- #1
Appleton
- 91
- 0
Homework Statement
Prove that, if x is so small that
[itex]x^6[/itex] and higher powers of x may be neglected, then [itex]\frac{e^{2x}-1}{e^{2x}+1}\approx x-\frac{x^3}{3}+\frac{2x^5}{15}[/itex]
Homework Equations
The Attempt at a Solution
[/B]
[itex]\\
\frac{e^{2x}-1}{e^{2x}+1}\approx\frac{2x+\frac{(2x)^2}{2}+\frac{(2x)^3}{3!}+
\frac{(2x)^4}{4!}+
\frac{(2x)^5}{5!}}{2+2x+\frac{(2x)^2}{2}+\frac{(2x)^3}{3!}+
\frac{(2x)^4}{4!}+
\frac{(2x)^5}{5!}}
\\
\approx
\frac{x+x^2+\frac{2x^3}{3}+
\frac{x^4}{3}+
\frac{2x^5}{15}}{1+x+x^2+\frac{2x^3}{3}+
\frac{x^4}{3}+
\frac{2x^5}{15}}
\\
\approx
\frac{x-\frac{x^3}{3}+\frac{2x^5}{15}+x^2+x^3+
\frac{x^4}{3}}{1+x+x^2+\frac{2x^3}{3}+
\frac{x^4}{3}+
\frac{2x^5}{15}}
\\
[/itex]
At this point I am unable to isolate
[itex]
x-\frac{x^3}{3}+\frac{2x^5}{15}
\\
[/itex]
The only other approaches that occur to me are:
[itex]
e^{2x}(e^{2x} +1)^{-1}-(e^{2x}+1)^{-1}
[/itex]
From which I could yield a series of ascending powers of [itex]e^{2x}[/itex], however this doesn't seem to lead anywhere useful.
Alternatively:
[itex]\frac{e^{2x}-1}{e^{2x}+1}\approx 1-\frac{1}{1+x+x^2+\frac{2x^3}{3}+
\frac{x^4}{3}+
\frac{2x^5}{15}}
\\
[/itex]
was another expression I looked at but didn't seem to go anywhere.
Part a) of this question asks me to express [itex]ln{\frac{1+x}{1-x}}[/itex] as a series of ascending powers of x up to and including [itex]x^5[/itex]. I got [itex]2(x+\frac{x^3}{3}+\frac{x^5}{5}).[/itex] Given that the two problems form 2 parts of the same question and the form of the expressions look so similar and ln and e are closely related, I would expect there to be some strong relationship between the route to finding the answers to both these questions.