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Fun Maths Questions (1 Viewer)

hyparzero

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Found the following questions quite interesting:

1. Find ii where i = root(-1)

2. The number a is randomly selected from the set {0,1,2,3,...,98,99}. The number b is selected from the same set.

What is the probabilty that the number 3a+7b is an 16 digit number?
 
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Trev

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Just having a go at it, i<sup>i</sup>:
Since e<sup>ix</sup>=cisx, when x=pi; e<sup>i.pi</sup> = -1.
ln[e<sup>i.pi</sup>] = ln[-1]
So i.pi = ln[-1].

Let k=i<sup>i</sup>
k<sup>2</sup>=i<sup>2i</sup>=(-1)<sup>i</sup>
ln[k<sup>2</sup>]=ln[(-1)<sup>i</sup>]
2ln[k]=i.ln[-1], since ln[-1]=i.pi then:
2ln[k]=i.i.pi
k=e<sup>-pi/2</sup>
&there4; i<sup>i</sup>=e<sup>-pi/2</sup>.
Yeah?
 

Bobness

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I couldn't resist it ...

Maths is not fun

That is all :eek:
 

hyparzero

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Trev said:
Just having a go at it, i<sup>i</sup>:
Since e<sup>ix</sup>=cisx, when x=pi; e<sup>i.pi</sup> = -1.
ln[e<sup>i.pi</sup>] = ln[-1]
So i.pi = ln[-1].

Let k=i<sup>i</sup>
k<sup>2</sup>=i<sup>2i</sup>=(-1)<sup>i</sup>
ln[k<sup>2</sup>]=ln[(-1)<sup>i</sup>]
2ln[k]=i.ln[-1], since ln[-1]=i.pi then:
2ln[k]=i.i.pi
k=e<sup>-pi/2</sup>
&there4; i<sup>i</sup>=e<sup>-pi/2</sup>.
Yeah?
Yup absolutely correct. However it seems e<sup>ix</sup>=cisx might have to be be proved using the power series beforehand.

Anyone want to take a shot at the question 2?
 

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