i^i

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AntiviruShahriar
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i^i

Unread post by AntiviruShahriar » Thu Jan 06, 2011 8:43 pm

$e^{ix} = cos x + i \cdot sin x $ ব্যবহার করলে আমরা $i^i$ এর মান বের করতে পারি। এবং মান পাই $i^i=e^{\frac{- \pi}{2}}$
$i$ একটা অবাস্তব সংখ্যা। একটা অবাস্তব সংখ্যার অবাস্তব তম ঘাত কিভাবে বাস্তব হইতে পারে??? :shock:
:o

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Moon
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Re: i^i

Unread post by Moon » Thu Jan 06, 2011 8:47 pm

হতে যে পারে সেইটা তো তুমি নিজেই প্রমাণ করলা! $e^{i\pi/2}=i \Longrightarrow i^i=e^{\frac{- \pi}{2}}$ :)
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AntiviruShahriar
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Re: i^i

Unread post by AntiviruShahriar » Thu Jan 06, 2011 8:59 pm

বুঝলাম না :?
$e^{i \frac{\pi}{2}}$ কি অবাস্তব না বাস্তব???এইটা নিয়া ও তো এখন সন্দেহ হইতাসে :(
সবই এখন সন্দেহজনক লাগতেসে। :cry: পড়তে যাই, আমার মাথা কেমন ভারি ভারি লাগতেসে।

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Moon
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Re: i^i

Unread post by Moon » Thu Jan 06, 2011 9:11 pm

বেশি সন্দেহ করার কিছু নাই।

তবে আমারও একটা প্রশ্ন আছে বড়দের প্রতি। $a^b$ এর মানে কী? $b \in \mathbb{Z}$ হইলে আমরা বলতে পারি যে $b$ সংখ্যক $a$ কে গুণ করা। কিন্তু অন্যান্য ক্ষেত্রে formal definition টা কী? :?:
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Re: i^i

Unread post by tanvirab » Thu Jan 06, 2011 11:52 pm

$i^i$ বাস্তব হইতে কোনো সমস্যা নাই। এবং বাস্তব যে সেইটাতো তুমি প্রমাণই করলা।

যেকোনো দুইটা জটিল সংখ্যা $a,b$ এর জন্য $a^b = e^{b \mbox{ } log(a)}$ যেইখানে exponential এর সংজ্ঞা দেয়া হইসে power series এর মাধ্যমে এবং logarithm এর সংজ্ঞা দেয়া হইসে exponential এর inverse হিসেবে।

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Re: i^i

Unread post by nayel » Fri Jan 07, 2011 12:10 am

The definition that I know of $x^a$ is $x^a=e^{a\ln x}=\exp(a\ln x)$, where $\exp$ is defined as
\[\exp(x)=\sum_{n=0}^\infty\frac{x^n}{n!}.\]
Note that this definition generalizes to complex powers as well.

However, from your point of view: If $a$ is a positive integer, we regard $x^a$ as $\underbrace{x\cdot x\cdots x}_{a\text{ times}}$. From here it's clear that
\[x^{m+n}=x^m\cdot x^n\qquad (*)\]
for positive integers $m,n$. We can define negative powers from $(*)$, for example $x^{n-1}=x^{-1}\cdot x^n$ so $x^{-1}=1/x$. From $(*)$ we can also see that
\[(x^m)^n=x^{mn}\qquad (**)\]
for integers $m,n$. Now, from $(**)$ we can define $x^{1/m}$ as $y$, which is the solution to $y^m=(x^{1/m})^m=x$. Thus, for rational $a$, i.e. $a=p/q$ we have $x^a=(x^p)^{1/q}$. But this stops here, because we can't define irrational powers in this way.

P.S. Tanvir bhai beat me :|
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Moon
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Re: i^i

Unread post by Moon » Fri Jan 07, 2011 12:17 am

ধন্যবাদ। ভাল একটা জিনিস শিখলাম। :)

তানভীর ভাই:
তবে এইখানে একটা পুরানো প্রশ্ন চলে আসছে। সেইটা হল জটিল সংখ্যার লগারিদম কেমন করে define করব? exponential এর inverse power series হিসেবে?
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Re: i^i

Unread post by Moon » Fri Jan 07, 2011 12:19 am

oups...sorry...I think that we have posted in Bangla in Olympiad forum. BTW I think that this topic is suitable for college forum. It has nothing to with Olympiad problems/NT. moving it. :)
"Inspiration is needed in geometry, just as much as in poetry." -- Aleksandr Pushkin

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Re: i^i

Unread post by tanvirab » Fri Jan 07, 2011 3:10 am

Logarithm is defined as the inverse of the exponential function. So, $log(z) = w$ such that $e^w = z$.

But, one needs to be careful since the exponential of complex numbers is not a one-to-one function. It's a many-to-one function and the inverse is not automatically a well defined map. So, we need to make a choice. (This is similar to the inverse trigonometric functions, where we need to choose a principle value)

The usual convention is to note that, if we represent complex numbers in polar form, $z = re^{i\theta}$ and restrict $\theta$ to $[0, 2\pi)$ then the above definition is equivalent to $log(z) = log(r) + i \theta$. ($r$ is positive real, the function is not defined for $z=0$)

This function can also be expressed as a power series, and it is the obvious generalization of the Taylor series for the usual logarithm function. But, there is a technical complication in doing this, known as branch cuts;

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