begin{align}
 \lim_{x \to 0}\left(x^k\ln(x)\right) &= \lim_{y \to \infty}\left(-\frac y k e^{-y}\right) \\
 &= -\frac 1 k \lim_{y \to \infty} \left(y e^{-y} \right) \end{align}</script> <br>
  Using the established limit for <script type="math/tex" id="MathJax-Element-194">x^ke^{-k}</script> with <script type="math/tex" id="MathJax-Element-195">k = 1</script>, we have <script type="math/tex" id="MathJax-Element-196">\lim \limits_{y \to \infty}\left(ye^{-y} \right)  = 0</script>  <br>
  Hence <script type="math/tex; mode=display" id="MathJax-Element-197">\lim_{x\to 0}{\left(x^k \ln(k)\right)}</script> </p>

  <p>For small values of <script type="math/tex" id="MathJax-Element-198">x</script>, <script type="math/tex" id="MathJax-Element-199">x^k \ln(k)</script> will be negative because <script type="math/tex" id="MathJax-Element-200">\ln(x) < 0 </script> for <script type="math/tex" id="MathJax-Element-201">0 < x < 1</script>. Hence the limit is approached from negative values.</p>

  <p>We then have <script type="math/tex" id="MathJax-Element-202">x^k \ln(x) \to 0- </script> when <script type="math/tex" id="MathJax-Element-203">x \to 0+ \quad (k > 0)</script></p>
</blockquote>


<h3 id="improper-integrals">Improper integrals</h3>

<p>Consider the integral <script type="math/tex" id="MathJax-Element-204">I_1 = \int_0^{\infty}\frac{1}{1+x^2} dx</script>. <br>
This integral gives the area of the region <script type="math/tex" id="MathJax-Element-205">R_1</script> in the first quadrant enclosed by the curve <script type="math/tex" id="MathJax-Element-206">y = \frac{1}{1+x^2}</script> the <script type="math/tex" id="MathJax-Element-207">x</script> axis and the <script type="math/tex" id="MathJax-Element-208">y</script> axis. <br>
It is unclear whether the area will have a finite value, as the upper bound of the integral is infinite.</p>

<p>To investigate this, the upper bound is replaced by <script type="math/tex" id="MathJax-Element-209">c</script>. <br>
We then have  <br>
<script type="math/tex; mode=display" id="MathJax-Element-210">\begin{align}I_1 &= \int_0^c \frac{1}{1+x^2}dx \\
&= \left[ \operatorname{atan}(x) \right]_0^c \\
&= \operatorname{atan}(c)\end{align}</script> </p>

<p>Now let <script type="math/tex" id="MathJax-Element-211">c \to \infty</script> and it can be seen that <script type="math/tex" id="MathJax-Element-212">I_1 \to \frac{\pi}{2} </script> because <script type="math/tex" id="MathJax-Element-213">\operatorname{atan}(c) \to \frac{\pi}{2}</script>. <br>
The area of the region <script type="math/tex" id="MathJax-Element-214">R_1</script> is therefore <script type="math/tex" id="MathJax-Element-215">\frac{\pi}{2}</script>.</p>

<p>Next consider the integral <script type="math/tex" id="MathJax-Element-216">I_2 = \int_0^4 \frac{1}{x^{\frac 1 2}} dx</script>. This gives the area of the region <script type="math/tex" id="MathJax-Element-217">R_2</script> bounded by the curve <script type="math/tex" id="MathJax-Element-218">y = \frac{1}{x^{\frac 1 2}}</script> the <script type="math/tex" id="MathJax-Element-219">x</script> axis, the <script type="math/tex" id="MathJax-Element-220">y</script> axis and the ordinate <script type="math/tex" id="MathJax-Element-221">x=4</script>.</p>

<p><script type="math/tex" id="MathJax-Element-222">R_2</script> is an unbounded region because <script type="math/tex" id="MathJax-Element-223">y \to \infty</script> as <script type="math/tex" id="MathJax-Element-224">x \to 0</script>. <br>
To determine whether <script type="math/tex" id="MathJax-Element-225">R_2</script> is finite we replace the lower limit with <script type="math/tex" id="MathJax-Element-226">c</script>.</p>


<p><script type="math/tex; mode=display" id="MathJax-Element-227">\begin{align}I_2 &= \int_c^4 \frac{1}{x^{\frac 1 2}} dx \\
&= \left[2x^{\frac 1 2} \right]_c^0 \\
&= 4 - 2c^{\frac 1 2} \end{align}</script></p>

<p>Now let <script type="math/tex" id="MathJax-Element-228">c \to 0</script> and it can be seen that <script type="math/tex" id="MathJax-Element-229">I_2 \to 4</script>. The area of <script type="math/tex" id="MathJax-Element-230">R_2</script> is therefore equal to 4.</p>

<p>The formal definition of an improper integral is as follows</p>

<p>The integral <script type="math/tex" id="MathJax-Element-231">\int _a^b f(x)\ dx</script> is said to be improper if:</p>

<ul>
<li>The interval of integration is infinite</li>
<li>The integrand is not defined at either of the limits</li>
<li>The integrand is not defined at one or more of the points bounded by the limits of the integral</li>
</ul>


<h2 id="polar-coordinates">Polar coordinates</h2>

<p>Let <script type="math/tex" id="MathJax-Element-232">O</script>  be a fixed point and <script type="math/tex" id="MathJax-Element-233">OL</script> a fixed line in the plane. For any point <script type="math/tex" id="MathJax-Element-234">P</script>, let the distance of <script type="math/tex" id="MathJax-Element-235">P</script> from <script type="math/tex" id="MathJax-Element-236">O</script> be <script type="math/tex" id="MathJax-Element-237">r</script> and the angle that <script type="math/tex" id="MathJax-Element-238">OP</script> makes with <script type="math/tex" id="MathJax-Element-239">OL</script> be <script type="math/tex" id="MathJax-Element-240">\theta</script>.</p>

<p><script type="math/tex" id="MathJax-Element-241">r</script> and <script type="math/tex" id="MathJax-Element-242">\theta</script> are then the polar coordinates of <script type="math/tex" id="MathJax-Element-243">P</script>.</p>

<p>The point <script type="math/tex" id="MathJax-Element-244">O</script> is called the pole and <script type="math/tex" id="MathJax-Element-245">OL</script> is called the initial line. <br>
The angle <script type="math/tex" id="MathJax-Element-246">\theta</script> is measured in radians. <br>
Positive values of <script type="math/tex" id="MathJax-Element-247">\theta</script> correspond to an anticlockwise rotation from <script type="math/tex" id="MathJax-Element-248">OL</script>, and negative values to a clockwise rotation.</p>

<p>The AQA specification defines <script type="math/tex" id="MathJax-Element-249">r</script> as being non-negative.</p>


<h3 id="restrictions-on-the-values-of-theta">Restrictions on the values of <script type="math/tex" id="MathJax-Element-250">\theta</script></h3>

<p>To ensure that each point in the plane, other than the pole <script type="math/tex" id="MathJax-Element-251">O</script>, has only one pair of polar coordinates, the values that <script type="math/tex" id="MathJax-Element-252">\theta</script> can take will sometimes be restricted. <br>
The pole is an exception as it is defined by <script type="math/tex" id="MathJax-Element-253">r = 0 </script> without reference to <script type="math/tex" id="MathJax-Element-254">\theta</script>.</p>


<h3 id="the-relationship-between-cartesian-and-polar-coordinates">The relationship between Cartesian and polar coordinates</h3>

<p>We have </p>

<p><script type="math/tex" id="MathJax-Element-255">x = r\cos(\theta)</script></p>

<p><script type="math/tex" id="MathJax-Element-256">y = r\sin(\theta)</script></p>

<p><script type="math/tex" id="MathJax-Element-257">r^2 = x^2 + y^2</script></p>

<p><script type="math/tex" id="MathJax-Element-258">\tan(\theta) = \frac y x</script></p>


<h3 id="example-1">Example</h3>

<blockquote>
  <p>The values of A and B have polar coordinates <script type="math/tex" id="MathJax-Element-259">\left(3, \frac \pi 6 \right)</script> and <script type="math/tex" id="MathJax-Element-260">\left(4, -\frac \pi 3 \right)</script> respectively.</p>

  <p>Show that AB = 5</p>

  <p><script type="math/tex" id="MathJax-Element-261">AB = \sqrt{(x_B - x_A)^2 + (y_B - y_A)^2}</script> <br>
    We have</p>

  <p><script type="math/tex" id="MathJax-Element-262">x_B = 4\times \cos(-\frac \pi 3) \quad x_A = 3 \times \cos(\frac \pi 6)</script> <br>
    <script type="math/tex" id="MathJax-Element-263">y_B = 4 \times \sin(-\frac \pi 3) \quad y_A = 3 \times \sin(\frac \pi 6)</script> </p>

  <p><script type="math/tex" id="MathJax-Element-264">AB</script> is then <script type="math/tex" id="MathJax-Element-265">\sqrt{25} = 5</script></p>

  <p>Alternatively, we can find the distance without converting to cartesian form given that the distance between two polar points <script type="math/tex" id="MathJax-Element-266">(r_1, \theta_1)</script> and <script type="math/tex" id="MathJax-Element-267">(r_2, \theta_2)</script> is given by <br>
     <script type="math/tex" id="MathJax-Element-268">\sqrt{r_1^2 + r_2^2 - 2r_1r_2\cos(\theta_1 - \theta_2)}</script> <br>
     Which also gives <script type="math/tex" id="MathJax-Element-269">AB = 5</script>.</p>
</blockquote>


<h3 id="representations-of-curves-in-polar-form">Representations of curves in polar form</h3>

<p>If a point <script type="math/tex" id="MathJax-Element-270">P</script> with Cartesian coordinates <script type="math/tex" id="MathJax-Element-271">(x, y)</script> moves on a circle of radius <script type="math/tex" id="MathJax-Element-272">a</script> and centre <script type="math/tex" id="MathJax-Element-273">C(a, 0)</script>, then, for all positions of <script type="math/tex" id="MathJax-Element-274">P</script>,</p>

<p><script type="math/tex" id="MathJax-Element-275">(x-a)^2 + y^2 = a^2</script></p>

<p>This is the Cartesian equation of the circle.</p>

<p>Now let <script type="math/tex" id="MathJax-Element-276">P</script> have polar coordinates <script type="math/tex" id="MathJax-Element-277">(r, \theta)</script>. The circle intersects the positive <script type="math/tex" id="MathJax-Element-278">x</script> axis at the point <script type="math/tex" id="MathJax-Element-279">A(2a, 0)</script>.  <br>
Hence from the triangle <script type="math/tex" id="MathJax-Element-280">OAP</script> we have <script type="math/tex" id="MathJax-Element-281">r = 2a\cos(\theta)</script>.</p>

<p>This is the polar equation of the circle. Note that the equation is valid for negative values of <script type="math/tex" id="MathJax-Element-282">\theta</script> because <script type="math/tex" id="MathJax-Element-283">\cos(-\theta)=\cos(\theta)</script>.</p>

<p>There are many examples of curves whose properties are more easily investigated using a polar equation rather than a Cartesian equation.</p>

<ul>
<li>The equation <script type="math/tex" id="MathJax-Element-284">r=a</script> which represents a circle centred at <script type="math/tex" id="MathJax-Element-285">O</script> and of radius <script type="math/tex" id="MathJax-Element-286">a</script>   </li>
<li>The equation <script type="math/tex" id="MathJax-Element-287">\theta = \alpha</script> represents a semi-infinite straight line <script type="math/tex" id="MathJax-Element-288">OA</script> radiating from the origin and making an angle of <script type="math/tex" id="MathJax-Element-289">\alpha</script> with the initial line <script type="math/tex" id="MathJax-Element-290">OL</script>.</li>
</ul>


<h1 id="introduction-to-differential-equations">Introduction to differential equations</h1>


<h2 id="order-and-linearity">Order and linearity</h2>

<p>When only the first order derivative <script type="math/tex" id="MathJax-Element-291">\frac{dy}{dx}</script> is involved, the differential equation is said to be of first order. When the second order derivative <script type="math/tex" id="MathJax-Element-292">\frac{d^2y}{dx^2}</script> is involved the derivative is of second order. <br>
An order <script type="math/tex" id="MathJax-Element-293">n</script> derivative involves <script type="math/tex" id="MathJax-Element-294">\frac{d^ny}{dx^n}</script>.</p>

<p>A differential equation is said to be linear if it is linear in the dependent variable <script type="math/tex" id="MathJax-Element-295">y</script> and the derivatives of <script type="math/tex" id="MathJax-Element-296">y</script>.</p>

<p>This can also be defined by the statement: ‘a differential equation is linear if the highest order derivative of the dependent variable <script type="math/tex" id="MathJax-Element-297">y</script> can be expressed as a linear function of <script type="math/tex" id="MathJax-Element-298">y</script> and the lower order derivatives.’. <br>
Hence, for a second order differential equation to be linear, it must be possible to express <script type="math/tex" id="MathJax-Element-299">\frac{d^2y}{dx^2}</script> in the form</p>


<p><script type="math/tex; mode=display" id="MathJax-Element-300">\frac{d^2y}{dx^2} = f(x) + g(x)y + h(x)\frac{dy}{dx}</script></p>

<p>where <script type="math/tex" id="MathJax-Element-301">f</script>, <script type="math/tex" id="MathJax-Element-302">g</script>, and <script type="math/tex" id="MathJax-Element-303">h</script> are functions of <script type="math/tex" id="MathJax-Element-304">x</script> only.</p>


<h3 id="example-2">Example</h3>

<blockquote>
  <p>State order of each of the following differential equations and whether they are linear <br>
  a) <script type="math/tex" id="MathJax-Element-305">x\frac{dy}{dx} + y = x^2</script> <br>
  Order 1, linear</p>

  <p>b) <script type="math/tex" id="MathJax-Element-306">\frac{d^2y}{dx^2} + \frac{dy}{dx} + y = 0</script> <br>
  Order 2, linear</p>

  <p>c) <script type="math/tex" id="MathJax-Element-307">\frac{dy}{dx} = x^3 + y^3</script>  <br>
  Order 1, non-linear</p>

  <p>d) <script type="math/tex" id="MathJax-Element-308">x\left(\frac{dy}{dx}\right)^2 + y = 1</script> <br>
  Order 1, non-linear</p>
</blockquote>


<h2 id="families-of-solutions-general-solutions-and-particular-solutions">Families of solutions, general solutions, and particular solutions</h2>

<p>A differential equation of the form <script type="math/tex" id="MathJax-Element-309">\frac{dy}{dx} = f(x)</script> is solved by integrating each side, to give a function <script type="math/tex" id="MathJax-Element-310">y(x) = \int f(x)\ dx</script>.</p>

<p>A differential equation of the form <script type="math/tex" id="MathJax-Element-311">\frac{dy}{dx} = f(x)g(y)</script> can be solved by separation of variables, which involves rewriting the equation as <script type="math/tex" id="MathJax-Element-312">\frac{1}{g(y)} \frac{dy}{dx} = f(x)</script>, integrating both sides with respect to <script type="math/tex" id="MathJax-Element-313">x</script>, and solving for <script type="math/tex" id="MathJax-Element-314">y</script>.</p>

<p>The set of all possible solutions for a differential equation is said to form a family of solutions. <br>
Particular members of the family of solutions are solutions with a particular value for the constant of integration.</p>

<p>Solutions which involve arbitrary constants are called general solutions because they represent the whole family of possible solutions. <br>
A solution which satisfies the differential equation but contains no arbitrary constants is called a particular solution. <br>
General solutions of first order differential equations always contain exactly one arbitrary constant.</p>

<p>In most of the applications of differential equations, a particular solution valid over some specified interval is required. The required solution of a first order differential equation is often chosen in order to satisfy a given condition at an end point of an interval under consideration.</p>


<h3 id="example-3">Example</h3>

<blockquote>
  <p>The function <script type="math/tex" id="MathJax-Element-315">y(x)</script> satisfies the differential equation  <br>
  <script type="math/tex; mode=display" id="MathJax-Element-316">x^2 \frac{dy}{dx} - y^2 = 0, \quad x \geq 1</script></p>

  <p>a) Find the general solution for <script type="math/tex" id="MathJax-Element-317">y(x)</script></p>

  <p>Writing the differential equation as <script type="math/tex" id="MathJax-Element-318">\frac{dy}{dx} = \frac{x^2}{y^2}</script> and separating the variables we have <br>
    <script type="math/tex; mode=display" id="MathJax-Element-319">\int \frac{dy}{y^2} = \int \frac{dx}{x^2}</script> <br>
    giving  <br>
    <script type="math/tex; mode=display" id="MathJax-Element-320">-\frac 1 y = -\frac 1 x + C</script> <br>
    Hence  <br>
    <script type="math/tex; mode=display" id="MathJax-Element-321">y = \frac{x}{1- Cx}</script></p>

  <p>b) Hence find the particular solution satisfying the boundary condition <script type="math/tex" id="MathJax-Element-322">y(1) = \frac 1 2</script> </p>

  <p>Applying the boundary condition gives <br>
    <script type="math/tex; mode=display" id="MathJax-Element-323">\frac 1 2 = \frac{1}{1-C}</script> <br>
    Hence <script type="math/tex" id="MathJax-Element-324">C = -1</script>. The required particular solution is therefore</p>

  <p><script type="math/tex; mode=display" id="MathJax-Element-325">y = \frac{x}{1+x}</script></p>
</blockquote>


<h1 id="numerical-methods-for-the-solution-of-first-order-differential-equations">Numerical methods for the solution of first order differential equations</h1>

<p>These formula are concerned with differential equations of the form </p>

<p><script type="math/tex" id="MathJax-Element-326">\frac{dy}{dx} = f(x, y)</script> subject to the boundary conditions of the form <script type="math/tex" id="MathJax-Element-327">y(x_0) = y_0</script>.</p>

<p>In each case, the solution <script type="math/tex" id="MathJax-Element-328">y(x)</script>, valid for <script type="math/tex" id="MathJax-Element-329">x \geq x_0</script> is required.</p>

<p>The basic idea of all the numerical methods is to use a step-by-step procedure to obtain approximations to the values of <script type="math/tex" id="MathJax-Element-330">y_1, y_2, y_3,\ \ ...</script> successively. <br>
The interval, <script type="math/tex" id="MathJax-Element-331">h</script>, between successive x-values is called the step length. <br>
Some of the methods of obtaining these approximations are suggested by geometrical considerations of the graph.</p>


<h2 id="eulers-formula">Euler’s formula</h2>

<p>One way of obtaining an approximation to the value of <script type="math/tex" id="MathJax-Element-332">y_1</script> is to assume that the part of the curve between <script type="math/tex" id="MathJax-Element-333">P_0</script> and <script type="math/tex" id="MathJax-Element-334">P_1</script> is a straight line segment with a gradient equal to the gradient of the curve at <script type="math/tex" id="MathJax-Element-335">P_0</script>.</p>

<p>Since <script type="math/tex" id="MathJax-Element-336">\frac{dy}{dx} = f(x, y)</script> , the gradient of the curve at <script type="math/tex" id="MathJax-Element-337">P_0</script> is <script type="math/tex" id="MathJax-Element-338">f(x_0, y_0)</script>. Hence with this approximation</p>


<p><script type="math/tex; mode=display" id="MathJax-Element-339">\frac{y_1 - y+0}{h} = f(x_0, y_0)</script></p>


<p><script type="math/tex; mode=display" id="MathJax-Element-340">y_1 = y_0 + hf(x_0, y_0)</script> </p>

<p>Using this approximation to obtain the value of <script type="math/tex" id="MathJax-Element-341">y</script> at <script type="math/tex" id="MathJax-Element-342">P_1</script>, the process can be repeated, assuming that the part of the curve between <script type="math/tex" id="MathJax-Element-343">P_1</script> and <script type="math/tex" id="MathJax-Element-344">P_2</script> is a straight line segment with gradient equal to the value of <script type="math/tex" id="MathJax-Element-345">\frac{dy}{dx}</script> at <script type="math/tex" id="MathJax-Element-346">P_1</script>. <br>
This gives <script type="math/tex" id="MathJax-Element-347">y_2 = y_1 + hf(x_1, y_1)</script>. <br>
Continuing in this manner we have the general form</p>


<p><script type="math/tex; mode=display" id="MathJax-Element-348">y_{r+1} = y_r + hf(x_r, y_r), \quad r \in \mathbb{Z}^+_0</script></p>

<p>This is Euler’s formula. Successive calculation of values of <script type="math/tex" id="MathJax-Element-349">y</script> using this formula is known as Euler’s method. It is clear from the nature of the linear approximation on which this method is based that the step length <script type="math/tex" id="MathJax-Element-350">h</script> needs to be relatively small in order to achieve reasonable accuracy.</p>


<h3 id="example-4">Example</h3>

<blockquote>
  <p>The function <script type="math/tex" id="MathJax-Element-351">y(x)</script> satisfies the differential equation <script type="math/tex" id="MathJax-Element-352">\frac{dy}{dx} = \sin(x + y)</script> and the condition <script type="math/tex" id="MathJax-Element-353">y(1) = 0</script>. <br>
  Use Euler’s formula to estimate the values of <script type="math/tex" id="MathJax-Element-354">y(1.1), y(1.2), y(1.3)</script>, giving the answers to four decimal places.</p>

  <p>In this case <script type="math/tex" id="MathJax-Element-355">x_0 = 1</script>, <script type="math/tex" id="MathJax-Element-356">y_0 = 0</script>, <script type="math/tex" id="MathJax-Element-357">h = 0.1</script> and <script type="math/tex" id="MathJax-Element-358">f(x, y) = \sin(x + y)</script>. <br>
   Also <script type="math/tex" id="MathJax-Element-359">y(1.1) = y_1</script>, <script type="math/tex" id="MathJax-Element-360">y(1.2) = y_2</script>, and <script type="math/tex" id="MathJax-Element-361">y(1.3) = y_3</script>. <br>
   Using Euler’s formula <br>
   <script type="math/tex; mode=display" id="MathJax-Element-362">\begin{align}y_1 &= y_0 + h\sin(x_0 + y_0) \\
    &= 0.1\sin(1) \\
    &\approx 0.08414709848\\
    &\approx 0.0841\ (\text{4 d.p})\end{align}</script> <br>
  <script type="math/tex; mode=display" id="MathJax-Element-363">\begin{align}y_2 &= y_1+ h\sin(x_1 + y_1) \\
    &= 0.08414709848 +0.1\sin(1.1 + 0.08414709848) \\
    &\approx 0.1767648775\\
    &\approx 0.1768\ (\text{4 d.p})\end{align}</script> <br>
  <script type="math/tex; mode=display" id="MathJax-Element-364">\begin{align}y_3 &= y_2+ h\sin(x_2 + y_2) \\
    &= 0.1767648775 +0.1\sin(1.2 + 0.1767648775) \\
    &\approx 0.2748883657\\
    &\approx 0.2749\ (\text{4 d.p})\end{align}</script></p>
</blockquote>

<p>Note that at each stage after the first, the previous value of <script type="math/tex" id="MathJax-Element-365">y</script> should be used to the maximum available precision.</p>


<h2 id="the-mid-point-formula">The Mid-Point formula</h2>

<p>Consider three points, <script type="math/tex" id="MathJax-Element-366">P_{r-1}</script>, <script type="math/tex" id="MathJax-Element-367">P_r</script>, and <script type="math/tex" id="MathJax-Element-368">P_{r+1}</script> on part of a curve representing the solution of the differential equation <script type="math/tex" id="MathJax-Element-369">\frac{dy}{dx} = f(x,y)</script>. <br>
Provided that the three points are reasonably close together, the gradient of the line segment joining <script type="math/tex" id="MathJax-Element-370">P_{r-1}</script> and <script type="math/tex" id="MathJax-Element-371">P_{r+1}</script> will be approximately the same as the gradient of the tangent to the curve at <script type="math/tex" id="MathJax-Element-372">P_r</script>.</p>

<p>Since <script type="math/tex" id="MathJax-Element-373">\frac{dy}{dx} = f(x_r, y_r)</script> at <script type="math/tex" id="MathJax-Element-374">P_r</script>, this approximation gives</p>


<p><script type="math/tex; mode=display" id="MathJax-Element-375">\frac{y_{r+1} - y_{r-1}}{2h} = f(x_r, y_r)</script></p>


<p><script type="math/tex; mode=display" id="MathJax-Element-376">y_{r+1} = y_{r-1} + 2hf(x_r, y_r)</script></p>

<p>Since only the value of <script type="math/tex" id="MathJax-Element-377">y_{r-1}</script>  is known, it is necessary t calculate <script type="math/tex" id="MathJax-Element-378">y_r</script> by another means.</p>


<h3 id="example-5">Example</h3>

<blockquote>
  <p>The function <script type="math/tex" id="MathJax-Element-379">y(x)</script> satisfies the differential equation <script type="math/tex" id="MathJax-Element-380">\frac{dy}{dx} = 1 + \sqrt{xy}</script> and the condition <script type="math/tex" id="MathJax-Element-381">y(0) = 1</script>. <br>
  Use the mid-point formula with a step length of <script type="math/tex" id="MathJax-Element-382">0.25</script> to obtain an approximate value for <script type="math/tex" id="MathJax-Element-383">y(1)</script> to three decimal places. </p>

  <p><script type="math/tex" id="MathJax-Element-384">x_0 = 0,\ y_0 =1,\ h=0.25,\ f(x, y) = 1 + \sqrt{xy}</script> </p>

  <p>Euler’s formula gives <br>
   <script type="math/tex; mode=display" id="MathJax-Element-385">\begin{align}y(0.25) = y_1 &= y_0 +h(1 + \sqrt{x_0 y_0}) \\
   &= 1 + 0.25 \times 1 \\
   &= 1.25\end{align}</script> <br>
  The mid-point formula gives <br>
  <script type="math/tex; mode=display" id="MathJax-Element-386"> \begin{alignat*}{3}
    &y(0.5) &&=y_2 &&&=y_1 + 2h(1 + \sqrt{x_1 y_1}) &&&&\\
		    &\ &&\  &&&= 1+ 0.5(1 + \sqrt{0.25 \times 1.25}) &&&&\\
		    &\ &&\ &&&\approx 1.77951 &&&&\\
    &y(0.75) &&=y_3 &&&= 2h(1+ \sqrt{x_2 y_2}) &&&&\\
		    &\ &&\  &&&= 1.25 + 0.5(1 + \sqrt{0.5 \times 1.77951}) &&&& \\
		    &\ &&\ &&&\approx 2.22163 &&&&\\
    &y(1) &&=y_4 &&&= y_2 + 2h(1 + \sqrt{x_3 y_3}) &&&& \\
            &\ &&\  &&&= 1.77951 + 0.5(1 + \sqrt{0.75 \times 2.22163})&&&& \\
		    &\ &&\ &&&\approx 2.92492 &&&&\\
\end{alignat*}</script> </p>
</blockquote>


<h2 id="the-improved-euler-formula">The Improved Euler formula</h2>

<p>Another formula can be obtained by assuming that the gradient of <script type="math/tex" id="MathJax-Element-387">P_r P_{r+1}</script> is equal to the average of the values of <script type="math/tex" id="MathJax-Element-388">\frac{dy}{dx}</script> at <script type="math/tex" id="MathJax-Element-389">P_r</script> and <script type="math/tex" id="MathJax-Element-390">P_{r+1}</script>.</p>

<p>With this assumption</p>


<p><script type="math/tex; mode=display" id="MathJax-Element-391">\frac{y_{r+1} - y_r}{h} = \frac 1 2 [f(x_r, y_r) + f(x_{r+1}, y_{r+1})]</script></p>

<p>Hence </p>


<p><script type="math/tex; mode=display" id="MathJax-Element-392">y_{r+1} = y_r + \frac h 2 [f(x_r, y_r) + f(x_{r+1}, y_{r+1}^*)]</script></p>

<p>Where <script type="math/tex" id="MathJax-Element-393">y_{r+1}^*</script> is an approximation for <script type="math/tex" id="MathJax-Element-394">y_{r+1}</script>, often obtained by Euler’s formula. </p>


<h3 id="example-6">Example</h3>

<blockquote>
  <p>The function <script type="math/tex" id="MathJax-Element-395">y(x)</script> satisfies the differential equation <script type="math/tex" id="MathJax-Element-396">\frac{dy}{dx} = \ln(x+y), \quad x \geq 1</script>. <br>
  The boundary condition is <script type="math/tex" id="MathJax-Element-397">y(1) = 1</script>. Use the improved Euler formula with a step length of <script type="math/tex" id="MathJax-Element-398">0.25</script> to estimate the value of <script type="math/tex" id="MathJax-Element-399">y(1.5)</script></p>

  <p><script type="math/tex" id="MathJax-Element-400">x_0 = 1,\ y_0 = 1, \ h = 0.25, \text{and}\ f(x,y) = \ln(x+y)</script>  <br>
   the first estimate of <script type="math/tex" id="MathJax-Element-401">y_1</script> is obtained by Euler’s formula <br>
   <script type="math/tex; mode=display" id="MathJax-Element-402">\begin{align}y_1^* &= y_0 + h\ln(x_0 + y_0) \\
    &= 1 + 0.25\ln(2) \\
    &\approx 1.17329\end{align}</script> <br>
  For <script type="math/tex" id="MathJax-Element-403">y_{1}</script> we then have <br>
  <script type="math/tex; mode=display" id="MathJax-Element-404">\begin{align}y_1 &= y_0 + \frac h 2 [\ln(x_0 + y_0) + \ln(x_1 + y_1)] \\
  &= 1 + 0.125[\ln(2) + \ln(1.25 + 1.19728)] \\
  &\approx 1.42102\end{align}</script> <br>
  We can now use Euler’s formula with the improved value of <script type="math/tex" id="MathJax-Element-405">y_1</script> to give <br>
  <script type="math/tex; mode=display" id="MathJax-Element-406">\begin{align}y_2^* &= y_1 + h\ln(x_1 +y_1) \\
    &= 1.19728 + 0.25\ln(1.25 + 1.19728) \\
    &\approx 1.42102\end{align}</script> <br>
  And finally  <br>
  <script type="math/tex; mode=display" id="MathJax-Element-407">\begin{align}y_2 &= y_1 + \frac h 2 [\ln(x_1 + y_1) + \ln(x_2 + y_2^*)] \\
   &= 1.19728 + 0.125[\ln(1.25 + 1.19728) + \ln(1.5  1.42102)] \\
   &\approx 1.443\end{align}</script></p>
</blockquote>


<h2 id="alternative-calculation-for-the-improved-euler-method">Alternative calculation for the improved Euler method</h2>

<p>By substituting the the expression for <script type="math/tex" id="MathJax-Element-408">y^*_{r+1}</script> into the expression for <script type="math/tex" id="MathJax-Element-409">y_{r+1}</script> we achieve</p>


<p><script type="math/tex; mode=display" id="MathJax-Element-410">y_{r+1} = y_r + \frac h 2 [f(x_r, y_r) + f(x_{r+1}, y_r + hf(x_r, y_r))]</script></p>

<p>Noting that <script type="math/tex" id="MathJax-Element-411">x_{r+1} = x_r +h </script> this can be expressed as</p>


<p><script type="math/tex; mode=display" id="MathJax-Element-412">y_{r+1} = y_r + \frac 1 2(k_1 + k_2)</script></p>

<p>where <script type="math/tex" id="MathJax-Element-413">k_1 = hf(x_r, y_r)</script> <br>
and <script type="math/tex" id="MathJax-Element-414">k_2 = hf(x_r + h, y_r + k_1)</script></p>


<h3 id="example-7">Example</h3>

<blockquote>
  <p>The function <script type="math/tex" id="MathJax-Element-415">y(x)</script> satisfies the differential equation <script type="math/tex" id="MathJax-Element-416">\frac{dy}{dx} = f(x, y)</script> where <script type="math/tex" id="MathJax-Element-417">f(x, y) = 2x^3 + 3 -2y, \quad x \geq 0</script>.  <br>
  The boundary condition is <script type="math/tex" id="MathJax-Element-418">y(0) = 5</script> </p>

  <p>a) Use the improved Euler method, with <script type="math/tex" id="MathJax-Element-419">h = 0.25</script>, to calculate approximations to the values of <script type="math/tex" id="MathJax-Element-420">y(0.25)</script> and <script type="math/tex" id="MathJax-Element-421">y(0.5)</script>. </p>

  <p>Setting out the solution in a tabular form</p>

  <table>
<thead>
<tr>
  <th><script type="math/tex" id="MathJax-Element-422">r</script></th>
  <th><script type="math/tex" id="MathJax-Element-423">x_r</script></th>
  <th><script type="math/tex" id="MathJax-Element-424">y_r</script></th>
  <th><script type="math/tex" id="MathJax-Element-425">k_1</script></th>
  <th><script type="math/tex" id="MathJax-Element-426">x_r + h</script></th>
  <th><script type="math/tex" id="MathJax-Element-427">y_r + k_1</script></th>
  <th><script type="math/tex" id="MathJax-Element-428">k_2</script></th>
  <th><script type="math/tex" id="MathJax-Element-429">y_{r+1}</script></th>
</tr>
</thead>
<tbody><tr>
  <td><script type="math/tex" id="MathJax-Element-430">0</script></td>
  <td><script type="math/tex" id="MathJax-Element-431">0</script></td>
  <td><script type="math/tex" id="MathJax-Element-432">5</script></td>
  <td><script type="math/tex" id="MathJax-Element-433">-1.75</script></td>
  <td><script type="math/tex" id="MathJax-Element-434">0.25</script></td>
  <td><script type="math/tex" id="MathJax-Element-435">3.25</script></td>
  <td><script type="math/tex" id="MathJax-Element-436">-0.84375</script></td>
  <td><script type="math/tex" id="MathJax-Element-437">3.70125</script></td>
</tr>
<tr>
  <td><script type="math/tex" id="MathJax-Element-438">1</script></td>
  <td><script type="math/tex" id="MathJax-Element-439">0.25</script></td>
  <td><script type="math/tex" id="MathJax-Element-440">3.703125</script></td>
  <td><script type="math/tex" id="MathJax-Element-441">-1.0703125</script></td>
  <td><script type="math/tex" id="MathJax-Element-442">0.5</script></td>
  <td><script type="math/tex" id="MathJax-Element-443">2.638125</script></td>
  <td><script type="math/tex" id="MathJax-Element-444">-0.44140625</script></td>
  <td><script type="math/tex" id="MathJax-Element-445">2.947266</script></td>
</tr>
</tbody></table>


  <p>b) The exact solution to the differential equation is <script type="math/tex" id="MathJax-Element-446">y(x) = 3e^{-2x} + x^2 -x + 2</script>. <br>
  Calculate, to two significant figures, the percentage error in the value of <script type="math/tex" id="MathJax-Element-447">y(0.5)</script> obtained in part (a).</p>

  <p><script type="math/tex" id="MathJax-Element-448">y(0.5) = 3e^{-1} + 0.5^2 - 0.5 + 2 \approx 2.853638</script>  <br>
   Hence the error is  <br>
   <script type="math/tex" id="MathJax-Element-449">\dfrac{2.942766 - 2.853638}{2.853638} \times 100\% \approx 3.3 \%</script></p>
</blockquote>

<h2 id="error-analysis">Error analysis</h2>

<p>The three numerical formulae were derived by geometrical considerations.  <br>
They can also be derived analytically.</p>

<p>The Maclaurin series for <script type="math/tex" id="MathJax-Element-452">y(h)</script> is</p>


<p><script type="math/tex; mode=display" id="MathJax-Element-490">y(h) = y(0) + hy'(0) + \frac{h^2}{2!}y''(0) + \frac{h^3}{3!}y'''(0) +\ ...,</script></p>

<p>If the origin is transferred to the point <script type="math/tex" id="MathJax-Element-491">x_r</script>, this becomes</p>


<p><script type="math/tex; mode=display" id="MathJax-Element-521">y(x_r + h) = y(x_r) + hy'(x_r) + \frac{h^2}{2!}y''(x_r) + \frac{h^3}{3!}y'''(x_r) +\ ...,</script></p>

<p>which is Taylor’s series.</p>

<p>Noting that <script type="math/tex" id="MathJax-Element-522">y'(x_r) = f(x_r, y_r)</script>, the Taylor series may be expressed as</p>


<p><script type="math/tex; mode=display" id="MathJax-Element-760">y_{r+1} = y_r + hf(x_r, y_r) + \frac{h^2}{2!}y''(x_r) + \frac{h^3}{3!}y'''(x_r) +\ ..., </script></p>

<p>If <script type="math/tex" id="MathJax-Element-761">h</script> is assumed to be sufficiently small for terms in <script type="math/tex" id="MathJax-Element-762">h^2,\ h^3,\ ...</script> to be negligible, then the series reduces to </p>

<p><script type="math/tex" id="MathJax-Element-763">y_{r+1} = y_r + hf(x_r, y_r)</script></p>

<p>which is Euler’s formula.</p>

<p>Taylor’s series can also be used to step backwards from <script type="math/tex" id="MathJax-Element-764">x_r</script> to <script type="math/tex" id="MathJax-Element-765">x_{r-1}</script>, <script type="math/tex" id="MathJax-Element-766">(r \geq 1)</script>, by replacing <script type="math/tex" id="MathJax-Element-767">h</script> with <script type="math/tex" id="MathJax-Element-768">-h</script>.</p>

<p>This gives</p>


<p><script type="math/tex; mode=display" id="MathJax-Element-812">y_{r-1} = y_r -hf(x_r, y_r) + \frac{h^2}{2!}y''(x_r) - \frac{h^3}{3!}y'''(x_r)+\ ..., </script> </p>

<p>Subtracting this from the series for <script type="math/tex" id="MathJax-Element-813">y_{r+1}</script> and neglecting all terms in $h^3£ and higher powers gives</p>


<p><script type="math/tex; mode=display" id="MathJax-Element-842">y_{r+1} = y_{r-1} + 2hf(x_r, y_r), \quad (r \geq 1)</script></p>

<p>which is the mid-point formula.</p>

<p>Finally, to obtain the improved Euler formula, first replace <script type="math/tex" id="MathJax-Element-843">y</script> by <script type="math/tex" id="MathJax-Element-844">y'</script> in the Taylor series, giving</p>


<p><script type="math/tex; mode=display" id="MathJax-Element-874">y'(x_r +h) = y'(x_r) + hy''(x_r) + \frac{h^2}{2!}y'''(x_r)+\ ...</script></p>

<p>which may be expressed as </p>


<p><script type="math/tex; mode=display" id="MathJax-Element-910">f(x_{r+1}, y_{r+1}) = f(x_r, y_r) + hy''(x_r) + \frac{h^2}{2!}y'''(x_r)+\ ...</script></p>

<p>Multiplying each side by <script type="math/tex" id="MathJax-Element-911">\frac h 2</script> and subtracting the result from the Taylor series for <script type="math/tex" id="MathJax-Element-912">y_{r+1}</script> gives</p>


<p><script type="math/tex; mode=display" id="MathJax-Element-951">y_{r+1} - \frac h 2 f(x_{r+1}, y_{r+1}) = y_r + \frac h 2 f(x_r, y_r) + \text{terms  in } h^3 \text{ and higher powers}</script></p>

<p>Neglecting the terms in <script type="math/tex" id="MathJax-Element-952">h^3</script> and higher powers we have</p>


<p><script type="math/tex; mode=display" id="MathJax-Element-1264">y_{r+1} = y_r + \frac h 2 [f(x_r, y_r), + f(x_{r+1}, y_{r+1})]</script></p>

<p>The error incurred by neglecting terms is called the truncation error. <br>
The derivations show that the truncation error in Euler’s formula is greater than that in the mid-point formula and the improved Euler formula because the extra term <script type="math/tex" id="MathJax-Element-1265">h^2</script> is neglected.</p>

<p>In addition to the truncation area, for all values of <script type="math/tex" id="MathJax-Element-1266">r \geq 1</script>, another error will be incurred in the calculation of <script type="math/tex" id="MathJax-Element-1267">y_{r+1}</script> because an approximate value of <script type="math/tex" id="MathJax-Element-1268">y_r</script> has to be used in the evaluation of <script type="math/tex" id="MathJax-Element-1269">f(x_r, y_r)</script>. As the calculation of <script type="math/tex" id="MathJax-Element-1270">y</script> values progresses, errors of this type sometimes decay, however they can also become increasingly larger, in which case the method is said to be unstable.</p>

<p>To estimate the order of accuracy of a solution, the usual procedure is to reduce the step length to half its previous value and repeat the calculations. It is usually safe to assume that the values are accurate to the number of decimal places which are the same in the two values.</p>

<h1 id="second-order-differential-equations">Second order differential equations</h1>

<h2 id="complex-numbers">Complex numbers</h2>

<p>The set of all complex numbers is denoted by <script type="math/tex" id="MathJax-Element-1358">\mathbb{C}</script>. When the coefficient of the imaginary component is 0, <script type="math/tex" id="MathJax-Element-1359">a +  \mathrm{i}b</script> reduces to the real value <script type="math/tex" id="MathJax-Element-1360">a</script>, so <script type="math/tex" id="MathJax-Element-1361">\mathbb{C}</script> includes the set <script type="math/tex" id="MathJax-Element-1362">\mathbb{R}</script> of real numbers.</p>

<h3 id="eulers-identity">Euler’s identity</h3>

<p><script type="math/tex; mode=display" id="MathJax-Element-1388">e^{\mathrm{i}x} = \cos(x) + \mathrm{i}\sin(x), \quad x \in \mathbb{R}</script></p>

<p>The identity can be proved by series expansions.</p>


<p><script type="math/tex; mode=display" id="MathJax-Element-1573">e^{\mathrm{i}x} = 1 + \mathrm{i}x + \frac{(\mathrm{i}x)^2}{2!} + \frac{(\mathrm{i}x)^3}{3!} + \frac{(\mathrm{i}x)^4}{4!} + \frac{(\mathrm{i}x)^5}{5!} +\ ...</script></p>

<p><script type="math/tex" id="MathJax-Element-1574">\mathrm{i}^2 = -1,\ \mathrm{i}^3 = -\mathrm{i},\ \mathrm{i}^4 = 1,\ \mathrm{i}^5 = \mathrm{i}</script> and the same sequence of results <script type="math/tex" id="MathJax-Element-1575">(-1,\ -\mathrm{i},\ 1,\ \mathrm{i})</script> is generated repeatedly for all higher integer powers. <br>
Substituting these expressions above and collecting together real and imaginary values gives</p>


<p><script type="math/tex; mode=display" id="MathJax-Element-1612">e^{\mathrm{i}x} = \left(1 - \frac{x^2}{2!} + \frac{x^4}{4!} -\ ... \right) + \mathrm{i}\left(x - \frac{x^3}{3!} + \frac{x^5}{5!} -\ ... \right) </script></p>

<p>Hence</p>


<p><script type="math/tex; mode=display" id="MathJax-Element-1623">e^{\mathrm{i}x} \equiv \cos(x) + \mathrm{i}\sin(x), \quad x \in \mathbb{R}</script></p>

<h4 id="example-8">Example</h4>

<blockquote>
  <p>The function <script type="math/tex" id="MathJax-Element-2638">y(x)</script> is given by</p>

  <p><script type="math/tex; mode=display" id="MathJax-Element-2639">y(x) = Ae^{(2 + 3\mathrm{i})x} + Be^{(2-3\mathrm{i})x}</script> <br>
   where <script type="math/tex" id="MathJax-Element-2640">A</script> and <script type="math/tex" id="MathJax-Element-2641">B</script> are constants. Show that this can be expressed as  <br>
   <script type="math/tex; mode=display" id="MathJax-Element-2642">y(x) = e^{2x}(C\cos(3x) + D\sin(3x))</script> <br>
   where <script type="math/tex" id="MathJax-Element-2643">C</script> and <script type="math/tex" id="MathJax-Element-2644">D</script> are constants.</p>

  <p><script type="math/tex; mode=display" id="MathJax-Element-2645">\begin{align}y(x) &= A\left(Ae^{2x} \times e^{3\mathrm{i}x} \right) + B\left(Ae^{2x} \times e^{-3\mathrm{i}x} \right) \\
   &= Ae^{2x}\left(\cos(3x) + \mathrm{i}\sin(3x) \right) + Be^{2x}\left(\cos(3x) - \mathrm{i}\sin(3x) \right) \\
   &= e^{2x} \left[(A+B)\cos(3x) + (A-B)\mathrm{i}\sin(3x) \right]\end{align}</script> <br>
  Writing <script type="math/tex" id="MathJax-Element-2646">A + B = C</script> and <script type="math/tex" id="MathJax-Element-2647">(A-B)\mathrm{i} = D</script>, this becomes </p>

  <p><script type="math/tex; mode=display" id="MathJax-Element-2648">y(x) = e^{2x}(C\cos(3x) + D\sin(3x))</script></p>
</blockquote></div></body>
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