How to find $P(-2)$ here?

Question

Where $$ P(n)=\sum_{i=1}^n i^{10}$$ And $n=1,2...$, $P:R\to R$.

Maybe there is a easier way of solving this without actually calculating the sum, which is not that pleasing.

Answer 1

Let $P(x)$ be any polynomial such that

$$P(n) = \sum_{k=1}^n k^{10}\quad\text{ for } n \in \mathbb{N}$$

By its definition, we have $P(0) = 0$ and

$$P(n) - P(n-1) = n^{10}\quad\text{ for } n \in \mathbb{Z}_{+}$$ This implies the polynomial $P(x) - P(x-1) - x^{10}$ has infinitely many roots and hence identically zero. From this, we can deduce

$$P(0) - P(-1) = 0^{10} = 0 \implies P(-1) = P(0) = 0$$ and hence $$P(-1) - P(-2) = (-1)^{10} = 1 \implies P(-2) = P(-1) - 1 = -1$$

Answer 2

In my book, you'll get the empty sum.

By definition, the empty sum in a commutative ring is the zero element of the ring.

The empty product in a commutative ring is the unit element of the ring.

Answer 3

Idea:

Clearly $P(n)$ is some polynomial in $n$ of the degree $11$. Find this polynomial and then calculate $P(-2)$.

Like if $$Q(n)=\sum_{i=1}^n i$$ We know it is $$Q(n)= {n(n+1)\over 2}$$ so $Q(-2)=...$

Answer 4

I'm not sure how $n = -2$ works under the summation definition. But there do exist closed forms for sums of these types. In the case of $10$, we have the formula $$P(n) = \frac{1}{66}(6n^{11} + 33 n^{10} + 55n^9 - 66n^7 + 66n^5 - 33n^3 + 5n).$$ In this case, $n=-2$ does make sense and can simply be substituted into the formula. See here for more information: http://mathworld.wolfram.com/PowerSum.html

Answer 5

If you accept that the sum just means sum from $n=1$ to $n=-2$ there are only four terms, so $P(-2)=1^{10}+0^{10}+(-1)^{10}+(-2)^{10}=1+0+1+1024=1026$ This is many fewer terms than the tenth power sum formula.

Whether one should accept a summation sign with the upper index less than the lower is a separate question. I think this is what is intended.

Answer 6

The question asks for the value of $\,P(-2)\,$ where $\,P(n) := \sum_{i=1}^n f(i)\,$ and $\,f(i) := i^{10}.\,$ The key part of the question is how to interpret $\,\sum_{i=1}^n f(i)\,$ when $\,n<1.\,$ One common approach is to bypass the need for interpretation by finding a formula for $\,P(n)\,$ which is valid for $\,n\ge0\,$ and then extending that by using it for $\,n<1\,$ also. This approach gives sensible answers and it can be a good method.

Another common approach is with the definition $$ \sum_{i=a}^b f(i) := f(a) + f(a+1) + \dots + f(b) $$ where the summation is undefined for $\,b<a\,$ and this interpretation makes sense especially if $\,f(i)\,$ is itself undefined for $\,i<a.\,$ In our case, it implies that $\,P(-2)\,$ is undefined and that is the correct answer according to this interpretation. A variation of this approach is to define the summation as $0$ for all $\,b\le a\,$ and this is also a valid interpretation. A less common variation is to reverse the limits of summation if the lower limit is greater than the upper limit.

Another interpretation of $\,\sum_{i=1}^n f(i)\,$ is to define it in a similar way as a definite integral using the first fundamental theorem of calculus. In this approach we find a function $\,F(n)\,$ such that $\,f(i) = F(i+1)-F(i)\,$ for all integer $\,i.\,$ Then define $\,\sum_{i=a}^b f(i) := F(b+1)-f(b)\,$ for all integer $\,a,b.\,$ One consequence of this definition is that $\,\sum_{i=a}^b f(i) = -\,\sum_{i=b+1}^{a-1} f(i).\,$ In our particular case, this implies that $$ P(-2) = -f(-1) - f(0) = -(1^{10}) - (0^{10}) = -1 -0 =-1. $$ This answer agrees with the formula approach as it should.

Notice that with this interpretation, and given that $\,f(i)>0\,$ for all $\,i,\,$ then the sum $\,S := \sum_{i=a}^b f(i)>0\,$ if $\,b>a\,$ but that $\,S<0\,$ if $b<a$ which seems counterintuitive but is a simple consequence of the definition.

To summarize, there are several interpretations of summation which do not all agree in at least some cases. The answer to your question depends on which interpretation is required to be used. Without knowing this, the question can not be correctly answered.