This kinda special limit

Let PaP_a and PbP_b be 2 polynomials with same degree, n, and same leading coefficient c, with c≠0c\neq 0 (otherwise the degree is less than n):

Pa(x)=cxn+an−1xn−1+∑i=0n−2aixiP_a(x)=cx^n+a_{n-1}x^{n-1}+\displaystyle\sum_{i=0}^{n-2}a_ix^i

Pb(x)=cxn+bn−1xn−1+∑i=0n−2bixiP_b(x)=cx^n+b_{n-1}x^{n-1}+\displaystyle\sum_{i=0}^{n-2}b_ix^i

Let y=lim⁡x→∞(Pa(x)n−Pb(x)n)y=\displaystyle\lim_{x\rightarrow\infty}\Big(\sqrt[n]{P_a(x)}-\sqrt[n]{P_b(x)}\Big). In this note I try to find the value of yy.

Let La(x)=∑i=0n−2aixiL_a(x)=\displaystyle\sum_{i=0}^{n-2}a_ix^i, and Lb(x)=∑i=0n−2bixiL_b(x)=\displaystyle\sum_{i=0}^{n-2}b_ix^i. Later, we'll see that both the value of La(x)L_a(x) and Lb(x)L_b(x) won't affect the value of the limit. We have:

y=lim⁡x→∞cn(xn+an−1cxn−1+La(x)cn−xn+bn−1cxn−1+Lb(x)cn)y=\displaystyle\lim_{x\rightarrow\infty}\sqrt[n]{c}\Bigg(\sqrt[n]{x^n+\dfrac{a_{n-1}}{c}x^{n-1}+\dfrac{L_a(x)}{c}}-\sqrt[n]{x^n+\dfrac{b_{n-1}}{c}x^{n-1}+\dfrac{L_b(x)}{c}}\Bigg)

y=cnlim⁡x→∞(xn+an−1cxn−1+La(x)c)−(xn+bn−1cxn−1+Lb(x)c)∑i=0n−1(Pa(x)cn)i(Pb(x)cn)n−1−iy=\sqrt[n]{c}\displaystyle\lim_{x\rightarrow\infty}\dfrac{\bigg(x^n+\dfrac{a_{n-1}}{c}x^{n-1}+\dfrac{L_a(x)}{c}\bigg)-\bigg(x^n+\dfrac{b_{n-1}}{c}x^{n-1}+\dfrac{L_b(x)}{c}\bigg)}{\displaystyle\sum_{i=0}^{n-1}\Bigg(\sqrt[n]{\frac{P_a(x)}{c}}\Bigg)^i\Bigg(\sqrt[n]{\frac{P_b(x)}{c}}\Bigg)^{n-1-i}}

y=cnclim⁡x→∞(an−1−bn−1)xn−1+(La(x)−Lb(x))∑i=0n−1(Pa(x)cn)i(Pb(x)cn)n−1−iy=\dfrac{\sqrt[n]{c}}{c}\displaystyle\lim_{x\rightarrow\infty}\dfrac{(a_{n-1}-b_{n-1})x^{n-1}+\big(L_a(x)-L_b(x)\big)}{\displaystyle\sum_{i=0}^{n-1}\Bigg(\sqrt[n]{\dfrac{P_a(x)}{c}}\Bigg)^i\Bigg(\sqrt[n]{\dfrac{P_b(x)}{c}}\Bigg)^{n-1-i}}

y=cnclim⁡x→∞an−1−bn−1+La(x)−Lb(x)xn−1∑i=0n−1(Pa(x)cxnn)i(Pb(x)cxnn)n−1−iy=\dfrac{\sqrt[n]{c}}{c}\displaystyle\lim_{x\rightarrow\infty}\dfrac{a_{n-1}-b_{n-1}+\dfrac{L_a(x)-L_b(x)}{x^{n-1}}}{\displaystyle\sum_{i=0}^{n-1}\Bigg(\sqrt[n]{\dfrac{P_a(x)}{cx^n}}\Bigg)^i\Bigg(\sqrt[n]{\dfrac{P_b(x)}{cx^n}}\Bigg)^{n-1-i}}

Since deg⁡(Pa)=deg⁡(Pb)=n\deg(P_a)=\deg(P_b)=n and both have leading coefficient c, then:

y=cnclim⁡x→∞an−1−bn−1∑i=0n−11i1n−1−iy=\dfrac{\sqrt[n]{c}}{c}\displaystyle\lim_{x\rightarrow\infty}\frac{a_{n-1}-b_{n-1}}{\displaystyle\sum_{i=0}^{n-1}1^{i}1^{n-1-i}}

y=cncan−1−bn−1ny=\dfrac{\sqrt[n]{c}}{c}\dfrac{a_{n-1}-b_{n-1}}{n}

y=an−1−bn−1ncn−1ny=\dfrac{a_{n-1}-b_{n-1}}{n\sqrt[n]{c^{n-1}}}

and we're done. We've found the value of yy. Take quite long, around 1 hour.

#Calculus

Note by Gian Sanjaya
5 years, 8 months ago

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Comments

You should state what you are trying to achieve at the start, so that others can understand what you are trying to say.

Calvin Lin Staff - 5 years, 8 months ago

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Oh okay, thanks for the suggestion. I'll edit it.

Gian Sanjaya - 5 years, 8 months ago

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Thanks. This clarifies the result :)

What happens if deg⁡Pa=n\deg P_a = n and deg⁡Pb=m \deg P_b = m ? What would be the conditions under which lim⁡Pan−Pbm \lim \sqrt[n]{P_a} - \sqrt[m]{P_b} is a constant? And under these conditions, what is the constant equal to?

Calvin Lin Staff - 5 years, 8 months ago

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@Calvin Lin – Let the leading coefficient of PaP_a and PbP_b be zaz_a and zbz_b, respectively. I would consider separating the root of polynomials in the limit to get:

lim⁡x→∞(Pa(x)n−Pb(x)m)=lim⁡x→∞(Pa(x)n−zanx)−lim⁡x→∞(Pb(x)m−zbmx)+lim⁡x→∞(zan−zbm)x\displaystyle \lim_{x\rightarrow\infty} \Big(\sqrt[n]{P_a(x)} - \sqrt[m]{P_b(x)}\Big)=\displaystyle \lim_{x\rightarrow \infty} \Big(\sqrt[n]{P_a(x)} - \sqrt[n]{z_a}x\Big) - \displaystyle \lim_{x\rightarrow \infty} \Big(\sqrt[m]{P_b(x)} - \sqrt[m]{z_b}x\Big) + \displaystyle \lim_{x\rightarrow\infty} \Big(\sqrt[n]{z_a} - \sqrt[m]{z_b}\Big)x

Applying what I've found, we would see that the first 2 limits are finite; the whole limit will be constant, not approaching either sides of infinity, if and only if the last limit is constant, which happens if and only if zan−zbm=0\sqrt[n]{z_a}-\sqrt[m]{z_b}=0, or equivalently zan=zbm\sqrt[n]{z_a}=\sqrt[m]{z_b}. Now, let za=cnz_a = c^n and zb=cmz_b=c^m for a positive real number c (so it holds for all values of m and n), and let PaP_a and PbP_b be defined as:

Pa(x)=cnxn+an−1xn−1+∑i=0n−2aixiP_a(x)=c^n x^n + a_{n-1} x^{n-1} + \displaystyle \sum_{i=0}^{n-2} a_i x^i

Pb(x)=cmxm+bm−1xm−1+∑i=0m−2bixiP_b(x)=c^m x^m + b_{m-1} x^{m-1} + \displaystyle \sum_{i=0}^{m-2} b_i x^i

We now have:

lim⁡x→∞(Pa(x)n−Pb(x)m)=an−1ncn−1−bm−1mcm−1\displaystyle \lim_{x\rightarrow\infty} \Big(\sqrt[n]{P_a(x)} - \sqrt[m]{P_b(x)}\Big)=\frac{a_{n-1}}{nc^{n-1}}-\frac{b_{m-1}}{mc^{m-1}}

Gian Sanjaya - 5 years, 8 months ago

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@Gian Sanjaya – That's a good analysis.

Pointers
1. Your end result disagrees with the previous result.
2. When you split out a limit (LHS) into 2 or more parts (RHS), it is not true that the LHS exists if and only if the RHS exists. If the RHS exists, then the LHS exists. But it is possible for the LHS to exist, while the RHS does not exist. Case in point:
lim⁡0≠lim⁡x+lim⁡(−x). \lim 0 \neq \lim x + \lim (-x).
3. I suggest doing something similar to what you did. Consider (anm−bnm)=(a−b)× ( a^{nm} - b^{nm}) = (a-b) \times stuff.

Calvin Lin Staff - 5 years, 8 months ago
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