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Chapter 8: Q. 70 (page 681)

Let fbe a function with an nth-order derivative at a point xoand let Pn(x)=鈭�k=0nf(k)x0k!x-x0k. Prove thatf(k)x0=Pn(k)x0 for every non-negative integerk鈮�n.

Short Answer

Expert verified

The equation Pn(k)x0=f(k)x0is true.

Step by step solution

01

Step 1. Given information

An function is given asPn(x)=鈭�k=0nf(k)x0k!x-x0k

02

Step 2. Verification

First we have to find the derivative Pn(x),

Pn'(x)=ddx鈭�k=0nf(k)x0k!x-x0k=鈭�k=0nf(k)x0k!ddxx-x0k=鈭�k=0nf(k)x0k!脳kx-x0k-1=鈭�k=0nf(k)x0(k-1)!x-x0k-1

Similarly,

Pn''(x)=ddx鈭�k=0nf(k)x0(k-1)!x-x0k-1=鈭�k=0nf(k)x0(k-2)!x-x0k-2

Implies that Pn(k)x0=f(k)x0

Hence proved.

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Most popular questions from this chapter

What is Taylor鈥檚 Theorem?

Complete Example 4 by showing that the power series 鈭�k=0鈭�鈥�(鈭�1)kk3k+1(2x鈭�3)kdiverges when x=2.

Exercise 64-68 concern with the bessel function.

What is the interval for convergence forJ0(x)?

Find the interval of convergence for power series:鈭�k=1鈭�1k2x+3k

If f(x)=4x3-5x2+6x+1andP3(x) is the third Taylor polynomial for f at 鈭�1, what is the third remainder R3(x)? What is R4(x)? (Hint: You can answer this question without finding any derivatives.)
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