Solve Exponential Function: e^x, e^(-1), Sin x, Cos x

Kenji Liew
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Homework Statement



This topic is under linear system differential equation.Solve the system by using exponential method. Just want to ask the expansion of exponential function

Homework Equations



e^x=1+x+(x^2)/2!+(x^3)/3!+...

The Attempt at a Solution


then how about the e^(-1)=?
Besides what is the function of sin x and cos x in continued function (such in e^x)?
Thanks!
 
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Kenji Liew said:

Homework Statement


The Attempt at a Solution


then how about the e^(-1)=?
Besides what is the function of sin x and cos x in continued function (such in e^x)?
Thanks!

I'm not really for sure what your question even is. What problem are you trying to solve? Are you asking for the power series for sine and cosine?
\begin{align*}<br /> \sin x &amp;= x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots \\<br /> \cos x &amp;= 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots<br /> \end{align*}

Or are you asking for the following identities?
\begin{align*}<br /> \sin x &amp;= \frac{e^{ix}-e^{-ix}}{2i} \\<br /> \cos x &amp;= \frac{e^{ix} + e^{-ix}}{2}<br /> \end{align*}
 
Kenji Liew said:

Homework Statement



This topic is under linear system differential equation.Solve the system by using exponential method. Just want to ask the expansion of exponential function

Homework Equations



e^x=1+x+(x^2)/2!+(x^3)/3!+...

The Attempt at a Solution


then how about the e^(-1)=?
Besides what is the function of sin x and cos x in continued function (such in e^x)?
Thanks!
e-1 = 1 + (-1) + (-1)2/2! + (-1)3/3! + ... + (-1)n/n! + ...
 
Also, you have posted what appears to be the same question twice, which is frowned upon in this and most other forums.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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