I tried looking for ways to solve this equation and came across something like Lambert's W function, which, by the way, I did not understand a bit, because I've never learned it nor do I have a decent mathematical background. I also came across one more method called the Newton's method, but never used it either. Can the procedure to solve this equation be made a little bit clearer? I just plugged in the value on my calculator and it solved it for me, I guess it must have used some trial and error. Please advise.
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Given that   $x^x = y$; and  given some value for $y$   is there a way to expressly solve that equation for $x$?
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I am trying to solve a central force problem and came to an equation like this:  $$\frac{d^2 y}{dx^2} + y = \frac{1}{y^3}$$  I can't find a decent method to solve it.
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I came across this example of how you could end up with an extraneous solution but I was wondering how it arose. We have the equation: $$x^2+x+1=0 $$ Since x=0 does not satisfy the equation, you can divide by x on both sides which yields: $$x+1+\frac{1}{x}=0$$ which is equivalent to our first equation. From our first eqution we can conclude that: $$-x^2=x+1$$ We now substitute that into the second equation to get:$$x^2=\frac{1}{x}$$ which results in $$x^3=1$$ which is equivalent to our previous equation since x cannot be 0. However, one solution from our last equation is x=1, which is not a solution to our original equation. I have a vague idea that it may have to do with the fact that you get a cubic equation and you began with a quadratic, and that steps imply the following and not vice versa, but can you provide a very detailed answer as to why it arises? Can you please provide more examples?
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Hello I have some problems concerning Taylor series.  Given the function $$f(x)=e^{\sin{x}} $$  I concluded that the Taylor series expansion would be   $$f(x) = \sum^\infty_{n=0}\frac{1}{n!}f^{(n)}(x)(x-x_0)^n$$ But Wolfram wrote a much simpler form   $$f(x)=\sum^\infty_{n=0}\frac{\sin^k{x}}{k!}$$  My question is: how come? Secondly, could somebody explain to me how  to get the second sum's radius of convergence?   Thank you very much for your time!
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Can anyone give a proof of the derivative of this type of function? Specifically showing that  $\dfrac{d(x^w)}{dx} = wx^{w-1}$ for a real $w$?  I tried to use the Taylor series expansion for $(x+dx)^w$ and got the correct result. However, the proof of the Taylor series requires knowledge of the derivative of these functions. So this is essentially circular reasoning. I know that the same series is also given by the binomial expansion, but that's not entirely satisfactory either, because where's the proof that the binomial expansion works for all reals (isn't it only apparent for integers)? So far all of the arguments I've come across involve circular reasoning. I was thinking of showing that the binomial expansion is true for all reals using some form of proof by induction e.g. something like this.  I'm really not sure.
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The question is:     When $f(x+1)-f(x)=f'(x)$, what are the solutions for $f(x)$?   The most obvious solution is a linear function of the form $f(x)=ax+b$. Is this the only solution?  Edit  I should add that $f:\mathbb R\to\mathbb R$ to the question.
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I was watching the said movie the other night, and I started thinking about the equation posed by Nash in the movie.  More specifically, the one he said would take some students a lifetime to solve (obviously, an exaggeration).  Nonetheless, one can't say it's a simple problem.  Anyway, here it is  $$V = \{F:\mathbb{R^3}/X\rightarrow \mathbb{R^3} \text{ so } \hspace{1mm}\nabla \times F=0\}$$ $$W = \{F = \nabla g\}$$ $$\dim(V/W) = \; 8$$  I haven't actually attempted a solution myself to be honest, but I thought it would be an interesting question to pose. I have done a quick search on this site and Google, but there were surprisingly few results.   In any case, I was curious if anyone knew the answer aside from the trivial.
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What would the steps be for someone new trying to learn how to solve this? I feel like I understand the concepts and what they mean but not how to prove it. Thanks
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I'm having a hard time finding the solution of the following equation:  $$0 = x\cos(x)+2$$  This is part of showing that $f(x):= x^2e^{\sin(x)}$ has 3 extremas in $[-\frac{\pi}{2},\frac{\pi}{2}]$.
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Here is an interesting topic.  It comes from evaluating $$\sum_{k=1}^{\infty}tan^{-1}\left(\frac{1}{k^{2}}\right)$$  I managed to dig up an old paper I have on the sum of arctans by Boros and Moll.  It is called the Method of Zeros.  It is located here:   on page 6-7   I will post, verbatim, what it says.  Perhaps someone can figure out how they arrive at the solution.  They skip over the details and I do not know how they arrived at the general form they establish in [2].  "Based on the factorization of the product  $$p_{n}:=\prod_{k=1}^{n}(a_{k}+ib_{k}), \;\ a_{k},b_{k}\in R$$  The argument of $p_{n}$ is given by $$\text{arg}(p_{n})=\sum_{k=1}^{n}\tan^{-1}\left(\frac{b_{k}}{a_{k}}\right)$$  up to an integral multiple of $\pi$.  This can be applied to the case of a polynomial with real coefficients given by  $$p_{n}(z)=\prod_{k=1}^{n}(z-z_{k})$$  Then,  $$\text{arg}(p_{n}(z))=\sum_{k=1}^{n}\tan^{-1}\left(\frac{x-x_{k}}{y-y_{k}}\right)$$  up to an integral multiple of $\pi$.  The special case $$p_{n}(z)=z^{n}-1$$ has roots at   $$z_{k}=\cos(\frac{2\pi k}{n})+i\sin\left(\frac{2\pi k}{n}\right)$$,  so we get:  $$\text{arg}(z^{n}-1)=\sum_{k=1}^{n}\tan^{-1}\left(\frac{x-\cos\left(\frac{2\pi k}{n}\right)}{y-\sin\left(\frac{2\pi k}{n}\right)}\right)$$  The classical factorization  $$\sin(\pi z)=\pi z\prod_{k=1}^{\infty}\left(1-\frac{z^{2}}{k^{2}}\right)....[1]$$  shows that one can think of $\sin(\pi z)$ as a polynomial in z of infinite degree.  Euler than derived $$\sum_{k=1}^{\infty}\frac{1}{k^{2}}=\frac{{\pi}^{2}}{6}$$ by comparing the cubic terms of [1].  This now yields $$\sum_{k=1}^{\infty}\tan^{-1}\left(\frac{2xy}{k^{2}-x^{2}+y^{2}}\right)=\tan^{-1}(y/x)-\tan^{-1}\left(\frac{\tanh(\pi y)}{\tan(\pi x)}\right)....[2]$$"  That is it.  They mention Euler's famous sum of reciprocal of squares then just skip to "this then yields...."  Does anyone see how?.   Besides that, didn't Euler compare the square terms and not the cubic ones?.  So, for the problem at hand, we would let $x=y=\frac{1}{\sqrt{2}}$ and arrive at:  $$\sum_{k=1}^{\infty}\tan^{-1}\left(\frac{1}{k^{2}}\right)=\frac{\pi}{4}-\tan^{-1}\left(\frac{\tanh\left(\frac{\pi}{\sqrt{2}}\right)}{\tan\left(\frac{\pi}{\sqrt{2}}\right)}\right)\approx 1.42474......$$
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I have the following question:          Find the value of:     $$ \frac {1} {\sqrt {4} + \sqrt {5}} +  \frac {1} {\sqrt {5} + \sqrt {6}} + ... +  \frac {1} {\sqrt {80} + \sqrt {81}}$$      The book already provide the answer for this question but it didn't include the ways. Hint or anything that will help is appreciated
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I can't get any of these on my own and I am attempting to do $x^x$ I had it explained to me for about an hour and I still can't do it on my own. I thought I was supposed to make it into $x(\ln x)$ which should be equivalent to the original term. Then the differentiation should be easy for most people from here.  $$1(\ln x)+x(\frac1x) $$ or $$\ln x+1$$  This is of course wrong but I do not know why.
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Question: Can anyone shed some light on this formula? I can't find any information on it.  It has three corollaries that I also need to understand:
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Show that Newton's method can be used to compute the square root function $\sqrt a$ using the formula   $$x_{n+1} = \frac{1}{2}\left(x_{n} + \frac{a}{x_{n}}\right)$$  show that the error is  $$\sqrt a - x_{n+1} = -\frac{1}{2x_{n}}\left(\sqrt a - x_{n}\right)^2$$  edit: As pointed out below $x^2-a$ has $\sqrt a$ as a root.  I have done as suggested below and plugged in $\sqrt a + \epsilon$ for $x_n$ giving me  $$x_{n+1} = \frac{2\epsilon\sqrt a + \epsilon^2}{2(\sqrt a + \epsilon)}$$ and once again not sure where to proceed.
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I've got a problem with a modified Newton's method. We've got a function $f \in C^{(k+1)}$ and $r$ which is it's multiple root of multiciplity $k$. Also $f^{(k)}(r) \neq 0$ and $f'(x) \neq 0 $ in the neighbourhood of $r$.  The modified Newton's method is $$x_{n+1} = x_n - k\frac{f(x_n)}{f'(x_n)}$$ How to prove that $x_n$ converges to $r$ quadratically?  I found one method using a function $G(x) = (x-r) f'(x) - kf(x)$ but then it's said that $ G^{(k+1)}(r) \neq 0$ but it comes from the fact that $f^{(k+1)}(r) \neq 0$, but it doesn't necessarily have to be true, am I right? We know that $f^{(k)}(r) \neq 0$, but we have no information about the $k+1$st derivative. I would appreciate if somebody explained to me precisely, why it quadratically converges.
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Okay I like have no idea how to do this. I tried integration by parts then everything came back uglier. I then put this in wolfram it says "No found in standard mathematical functions"   Is there a easy way to do this?   Maybe like FTOC II?
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So I was practicing for an exam, and through my studying I found this problem. I really don't know how to solve it although the answer must not be too complicated since the question that came with this one were fairly simple. I first thought that maybe I could use vectors to solve this problem. I tried using this equation:    I noticed that this wouldn't help since I would then have the following equation :. I would have two variables to solve for using only one equation so I couldnt solve it.   I also thought of using geometry thus making a triangle with two sides equal to sqrt(5) thus it would have an angle of 45 degrees and two of 67.5 degress. I could then proceed to use sine rule but this wouldn't work out for me since I am supposed to do the problem with calculator.   Can someone help me?