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7 votes
Accepted

Solving the system $\frac{xy}{ay+bx}=c$, $\frac{xz}{az+cx}=b$, $\frac{yz}{bz+cy}=a$ for $x$, $y$, $z$

From the given equations you have $ xy = bc x + a c y $ $ xz = bc x + ab z $ $ yz = a c y + ab z $ Divide the first equation by $xy$ , the second by $xz$ and the third by $yz$, you get $ 1 = bc \left( ...
Quadrics's user avatar
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6 votes

$\frac{1}{x} + \frac{1}{x-1} + \ldots + \frac{1}{x-n} = 0$ has only one root $(0;1)$

If it has two roots then its derivative will have to vanish at some point (by Rolle's Theorem). But the derivative is striclty negative at every point of $(0,1)$.
geetha290krm's user avatar
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6 votes
Accepted

Quickly finding $(-M)^{ATH}$ from $M+A+T+H=10$, $M-A+T+H=6$, $M+A-T+H=4$, $M+A+T-H=2$

$$\begin{align} M+A+T+H &=10 \tag1\\ M-A+T+H &=\;6 \tag2\\ M+A-T+H &=\;4 \tag3\\ M+A+T-H &=\;2 \tag4\\ \end{align}$$ Then the symmetry of the coefficients of the variables in each ...
Sahaj's user avatar
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5 votes
Accepted

How can a modulus produce a negative value? For example, why is $|y|=–y$ when $y<0$?

A minus sign in front of a variable doesn't mean it's negative. It means we "reverse" its sign. If $x$ is negative, then $-x$ is positive. In this case, both $y$ and $c^5$ are negative, so ...
Arthur's user avatar
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4 votes
Accepted

A system of equations of power sums for 3 variables

HINT If $\alpha = x + y + z$, $\beta = xy + xz + yz$ and $\gamma = xyz$, the proposed system of equations becomes: \begin{align*} \begin{cases} \alpha = 2\\\\ \alpha^{2} - 2\beta = 6\\\\ \alpha^{3} - ...
Átila Correia's user avatar
3 votes

Why is interpolating $y=g(x)$ then applying $h(y)$ not equivalent to interpolating $h(g(x))$?

Interpolation treats your function as if it's a line (i.e. linear function) between your two interpolation endpoints. If your function $h( )$ preserved "being a line" then you would get the ...
JonathanZ's user avatar
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3 votes

$\frac{1}{x} + \frac{1}{x-1} + \ldots + \frac{1}{x-n} = 0$ has only one root $(0;1)$

Using calculus, $\frac{d}{dx}\frac{1}{x-\alpha}=-\frac{1}{(x-\alpha)^2}$, so the derivative of the original function is always negative.Therefore, it can have only one real zero in the region $(0,1)$ ...
Zima's user avatar
  • 3,382
3 votes

Two numbers written on a board get replaced

The problem with considering the sum, is that this can either increase or decrease with each turn. If $a<b$, then the sum increases. If $a>b$, then the sum decreases. And so it is unclear ...
Adam Dougall's user avatar
2 votes

$\frac{1}{x} + \frac{1}{x-1} + \ldots + \frac{1}{x-n} = 0$ has only one root $(0;1)$

As Zima wrote in comment, you proved there is at least one root. To prove there is at most one root, note that every term monotonically decreases for $x \in (0, 1)$ - so the sum also monotonically ...
mihaild's user avatar
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2 votes

Quickly finding $(-M)^{ATH}$ from $M+A+T+H=10$, $M-A+T+H=6$, $M+A-T+H=4$, $M+A+T-H=2$

Yes. Let $L_{i}$ denote the $i$-th equation (counted from bottom to top). Notice that: $L_{2}=L_{1}-2A$ $L_{3}=L_{1}-2T$ $L_{4}=L_{1}-2H$ So, it's easy to see: $10-2A = 6$ $10-2T = 4$ $10-2H = 2$ ...
Almeida's user avatar
  • 595
2 votes

Quickly finding $(-M)^{ATH}$ from $M+A+T+H=10$, $M-A+T+H=6$, $M+A-T+H=4$, $M+A+T-H=2$

The sum of the last three equations has coefficient 1 on any of the ATH variables. This means $2=(2)+(3)+(4)-(1)= 2M$, that implies $M=1$. In particular, we are only interested in the parity of $ATH$, ...
Andrea Marino's user avatar
1 vote
Accepted

Why is interpolating $y=g(x)$ then applying $h(y)$ not equivalent to interpolating $h(g(x))$?

First, an objection: I don't think the title formulation matches the context provided. In the first case, you approximate $V(I)\approx V_L(I)$ via linear interpolation and then compute $V_L(I)/I$. In ...
Semiclassical's user avatar
1 vote

How can a modulus produce a negative value? For example, why is $|y|=–y$ when $y<0$?

the modulus is literally designed for removing the minus. You clearly don't know that the "minus" sign has three meanings: It can be used for indicating that a number is negative: if you ...
Dominique's user avatar
  • 2,527

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