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**Kenjiska****Guest**

Please,tell me the value of these-

i !

(-1) !

(1+i) !

(.1) !

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 102,917

Hi;

(-1) ! does not exist. All the negative integers of the factorial function are equal to infinity.

I would use a Taylor series for some of them. How much accuracy do you need.

Basically though you would use a computer or Wolfram Alpha to look those up. If your teacher wants to see some method then how is it you do not know that method? If I use series expansion he/she may not want that method used. So provide me with what method is to be used.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.** **A number by itself is useful, but it is far more useful to know how accurate or certain that number is.**

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**Kenjiska****Guest**

Show me using the taylor series

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 102,917

Hi;

Those are all difficult but here goes nothing!

Let's try (.1)! first. For that we use a series and a trick.

It is in nested or Horner form for fast computation. The series is best for values >=5 so we put in z = 5.1

We take the exponential of both sides and get.

Now to get .1! we use a simple relation:

So

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.** **A number by itself is useful, but it is far more useful to know how accurate or certain that number is.**

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**Kenjiska****Guest**

What about the others

(1+i)!

i !

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 102,917

For i! use this truncated series:

Substituting z = i you get

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.** **A number by itself is useful, but it is far more useful to know how accurate or certain that number is.**

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**zetafunc.****Guest**

Also, for (1+i)!, you can find that using bobbym's answer, since