Question

For $p, n \in N$, let $f(x) = 1-x^p$ and $g_n(x) = \frac{n}{\frac{1}{f(x)}+\frac{1}{f(2x)}+....+\frac{1}{f(nx)}}$. Find the value of $\lim_{x \to 0} \frac{1-g_n(x)}{x^p}$ at $n=5$ and $p=3$.

Answer 1

45

Answer 2

To find the value of the limit $\lim_{x \to 0} \frac{1-g_n(x)}{x^p}$, where $f(x) = 1-x^p$ and $g_n(x) = \frac{n}{\frac{1}{f(x)}+\frac{1}{f(2x)}+....+\frac{1}{f(nx)}}$, we will use the Taylor series expansion.

First, let's express $g_n(x)$ in terms of $f(kx)$: $g_n(x) = \frac{n}{\sum_{k=1}^{n} \frac{1}{f(kx)}}$

Substitute $f(kx) = 1-(kx)^p = 1-k^p x^p$: $g_n(x) = \frac{n}{\sum_{k=1}^{n} \frac{1}{1-k^p x^p}}$

We know the geometric series expansion for $\frac{1}{1-y}$ for small $y$: $\frac{1}{1-y} = 1+y+y^2+O(y^3)$

In our case, $y = k^p x^p$. As $x \to 0$, $k^p x^p \to 0$. So, we can write: $\frac{1}{1-k^p x^p} = 1+k^p x^p + (k^p x^p)^2 + O(x^{3p}) = 1+k^p x^p + k^{2p} x^{2p} + O(x^{3p})$

Let $D(x)$ be the denominator of $g_n(x)$: $D(x) = \sum_{k=1}^{n} \frac{1}{1-k^p x^p} = \sum_{k=1}^{n} (1+k^p x^p + k^{2p} x^{2p} + O(x^{3p}))$ $D(x) = \sum_{k=1}^{n} 1 + x^p \sum_{k=1}^{n} k^p + x^{2p} \sum_{k=1}^{n} k^{2p} + O(x^{3p}) \sum_{k=1}^{n} 1$ Let $S_1 = \sum_{k=1}^{n} k^p$ and $S_2 = \sum_{k=1}^{n} k^{2p}$. $D(x) = n + S_1 x^p + S_2 x^{2p} + O(x^{3p})$

Now, substitute $D(x)$ back into $g_n(x)$: $g_n(x) = \frac{n}{n + S_1 x^p + S_2 x^{2p} + O(x^{3p})}$

Next, we calculate $1-g_n(x)$: $1-g_n(x) = 1 - \frac{n}{n + S_1 x^p + S_2 x^{2p} + O(x^{3p})}$ $1-g_n(x) = \frac{(n + S_1 x^p + S_2 x^{2p} + O(x^{3p})) - n}{n + S_1 x^p + S_2 x^{2p} + O(x^{3p})}$ $1-g_n(x) = \frac{S_1 x^p + S_2 x^{2p} + O(x^{3p})}{n + S_1 x^p + S_2 x^{2p} + O(x^{3p})}$

Finally, we find the limit: $\lim_{x \to 0} \frac{1-g_n(x)}{x^p} = \lim_{x \to 0} \frac{\frac{S_1 x^p + S_2 x^{2p} + O(x^{3p})}{n + S_1 x^p + S_2 x^{2p} + O(x^{3p})}}{x^p}$ Divide the numerator by $x^p$: $\lim_{x \to 0} \frac{S_1 + S_2 x^p + O(x^{2p})}{n + S_1 x^p + S_2 x^{2p} + O(x^{3p})}$

As $x \to 0$, all terms containing $x$ will go to zero. So, the limit simplifies to: $\frac{S_1}{n} = \frac{\sum_{k=1}^{n} k^p}{n}$

Now, we apply the given values $n=5$ and $p=3$: The limit value is $\frac{\sum_{k=1}^{5} k^3}{5}$.

Calculate the sum $\sum_{k=1}^{5} k^3$: $\sum_{k=1}^{5} k^3 = 1^3 + 2^3 + 3^3 + 4^3 + 5^3$ $= 1 + 8 + 27 + 64 + 125$ $= 225$

Substitute this sum back into the limit expression: Limit $= \frac{225}{5} = 45$.

The final answer is $\boxed{45}$.

Explanation: The problem involves evaluating a limit of a function $g_n(x)$ which is defined in terms of a sum. The key insight is to use the Taylor series expansion for $\frac{1}{1-y} = 1+y+y^2+...$ for small $y$. By substituting $y=k^p x^p$ for each term in the sum, we can express the denominator of $g_n(x)$ as $n + x^p \sum_{k=1}^{n} k^p + O(x^{2p})$. Then, $1-g_n(x)$ simplifies to $\frac{x^p \sum_{k=1}^{n} k^p + O(x^{2p})}{n + O(x^p)}$. Dividing by $x^p$ and taking the limit as $x \to 0$ yields $\frac{\sum_{k=1}^{n} k^p}{n}$. Finally, substituting $n=5$ and $p=3$, we calculate the sum of the first five cubes and divide by 5, which gives 45.