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Kalkylator för Taylor- och Maclaurinserier (potensserier)

Bestäm Taylor-/Maclaurinserie steg för steg

Kalkylatorn hittar Taylor- (eller potens-)serieutvecklingen för den givna funktionen kring den givna punkten, med visade steg. Du kan ange ordningen för Taylorpolynomet. Om du vill ha Maclaurinpolynomet, sätt bara punkten till $$$0$$$.

Enter a function:

Enter a point:

For Maclaurin series, set the point to `0`.

Order `n=`

Evaluate the series and find the error at the point

The point is optional.

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Solution

Your input: calculate the Taylor series of $$$\ln{\left(x \right)}$$$ around $$$a=1$$$ up to $$$n=3$$$

A Taylor series is given by $$$f\left(x\right)=\sum\limits_{k=0}^{\infty}\frac{f^{(k)}\left(a\right)}{k!}\left(x-a\right)^k$$$

In our case, $$$f\left(x\right) \approx P\left(x\right) = \sum\limits_{k=0}^{n}\frac{f^{(k)}\left(a\right)}{k!}\left(x-a\right)^k=\sum\limits_{k=0}^{3}\frac{f^{(k)}\left(a\right)}{k!}\left(x-a\right)^k$$$

So, what we need to do to get the desired polynomial is to calculate the derivatives, evaluate them at the given point, and plug the results into the given formula.

$$$f^{(0)}\left(x\right)=f\left(x\right)=\ln{\left(x \right)}$$$

Evaluate the function at the point: $$$f\left(1\right)=0$$$

  1. Find the 1st derivative: $$$f^{(1)}\left(x\right)=\left(f^{(0)}\left(x\right)\right)^{\prime}=\left(\ln{\left(x \right)}\right)^{\prime}=\frac{1}{x}$$$ (steps can be seen here).

    Evaluate the 1st derivative at the given point: $$$\left(f\left(1\right)\right)^{\prime }=1$$$

  2. Find the 2nd derivative: $$$f^{(2)}\left(x\right)=\left(f^{(1)}\left(x\right)\right)^{\prime}=\left(\frac{1}{x}\right)^{\prime}=- \frac{1}{x^{2}}$$$ (steps can be seen here).

    Evaluate the 2nd derivative at the given point: $$$\left(f\left(1\right)\right)^{\prime \prime }=-1$$$

  3. Find the 3rd derivative: $$$f^{(3)}\left(x\right)=\left(f^{(2)}\left(x\right)\right)^{\prime}=\left(- \frac{1}{x^{2}}\right)^{\prime}=\frac{2}{x^{3}}$$$ (steps can be seen here).

    Evaluate the 3rd derivative at the given point: $$$\left(f\left(1\right)\right)^{\prime \prime \prime }=2$$$

Now, use the calculated values to get a polynomial:

$$$f\left(x\right)\approx\frac{0}{0!}\left(x-\left(1\right)\right)^{0}+\frac{1}{1!}\left(x-\left(1\right)\right)^{1}+\frac{-1}{2!}\left(x-\left(1\right)\right)^{2}+\frac{2}{3!}\left(x-\left(1\right)\right)^{3}$$$

Finally, after simplifying we get the final answer:

$$$f\left(x\right)\approx P\left(x\right) = \left(x-1\right)- \frac{1}{2}\left(x-1\right)^{2}+\frac{1}{3}\left(x-1\right)^{3}$$$

Answer: the Taylor series of $$$\ln{\left(x \right)}$$$ around $$$a=1$$$ up to $$$n=3$$$ is $$$\ln{\left(x \right)}\approx P\left(x\right)=\left(x-1\right)- \frac{1}{2}\left(x-1\right)^{2}+\frac{1}{3}\left(x-1\right)^{3}$$$


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Relaterade anteckningar: Taylor Polynomial , Maclaurin Polynomials of Common Functions