# Taylor series

#### Tranquillity

##### Member
Hello guys,

I am finding difficulties expanding log[(x+epsilon)^2 + y^2] as epsilon->0.

Could anyone help me with the expansion?

Kindest regards

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Hello guys,

I am finding difficulties expanding log[(x+epsilon)^2 + y^2] as epsilon->0.

Could anyone help me with the expansion?

Kindest regards
Welcome to MHB, Tranquillity!

You would have to expand around $x^2+y^2$.
To do so, it's easiest if you use:

$\log((x+\varepsilon)^2+y^2)=\log\left({x^2+y^2+2x \varepsilon +\varepsilon^2 \over x^2 + y^2}(x^2+y^2)\right) = \log\left(1 + {2x\varepsilon +\varepsilon^2 \over x^2 + y^2}\right) + \log(x^2+y^2)$

Do you know how to expand that?

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#### Tranquillity

##### Member
Welcome to MHB, Tranquillity!

You would have to expand around $x^2+y^2$.
To do so, it's easiest if you use:

$\log((x+\varepsilon)^2+y^2)=\log({x^2+y^2+2x \varepsilon +\varepsilon^2 \over x^2 + y^2}(x^2+y^2)) = \log(1 + {2x\varepsilon +\varepsilon^2 \over x^2 + y^2}) + \log(x^2+y^2)$

Do you know how to expand that?
I could use that log(1+a) = a-a^2/2+... for the first parenthesis right?

But then how should I expand log(x^2+y^2)?

Thanks for that!

#### Klaas van Aarsen

##### MHB Seeker
Staff member
I could use that log(1+a) = a-a^2/2+... for the first parenthesis right?

But then how should I expand log(x^2+y^2)?

Thanks for that!
Right!

You're not supposed to expand $\log(x^2+y^2)$.
It's supposed to be fixed.

#### Tranquillity

##### Member
Right!

You're not supposed to expand $\log(x^2+y^2)$.
It's supposed to be fixed.
So my overall expansion should be [(2*x*epsilon + epsilon^2) / (x^2+y^2) ] + log(x^2+y^2) ?

Thanks for all the help, it is greatly appreciated

#### Klaas van Aarsen

##### MHB Seeker
Staff member
So my overall expansion should be [(2*x*epsilon + epsilon^2) / (x^2+y^2) ] + log(x^2+y^2) ?

Thanks for all the help, it is greatly appreciated
In an expansion you get $\displaystyle\left[{2*x*\varepsilon + \varepsilon^2 \over x^2+y^2} - \frac 1 2 \left({2*x*\varepsilon + \varepsilon^2 \over (x^2+y^2)}\right)^2 + ...\right.$

At first order this simplifies to:

$\displaystyle{2*x*\varepsilon \over x^2+y^2} + ...$

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#### Tranquillity

##### Member
Thanks for all the help!

Kindest regards