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First of all, notice that $f$ is undefined at $x=0$ and has no derivatives there. But this is a removable singularity, writing $f$ in the equivalent form $$f(x)=\frac{2}{1-2x+\sqrt{1-8x+8x^2}}$$ allows us to avoid this. Now the first few coefficients are $f_0=1$, $f_1=3$, $f_2=11$ and $f_3=47$. Plugging these into the formula with $n=3$ starts the induction....

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You may have the right idea, but for the proof to be complete and rigorous you need to more clearly justify the following crucial inference: When we square a number, we merely repeat its factors, therefore $A^2$ and $B^2$ must also not share any factors". As it stands, your justification "when we square a number, we merely repeat its factors" could be ...

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I will mention one (easily corrected) logical error and one stylistic piece of advice that could make the proof more readable. But the upshot is that this is a well-argued proof by any standard, and especially impressive for a first effort. When you said that $A^2$ and $B^2$ share no factors aside from 1, that does not imply that $\frac{A^2}{B^2}$ is not ...

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Doing the common denominator and simplifying, I obtain: $\frac{a+\sqrt{ab}}{-(b+\sqrt{ab})}$. From this, I multiply the numerator and the denominator by $b-\sqrt{ab}$ and I get: $\frac{(a+\sqrt{ab})(b-\sqrt{ab})}{-(b+\sqrt{ab})(b-\sqrt{ab})}=\frac{ab-a\sqrt{ab}+b\sqrt{ab}-ab}{-(b^2-b\sqrt{ab}+b\sqrt{ab}-ab)}=\frac{\sqrt{ab}(b-a)}{-b(b-a)}=-\frac{\sqrt{ab}}{b}... 1 The rule that you used$x^{ab}=(x^a)^b$is true only when$x \ge 0$you can't use it when$x=-2$, and as Kavi Rama Murthy said$\sqrt x=x^{1/2} \ge 0$3 Notice that $$a-b = (\sqrt{a}+\sqrt{b})(\sqrt{a}-\sqrt{b})$$ so you get $$(\sqrt{a}-\sqrt{b}+\sqrt{b})/(\sqrt{a}-\sqrt{b}-\sqrt{a})=-\sqrt{a\over b}$$ 0 just simplify the equation:$ \Large x =\frac{\frac{\left(a-b\right)\left(\sqrt{a}-\sqrt{b}\right)}{\left(a-b\right)}+\sqrt{b}}{\frac{\left(a-b\right)\left(\sqrt{a}-\sqrt{b}\right)}{\left(a-b\right)}-\sqrt{a}}=\frac{\left(\sqrt{a}-\sqrt{b}\right)+\sqrt{b}}{\left(\sqrt{a}-\sqrt{b}\right)-\sqrt{a}}=\frac{\sqrt{a}}{-\sqrt{b}}=-\sqrt{\frac{a}{b}}$-1 One of the approache is use Euler form if Z is a complex numbers such that$ Z\, = \,r.e^{ix}$then Square root of Z will be$Z_{1}= \, \sqrt{r} . e^{iy} $Where angle y will be x/2 and (x + 2π)/2. 2 You made a mistake in your derivation. You should have$(x^2+y^2)^\color{red}2=a^2+b^2$. This follows from$a=x^2-y^2$and$b=2xy$(or from known properties of complex modulus). Thus,$a=x^2-\dfrac {b^2}{4x^2}$; solving this quadratic equation in$x^2$yields$x^2=\dfrac{a+\sqrt{a^2+b^2}}2$as the correct answer. 0 After squaring and rearranging we have: $$4x^6-12x^5+9x^4+3x^3-14x^2+8x=0$$ Since there is no constant term,$x=0$is a solution. Using the rational root theorem,$p$and$q$have to be coprime. Since the only factors of$4$and$8$are powers of$2$, then the only solutions are when either$1$is in the numerator or$1is the denominator: $$x = ±\frac{1}... 0 (x \le 3y \ne -x) Let a = \dfrac{2}{x^2} > 0 and b = \dfrac{3y}{x}. The system of equation becomes$$\left\{ \begin{align} \left(\frac{2}{a} - 1\right)^2 + 3 = \frac{2b \cdot \left(\dfrac{2}{a}\right)^3}{\dfrac{2}{a} + 2}\\ b - 1 = \sqrt{\frac{2a - b - b^2}{1 + b}} \end{align} \right.\iff \left\{ \begin{align} (2 - a)^2 + 3a^2 = \frac{8b}{a +... 1 Your convention is a little different from what I am accustomed to; I usually denote Pochhammer symbols (rising factorials) by(a)_n$, while falling factorials are$a^{(n)}. That is, \begin{align*} (a)_n&=\prod_{j=0}^{n-1}(a+j)\\ a^{(n)}&=\prod_{j=0}^{n-1}(a-j)=(-1)^n(-a)_n\end{align*} The general relation you want is \left(a+\frac12\right)... 0 Some years ago, I came up with a proof. Proof: By using the Cauchy-Schwarz inequality, we have \begin{align} &\frac{xy}{\sqrt{xy+yz}}+\frac{yz}{\sqrt{yz+zx}}+\frac{zx}{\sqrt{zx+xy}}\\ \le \ & \sqrt{(xy+yz+zx)\Big(\frac{xy}{xy+yz}+\frac{yz}{yz+zx}+\frac{zx}{zx+xy}\Big)}\\ = \ & \sqrt{xy + \frac{x^2z}{x+z} + yz + \frac{xy^2}{y+x} + xz + \frac{yz^2}... 0 You can continue so:x^2=10 + 2(\sqrt{6} + \sqrt{10} + \sqrt{15})=2+5+3+ 2(\sqrt{6} + \sqrt{10} + \sqrt{15})=(\sqrt2+\sqrt5+\sqrt3)^2.$$Can you end it now? 3 Let$$(\sqrt{x}+\sqrt{y}+\sqrt{z})^2=x+y+z+2\sqrt{xy}+2\sqrt{yz}+2\sqrt{xz}$$So$$\sqrt{x+y+z+2\sqrt{xy}+2\sqrt{yz}+2\sqrt{xz}}=\sqrt{x}+\sqrt{y}+\sqrt{z}$$For$$\sqrt{10+\sqrt{24}+\sqrt{40}+\sqrt{60}}=\sqrt{x}+\sqrt{y}+\sqrt{z}.$$We get x+y+z=10, 2\sqrt{xy}=\sqrt{24}, 2\sqrt{yz}=\sqrt{40}, 2 \sqrt{xz}=\sqrt{60}. So we get xy=6, yz=10, zx=15 \Rightarrow ... 8 From 6 = 2\times 3, 10 = 2\times 5, 15 = 3 \times 5, observe that 10 + 2(\sqrt6 + 2\sqrt{10} + 2\sqrt{15}) = (\sqrt2+\sqrt3+\sqrt5)^2. 2 This is a redo of a previously posted answer. This one attempts to be more approachable to readers without knowledge of finite field extensions. The \sqrt[3]{16} and \sqrt[3]{27} are distractions. Write these as 2\sqrt[3]{2} and 3. So you have$$\frac{-1+2\sqrt[3]{2}}{2+\sqrt[3]{2}+\sqrt[3]{4}}$$All five terms are in the form a\sqrt[3]{2^n}. (For ... 0 Let x=\sqrt[3]{2} then we have \frac{x^4-1}{x^2+x+3} Multiply by x-1 \frac{x^4-1}{x^2+x+3}\cdot\frac{x-1}{x-1}=\frac{(x^4-1)(x-1)}{x^3+2x-3}=\frac{(x^4-1)(x-1)}{2+2x-3}=\frac{(x^4-1)(x-1)}{2x-1} Multiply by 4x^2+2x+1 \frac{(x^4-1)(x-1)(4x^2+2x+1)}{(2x-1)(4x^2+2x+1)}=\frac{(x^4-1)(x-1)(4x^2+2x+1)}{8x^3-1}=\frac{(x^4-1)(x-1)(4x^2+2x+1)}{15}=\... 6 With the same notation as in the other answer, i.e. x = \sqrt[3] 2, noting that x^3+1=3, you can write your quantity as \begin{eqnarray} \frac{x^4-1}{x^2+x+3} &=& \frac{(x-1)(x^3+x^2+x+1)}{x^2+x+x^3+1}=\\ &=&x-1. \end{eqnarray} 9 Let x=\sqrt[3]2 then we have$${x^4-1\over x^2+x+3}={x^6-x^2\over x(x^3+x^2+3x)}={4-x^2\over x(2+x^2+3x)}= {(2-x)(2+x)\over x(x+2)(x+1) } = {2-x\over x^2+x}= {(2-x)(x-1)\over x(x+1)(x-1)}= {(2-x)(x-1)\over x^3-x}= {(2-x)(x-1)\over 2-x} = x-1$$Edit: but other solution is much nicer then this one. -3 It is so easy to solve x6−4x3−1=0 just let x3=t and refine the equation as t2-4t-1=0 after you find the t, you can bring back the x3 for example if t=5, then x3=5 and x=3√5 2 Consider points A(2,6), B(-4,3) and a point P(0,y) on y-axis. Then notice that given equation is actually$$AP+BP= AB$$so P is (by triangle inequality ) on a line AB so it is an intersection point of the line AB:$$ y= {1\over 2}x+5$$and y-axis, so y=5. 0 You can factor out 2 at the second summand.$$\sqrt{\underbrace{4+(y-6)^2}_{=5}}+2\cdot \sqrt{\underbrace{4+\left(\frac{y-3}2\right)^2}_{=5}}=\sqrt{5}+2\cdot\sqrt{5}$$Now we see that the following equations has to be true at the same time. (y-6)^2=1 \Rightarrow y_1=7,y_2=5 \left(\frac{y-3}2\right)^2=1\Rightarrow y_1=5,y_2=1 Thus the solution is ... 3 Set y-3=z, then square both sides and simplify to get$$z^2-3z-8=-\sqrt{4+(z-3)^2}\sqrt{16+z^2}.$$Squaring again and simplifying the right hand side gives$$(z^2-3z-8)^2=(z^2-6z+13)(16+z^2).$$Now let 16+z^2=w, then expand and simplify to get$$z^2+16z+64=5w.$$Substituting back for w, simplifying and factoring gives the quadratic$$(z-2)^2=0.$$Thus, ... 0 Hint: After squaring one times we get$$2\sqrt{4+(y-6)^2}\sqrt{16+(y-3)^2}=25-(y-6)^2-(y-3)^2$$squaring again and simplfying we get$$4 \left(46 y^2-450 y+975\right)=0$$1 If there exists a cubic polynomial of integer coefficients with roots:$$0,\sqrt[3] {A-15√3} , \sqrt[3] {A+15√3} $$then it will be of the form x^3-4x^2 + \sqrt[3]{A^2-675}x You just need to find A such that A^2-675 is a perfect cube. Trivially A^2 = 676 will do. 1 Let x=A-15\sqrt 3 and y=A+15\sqrt 3, then the equation becomes$$x^{1/3}+y^{1/3}=4.$$Taking cubes of both sides gives$$x+y+3(xy)^{1/3}(x^{1/3}+y^{1/3})=4^3.$$Now since x+y=2A, and x^{1/3}+y^{1/3}=4, the equation becomes$$2A+3(xy)^{1/3}(4)=4^3,$$which gives$$6(xy)^{1/3}=32-A.$$Now, cubing and substituting for xy=A^2-15^2\cdot 3 gives$$6^3(A^... 1 Letx,y$be the numbers$(A\pm 15\sqrt 3)^{1/3}$so that$x+y=4. hen we have: \begin{aligned} 2A &= x^3+y^3\\ &= (x+y)^3-3xy(x+y) \\ &= 4^3-3xy\cdot 4 \\ &=64 - 3(A^2-675)^{1/3}\cdot 4\ .\text{ So:} \\[3mm] 6^3(A^2-675) &=(32-A)^3\ . \end{aligned} This gives us an equation inA$, that we may solve (with bare hands or not). sage: ... 0 In fact it is not unproblematic to work with an expression$a^x$for$x \notin \mathbb Z$and$a <0$. See my answer to Why$(-2)^{2.5}$isn't equal to$((-2)^{25})^{1/10}$? The "universal definition" would be$a^x = e^{x\ln a}$, but is valid only for$a > 0$. Extending this to$a < 0$is possible, but involves the complex logarithm and doing so ... 0 I think it is$\sqrt[3]{x}=x^{\frac{1}{3}} \ne x^{\frac{2}{6}}$Consider this example$-1=(-1)^3=(-1)^{2.\frac{3}{2}} \ne ((-1)^2)^{\frac{3}{2}}=1$The idea is that we must write the numbers we use in their irreducible forms to remove any kind of ambiguity. 2 If$n$is odd,$x^n$is an invertible function from$\Bbb R$to itself; we denote the inverse either$\sqrt[n]{x}$or$x^{1/n}$. This is consistent with$\left(x^a\right)^b=x^{ab}$. We can now uniquely define$x^{p/q}\in\Bbb R$for any$x\in\Bbb R$with$x\ne0$(a restriction we can drop if$p/q\gt0$), and any integers$p,\,q$with odd$q>0$. It won't ... 1 Let$y=x-2$, then $$L=.\lim_{y\rightarrow 0} \frac{(1+3y)^{1/2}-(1+y)^{1/3}}{y}$$ Use binomial approximation$(1+z)^p \approx 1+pz$if$|z|<<1$. Then $$L=\lim_{y \rightarrow 0} \frac{(1+3y/2)-(1+y/3)}{y}= \lim_{y\rightarrow 0} \frac{7y}{6y}=7/6.$$ Since it is$0/0$form you may also use L'Hospital rule, differentiate up and down separately: $$L=\... 4 Hint: \displaystyle\lim_{x\to2}\frac{\sqrt{3x-5}-\sqrt[3]{x-1}}{x-2}=\lim_{x\to2}\frac{\sqrt{3x-5}-1}{x-2}-\lim_{x\to2}\frac{\sqrt[3]{x-1}-1}{x-2}. 1 As mentioned above: "An extension field F of a field K is a radical extension of K if F=K(u_1,…,u_n), some power of u_1 lies in K and for each i≥2, some power of u_i lies in K(u_1,…,u_i$$_-$$_1)". All what the definition is saying is that if we can find a sequence of the u's which satisfies the conditions [namely: some power of u_1 lies in K ... 0 The first step is wrong already. In \mathbb C we have \sqrt1=\{1,-1\}. Actually, in \mathbb R we have \sqrt1=1. 0 Square root is only a single valued function if you restrict yourself to positive real numbers. This is something that people are at least casually aware of when they solve quadratic equations: if you want to solve x^2=4, you "take the square root of both sides", but you get two roots, x=\pm 2. When you're not working in a situation where there is a ... 5 The only incorrect statement is that$$\sqrt{e^{i2\pi}}=e^{i\pi}$$Because we have that$$\sqrt{x^2}=|x|$$for real x. So we would have the answer as$$\sqrt{e^{i2\pi}}=|e^{i\pi}|=|-1|=1$$3 There is nothing wrong with the equality 1=\sqrt{(-1)(-1)}. The problem lies in the equality \sqrt{e^{2\pi i}}=e^{\pi i}, because e^{2\pi i}=1, and therefore \sqrt{e^{2\pi i}}=1\neq e^{\pi i}. 0 Just for sake of simplicity, it is possible to compute \sqrt[n]a_s very easily when a>1 or n is odd and a>0 using bisection, which doesn't involve any hard calculations (so this is something doable with a calculator that only has exponentiation, and perhaps some paper). As long as you have x\le\sqrt[n]a_s\le y, we can iteratively consider [(... 0 Calculate \sqrt [ 4 ]{ 136 } using interpololation. 3^4 = 81 4^4 = 256 256 - 81 = 175 136-81 = 55 \sqrt [ 4 ]{ 136 } \approx 3 + \frac{55}{175} We started with rough estimations, but now we're going to tenths: 34^4 = 1336336 35^4 = 1500625 1500625 - 1336336 = 164289 1360000 - 1336336 = 23664 \sqrt [ 4 ]{ 136 } \approx 3.4 + \frac{... 1 There is also an (individual) powerseries solution (Puisieux-series) for each n separately. Unfortunately that series have a little radius of convergence (if nonzero at all), but might be summable using Euler-summation. I'll give an example for n=3. (More examples are in my small treatize on my webspace) Let's define our basic function$$ v = f_3(u) = ... 2 If$\sqrt{n-1}+\sqrt{n+1}$is rational and$n>1$then$(\sqrt{n-1}+\sqrt{n+1})^2=2n+2\sqrt{n^2-1}$is rational. Then$\sqrt{n^2-1}$is rational. Then$n^2-1=\frac{p^2}{q^2}$, where$p,q \in \mathbb{N}$,$gcd(p,q)=1$. Then$q=1$. Then$n^2-1=p^2$. But$n^2-1=p^2$hasn't solutions. 2 The sum of two irrational numbers can be rational (e.g.$x,\,q-x$with$q\in\Bbb Q\not\owns x$). Their strategy was to note that, since$\sqrt{n-1}\pm\sqrt{n+1}\in\Bbb Q$, taking linear combinations thereof gives$\sqrt{n\pm 1}\in\Bbb Q$. 1 By Eisenstein's criterion and Gauss' lemma,$x^3+y^3+z^3$is irreducible in$k[x,y,z]$for any field$k$not of characteristic$3$. So, for example, it is irreducible in$\mathbb C[x,y,z]$. 2 One thing we can do is render the third superroot as an iteration of second superroots. Let$x^{x^x}=a$. Then$(x^x)^{(x^x)}=a^x$and we take two square superroots to get a fixed point iteration:$\color{blue}{x=\sqrt{\sqrt{a^x}_s}_s}$If we put$a>1$and$x=1$on the right side, we get$x=\sqrt{\sqrt{a}_s}_s$as our next iteration, and this will ... 0 $$\lim_{h\to0^+}\frac{\sqrt{\dfrac 1h+1}-\sqrt{\dfrac 1h}}{\sqrt h}=\lim_{h\to0}\frac{\sqrt{h+1}-1}h=\left.(\sqrt{x+1})'\right|_{x=0}=\frac12.$$ 1 $$\lim_{n \to \infty}\sqrt{n} \cdot (\sqrt{n+1} - \sqrt{n})= \lim_{n \to \infty}\sqrt{n} \cdot \frac{(\sqrt{n+1} - \sqrt{n})}{1}= \lim_{n \to \infty}\sqrt{n} \cdot \frac{(\sqrt{n+1} - \sqrt{n})(\sqrt{n+1} + \sqrt{n})}{(\sqrt{n+1} + \sqrt{n})}= \lim_{n \to \infty}\sqrt{n}\cdot \frac{n+1-n}{\sqrt{n+1}+\sqrt n}= \lim_{n \to \infty}\sqrt{n}\cdot \frac1{\sqrt{n+... 1 Hint$$\sqrt{n+1}-\sqrt n=\dfrac{n+1-n}{?}$$Now set 1/n=h,h\to0^+ Alternatively$$\sqrt n(\sqrt{n+1}-\sqrt n)=\lim_{h\to0}\dfrac{\sqrt{1+h}-1}{h}$$Set$\sqrt{1+h}=u$5 Hint:$\displaystyle\sqrt{n+1}-\sqrt n=\frac1{\sqrt{n+1}+\sqrt n}=\frac1{\sqrt n}\times\frac1{\sqrt{1+\frac1n}+1}.\$

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Your expression can be verified by the following identity: \begin{align} & \; \sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c} \\ = & \: \dfrac {(a+b+c)^3-27abc} {\left( \sqrt[3]{a^2}+\sqrt[3]{b^2}+\sqrt[3]{c^2}- \sqrt[3]{bc}-\sqrt[3]{ca}-\sqrt[3]{ab} \, \right) \left[ (a+b+c)^2+ 3(a+b+c)\sqrt[3]{abc}+ 9\...

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