calculussummation
In certain special series, we can use the sigma notation to obtain the sum of $n$ terms. For example;
$$1^3 + 2^3 + 3^3 + 4^3 + 5^3 +\cdots+ n^3 = \frac {n^2(n+1)^2}{4}$$
The sum can also be written as
$$\sum_{i=1}^n a_i^3$$
Why can we not simply integrate the function $n^3$? Using the aforementioned formula gives the correct answer. whereas integrating the function doesn't. Why is there a disparity between the two?
The exact expression for $\displaystyle H_n:=1+\frac{1}{2}+\frac{1}{3}+\cdots\ +\frac{1}{n} $ is not known, but you can estimate $H_n$ as below
Let us consider the area under the curve $\displaystyle \frac{1}{x}$ when $x$ varies from $1$ to $n$.
Now note that $\displaystyle H_{n}-\frac{1}{n}=1+\frac{1}{2}+\frac{1}{3}+\cdots\ +\frac{1}{n-1}$ is an overestimation of this area by rectangles. See below
And $\displaystyle H_n-1=\frac{1}{2}+\frac{1}{3}+\cdots\ +\frac{1}{n} $ is an underestimation of the area. See below
(source: uark.edu)
Hence $$\large H_n-1<\int_{1}^n\frac{1}{x}dx<H_n-\frac{1}{n}\\ \Rightarrow \ln n+\frac{1}{n}<H_n<\ln n+1$$
Also, Euler discovered this beautiful property of harmonic number $H_n$ that $$\large \lim_{n\rightarrow \infty}\left(H_n-\ln n\right)=\gamma\approx 0.57721566490153286060651209008240243104215933593992…$$ $\gamma$ is called the Euler-Mascheroni constant.
We have \begin{eqnarray*} f''(x)=-f(x) \cos^2(x) -f'(x) \tan(x). \end{eqnarray*} The first few terms of $-\cos^2(x)$ are \begin{eqnarray*} -\cos^2(x) = -1+ x^2+ b_4 x^4 + \cdots \end{eqnarray*} and further terms can be calculated by squaring the power series for $ \cos(x)$.
The first few terms of $-\tan(x)$ are \begin{eqnarray*} -\tan(x) = -x+ \frac{1}{3} x^3+ c_5 x^5 + \cdots \end{eqnarray*} and further terms can found in equation $(32)$ here ... http://mathworld.wolfram.com/Tangent.html
We want to calculate $f(x)$ and we shall need its first & second derivatives \begin{eqnarray*} f(x)&=&1+a_2 x^2 +a_4 x^4+ \cdots \\ f'(x)&=&2a_2 x +4a_4 x^3+ \cdots \\ f''(x)&=&2a_2 +12a_4 x^2+ \cdots. \\ \end{eqnarray*} You see where the terms in the red boxes come from now ?
Plugging all these into the first equation gives \begin{eqnarray*} 2a_2 +12a_4 x^2+ \cdots = \left( a_0+a_2 x^2 +a_4 x^4+ \cdots \right) \left( -1+ x^2+ b_4 x^4 + \cdots \right)+ \left( 2a_2 x +4a_4 x^3+ \cdots \right) \left( -x+ \frac{1}{3} x^3+ c_5 x^5 + \cdots \right) \end{eqnarray*} Now just expand the brackets and equate coefficients of $x$.
We have \begin{eqnarray*} \cos(\sin(x))=1 -\frac{x^2}{2} +\frac{5x^4}{24}+\cdots. \end{eqnarray*}
Edit : Equating the $x^0$ terms (i.e constant terms) gives \begin{eqnarray*} 2a_2=-a_0 \\ a_2= -\frac{1}{2} \end{eqnarray*} Now collect all the $x^2$ terms and we have \begin{eqnarray*} 12a_4=a_0 -a_2 -2a_2 \\ a_4= \frac{5}{24}. \end{eqnarray*} You will need to include higher order terms if you wish to calculate the next order ... Good luck $ \ddot \smile$
Best Answer
let's consider $1^3+2^3$ for now
when i use integration, i mean $1^3 + (1+dx)^3 + (1+2dx)^3 + ... + (2-dx)^3 + 2^3$, where $dx$ is a very small number
while the sigma notation means sum of 2 numbers only in this case, $1^3 + 2^3$.