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<!-- Header -->
<div class="header">
<button class="toc-btn" onclick="toggleTOC()">☰</button>
<span class="header-title">Exponentials</span>
<span class="page-indicator"><span id="currentPageNum">1</span> / 30</span>
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<!-- Content -->
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<div class="pages-container" id="pagesContainer">
<!-- Page 1: Title page -->
<div class="page" data-page="1">
<div class="title-page">
<h1>Exponentials and Logarithms</h1>
<div class="subtitle">part 1: what is \(e\,\)?</div>
<div class="author">Dr Brian Brooks<br><a href="http://www.mathsinsight.co.uk">www.mathsinsight.co.uk</a></div>
</div>
</div>
<!-- Page 2 -->
<div class="page" data-page="2">
<h2 class="chapter">Part 1: the number \(\mathbf{e}\)</h2>
<h2 class="section">what is \(\mathbf{e}\) anyway?</h2>
<div class="intro">Work on the question in yellow.
When you are ready, navigate to the next page using arrow keys,
swiping, or hovering near the bottom of the page.
Compare your work to the answer in green.
Don't go to the page too soon!
Work your way through the whole document this way,
but take your time!
</div>
<h2 class="section">what is \(\mathbf{e}\) anyway?</h2>
<div class="intro">\(e\) is a number. A very special number, to be sure, but just a number none the less. It is absolutely crucial to so many areas of mathematics: there is, for example, little in physics that we would be able to do without knowing about \(e\), and nothing whatsoever in studies of growth, decay, and populations.</div>
<h2 class="section">\(\mathbf{e}\) by exponential graphs</h2>
<div class="intro">
<p>There are many ways to start thinking about \(e\). Here's one:</p>
</div>
<div class="exercise">Sketch the graph \(y=2^{x}\).</div>
</div>
<!-- Page 3 -->
<div class="page" data-page="3">
<div class="answer">Here is the graph \(y=2^{x}\) with an added line.</div>
<figure><img src="images/image1.png" alt="Diagram"></figure>
<div class="exercise">What is the gradient of the red line?</div>
</div>
<!-- Page 4 -->
<div class="page" data-page="4">
<figure><img src="images/image1.png" alt="Diagram"></figure>
<div class="answer">
<p>\[\begin{aligned}
\text{gradient}&=\frac{y-\text{step}}{x-\text{step}}\\[1em]
&=\frac{2^{h}-1}{h}
\end{aligned}\]</p>
</div>
</div>
<!-- Page 5 -->
<div class="page" data-page="5">
<div class="answer">Here is the graph \(y=2^{x}\) with a tangent added at the red cross.</div>
<figure><img src="images/image2.png" alt="Diagram"></figure>
<div class="exercise">Use the video below to help you understand and explain
why the gradient of the tangent at \(\left(0,1\right)\) is
\(\displaystyle{\lim_{h\to0}\frac{2^{h}-1}{h}}\)</div>
<figure>
<video autoplay loop muted playsinline><source src="images/gif1.mp4" alt="Diagram"></video>
</figure>
</div>
<!-- Page 6 -->
<div class="page" data-page="6">
<div class="exercise">Explain why the gradient of the tangent at \(\left(0,1\right)\) is
\(\displaystyle{\lim_{h\to0}\frac{2^{h}-1}{h}}\)</div>
<figure>
<video autoplay loop muted playsinline><source src="images/gif1.mp4" alt="Diagram"></video>
</figure>
<div class="answer">\[\begin{aligned}
\text{As }h\to0,&\text{ green dot approaches red cross}\\[1em]
\Rightarrow\;&\text{red line approaches blue tangent}\\[1em]
\Rightarrow\;&\text{red gradient}\to\text{gradient of tangent} \\[1em]
\Rightarrow\;&\frac{2^{h}-1}{h}\to\text{ gradient of tangent} \\[1em]
\Rightarrow\;&\text{gradient of tangent}=\lim_{h\to0}\frac{2^{h}-1}{h}
\end{aligned}\]</div>
</div>
<!-- Page 7 -->
<div class="page" data-page="7">
<div class="exercise">
<p>Use a calculator and increasingly small values of \(h\) to find an approximate
value for \(\displaystyle{\lim_{h\to0}\frac{2^{h}-1}{h}}\)</p>
<p>Use this to find an approximate value for the gradient of the tangent
to the curve \(y=2^x\) at the point \((0,\,1)\)
</p>
</div>
</div>
<!-- Page 8 -->
<div class="page" data-page="8">
<div class="exercise">
<p>Use a calculator and increasingly small values of \(h\) to find an approximate
value for \(\displaystyle{\lim_{h\to0}\frac{2^{h}-1}{h}}\)</p>
<p>Use this to find an approximate value for the gradient of the tangent
to the curve \(y=2^x\) at the point \((0,\,1)\)
</p>
</div>
<div class="answer">
<p>when \(\displaystyle{h=0.01,\;\frac{2^{h}-1}{h}\approx 0.696}\)</p>
<p>when \(\displaystyle{h=0.0001,\;\frac{2^{h}-1}{h}\approx 0.693}\)</p>
<p>when \(\displaystyle{h=0.0000001,\;\frac{2^{h}-1}{h}\approx 0.693}\)</p>
<p>\[\Rightarrow \lim_{h\to 0}\frac{2^{h}-1}{h}\approx 0.693\]</p>
<p>\[\Rightarrow \text{gradient of tangent at }(0,\,1)\approx 0.693\]</p>
</div>
</div>
<!-- Page 9 -->
<div class="page" data-page="9">
<div class="exercise">Here are the graphs \(y=2^{x}\) and \(y=3^{x}\). Which is which, and why do they
cross over each other on the \(y\) axis?
</div>
<figure><img src="images/image3.png" alt="Diagram"></figure>
<div class="exercise">Draw a tangent to the curve \(y=3^{x}\) at the point \(\left(0,1\right)\).</div>
<div class="exercise">What is the formula for the gradient of this tangent?</div>
</div>
<!-- Page 10 -->
<div class="page" data-page="10">
<figure>
<div class="row">
<div class="column">
<img src="images/image21.png" alt="image">
</div>
<div class="column">
<img src="images/image22.png" alt="image">
</div>
</div>
</figure>
<div class="answer">gradient\(=\displaystyle{\lim_{h\to0}\frac{3^{h}-1}{h}}\)</div>
<div class="exercise">Use your calculator to find an approximate value of the gradient
of the tangent to the curve \(y=3^x\) at the point \((0,\,1)\).</div>
</div>
<!-- Page 11 -->
<div class="page" data-page="11">
<div class="exercise">Use your calculator to find an approximate value of the gradient
of the tangent to the curve \(y=3^x\) at the point \((0,\,1)\).</div>
<div class="answer">
<p>when \(\displaystyle{h=0.01,\;\frac{3^{h}-1}{h}\approx 1.1047}\)</p>
<p>when \(\displaystyle{h=0.0001,\;\frac{3^{h}-1}{h}\approx 1.099}\)</p>
<p>when \(\displaystyle{h=0.0000001,\;\frac{3^{h}-1}{h}\approx 1.099}\)</p>
<p>\[\Rightarrow \lim_{h\to 0}\frac{3^{h}-1}{h}\approx 1.099\]</p>
</div>
<div class="intro">
<p>At the point \(\left(0,1\right)\), the gradient of the curve \(y=2^{x}\) is less than \(1\), whereas the gradient of the curve \(y=3^{x}\) is more than \(1\).</p>
<p>So somewhere between \(2\) and \(3\) there should be a number \(a\) that makes the gradient of the curve \(y=a^{x}\) equal to \(1\). This is the number that we call \(e\).</p>
</div>
<div class="exercise">What is the gradient of the tangent to the curve
\(y=a^{x}\) at the point \((0,\,1)\) in terms of \(a\)?</div>
</div>
<!-- Page 12 -->
<div class="page" data-page="12">
<div class="exercise">What is the gradient of the tangent to the curve
\(y=a^{x}\) at the point \((0,\,1)\) in terms of \(a\)?</div>
<div class="answer">The same process that we used when \(a=2\)
and \(a=3\) will show that
gradient\(=\displaystyle{\lim_{h\to0}\frac{a^{h}-1}{h}}\)</div>
<div class="exercise">Experiment with your calculator to find the value of \(a\) that makes this gradient as close to \(1\) as you can.</div>
</div>
<!-- Page 13 -->
<div class="page" data-page="13">
<div class="exercise">Experiment with your calculator to find the
value of \(a\) that makes this gradient \(=\displaystyle{\lim_{h\to0}\frac{a^{h}-1}{h}}\)
as close to \(1\) as you can.</div>
<div class="answer">
<p>taking \(\displaystyle{h=0.0000001}\)</p>
<p>when \(\displaystyle{a=2.7,\;\frac{a^{h}-1}{h}\lt 1}\)</p>
<p>when \(\displaystyle{a=2.71,\;\frac{a^{h}-1}{h}\lt 1}\)</p>
<p>when \(\displaystyle{a=2.72,\;\frac{a^{h}-1}{h}\gt 1}\)</p>
<p>\[\Rightarrow 2.71\lt a\lt 2.72\]</p>
<p>when \(\displaystyle{a=2.717,\;\frac{a^{h}-1}{h}\lt 1}\)</p>
<p>when \(\displaystyle{a=2.718,\;\frac{a^{h}-1}{h}\lt 1}\)</p>
<p>when \(\displaystyle{a=2.719,\;\frac{a^{h}-1}{h}\gt 1}\)</p>
<p>\[\Rightarrow 2.718\lt a\lt 2.719\]</p>
</div>
</div>
<!-- Page 14 -->
<div class="page" data-page="14">
<figure><img src="images/image6.png" alt="Diagram"></figure>
<div class="key-concept">
<p>\(e\) is the number for which the gradient of
the curve \(y=e^{x}\) at the point \(\left(0,1\right)\) is \(1\).</p>
<p>Correct to 30 decimal places, \[e=2.718281828459045235360287471352\]</p>
<p>But \(e\) is irrational, so any attempt to write it as a decimal
can only be an approximation.</p>
<p>This property that the gradient of \(y=e^x\) at the point
\((0,\,1)\) is \(1\) may not seem that important. In fact, however,
it is this very property that will turn out to make \(e\) such
a crucial number in all physics (and other disciplines, too).</p>
<p>Next, we will begin to investigate why this is.</p>
</div>
</div>
<!-- Page 15 -->
<div class="page" data-page="15">
<h2 class="chapter">the exponential function</h2>
<div class="intro">
<p>Now we know a bit about the number \(e\),
it's time to think more deeply about the exponential function \(f(x)=e^{x}\), sometimes written \(\exp(x)\).</p>
<p>We started with graphs of the form \(y=a^{x}\), and we defined \(e\) as the number for which the gradient of \(y=e^{x}\) is \(1\) at the point \((0,1)\).</p>
<p>This led us to the definition of \(e\) as the number for which
\[\lim_{h\to0}\frac{e^{h}-1}{h}=1\]</p>
<p>Now, we are going to go back to graphs and use this
property to show that, if \(f(x)=e^{x}\), then \[f'(x)=f(x)\]
for all values of \(x\), not only when \(x=0\).</p>
</div>
</div>
<!-- Page 16 -->
<div class="page" data-page="16">
<div class="exercise">Here is the graph \(y=e^x\). What is the gradient of the red line?</div>
<figure><img src="images/part2_image1.png" alt="Diagram showing connections"></figure>
</div>
<!-- Page 17 -->
<div class="page" data-page="17">
<div class="exercise">Here is the graph \(y=e^x\). What is the gradient of the red line?</div>
<figure><img src="images/part2_image1.png" alt="Diagram showing connections"></figure>
<div class="answer">gradient\(=\displaystyle{\frac{e^{x+h}-e^{x}}{h}}\)</div>
<div class="exercise">Factorize the top of the fraction \(\displaystyle{\frac{e^{x+h}-e^{x}}{h}}\).</div>
</div>
<!-- Page 18 -->
<div class="page" data-page="18">
<div class="exercise">Factorize the top of the fraction \(\displaystyle{\frac{e^{x+h}-e^{x}}{h}}\).</div>
<div class="answer">\[\begin{aligned}
\text{gradient}&=\frac{e^{x+h}-e^{x}}{h}\\[2em]
&=\frac{e^x e^h-e^{x}}{h}\\[2em]
&=\frac{e^x \left(e^h-1\right)}{h}\\[2em]
&={e^{x}\cdot\frac{e^{h}-1}{h}}
\end{aligned}\]</div>
<div class="exercise">What is the gradient of the tangent at the point \((x,e^{x})\) as a limit?</div>
</div>
<!-- Page 19 -->
<div class="page" data-page="19">
<div class="exercise">What is the gradient of the tangent at the point \((x,e^{x})\) as a limit?</div>
<div class="answer">
\[\begin{aligned}
\text{gradient}&=\lim_{h\to 0}{e^{x}\cdot\frac{e^{h}-1}{h}}
\end{aligned}\]
</div>
<div class="exercise">
<p>Explain why this is the same as \[{e^{x}\lim_{h\to0}\frac{e^{h}-1}{h}}\]</p>
</div>
</div>
<!-- Page 20 -->
<div class="page" data-page="20">
<div class="exercise">
<p>Explain why \[\lim_{h\to 0}e^{x}\cdot\frac{e^{h}-1}{h}\] is the same as \[{e^{x}\lim_{h\to0}\frac{e^{h}-1}{h}}\]</p>
</div>
<div class="answer">
As \(h\) changes, \(\frac{e^{h}-1}{h}\) changes. If I multiply
this fraction by a number and then take the limit as \(h\to0\),
this is the same as taking the limit first and then multiplying by
a number. But, as far as \(h\) is concerned, \(e^x\) is just a number.
</div>
<div class="exercise">
If \(f(x)=e^{x}\), what is \(f'(x)\)?</div>
</div>
<!-- Page 21 -->
<div class="page" data-page="21">
<div class="exercise">
If \(f(x)=e^{x}\), what is \(f'(x)\)?</div>
<div class="answer">\[\begin{aligned}
f'(x)&=\lim_{h\to0}\frac{e^{x+h}-e^{x}}{h}\\[6pt]
&=e^{x}\lim_{h\to0}\frac{e^{h}-1}{h}\\[6pt]
&=e^{x}\times1\\[6pt]
&=e^{x}
\end{aligned}\]</div>
</div>
<!-- Page 22 -->
<div class="page" data-page="22">
<div class="key-concept">
<p>This gives us another characterization of \(e\):</p>
<p>It's the number for which the gradient of the curve \(y=e^{x}\) is always equal to the value of \(y\).</p>
</div>
<figure><img src="images/part2_image1.png" alt="Updated diagram"></figure>
</div>
<!-- Page 23 -->
<div class="page" data-page="23">
<h2 class="chapter">\(\mathbf{e}\) as a limit</h2>
<div class="exercise">
Use your calculator to find:
\[\begin{aligned}
\left(1+\frac{1}{2}\right)^{2}\\[2em]
\left(1+\frac{1}{3}\right)^{3}\\[2em]
\left(1+\frac{1}{4}\right)^{4}\\[2em]
\left(1+\frac{1}{5}\right)^{5}\\[2em]
\end{aligned}\]</div>
</div>
<!-- Page 24 -->
<div class="page" data-page="24">
<div class="answer">
\[\begin{aligned}
\left(1+\frac{1}{2}\right)^{2}=2.25\\[2em]
\left(1+\frac{1}{3}\right)^{3}\approx2.3704\\[2em]
\left(1+\frac{1}{4}\right)^{4}\approx2.4414\\[2em]
\left(1+\frac{1}{5}\right)^{5}\approx2.5216\\[2em]
\end{aligned}\]
</div>
<div class="exercise">On your calculator, investigate this expression as \(n\) gets larger:
\[\left(1+\frac{1}{n}\right)^{n}\]</div>
</div>
<!-- Page 25 -->
<div class="page" data-page="25">
<div class="exercise">On your calculator, investigate this expression as \(n\) gets larger:
\[\left(1+\frac{1}{n}\right)^{n}\]</div>
<div class="answer">
\[\begin{aligned}
\left(1+\frac{1}{10^4}\right)^{10^4}\approx2.7181\\[2em]
\left(1+\frac{1}{10^9}\right)^{10^4}\approx2.71828\\[2em]
\left(1+\frac{1}{4}\right)^{4}\approx2.4414\\[2em]
\end{aligned}\]
<p>and it looks as though the expression is getting increasingly close to \(e\)</p>
</div>
<div class="intro">
Next, we will figure out why it is that
\[\text{as }n\to\infty,\;\left(1+\frac{1}{n}\right)^{n}\to e\]
</div>
</div>
<!-- Page 26 -->
<div class="page" data-page="26">
<div class="exercise">
<p>Show that, if \(\displaystyle{\lim_{h\to0}\frac{e^{h}-1}{h}=1}\),
then for small values of \(h\),
\[e\approx\left(1+h\right)^{\frac{1}{h}}\]</p>
<p>Use this to show that
\[e\approx\left(1+\frac{1}{n}\right)^{n}\]and hence that
\[e=\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^{n}\]</p>
</div>
</div>
<!-- Page 27 -->
<div class="page" data-page="27">
<div class="exercise">
<p>Show that, if \(\displaystyle{\lim_{h\to0}\frac{e^{h}-1}{h}=1}\),
then for small values of \(h\),
\(e\approx\left(1+h\right)^{\frac{1}{h}}\)</p>
</div>
<div class="answer">\[\begin{aligned}
\text{For small values of }h,\;&\frac{e^{h}-1}{h}\approx1\\[1em]
\Rightarrow \;&e^{h}\approx1+h\\[1em]
\Rightarrow \;&e\approx\left(1+h\right)^{\frac{1}{h}}\\[1em]
\end{aligned}\]
</div>
<div class="exercise">
<p>Show that
\(e\approx\left(1+\frac{1}{n}\right)^{n}\) and hence that
\[e=\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^{n}\]</p>
</div>
<div class="answer">
\[\begin{aligned}
\text{Put } n=\frac{1}{h},\quad &h=\frac{1}{n}\quad \text{so that as }h\to 0,\;n\to\infty\\[1em]
\Rightarrow \;&e\approx\left(1+\frac{1}{n}\right)^{n}\\[1em]
\Rightarrow \;&e=\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^{n}
\end{aligned}\]</div>
<div class="intro">
<p>Now we are going to extend this to a similar result for \(e^{x}\).</p>
</div>
</div>
<!-- Page 28 -->
<div class="page" data-page="28">
<div class="exercise">
Remembering that \[e=\lim_{h\to 0}(1+h)^\frac{1}{h}\]
and using the substitution \(k=\frac{h}{x}\), show that
\[e^{x}=\lim_{h\to 0}(1+xh)^\frac{1}{h}\] and hence that
\[e^{x}=\lim_{n\to\infty}\left(1+\frac{x}{n}\right)^{n}\]</div>
</div>
<!-- Page 29 -->
<div class="page" data-page="29">
<div class="exercise">
Remembering that \(\displaystyle{e=\lim_{h\to 0}(1+h)^\frac{1}{h}}\)
and using the substitution \(k=\frac{h}{x}\),
\[\text{show that }e^{x}=\lim_{h\to 0}(1+xh)^\frac{1}{h}\text{ and hence that}\]
\[e^{x}=\lim_{n\to\infty}\left(1+\frac{x}{n}\right)^{n}\]</div>
<div class="answer">\[\begin{aligned}
e&=\lim_{h\to 0}(1+h)^\frac{1}{h}\\[1em]
\Rightarrow e^x&=\lim_{h\to 0}(1+h)^\frac{x}{h}\\[1em]
&=\lim_{h\to 0}(1+xk)^\frac{1}{k}\\[1em]
&=\lim_{k\to 0}(1+xk)^\frac{1}{k}\\[1em]
&=\lim_{n\to\infty}\left(1+\frac{x}{n}\right)^{n}
\end{aligned}\]</div>
</div>
<!-- Page 30: End -->
<div class="page" data-page="30">
<div class="title-page">
<h1>End of Part 1</h1>
<div class="subtitle">what is \(e\,\)?</div>
<div class="author">Dr Brian Brooks<br><a href="http://www.mathsinsight.co.uk">www.mathsinsight.co.uk</a></div>
</div>
</div>
</div>
</div>
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<h2>Contents</h2>
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<div class="toc-list" id="tocList">
<div class="toc-item" onclick="goToPage(1)">1. Title</div>
<div class="toc-item" onclick="goToPage(2)">2. What is e?</div>
<div class="toc-item" onclick="goToPage(3)">3. Graph y = 2^x</div>
<div class="toc-item" onclick="goToPage(4)">4. Gradient formula</div>
<div class="toc-item" onclick="goToPage(5)">5. Tangent at (0,1)</div>
<div class="toc-item" onclick="goToPage(6)">6. Limit explanation</div>
<div class="toc-item" onclick="goToPage(7)">7. Calculator exercise</div>
<div class="toc-item" onclick="goToPage(8)">8. Gradient approximation</div>
<div class="toc-item" onclick="goToPage(9)">9. Graphs 2^x and 3^x</div>
<div class="toc-item" onclick="goToPage(10)">10. Gradient of 3^x</div>
<div class="toc-item" onclick="goToPage(11)">11. Finding e</div>
<div class="toc-item" onclick="goToPage(12)">12. General formula</div>
<div class="toc-item" onclick="goToPage(13)">13. Approximating e</div>
<div class="toc-item" onclick="goToPage(14)">14. Definition of e</div>
<div class="toc-item" onclick="goToPage(15)">15. Exponential function</div>
<div class="toc-item" onclick="goToPage(16)">16. Graph y = e^x</div>
<div class="toc-item" onclick="goToPage(17)">17. Gradient of e^x</div>
<div class="toc-item" onclick="goToPage(18)">18. Factorization</div>
<div class="toc-item" onclick="goToPage(19)">19. Tangent gradient</div>
<div class="toc-item" onclick="goToPage(20)">20. Limit properties</div>
<div class="toc-item" onclick="goToPage(21)">21. Derivative of e^x</div>
<div class="toc-item" onclick="goToPage(22)">22. Key result</div>
<div class="toc-item" onclick="goToPage(23)">23. e as a limit</div>
<div class="toc-item" onclick="goToPage(24)">24. Sequence values</div>
<div class="toc-item" onclick="goToPage(25)">25. Limit investigation</div>
<div class="toc-item" onclick="goToPage(26)">26. Proof exercise</div>
<div class="toc-item" onclick="goToPage(27)">27. Limit derivation</div>
<div class="toc-item" onclick="goToPage(28)">28. e^x limit exercise</div>
<div class="toc-item" onclick="goToPage(29)">29. e^x limit proof</div>
<div class="toc-item" onclick="goToPage(30)">30. End</div>
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