Year 12s: what will be the first way you research your future options? >>

Year 12s: what will be the first way you research your future options? >>

Year 12s: what will be the first way you research your future options? >>

Simple Inversing of a Function

Yeah I know this is trivial, but it's always the trivial stuff I get wrong. I want to know if I did this right. If I didn't, I'd like some correction please :smile:

:smile:

y = 255(\frac{x}{255})^\gamma

x = 255(\frac{y}{255})^\gamma

\frac{x}{255} = (\frac{y}{255})^\gamma

ln(\frac{x}{255}) = \gamma ln(\frac{y}{255})

\frac{1}{\gamma} ln(\frac{x}{255}) = ln(\frac{y}{255})

e^{\frac{1}{\gamma} ln(\frac{x}{255})} = \frac{y}{255}

\frac{x}{255}e^{\frac{1}{\gamma}} = \frac{y}{255}

y = xe^{\frac{1}{\gamma}}

(edited 13 years ago)

Original post by strawberry

Yeah I know this is trivial, but it's always the trivial stuff I get wrong. I want to know if I did this right. If I didn't, I'd like some correction please :smile:

:smile:

y = 255(\frac{x}{255})^\gamma

x = 255(\frac{y}{255})^\gamma

\frac{x}{255} = (\frac{y}{255})^\gamma

ln(\frac{x}{255}) = \gamma ln(\frac{y}{255})

\frac{1}{\gamma} ln(\frac{x}{255}) = ln(\frac{y}{255})

e^{\frac{1}{\gamma} ln(\frac{x}{255})} = \frac{y}{255}

\frac{x}{255}e^{\frac{1}{\gamma}} = \frac{y}{255}

y = xe^{\frac{1}{\gamma}}

There's a mistake going from the third line from the bottom to the next line.
That should be:

Unparseable latex formula:

\frac{1}{\gamma}\ln (\frac{x}{255})} = \ln(\frac{y}{255})\implies (\frac{x}{255})^{\frac{1}{ \gamma }} = \frac{y}{255}

.

(edited 13 years ago)

OP

Farhan.Hanif93

x

thanks!

y = 255(\frac{x}{255})^\gamma

x = 255(\frac{y}{255})^\gamma

\frac{x}{255} = (\frac{y}{255})^\gamma

ln(\frac{x}{255}) = \gamma ln(\frac{y}{255})

\frac{1}{\gamma} ln(\frac{x}{255}) = ln(\frac{y}{255})

e^{\frac{1}{\gamma} ln(\frac{x}{255})} = \frac{y}{255}

(\frac{x}{255})^\frac{1}{\gamma} = \frac{y}{255}

y = 255(\frac{x}{255})^{\frac{1}{\gamma}}

(edited 13 years ago)

The kind of problems you people dismiss as trivial give me aneurysms. Consider yourselves fortunate.

Original post by strawberry

thanks!

y = 255(\frac{x}{255})^\gamma

x = 255(\frac{y}{255})^\gamma

\frac{x}{255} = (\frac{y}{255})^\gamma

ln(\frac{x}{255}) = \gamma ln(\frac{y}{255})

\frac{1}{\gamma} ln(\frac{x}{255}) = ln(\frac{y}{255})

e^{\frac{1}{\gamma} ln(\frac{x}{255})} = \frac{y}{255}

\frac{x}{255}^(\frac{1}{\gamma}) = \frac{y}{255}

y = 255(\frac{x}{255})^{\frac{1}{\gamma}}

:yep:

OP

Original post by Eldedu

The kind of problems you people dismiss as trivial give me aneurysms. Consider yourselves fortunate.

I meant trivial as in "everybody should be able to do these right the first time".

Original post by strawberry

y = 255(\frac{x}{255})^{\frac{1}{\gamma}}

Doing gamma correction for an 8 bit display, by any chance? :smile:

:smile:

It's simplest to normalise x and y to the range [0, 1].

That is, let

X = \frac{x}{255}, Y = \frac{y}{255}

, then

Y = X^\gamma

So

X = Y^{1/\gamma}

so

\frac{x}{255} = \left(\frac{y}{255}\right)^{1/\gamma}

so

x = 255 \left(\frac{y}{255}\right)^{1/\gamma}

Edit: note that a key check here is to make sure you get the right answers when y = 0 and y = 255.

OP

lol why YES! image processing :smile:

:smile:

Original post by strawberry

lol why YES! image processing :smile:

:smile:

I work in the field.

OP

I'm interested in DSP applications and stuff, but more towards the communications side of things, not image processing - though I think it's cool

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