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Find all functions, \(f : \mathbb{R} \rightarrow \mathbb{R},\) such that

\[f(x+y) - 2f(x-y) + f(x) - 2f(y) = y-2.\]

Note by Victor Loh 3 years, 2 months ago

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Let \(P(x,y)\) be the statement \(f(x+y)-2f(x-y)+f(x)-2f(y)=y-2\).

\(P(0,0)\implies f(0)=1\)

\(P(x,x)\implies f(2x)=f(x)+x\)

\(P(x,-x)\implies f(x)+2f(-x)=3-x\stackrel{x\to -x}\implies f(-x)+2f(x)=3+x\)

\(\implies f(x)+f(-x)=2\)

The last one was got by adding the 2 previous equations.

\((2-f(x))+2f(x)=3+x\implies f(x)=x+1\)

Thus \(f(x)=x+1\) is the only possible solution. After checking it we see it works.\(\square\)

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Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

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TopNewestLet \(P(x,y)\) be the statement \(f(x+y)-2f(x-y)+f(x)-2f(y)=y-2\).

\(P(0,0)\implies f(0)=1\)

\(P(x,x)\implies f(2x)=f(x)+x\)

\(P(x,-x)\implies f(x)+2f(-x)=3-x\stackrel{x\to -x}\implies f(-x)+2f(x)=3+x\)

\(\implies f(x)+f(-x)=2\)

The last one was got by adding the 2 previous equations.

\((2-f(x))+2f(x)=3+x\implies f(x)=x+1\)

Thus \(f(x)=x+1\) is the only possible solution. After checking it we see it works.\(\square\)

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