3 years ago

Pleaseeee help What one are equivalent?

Mathematics

1 answer:

PSYCHO15rus [73]3 years ago

7 0

2, 3, 5. I just solved them and they look equivalent.

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use Taylor's Theorem with integral remainder and the mean-value theorem for integrals to deduce Taylor's Theorem with lagrange r

Vadim26 [7]

Answer:

As consequence of the Taylor theorem with integral remainder we have that

$f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \cdots + \frac{f^{(n)}(a)}{n!}(x-a)^n + \int^a_x f^{(n+1)}(t)\frac{(x-t)^n}{n!}dt$

If we ask that $f$ has continuous $(n+1)$ th derivative we can apply the mean value theorem for integrals. Then, there exists $c$ between $a$ and $x$ such that

$\int^a_x f^{(n+1)}(t)\frac{(x-t)^k}{n!}dt = \frac{f^{(n+1)}(c)}{n!} \int^a_x (x-t)^n d t = \frac{f^{(n+1)}(c)}{n!} \frac{(x-t)^{n+1}}{n+1}\Big|_a^x$

Hence,

$\int^a_x f^{(n+1)}(t)\frac{(x-t)^k}{n!}d t = \frac{f^{(n+1)}(c)}{n!} \frac{(x-t)^{(n+1)}}{n+1} = \frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1} .$

Thus,

$\int^a_x f^{(n+1)}(t)\frac{(x-t)^k}{n!}d t = \frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}$

and the Taylor theorem with Lagrange remainder is

$f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \cdots + \frac{f^{(n)}(a)}{n!}(x-a)^n + \frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}$ .

Step-by-step explanation:

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How to do this in math

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Multiply both sides by 5

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Solve for y<br> 1.02 = -3.6 + 3.9

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Answer:

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Step-by-step explanation:

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Program the VR Robot to drive forward 6 grid squares, turn around, and drive back 6 grid squares. How many degrees did you have

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