Looking for reliable answers? Westonci.ca is the ultimate Q&A platform where experts share their knowledge on various topics. Ask your questions and receive accurate answers from professionals with extensive experience in various fields on our platform. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts.
Sagot :
Answer:
[tex]\frac{x+6}{2x^2-9x+5}=-\sum_{n=0}^{\infty} [(-2)^{n}x^{n} + \frac{x^{n}}{5^{n+1}}][/tex]
when:
[tex]|x|<\frac{1}{2}[/tex]
Step-by-step explanation:
In order to solve this problem, we must begin by splitting the function into its partial fractions, so we must first factor the denominator.
[tex]\frac{x+6}{2x^2-9x+5}=\frac{x+6}{(2x+1)(x-5)}[/tex]
Next, we can build our partial fractions, like this:
[tex]\frac{x+6}{(2x+1)(x-5)}=\frac{A}{2x+1}+\frac{B}{x-5}[/tex]
we can then add the two fraction on the right to get:
[tex]\frac{x+6}{(2x+1)(x-5)}=\frac{A(x-5)+B(2x+1)}{(2x+1)(x-5)}[/tex]
Since we need this equation to be equivalent, we can get rid of the denominators and set the numerators equal to each other, so we get:
[tex]x+6=A(x-5)+B(2x+1)[/tex]
and expand:
[tex]x+6=Ax-5A+2Bx+B[/tex]
we can now group the terms so we get:
[tex]x+6=Ax+2Bx-5A+B[/tex]
[tex]x+6=(Ax+2Bx)+(-5A+B)[/tex]
and factor:
[tex]x+6=(A+2B)x+(-5A+B)[/tex]
so we can now build a system of equations:
A+2B=1
-5A+B=6
and solve simultaneously, this one can be solved by substitution, so we get>
A=1-2B
-5(1-2B)+B=6
-5+10B+B=6
11B=11
B=1
A=1-2(1)
A=-1
So we can use these values to build our partial fractions:
[tex]\frac{x+6}{(2x+1)(x-5)}=\frac{A}{2x+1}+\frac{B}{x-5}[/tex]
[tex]\frac{x+6}{(2x+1)(x-5)}=-\frac{1}{2x+1}+\frac{1}{x-5}[/tex]
and we can now use the partial fractions to build our series. Let's start with the first fraction:
[tex]-\frac{1}{2x+1}[/tex]
We can rewrite this fraction as:
[tex]-\frac{1}{1-(-2x)}[/tex]
We can now use the following rule to build our power fraction:
[tex]\sum_{n=0}^{\infty} ar^{n} = \frac{a}{1-r}[/tex]
when |r|<1
in this case a=1 and r=-2x so:
[tex]-\frac{1}{1-(-2x)}=-\sum_{n=0}^{\infty} (-2x)^n[/tex]
or
[tex]-\frac{1}{1-(-2x)}=-\sum_{n=0}^{\infty} (-2)^{n} x^{n}[/tex]
for: |-2x|<1
or: [tex] |x|<\frac{1}{2} [/tex]
Next, we can work with the second fraction:
[tex]\frac{1}{x-5}[/tex]
On which we can factor a -5 out so we get:
[tex]-\frac{1}{5(1-\frac{x}{5})}[/tex]
In this case: a=-1/5 and r=x/5
so our series will look like this:
[tex]-\frac{1}{5(1-\frac{x}{5})}=-\frac{1}{5}\sum_{n=0}^{\infty} (\frac{x}{5})^n[/tex]
Which can be simplified to:
[tex]-\frac{1}{5(1-\frac{x}{5})}=-\sum_{n=0}^{\infty} \frac{x^n}{5^(n+1)}[/tex]
when:
[tex]|\frac{x}{5}|<1[/tex]
or
|x|<5
So we can now put all the series together to get:
[tex]\frac{x+6}{2x^2-9x+5}=-\sum_{n=0}^{\infty} [(-2)^{n}x^{n} + \frac{x^{n}}{5^{n+1}}}[/tex]
when:
[tex]|x|<\frac{1}{2}[/tex]
We use the smallest interval of convergence for x since that's the one the whole series will be defined for.
Thanks for using our platform. We're always here to provide accurate and up-to-date answers to all your queries. Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. Stay curious and keep coming back to Westonci.ca for answers to all your burning questions.
