New answers tagged

0

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....


0

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 ...


4

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 ...


0

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 $1$ is 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

Let $x,y$ 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 in $A$, 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: $\displaystyle\sqrt{n+1}-\sqrt n=\frac1{\sqrt{n+1}+\sqrt n}=\frac1{\sqrt n}\times\frac1{\sqrt{1+\frac1n}+1}.$


3

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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