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Taylor's theorem states that

$$f(x)-\sum_{k=0}^n\frac{f^{(k)}(a)}{k!}(x-a)^k = \int_a^x \frac{f^{(n+1)} (t)}{n!} (x - t)^n \, dt $$

We can use this to evaluate integrals. For example, consider $f(x)=\frac{b!x^{b+n+1}}{(b+n+1)!}$. This has $f^{(k)}(0)=0$ for $k\leq n$ and $f^{(n+1)}(x)=x^b$. Hence, by Taylor's theorem,

$$f(1)-\sum_{k=0}^n\frac{f^{(k)}(0)}{k!} = \int_0^1 \frac{f^{(n+1)} (t)}{n!} (1 - t)^n \, dt $$ $$\implies \frac{b!}{(b+n+1)!}=\frac1{n!}\int_0^1t^b(1-t)^n\,dt$$ $$\implies\beta(b+1,n+1)=\int_0^1t^b(1-t)^n\,dt=\frac{b!\, n!}{(b+n+1)!}$$

So we have determined an explicit expression for the Beta function very efficiently using Taylor's theorem.

Taylor's theorem could also be used to evaluate partial sums using knowledge of the infinite sum; for example, considering $f(x)=\sum\limits_{k=0}^\infty kx^k=\frac x{(1-x)^2}=\frac1{(1-x)^2}-\frac1{1-x}$ for $|x|<1$, and noting that $f^{(n+1)}(x)=(n+2)!(1-x)^{-(n+3)}-(n+1)!(1-x)^{-(n+2)}$, we can evaluate its partial sum as

$$\sum_{k=0}^nkx^k=\frac x{(1-x)^2}-(n+2)(n+1)\int_0^x(1-t)^{-(n+3)} (x - t)^n \, dt$$ $$-(n+1)\int_0^x(1-t)^{-(n+2)} (x - t)^n \, dt ={\frac {x ( 1-{x}^{n} ) }{ \left( 1-x \right) ^{2}}}-{ \frac {n{x}^{n+1}}{1-x}} $$ after evaluating the integrals and simplifying. In fact, this holds for all $x$ except $x=1$, and if a limit towards $1$ is taken, we correctly obtain $\frac{n(n+1)}{2}$. What other sort of sums and integrals can be evaluated using Taylor's Theorem? Is this an effective method for summing or integrating? Are there certain limitations we should be aware of with this method?

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    $\begingroup$ It's an effective method for many well-behaved functions within the radius of convergence of the series about the point $a$ at which you're evaluating the derivatives. It is not in general a particularly effective method of integrating many functions of practical interest, hence the vast body of research in quadrature methods. See this link and in particular, the caveat on page 5: math.caltech.edu/~syye/teaching/courses/Ma8_2015/… $\endgroup$ Commented May 7, 2019 at 16:37

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For what concerns the first question, it would be a nice exercise trying to evaluate the following integral $$\int_{0}^{1}x^x\textrm{d}x.$$ For what concerns the second question, there are different necessary and sufficient conditions actually i.e. If you are integrating over an interval, the uniform convergence of Taylor series is sufficient to expand the function and interchange integral with series.

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  • $\begingroup$ Is it right? $$\int_0^1 x^x \, dx=\sum _{n=1}^{\infty } \frac{(-1)^{n+1}}{n^n}$$ $\endgroup$
    – Raffaele
    Commented Dec 19, 2020 at 21:38
  • $\begingroup$ Yes, it is right! $\endgroup$
    – Gauge_name
    Commented Dec 19, 2020 at 22:58

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