exercise:F8501d943e: Difference between revisions
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gives a different form for the remainder | gives a different form for the remainder | ||
is the following: | is the following: | ||
Let <math>f</math> be a function with continuous | |||
<math>(n+1)\mathrm{st | <math>(n+1)\mathrm{st}</math> derivative at every point | ||
of the interval <math>[a,b]</math>. Then | of the interval <math>[a,b]</math>. Then | ||
| Line 48: | Line 48: | ||
(t) \; dt | (t) \; dt | ||
. | . | ||
</math> | </math> | ||
<ul style{{=}}"list-style-type:lower-alpha" | <ul style{{=}}"list-style-type:lower-alpha"> | ||
<li> | <li> | ||
Using integration by parts, show that | Using integration by parts, show that | ||
| Line 63: | Line 63: | ||
. | . | ||
</math></li> | </math></li> | ||
<li>Using induction on <math>n</math> and the result of part | <li>Using induction on <math>n</math> and the result of part (a), prove the above form of | ||
Taylor's Theorem in which the remainder | Taylor's Theorem in which the remainder | ||
appears as an integral.</li> | appears as an integral.</li> | ||
</ul> | </ul> | ||
Latest revision as of 03:26, 25 November 2024
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[/math]
Another statement of Taylor's Theorem which gives a different form for the remainder is the following: Let [math]f[/math] be a function with continuous [math](n+1)\mathrm{st}[/math] derivative at every point of the interval [math][a,b][/math]. Then
[[math]]
f(b) = f(a) + f^\prime(a)(b-a) + \cdots +
\frac1{n!} f^{(n)} (a)(b-a)^n
[[/math]]
[[math]]
+ \int_a^b \frac{(b-t)^n}{n!} f^{(n+1)}
(t) \; dt
.
[[/math]]
-
Using integration by parts, show that
[[math]] \int_a^b \frac{(b-t)^n}{n!} f^{(n+1)} (t) \; dt [[/math]][[math]] = - \frac1{n!} f^{(n)} (a)(b-a)^n + \int_a^b \frac{(b-t)^{n-1}}{(n-1)!} f^{(n)} (t) \; dt . [[/math]]
- Using induction on [math]n[/math] and the result of part (a), prove the above form of Taylor's Theorem in which the remainder appears as an integral.