0

Solve $U_{n+1}=U_n+2n^2+3n+4, U_1=2$

My problem occurs when trying to find the particular equation, as the variable start to cancel out. This is my working:

Reduced form: $U_{n+1}=U_n$

Auxiliary equation: $r^{n+1}=r^n$, so $r=1$.

$C(n)=A(1^n)=A$

$P(n)=an^2+bn+c$

$a(n+1)^2+b(n+1)+c=an^2+bn+c+2n^2+3n+4$

$an^2+2an+a+bn+b+c=an^2+bn+c+2n^2+3n+4$

Considering coefficients of $n^2$: $a=a+2$ (no solutions)

Considering coefficients of $n$: $2a+b=b+3$, so $a=\frac{3}{2}$.

Considering coefficients of constants: $a+b+c=c+4$, so $b=\frac{5}{2}$

Hence, $P(n)=\frac{3}{2}n^2+\frac{5}{2}n$

General solution: $U_n=A+\frac{3}{2}n^2+\frac{5}{2}n$

Since $U_1=2$, $2=A+\frac{3}{2}+\frac{5}{2}$, so $A=-2$

Final solution: $U_n=\frac{3}{2}n^2+\frac{5}{2}n-2$.

However, this final solution does not correspond to the sequence generated by the initial inductive definition. Could any help me by explaining where my method above went wrong?

Jamminermit
  • 1,923

1 Answers1

1

$$U_{n+1}-U_n=2n^2+3n+4$$ implies that $U_n=An^3+Bn^2+Cn+D$, then $$U_{n+1}-U_n=3An^2+(3A+2B)n+(A+B+C)=2n^2+3n+4$$ $$\implies A=2/3, 3A+2B=3,A+B+C=4, \implies B=1/2,C=17/6$$ Next given that $U_1=2=A+B+C+D \implies D=-2$. Finally we get $$U_n=\frac{2n^2}{3}+\frac{n^2}{2}+\frac{17n}{6}-2$$

Z Ahmed
  • 43,235
  • Why does $U_{n+1}-U_n=2n^2+3n+4$ imply that $U_n=An^3+Bn^2+Cn+D$? – Jamminermit Jan 12 '20 at 10:41
  • It is just like this: $ dy/dx=Dy=ax^2+bx+c \implies y=ax^3/3+bx^2/2+cx+g$ (at most a cubic). Similarly, $U_{n+1}-U_n=\Delta U_n=an^2+bn+c$ means $U_n$ is at most a cubic of $n$. Interestingly if you take $U_n=pn^4+qn^3+ r n^2+sn+t $ you will get $p=0, q=2/3, r=1/2, s=17/6, t=-2.$ try out! – Z Ahmed Jan 12 '20 at 11:06
  • Thank you so much! I understand it now. – Jamminermit Jan 12 '20 at 11:46
  • Welcome, why not an upvote too!. Also a first order Rec Rel. means you have the first order difference on LHS that is $\Delta=\Delta^{1}=U_{n+1}-U_n$, similarly $\Delta^2=U_{n+2}-2 U_{n+1}+U_n$ if this equals a quadratic, then $U_n$ will be at most a fourth degree polynomial. – Z Ahmed Jan 12 '20 at 12:07