This is what I thought about when I was 17 year old. Because when I read about *Maclaurin* expansion, which is *Taylor* expansion at zero.

,

however, this equality is valid for every .

Because of this convergent series, I got to know that:

Though there is also a elementary way to calculate if we know .

We shall ask if we need to calculate other series,

.

Certainly we can use calculus, it is pretty simple if we treat them as a series function’s calculus’ expansion.

This technique requires the primary function’s explicit form…it is very difficult.

At that time, I gave my method of substituting by a complex number, and depart the Re and Im. It really works with low degree .

If I replace by , where satifies . We need prove the validity of the* Maclaurin* Formula.

We first define the multi-value function , however we can restrict to one branch.

Then we have to know whether its expansion is convergent, because the radius of convergence is 1 actually.

**BY ABEL TEST FOR CONVERGENCE, **

,

.

- is monotonic decreasing to zero.
- is convergent whenever .

Thus the expansion of is valid now for .

Surely we can set or which to see what will happen. First we set ,

, and

.

Then it is done.

If we set it to be cubic root of 1, it is more complicated, but still calculable.

The process we shall omit, and we give the answer directly.

.

, thus the formula turns into:

And ,

Therefore,

,

,

.

Solve it and , , .

It is done. However, for higher degree, we could not always find a explicit answer to it (due to Guass’ proof), but if the root can be explicitly written, then we can solve it.

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Traveller, in math.
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