Proof that |arctan⁡(x)−π4−(x−1)2|≤(x−1)22I'm trying to prove that, for every x≥1 :|arctan⁡(x)−π4−(x−1)2|≤(x−1)22I could do it graphically on R, but how to make a formal algebraic proof?

We have that for x≥1arctan⁡x−π4−(x−1)2≤0⇒|arctan⁡x−π4−(x−1)2|=π4+(x−1)2−arctan⁡xthen considerf(x)=(x−1)22+arctan⁡x−π4−(x−1)2⇒f′(x)=1x2+1+x−32≥0and since f(1)=0 we have that f(x)≥0 and the inequality is proved.

Hint: For each x&gt;1, Taylor gives arctan⁡(x)=π4+(x−1)2+arctan⁡(cx)2(x−1)2 where 1&lt;cx&lt;x Thus all you need to show is that |arctan⁡(c)|≤1 for all c≥1

We have that for x≥1arctan⁡x−π4−(x−1)2≤0⇒|arctan⁡x−π4−(x−1)2|=π4+(x−1)2−arctan⁡xthen considerf(x)=(x−1)22+arctan⁡x−π4−(x−1)2⇒f′(x)=1x2+1+x−32≥0and since f(1)=0 we have that f(x)≥0 and the inequality is proved.

Hint: For each x&gt;1, Taylor gives arctan⁡(x)=π4+(x−1)2+arctan⁡(cx)2(x−1)2 where 1&lt;cx&lt;x Thus all you need to show is that |arctan⁡(c)|≤1 for all c≥1

"Proof that |arctan⁡(x)−π4−(x−1)2|≤(x−1)22I'm trying to prove that, for..." was answered by students like you

mairie0708zl

Answered question

2022-01-26

Proof that

| \arctan (x) - \frac{π}{4} - \frac{(x - 1)}{2} | \leq \frac{{(x - 1)}^{2}}{2}

I'm trying to prove that, for every

x \geq 1

| \arctan (x) - \frac{π}{4} - \frac{(x - 1)}{2} | \leq \frac{{(x - 1)}^{2}}{2}

I could do it graphically on

R

, but how to make a formal algebraic proof?

Answer & Explanation

chaloideq1

Beginner2022-01-27Added 11 answers

We have that for

x \geq 1

\arctan x - \frac{π}{4} - \frac{(x - 1)}{2} \leq 0 \Rightarrow | \arctan x - \frac{π}{4} - \frac{(x - 1)}{2} | = \frac{π}{4} + \frac{(x - 1)}{2} - \arctan x

then consider

f (x) = \frac{{(x - 1)}^{2}}{2} + \arctan x - \frac{π}{4} - \frac{(x - 1)}{2} \Rightarrow f^{'} (x) = \frac{1}{x^{2} + 1} + x - \frac{3}{2} \geq 0

and since f(1)=0 we have that

f (x) \geq 0

and the inequality is proved.

euromillionsna

Beginner2022-01-28Added 16 answers

Hint: For each

x > 1

, Taylor gives

\arctan (x) = \frac{π}{4} + \frac{(x - 1)}{2} + \frac{\arctan (c_{x})}{2} {(x - 1)}^{2}

where

1 < c_{x} < x

Thus all you need to show is that

| \arctan (c) | \leq 1

for all

c \geq 1

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