Function Challenge

For the function listed below, try expressing them in a single formula.

Example :

f(x)={xif x0xif x<0 f(x) = \begin{cases} -x & \quad \text{if } x \geq 0\\ x & \quad \text{if } x<0\\ \end{cases}

Answer : f(x)=xf(x) = -|x|

Challenge 1 :

f(x)={1if x<01if x>0 f(x) = \begin{cases} 1 & \quad \text{if } x < 0\\ -1 & \quad \text{if } x>0\\ \end{cases}

Challenge 2 :

f(x)={0if x0xif x>0 f(x) = \begin{cases} 0 & \quad \text{if } x\leq 0\\ x & \quad \text{if } x>0\\ \end{cases}

Challenge 3 :

f(x,y)={xyif x>yx+yif x<y f(x,y) = \begin{cases} x-y & \quad \text{if } x > y\\ x+y & \quad \text{if } x < y\\ \end{cases}

Comment your answer and vote up good functions!

Note by Christopher Boo
5 years, 6 months ago

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1 vote

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1)xx-\dfrac{x}{|x|} 2)x2+x2\left|\dfrac{x}{2}\right|+\dfrac{x}{2} 3)(xy)(x+y)(x+xyyxy)\dfrac{(x-y)(x+y)}{\left(x+\dfrac{|x-y|y}{x-y}\right)}

Aareyan Manzoor - 5 years, 6 months ago

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  1. f1(x)=xxf_1(x)=-\frac {|x|}{x}

  2. f2(x)=x+x2f_2(x)=\frac {|x|+x}{2}

  3. We can find a solution by using the solutions for challenge 1 and 2:

I don't know how to type arrays so let 1,21,2 denote the conditions yx<0,yx>0y-x<0, y-x>0 respectively.

Note that f2(yx)+x=1.x,2.yf_2(y-x)+x=1.x, 2.y and f2(yx)y=1.y,2.xf_2(y-x)-y=1.-y, 2.-x.

Further, f1(yx)(f2(yx)y)=1.y,2.xf_1(y-x)(f_2(y-x)-y)=1.-y, 2.x

Hence f(x)=f2(yx)+x+f1(yx)(f2(yx)y)=1.xy,2.x+yf(x)=f_2(y-x)+x+f_1(y-x)(f_2(y-x)-y)=1.x-y, 2.x+y

Another way is by considering 1+f1(yx)2=1.1,2.0,1f1(yx)2=1.0,2.1\frac {1+f_1(y-x)}{2}=1.1, 2.0, \frac {1-f_1(y-x)}{2}=1.0, 2.1.

Thus, f(x)=(xy)1+f1(yx)2+(x+y)1f1(yx)2f(x)=(x-y)\frac {1+f_1(y-x)}{2}+(x+y)\frac {1-f_1(y-x)}{2}

Xuming Liang - 5 years, 6 months ago

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For the first function, f(x)=1{xx<0}1{xx>0}f(x) = 1_{\{x|x<0\}} - 1_{\{x|x>0\}} where f:R{0}Rf : \mathbb{R} \setminus \{0\} \to \mathbb{R}. Because 1S1_S is the characteristic function that returns 11 if the argument is in SS and 00 otherwise.

If you want me to nitpick even more, those given aren't functions since you don't specify the domain (and the range, but the range usually can be inferred assuming it fits the whole domain). If I assume the domain of the first function is {1}\{1\}, then my "formula" can be as simple as f(x)=1f(x) = -1 (because it only needs to work for the single number x=1x = 1 in the domain).

A more interesting question is why you want to look specifically for just a single formula, and not just leave the function piecewise which is much easier to understand compared to the single formula forms?

Ivan Koswara - 5 years, 6 months ago

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Compute the value of

sin10sin30sin50sin70\sin 10^\circ \sin 30^\circ \sin 50^\circ \sin 70^\circ

without using a calculator.

Why shouldn't we use a calculator when we can? Isn't using a calculator faster and is less likely to have errors?

Christopher Boo - 5 years, 6 months ago

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You can't use calculator there because the result is rational. A calculator might round/truncate the result midway. I did throw it to WolframAlpha anyway, though.

Ivan Koswara - 5 years, 6 months ago

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@Ivan Koswara @Ivan Koswara, imagine you are solving a hard equation. yo just spend a huge time on it and are feeling proud of what you did! You want to show other people this cool eqn but when you show this to yor classmates, who suck at maths, say "waste of time, i solved it in a second by putting it in wolfram alpha. i am smarter then you."

But some stuff need analytical approach despite being easily solvable by numerical/computational approach. please think about this.

the "imagine" is taken from real life!

Aareyan Manzoor - 5 years, 6 months ago

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