The sum of power series and functions and the sum of union series

The sum function of power series and the sum of union series

The first thing to note is that for -1

In order to help everyone understand better, let’s integrate item by item to get the following results: -ln(1-x) = ∑{0 ≤ n} x^(n 1)/(n 1) = ∑{1 ≤ n} x^n/n (The integration constant can be determined from ln(1) = 0) . With this formula, we can calculate the value of -ln(1-x), which helps solve some mathematical problems. Hope this method is helpful to everyone!

In order to help players who have not yet solved the puzzle, let us learn about the specific methods of solving the puzzle. The key step is to transform the equation into the form "then -x·ln(1-x) = ∑{1 ≤ n} x^(n 1)/n = ∑{2 ≤ n} x^n/(n -1)" equation. The key to this step is to use series expansion to get the right side of the equation by summing the power series.

In order to help everyone understand better, let us interpret the specific meaning of this mathematical formula: $\ln(1-x)/x = \sum_{1 \leq n} \frac{x^{n- 1}}{n} = \sum_{0 \leq n} \frac{x^n}{n 1} = 1 \frac{x}{2} \sum_{2 \leq n} \frac{x^n }{n 1}.

In order to help everyone better understand this formula, we can prove its correctness through derivation and calculation. The specific steps are as follows: First, we can expand the series on the right-hand side into an infinite series. This series can be expressed by expanding the coefficients of each term into a geometric sequence. Next, we can simplify the expression on the left. Using the properties of series, we can express it as a fraction. Then, we can pass

To help everyone understand better, we can simplify the equation to ln(1-x)/x 1 x/2-x·ln(1-x) = 2·∑{2 ≤ n} x ^n/(n²-1). This way we can see the structure and relationships of the equation more clearly.

∑{2 ≤ n} x^n/(n²-1) = ln(1-x)/(2x) 1/2 x/4 - x·ln(1-x)/2 This series is It converges uniformly in the closed interval (-1,1).

After substituting x = 1/2, we get the result of ∑{2 ≤ n} 1/((n²-1)2^n) = 5/8-3ln(2)/4. This result can help us solve specific problems.

Power series and function issues

1, let an=x^n/n(n-1)

According to the given formula, we can deduce the following conclusion: when x=1, an=1/n(n-1)=1/(n-1)-1/n, this series is convergent of. When x=-1, an=(-1)^n*(1/(n-1)-1/n) is also convergent. This is a staggered series.

So the convergence interval is [-1,1]

2. This question should go from item 2 to infinity, right? Otherwise it is meaningless.

Since an=x^n/n(n-1)=x^n[1/(n-1)-1/n]=x^n/(n-1)-x^n/n

In order to help players who have not passed the level yet, let us learn about the specific puzzle solving methods. During the puzzle solving process, please note that the sum is calculated starting from n=2, and the second term of the formula is -x-ln(1-x). Furthermore, the first term can be written as (x^(n-1))*x/(n-1), which then leads to -xln(1-x). I hope these tips can help you solve the problem smoothly.

In order to help those players who have not solved the puzzle yet, let us take a look at the specific methods of solving the puzzle. The key to solving the puzzle is to convert the entire series sum into a simpler form. The specific calculation process is as follows: the entire series sum is -xln(1-x)-(-x-ln(1-x))=(1 -x)ln(1-x) x. This way it will be easier for you to understand and solve the puzzles.

Sum function of power series

Solution: [Use [.]' to indicate the derivative of x].

Let’s take a look at how to parse this expression: the original formula is ∑[(-1)^n]x^(2n) 2∑{[(-1)^n]/[2n(2n-1 )]}x^(2n). Now let’s explain in detail how to solve the puzzle.

To help everyone understand better, let’s discuss the summation formula in the convergence domain: ∑[(-1)^n]x^(2n)=(-x^2)/(1 x^2 ).

Suppose S=∑{[(-1)^n]/[2n(2n-1)]}x^(2n), and derive x with respect to S'=∑{[(-1)^n] /(2n-1)}x^(2n-1). Then derive the derivative of x to get S''=∑[(-1)^n]x^(2n-2)=-1/(1 x^2).

According to the problem-solving process, we got the final result: S = -xarctanx (1/2)ln(1 x^2) C. Among them, C is a constant. In addition, according to the conditions given in the question, we can determine that the value of C is 0.

The following is the original puzzle method for reference: We can use some mathematical formulas and properties to simplify and solve this expression. First, we can use the relationship of trigonometric functions to convert -arctan(x) into -ln(cos(arctan(x))). We can then combine -arctan(x) and ln(1 x^2) into a logarithmic function ln((1 x^2)/cos(arctan(x))). Next, we can combine -ln(cos(arctan(x))) and -ln((1 x^2)/cos(arctan(x))) into a logarithmic function

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