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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/… $\endgroup$ – Sharat V Chandrasekhar May 7 at 16:37

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