Question

312
series
write
Taylor
for f(x)= (1-x)^5/2 with lagrange's form
remainder up to 3 terms in the interval [0, 1]
of​

Answer 1

Step-by-step explanation:

expand the given maclaurin series

Answer 2

Answer:

The series of Taylor for given equation with Lagrange's form remainder up to 3 terms in the interval [0,1] is 11/36.

Step-by-step explanation:

As per the data given in the questions

we have to calculate Taylor series.

As per the questions it is given that Taylor

for f(x)= (1-x)^5/2 with Lagrange's form

remainder up to 3 terms in the interval [0, 1] .

so, f(x)= (1-x)^5/2

⇒f(1) = 0 (at x=1)

⇒f(x) = f(0) + $f^{-}$(0)x+($f^{"}$(0)x^1)/2+ .....

0= 1+(-2.5)+3.75/2-$\frac{15}{8\sqrt{1-c} }$

= $\frac{15}{48\sqrt{1-c} }$ = 0.375

= $\sqrt{1-c}=\frac{5}{6}$

= c= $\frac{11}{36}$

Hence, the series of Taylor for given equation with Lagrange's form remainder up to 3 terms in the interval [0,1] is 11/36.

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https://brainly.in/question/35963231

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