Science:Math Exam Resources/Courses/MATH103/April 2017/Question 08 (c)

MATH103 April 2017

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Question 08 (c)
Suppose the Taylor series $y=\sum _{n=0}^{\infty }a_{n}x^{n}$ solves the differential equation ${\frac {dy}{dx}}+2y=x^{2}$ with the initial condition $y(0)=4$. (c) Find the closed formula $a_{n}=f(n)$ for $n\geq 4$.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Use your answers for parts a) and b), and recall the definition of factorial. Try to find a formula for the product of all integers from 4 to n using the factorial function.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Recall that factorial is defined as follows. $0!=1$ $1!=1$ $2!=2\times 1=1$ $3!=3\times 2\times 1=6$ $4!=4\times 3\times 2\times 1=24$ $\dots$ and so on. From part b), we know that for $n\geq 4,$ $a_{n}={\frac {-2}{n}}a_{n-1}.$ So for all such $n,$ $a_{n}={\frac {-2}{n}}({\frac {-2}{n-1}}({\frac {-2}{n-2}}(\dots ({\frac {-2}{4}}a_{3})\dots )))={\frac {(-2)^{n-4+1}}{n!/3!}}a_{3}$ From part a), we know that $a_{3}=-5,$ so $a_{n}={\frac {(-2)^{n-4+1}}{n!/3!}}\times -5={\frac {-30\times (-2)^{n-3}}{n!}}$ Hence, the closed formula is $\color {blue}a_{n}={\frac {-30\times (-2)^{n-3}}{n!}}{\text{ for }}n\geq 4$