tag:blogger.com,1999:blog-12436121906667373382020-06-29T02:41:09.621+03:00Various notes on math and programming...petrovhttp://www.blogger.com/profile/14750572373308546381noreply@blogger.comBlogger10125tag:blogger.com,1999:blog-1243612190666737338.post-91395638961041779782020-06-20T13:55:00.044+03:002020-06-29T02:41:09.503+03:00Mean value theorem proof illustratedI was rereading recently the proof of the <b><a href="https://en.wikipedia.org/wiki/Mean_value_theorem" target="_blank">mean value theorem</a></b> from math real analysis.<div><br /></div><div>This led me to the idea to generate some drawing which nicely illustrates the idea of the proof. Let's first restate the <b>mean value theorem</b>.</div><div><br /></div><div><b>Theorem:</b> If the function $f$ is defined and continuous in the closed interval $[a,b]$ and is differentiable in the open interval $(a,b)$, then there exists a point $\theta$ which is strictly between $a$ and $b$ i.e. $a \lt \theta \lt b$, such that&nbsp;</div><div><br /></div><div>$$f'(\theta) = \frac{f(b)-f(a)}{b-a} \tag{1}$$</div><div><br /></div><div>The proof constructs a function and I was wondering what the idea is behind that function. So I finally understood that and wanted to illustrate it here via some nice drawing. It took me some time to find a good looking function but OK... I finally picked this one:&nbsp;</div><div><div><br /></div><div>$$f(x) = \sin(t) - \sin^2(t/2) + \cos^3(t/5) \tag{2}$$</div></div><div><br /></div><div>Here is the drawing I generated.&nbsp;</div><div class="separator" style="clear: both; text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-4zlQ4ZPagQg/XvH0esN5y8I/AAAAAAAABUU/pjGaUlXABVAUinvBC3gXFphnifmTiM2fwCK4BGAsYHg/s1920/Figure_1.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="1103" data-original-width="1920" src="https://1.bp.blogspot.com/-4zlQ4ZPagQg/XvH0esN5y8I/AAAAAAAABUU/pjGaUlXABVAUinvBC3gXFphnifmTiM2fwCK4BGAsYHg/d/Figure_1.png" /></a></div><div class="separator" style="clear: both; text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: center;"><br /></div><div>The given function $f(x)$ is shown in red in the figure above.</div><div><br /></div><div>The proof of the theorem is quite nice. Main role in it plays the the line which connects the two endpoints of the graph of $f(x)$ i.e. the points $(a,f(a))$ and $(b,f(b))$. This line is represented by the function</div><div><br /></div><div>$$g(x) = f(a) + \frac{f(b)-f(a)}{b-a}(x-a) \tag{3}$$&nbsp;</div><div><br /></div><div>The line is shown in green in the figure above.</div><div><br /></div><div>The proof then goes on to construct the function $h(x) = f(x)-g(x)$ and for this one it can be easily seen that $h(a)=h(b)=0$. The proof then applies <b><a href="https://en.wikipedia.org/wiki/Rolle%27s_theorem" target="_blank">Rolle's theorem</a>&nbsp;</b>to the function $h$ to get the desired result.&nbsp;</div><div><br /></div><div>So I was wondering how this function $h(x)$looks like. It is shown in deep sky blue in the figure above. The interesting thing about this function $h(x)$ is that it measures (at each point $x$) what the difference is between $f(x)$ and $g(x)$.&nbsp;</div><div><br /></div><div>In simple words this can be formulated in 2 different ways:&nbsp;&nbsp;</div><div>a) At any value of $x$ the point $(x,h(x))$ is <b>as far from </b>the X axis, <b>as </b>$(x,f(x))$ <b>is from</b> $(x,g(x))$.&nbsp;</div><div>This follows from the fact that: $h(x) = f(x) - g(x)$ which can be informally stated as <i style="font-weight: bold;"><font color="#00bfff">blue</font> = <font color="#d52c1f">red</font> - <font color="#41b375">green</font></i>.<b>&nbsp;</b></div><div>b) At any value of $x$ the point $(x,f(x))$ is <b>as far from&nbsp;</b>$(x,h(x))$,&nbsp;<b>as</b> $(x,g(x))$ <b>is from </b>the&nbsp;X axis.</div><div>This follows from the fact that: $f(x) - h(x) = g(x)$ which can be informally stated as&nbsp;<i style="font-weight: bold;"><font color="#d52c1f">red&nbsp;</font></i><i style="font-weight: bold;">-&nbsp;</i><i style="font-weight: bold;"><font color="#00bfff">blue</font>&nbsp;=&nbsp;<font color="#41b375">green</font></i>.</div><div><br /></div><div><br /></div><div><br /></div>petrovhttp://www.blogger.com/profile/14750572373308546381noreply@blogger.com0tag:blogger.com,1999:blog-1243612190666737338.post-666865608513635462020-05-16T15:23:00.142+03:002020-05-17T15:12:18.345+03:00The function $f(x) = x^x$Let us look at this function&nbsp;<div><br /></div><div>$$f(x) = x^x \tag{1}$$</div><div><br /></div><div>and try to find its derivative for real values of $x$.&nbsp;</div><div><br /></div><div>Before this we should mention that this function is well defined only when $x \gt 0$.&nbsp;</div><div><br /></div><div>In other words... to avoid any complications we are looking at this function only for positive values of $x$.&nbsp;</div><div><br /></div><div><b>I. How do we go about finding the derivative $f'(x) = (x^x)'$ ?&nbsp;</b></div><div><br /></div><div>Well, we will use the following identity&nbsp;</div><div><br /></div><div>$$a^x = e^{x \cdot \ln a} \tag{2}$$</div><div><br /></div><div>which is well-known from high school math and which holds true for any $a \gt 0$ and any $x \in \mathbb{R}$.&nbsp;</div><div><br /></div><div>Substituting $a = x$ in this identity we subsequently get&nbsp;</div><div><br /></div><div>$$f(x) = x^x = e ^ {x \cdot \ln x}$$&nbsp;</div><div><br /></div><div>$$f'(x) = e ^ {x \cdot \ln x} \cdot (x \cdot \ln x)'$$&nbsp;</div><div><br /></div><div>$$f'(x) = e ^ {x \cdot \ln x} \cdot (1 \cdot \ln x + x \cdot \frac{1}{x})$$&nbsp;</div><div><br /></div><div>$$f'(x) = e ^ {x \cdot \ln x} \cdot (\ln x + 1)$$&nbsp;</div><div><br /></div><div>$$f'(x) = x ^ x \cdot (\ln x + 1) \tag{3}$$&nbsp;</div><div><br /></div><div>The last equation $(3)$ gives us the derivative which we wanted to find.&nbsp;</div><div><br /></div><div>In the above derivation we used several simple facts from math analysis.&nbsp;</div><div><br /></div><div>$$(e^x)' = e^x$$</div><div><br /></div><div>$$(f(g(x)))' = f'(g(x)) \cdot g'(x)$$</div><div><br /></div><div>$$(u(x) \cdot v(x))' = u'(x) \cdot v(x) + u(x) \cdot v'(x)$$&nbsp;</div><div><br /></div><div>Finally let us restate the above established formula.&nbsp; &nbsp;</div><div><br /></div><div>$$\large { (x^x)' = x ^ x \cdot (\ln x + 1) } \tag{4}$$&nbsp;</div><div><br /></div><div><b>II. How do we find&nbsp;$\displaystyle{ \lim_{x \to 0^{+}} f(x) } = \displaystyle{ \lim_{x \to 0^{+}} x^x }$ ?</b></div><div><b><br /></b></div><div>I think the easiest way of finding this limit which I have seen is by letting $x=e^{-t}$ where $t$&nbsp;is some very large positive number. Then we easily get the following.&nbsp;</div><div><br /></div><div><div>$$\lim_{x \to 0^{+}} f(x) = \lim_{x \to 0^{+}} x^x = \lim_{x \to 0^{+}} e ^ {x \cdot \ln x} = \lim_{x \to 0^{+}} e ^ {(\ln x) \cdot x} =&nbsp;$$</div></div><div><br /></div><div>$$= \lim_{t \to \infty} e^{(-t) \cdot e^{-t} } = \lim_{t \to \infty} e^{ \frac{(-t)}{e^{t}} } = e ^ {\lim_{t \to \infty} \frac{(-t)}{e^{t} } } = e^0 = 1&nbsp;$$</div><div><br /></div><div>Here the main fact which we used was that&nbsp;&nbsp;</div><div><br /></div><div>$$\lim_{t \to \infty} \frac{(-t)}{e^{t} }&nbsp; = 0$$&nbsp;</div><div><br /></div><div>which is quite obvious given that the numerator is a polynomial of $t$ and the denominator is the exponential function $e^t$.&nbsp;&nbsp;</div><div><br /></div><div><div>This way we have just calculated this quite remarkable limit&nbsp;</div><div><br /></div><div>$$\large \lim_{x \to 0^{+}} x^x = 1 \tag{5}$$&nbsp;</div><div><br /></div></div><div>Finally... here is a video demonstrating nicely an informal numerical approach to finding the same limit&nbsp;<a href="https://youtu.be/r0_mi8ngNnM">What is 0 to the power of 0?</a></div><div><br /></div>petrovhttp://www.blogger.com/profile/14750572373308546381noreply@blogger.com0tag:blogger.com,1999:blog-1243612190666737338.post-24092719862507272912020-04-27T19:29:00.000+03:002020-04-28T01:32:07.609+03:00Euler's identity for $\frac{sin(x)}{x}$<div dir="ltr" style="text-align: left;" trbidi="on"><br />Here is this famous identity due to Euler.<br /><br />$$\frac{\sin x}{x} = \prod_{k=1}^{\infty} \cos \left(\frac{x}{2^k} \right) \tag{1}$$<br /><br />This holds true for every $x \ne 0$.<br /><br />Let's prove it.<br /><br /><br />Denote:<br /><br />$$S(n) = \prod_{k=1}^{n} \cos \left(\frac{x}{2^k} \right) \tag{2}$$<br /><br />Multiplying the two sides by $sin{\frac{x}{2^n}}$ we sequentially get:<br /><br />$$sin{\frac{x}{2^n}} \cdot S(n) = sin{\frac{x}{2^{n}}} \cdot \prod_{k=1}^{n} \cos \left(\frac{x}{2^k} \right)$$<br />$$sin{\frac{x}{2^n}} \cdot S(n) = sin{\frac{x}{2^{n-1}}} \cdot \frac{1}{2^1} \cdot&nbsp;\prod_{k=1}^{n-1} \cos \left(\frac{x}{2^k} \right)$$<br />$$sin{\frac{x}{2^n}} \cdot&nbsp;S(n) = sin{\frac{x}{2^{n-2}}} \cdot \frac{1}{2^2} \cdot&nbsp;\prod_{k=1}^{n-2} \cos \left(\frac{x}{2^k} \right)$$<br />$$...$$<br />$$sin{\frac{x}{2^n}} \cdot&nbsp;S(n) = sin{\frac{x}{2^{1}}} \cdot \frac{1}{2^{n-1}} \cdot&nbsp;\prod_{k=1}^{1} \cos \left(\frac{x}{2^k} \right)$$<br /><br />The last one obviously gives us:<br /><br />$$sin{\frac{x}{2^n}} \cdot&nbsp;S(n) = \frac{1}{2^{n}} \cdot sin(x) \tag{2}$$<br /><br />which can be easily reworked to:<br /><br />$$\large{ \frac{sin{\frac{x}{2^n}}}{\frac{x}{2^n}} \cdot&nbsp;S(n) = \frac{sin(x)}{x}} \tag{3}$$<br /><br />Now in $(3)$ when we take the limit as ${n\to\infty}$ while using that<br /><br />$$\lim_{u \to 0} \frac{sin(u)}{u} = 1$$<br /><br />we get equality $(1)$ which is what we wanted to prove.<br /><br /><br /></div>petrovhttp://www.blogger.com/profile/14750572373308546381noreply@blogger.com0tag:blogger.com,1999:blog-1243612190666737338.post-74748361156559585032020-04-18T15:00:00.003+03:002020-04-28T01:23:30.873+03:00Trigonometric identities<div dir="ltr" style="text-align: left;" trbidi="on">1) Even/Odd function identities<br /><br />$$\sin(-\alpha) = -\sin\alpha \tag{1.1}$$<br />$$\cos(-\alpha) = \cos\alpha \tag{1.2}$$<br />$$\tan(-\alpha) = -\tan\alpha&nbsp;\tag{1.3}$$<br />$$\cot(-\alpha) = -\cot\alpha&nbsp;\tag{1.4}$$<br /><br />2) Sum/Addition identities<br /><br />$$\sin(\alpha + \beta) = \sin\alpha\cos\beta + \cos\alpha\sin\beta \tag{2.1}$$<br />$$\sin(\alpha - \beta) = \sin\alpha\cos\beta - \cos\alpha\sin\beta \tag{2.2}$$<br />$$\cos(\alpha + \beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta \tag{2.3}$$<br />$$\cos(\alpha - \beta) = \cos\alpha\cos\beta + \sin\alpha\sin\beta \tag{2.4}$$<br /><br />$$\tan(\alpha + \beta) = \frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}\tag{2.5}$$<br /><br />$$\cot(\alpha + \beta) = \frac{\cot\alpha\cot\beta-1}{\cot\alpha+\cot\beta}\tag{2.6}$$<br /><br />3) Sum to product identities<br /><br />$$\sin\alpha + \sin\beta = 2 \sin\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2} \tag{3.1}$$<br />$$\sin\alpha - \sin\beta = 2 \sin\frac{\alpha-\beta}{2}\cos\frac{\alpha+\beta}{2}&nbsp;\tag{3.2}$$<br /><br />$$\cos\alpha + \cos\beta = 2 \cos\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2}&nbsp;\tag{3.3}$$<br /><br />$$\cos\alpha - \cos\beta = -2 \sin\frac{\alpha+\beta}{2}\sin\frac{\alpha-\beta}{2}&nbsp;\tag{3.4}$$<br /><br />4) Product to sum identities<br /><br />$$\sin\alpha\sin\beta = \frac{1}{2}[\cos(\alpha-\beta) - \cos(\alpha+\beta)] \tag{4.1}$$<br /><br />$$\cos\alpha\cos\beta = \frac{1}{2}[\cos(\alpha-\beta) + \cos(\alpha+\beta)] \tag{4.2}$$<br /><br />$$\sin\alpha\cos\beta = \frac{1}{2}[\sin(\alpha+\beta) + \sin(\alpha-\beta)] \tag{4.3}$$<br /><br />5) Double-angle identities<br /><br />$$\sin 2 \alpha&nbsp; = 2 \sin\alpha \cos\alpha \tag{5.1}$$<br /><br />$$\cos 2 \alpha&nbsp; = \cos^2\alpha - \sin^2\alpha \tag{5.2}$$<br /><br /><br /></div>petrovhttp://www.blogger.com/profile/14750572373308546381noreply@blogger.com0tag:blogger.com,1999:blog-1243612190666737338.post-76931441096219565082020-04-16T15:27:00.003+03:002020-04-17T03:34:27.675+03:00Fun problem on Ceva's theorem<div><b>Problem:&nbsp;</b><br /><b><br /></b>Given is $\Delta ABC$. The points $D \in BC$, $E \in CA$, $F \in AB$ are such that the lines $AD,BE,CF$ are concurrent.&nbsp;</div><div><br /></div><div><div>$A'$ - midpoint of $BC$&nbsp; &nbsp;</div><div>$B'$ - midpoint of $CA$&nbsp; &nbsp;</div><div>$C'$ - midpoint of $AB$&nbsp; &nbsp;</div><div>&nbsp; &nbsp;</div><div>$D'$ - midpoint of $AD$&nbsp;&nbsp;</div><div>$E'$ - midpoint of $BE$&nbsp;&nbsp;</div><div>$F'$ - midpoint of $CF$&nbsp;&nbsp;</div></div><div><br /></div><div>Prove that the lines&nbsp;</div><div><br /></div><div>$A'D', B'E', C'F'$ are also concurrent (i.e. that they pass through a common point).</div><div><br /></div><div><br /><br /></div><div><b>Solution:</b><br /><b><br /></b>It is crucial to come up with a nice realistic drawing here.<br /><br />Also, it is important to realize that:<br />a) $A', F', B'$ are on one line<br />b) $B', D', C'$ are on one line<br />c) $C', E', A'$ are on one line<br /><br /><div class="separator" style="clear: both; text-align: center;"></div><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-zZ47OSpB-c4/XphozOjx8rI/AAAAAAAAA20/1Bon4Oq-bEMVuu1K7-DU2lwDCWdv5QdGgCLcBGAsYHQ/s1600/Draw4.jpg" imageanchor="1" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img border="0" data-original-height="1063" data-original-width="1600" height="265" src="https://1.bp.blogspot.com/-zZ47OSpB-c4/XphozOjx8rI/AAAAAAAAA20/1Bon4Oq-bEMVuu1K7-DU2lwDCWdv5QdGgCLcBGAsYHQ/s400/Draw4.jpg" width="400" /></a></div><div class="separator" style="clear: both; text-align: center;"></div><br />From Ceva's theorem for triangle $ABC$ we get:<br />$$\frac{AF}{FB}\frac{BD}{DC}\frac{CE}{EA} = 1 \tag{1}$$<br /><br />Now the trick is to realize that:<br />$$\frac{B'F'}{F'A'} = \frac{AF}{FB} \tag{2}$$&nbsp; &nbsp; <br />$$\frac{C'D'}{D'B'} = \frac{BD}{DC} \tag{3}$$&nbsp; <br />$$\frac{A'E'}{E'C'} = \frac{CE}{EA} \tag{4}$$&nbsp; <br /><br />Why is this so?&nbsp; <br /><br />Because $B'C' || BC$&nbsp; &nbsp;, $C'A' || CA$ and&nbsp; $A'B' || AB$&nbsp; <br />so these relations follow from the Intercept theorem.&nbsp; &nbsp; <br /><br />Multiplying the last 3 equations and using $(1)$ we get:<br /><br />$$\frac{B'F'}{F'A'}\frac{C'D'}{D'B'}\frac{A'E'}{E'C'} = \frac{AF}{FB}\frac{BD}{DC}\frac{CE}{EA} = 1$$&nbsp; <br /><br />Thus:<br /><br />$$\frac{B'F'}{F'A'}\frac{A'E'}{E'C'}\frac{C'D'}{D'B'} = 1 \tag{5}$$<br /><br />Now using the converse Ceva's theorem (for the triangle $A'B'C'$ and for the points $D', E', F'$), we can conclude from $(5)$ that the three lines&nbsp; $A'D', B'E', C'F'$ intersect at a single/common point. This is what we had to prove hence the problem is solved.&nbsp; &nbsp; &nbsp;</div><div><br /></div><div><br /></div>petrovhttp://www.blogger.com/profile/14750572373308546381noreply@blogger.com0tag:blogger.com,1999:blog-1243612190666737338.post-23068876946091682532020-04-16T15:14:00.004+03:002020-04-28T01:28:02.095+03:00Klein bottle mystery<div dir="ltr" style="text-align: left;" trbidi="on">In the book "Basic Topology" by M.A.Armstrong I found an explanation about how to construct a Klein bottle. I had to reread it 5 times and I was still not quite convinced. I am retelling it here.<br /><br /><i>Begin with a sphere, remove two discs from it, and add a Möbius strip in their places.&nbsp;</i><br /><i>A Möbius strip has after all a single circle as boundary, and all that we are asking&nbsp;</i><br /><i>is that the points of its boundary circle be identified with those of the boundary&nbsp;</i><br /><i>circle of the hole in the sphere. One must imagine this identification taking place&nbsp;</i><br /><i>in some space where there is plenty of room (euclidean four-dimensional space will do).&nbsp;</i><br /><i>This cannot be realized in three dimensions without having each Möbius strip&nbsp;</i><br /><i>intersect itself. The resulting closed surface is called the Klein bottle.&nbsp;</i><br /><br />I was scratching my head around how this procedure actually produces a Klein bottle until I found this question in <b>MathSE</b>.<br /><br /><a href="https://math.stackexchange.com/questions/907176/klein-bottle-as-two-m%C3%B6bius-strips">Klein-bottle-as-two-Möbius-strips</a><br /><br />This picture there in one of the answers is really really nice, it really shows what happens if we cut a Klein bottle in half - we really get two Möbius strips as a result. The cut is done by a plane "parallel to the handle" which cuts the bottle in two symmetric parts.<br /><br /><br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-LdKzTrDVO1o/XphLbYQrQYI/AAAAAAAAA18/jB2UO88NnSURS4d2M37H1T9olYri6aoFgCLcBGAsYHQ/s1600/Klein1.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="400" data-original-width="629" height="203" src="https://1.bp.blogspot.com/-LdKzTrDVO1o/XphLbYQrQYI/AAAAAAAAA18/jB2UO88NnSURS4d2M37H1T9olYri6aoFgCLcBGAsYHQ/s320/Klein1.png" width="320" /></a></div><br /><br />So... it's really for a reason they say "a picture is worth a thousand words".<br /><br /><br /><br /><br /><br /></div>petrovhttp://www.blogger.com/profile/14750572373308546381noreply@blogger.com0tag:blogger.com,1999:blog-1243612190666737338.post-45951919470948907092020-04-03T01:15:00.002+03:002020-04-16T15:37:46.595+03:00How to solve a quadratic equation?This post comes to demonstrate the support for $\LaTeX$ available in Blogger.<br /><br />A quadratic equation is an equation of the form<br />$$ax^2 + bx + c = 0&nbsp; &nbsp;\tag{1}$$<br />where<br />$a,b,c \in \mathbb{R}$ and $a \ne 0$.<br />Let us assume that we are trying to solve this equation for real numbers only.<br />The value<br />$$D = b^2 - 4ac&nbsp; \tag{2}$$<br />is called discriminant of the equation $(1)$.<br />Case #1:<br />When $D \gt 0$, there are two distinct real solutions to $(1)$ and they are:<br />$$x_{1,2} = {-b \pm \sqrt{b^2-4ac} \over 2a} \tag{3}$$<br />Case #2:<br />When $D = 0$, there is one real solution to $(1)$ which is:<br />$$x = {-b \over 2a} \tag{4}$$<br />Case #3:<br />When $D \lt 0$, there are no real solutions to $(1)$.<br /><div><br /></div>petrovhttp://www.blogger.com/profile/14750572373308546381noreply@blogger.com0tag:blogger.com,1999:blog-1243612190666737338.post-63358186081927109882020-04-03T01:03:00.001+03:002020-04-16T15:37:41.833+03:00How to solve a linear equation?<br />This post comes to demonstrate the support for $\LaTeX$ available in Blogger.<br /><br />&nbsp;A linear equation is an equation of the form<br /><br />$$ax + b = 0&nbsp; \tag{1}$$<br /><br />where<br /><br />$$a,b \in \mathbb{R}, a \ne 0$$<br /><br />Let us assume that we are trying to solve this equation for real numbers only i.e. we are trying to find the real roots of the equation.<br /><br />Solving this equation is simple, it always has a single real solution which is<br /><br />$$x = - {b \over a}&nbsp; \tag{2}$$<br /><br /><br /><br /><br />petrovhttp://www.blogger.com/profile/14750572373308546381noreply@blogger.com0tag:blogger.com,1999:blog-1243612190666737338.post-28586826891725820512020-04-02T12:21:00.004+03:002020-04-16T15:16:05.560+03:00Calculate $\lim_{x \to \infty} \sqrt[n]{(x+a_1)(x+a_2) ... (x+a_n)} - x$ This is a problem which I found in a math textbook some time ago. I tried solving it but I did not manage to solve it as quickly as I wanted and so I lost patience. Therefore I posted it in <b>MathSE</b>. I liked quite a lot one of the solutions I got, so I wrote it on a sheet of paper and kept it.<br /><br />Here is the full story...<br /><br /><b>Problem:</b><br /><br />Find $\lim_{x \to \infty} \sqrt[n]{(x+a_1)(x+a_2) ... (x+a_n)} - x$,<br /><br />where $x$ is a real variable and $n$ is a fixed natural number $n \ge 1$.<br /><br /><b>Solution:</b><br /><br />Let<br />$f(x)=\sqrt[n]{(x+a_1)(x+a_2)\cdot (x+a_n)}$<br /><br />We now calculate:<br /><br />$$\lim_{x\to \infty} \frac{f(x)}{x}=\lim_{x\to \infty} \sqrt[n]{(1+\frac{a_1}{x})(1+\frac{a_2}{x}) \cdot (1+\frac{a_n}{x})}= \sqrt[n]{1} = 1 \tag{1}$$<br /><br />We now form the below difference and we rework it a bit:<br /><br />$$f(x)-x=\frac{f(x)^n-x^n}{\sum_{k=0}^{n-1} f(x)^{n-1-k} x^{k}}= \frac{f(x)^n-x^n}{f(x)^{n-1}+f(x)^{n-2}x+\dots+f(x)x^{n-2}+x^{n-1}}$$<br /><br />We note here that the numerator $P(x) = f(x)^n-x^n$ is a polynomial of degree $n-1$ and it has a leading coefficient of $c = a_1+a_2+\dots+a_n$. The denominator is composed of the sum of $n$ terms of the form $f(x)^{n-1-k}x^k$ (for $k=0,1, \dots, n-1$).<br /><br />So we divide the numerator and the denominator by $x^{n-1}$ and we get:<br /><br />\begin{align*}<br />\smash<br />\lim_{x\to \infty}f(x)-x &amp; = \lim_{x\to \infty} \frac {\frac{f(x)^n-x^n}{x^{n-1}}}{\frac{f(x)^{n-1}}{x^{n-1}}+\frac{f(x)^{n-2}}{x^{n-2}}+\dots+\frac{f(x)}{x}+1} \\[10pt]<br />&amp;= \lim_{x\to \infty} \frac {\frac{P(x)}{x^{n-1}}}{\frac{f(x)^{n-1}}{x^{n-1}}+\frac{f(x)^{n-2}}{x^{n-2}}+\dots+\frac{f(x)}{x}+1} \\[10pt]<br />&amp;= \frac{c}{\underbrace{1 + 1 + \dots + 1}_{n\text{ times}}} \\[10pt]<br />&amp;= \frac{a_1+a_2+\dots+a_n}{n}\end{align*}<br /><br />The cool thing here is that the limit value happens to be the arithmetic mean of the numbers $a_1, a_2, \dots a_n$.<br /><br /><br />See also&nbsp;<a href="https://math.stackexchange.com/a/3491452/116591">Math SE question/solution here</a><br /><br /><br /><br /><br />petrovhttp://www.blogger.com/profile/14750572373308546381noreply@blogger.com0tag:blogger.com,1999:blog-1243612190666737338.post-81559302195021210352020-04-02T00:59:00.002+03:002020-04-02T16:30:20.736+03:00Derivative of $f(x)=x^n$What is the derivative of $$f(x)=x^n$$ when $n$ is a fixed integer and $x$ is a real variable?<br /><br />As we all know this derivative is<br /><br />$$f'(x)=nx^{n-1}$$<br /><br />But for which values of $n$ and $x$ does this hold true?<br /><br />Well...<br /><br />1) this is true for integers $n \gt 0$<br />2) this is also true for integers $n \lt 0$ if $x \ne 0$<br /><br />Note that when $x=0$ the symbols $x^0, x^{-1}, x^{-2}, x^{-3}, \dots$ are usually treated as undefined. That is why for 2) we need the additional restriction that $x \ne 0$.<br /><br /><br /><br />petrovhttp://www.blogger.com/profile/14750572373308546381noreply@blogger.com0