Sum of Functions

Let $f(x)=x-\dfrac{1}{x}$.

(a) If $f(x)+f(y)=0$ , then prove or disprove that $f(x^n)+f(y^n)=0$ for all integers $n$ .

(b) If $f(x)+f(y)+f(z)=0$ , then prove or disprove that $f(x^n)+f(y^n)+f(z^n)=0$ for all integers $n$ .

(c) Do the above two problems with the added restriction that $x,y,z$ are positive.

#Algebra #Functions #Proofs #Zero #Powers

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
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Comments

(a) Counterexample: $y=-x$ and $n=2a$ for some positive integer $a$ . $f(x)+f(-x)=x-\dfrac{1}{x}-x+\dfrac{1}{x}=0$ , but $f(x^n)+f((-x)^n)=f(x^{2a})+f(x^{2a})=x^{2a}-\dfrac{1}{x^{2a}}+x^{2a}-\dfrac{1}{x^{2a}}\neq0$ .

(b) Again, the counterexample is $y=-x$ , $z=1$ , and $n=2a$ for some positive integer $a$ . $f(x)+f(-x)+f(1)=x-\frac{1}{x}-x+\frac{1}{x}+1-1=0$ , but $f(x^n)+f((-x)^n)+f(1^n)=f(x^{2a})+f(x^{2a})=f(1)=x^{2a}-\dfrac{1}{x^{2a}}+x^{2a}-\dfrac{1}{x^{2a}}+1-1\neq0$ .

(c) (part 1): Since $x,y$ are positive and $f(x)+f(y)=0$ , $x-\dfrac{1}{x}+y-\dfrac{1}{y}=0$ , and $x+y=\dfrac{1}{x}+\dfrac{1}{y}$ . Adding the fractions on the RHS gives $x+y=\dfrac{x+y}{xy}$ so $1=\dfrac{1}{xy}$ or $xy=1$ . Therefore, $f(x^n)+f(y^n)=x^n-\dfrac{1}{x^n}+y^n-\dfrac{1}{y^n}=x^n-\dfrac{1}{x^n}+\dfrac{1}{x^n}-x^n=0$ .

Jeffery Li - 6 years, 11 months ago

Sorry, your answer for part (a) is wrong. Also, the restriction of part (c) actually matters.

EDIT: thanks, fixed!

Daniel Liu - 6 years, 11 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$