User:Egm6321.f10.team2.oztekin/exp

=Exponantial Of Matrix=

Taylor series are consist of infinite components of function and its derivatives.The series are centered on any number that is interested in.


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$$  \displaystyle f(a)+\frac{{f}'(a)}{1!}\left ( x-a \right )+\frac{{f}''(a)}{2!}\left ( x-a \right )^{2}+\frac{f^{3}(a)}{3!}\left ( x-a \right )^{3}........... $$     (N)
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For f(x)=expx can be characterized as the same way.


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$$  \displaystyle e^{x}=\sum_{n=0}^{\infty }\frac{x^{n}}{n!}=1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+........... $$     (N)
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And if we substitute matrix A into x we concluded that:


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$$  \displaystyle e^{A}_{n\times n}=\sum_{n=0}^{\infty }\frac{x^{n}}{n!}=I_{n\times n}+A_{n\times n}+\frac{A_{n\times n}^{2}}{2!}+\frac{A_{n\times n}^{3}}{3!}+........... $$     (N)
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=Problem 6=

Given
Example given in MT15-3 as


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$$  \displaystyle x({y}')^{2}+y{y}'+xy{y}''=0 $$     (N)
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$$  \displaystyle g(x,y,p)=x({y}')^{2}+y{y}' $$     (N)
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$$  \displaystyle f(x,y,p)=xy $$     (N)
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$$  \displaystyle h_{x}+h_{y}p=0 $$     (N)
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Find
Without assuming h=const. discuss other solutions.

Solution
For the condition given Eq. both terms must be zero. So


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$$  \displaystyle h_{x}=0 $$     (N)
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$$  \displaystyle h_{y}p=0 $$     (N)
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And we know that h=h(x,y).Possible outcomes for the derivations;


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$$  \displaystyle h_{x}=h(x,y) $$     (N)
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$$  \displaystyle h_{y}=h(x,y) $$     (N)
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By this way we can not find common h for stisfy Eq.