Calculating a function given a certain x range.

Question

I have the function $f(x) = x^3 + 1$

and I want to calculate the following:

$f(x) = (\frac{a}{25})^3 + 1 $

Where "$a$" is 1,2,3,4,..,50.

1 through 50

Basically I'm adding up $f(1) + f(2) + f(3) + f(4),+..,f(50)$

I want to know if there a easier way of doing this instead of just adding each number on your calculator.

score 2 · Accepted Answer · answered Apr 05 '16 at 05:31

2

It turns out that $$ \sum_{n=1}^Nn^3=\frac{N^2(N+1)^2}{4} $$ for all natural numbers $N$, hence $$ \sum_{n=1}^{50}f\Big(\frac{n}{25}\Big)=\sum_{n=1}^{50}\Big[\frac{n^3}{25^3}+1\Big]=50+\frac{1}{25^3}\sum_{n=1}^{50}n^3=50+\frac{50^2\cdot 51^2}{4\cdot 25^3}$$

answered Apr 05 '16 at 05:31

carmichael561

53,688

By the way, if you're interested in the general formula for sums of the form $\sum_{n=1}^Nn^k$, look up Faulhaber's formula. – carmichael561 Apr 05 '16 at 05:36
Thanks man, wish I can give you both correct answer and I'll learn that for sure, helps a lot – shawn edward Apr 05 '16 at 05:48

score 1 · Answer 2 · answered Apr 05 '16 at 05:32

1

There is an identity $$\sum_{a=1}^n a^3=\left(\frac{n(n+1)}{2}\right)^2.$$ We can apply that here: $$\sum_{a=1}^{50}\left(\frac{a}{25}\right)^3=50+\frac{1}{25^3}\sum_{a=1}^{50} a^3=50+\frac{1}{25^3}\left(\frac{50\times 51}{2}\right)^2=50+\frac{51^2}{25}.$$ You can put that in your calculator, and do just one calculation.

answered Apr 05 '16 at 05:32

Plutoro

22,730
3
41
71

Thanks man, wish I can give you both correct answer. – shawn edward Apr 05 '16 at 05:48
Give it to carmichael. He was a few seconds faster than I. – Plutoro Apr 05 '16 at 05:49
alright, but thanks again, really really appreciate the help. – shawn edward Apr 05 '16 at 05:51

Calculating a function given a certain x range.

2 Answers2