What is expand (x-pi/2)^4 ?

The solution to expand (x-pi/2)^4 is x^4-2pix^3+(3pi^2x^2)/2-(pi^3x)/2+(pi^4)/(16)

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expand (x-pi/2)^4

expand

(x - \frac{π}{2})^{4}

x^{4} - 2 π x^{3} + \frac{3 π ^{2} x ^{2}}{2} - \frac{π ^{3} x}{2} + \frac{π ^{4}}{16}

Solution steps

(x - \frac{π}{2})^{4}

Apply binomial theorem:

(a + b)^{n} = i = 0 \sum n (i n) a^{(n - i)} b^{i}

a = x, b = - \frac{π}{2}

= i = 0 \sum 4 (i 4) x^{(4 - i)} (- \frac{π}{2})^{i}

Expand summation

= \frac{4 !}{0 ! ( 4 - 0 ) !} x^{4} (- \frac{π}{2})^{0} + \frac{4 !}{1 ! ( 4 - 1 ) !} x^{3} (- \frac{π}{2})^{1} + \frac{4 !}{2 ! ( 4 - 2 ) !} x^{2} (- \frac{π}{2})^{2} + \frac{4 !}{3 ! ( 4 - 3 ) !} x^{1} (- \frac{π}{2})^{3} + \frac{4 !}{4 ! ( 4 - 4 ) !} x^{0} (- \frac{π}{2})^{4}

Simplify

\frac{4 !}{0 ! ( 4 - 0 ) !} x^{4} (- \frac{π}{2})^{0} : x^{4}

Simplify

\frac{4 !}{1 ! ( 4 - 1 ) !} x^{3} (- \frac{π}{2})^{1} : - 2 π x^{3}

Simplify

\frac{4 !}{2 ! ( 4 - 2 ) !} x^{2} (- \frac{π}{2})^{2} : \frac{3 π ^{2} x ^{2}}{2}

Simplify

\frac{4 !}{3 ! ( 4 - 3 ) !} x^{1} (- \frac{π}{2})^{3} : - \frac{π ^{3} x}{2}

\frac{4 !}{4 ! ( 4 - 4 ) !} x^{0} (- \frac{π}{2})^{4} = \frac{π ^{4}}{16}

= x^{4} - 2 π x^{3} + \frac{3 π ^{2} x ^{2}}{2} - \frac{π ^{3} x}{2} + \frac{π ^{4}}{16}

Combine the fractions

\frac{3 π ^{2} x ^{2}}{2} - \frac{π ^{3} x}{2} : \frac{3 π ^{2} x ^{2} - π ^{3} x}{2}

= x^{4} - 2 π x^{3} + \frac{3 π ^{2} x ^{2} - π ^{3} x}{2} + \frac{π ^{4}}{16}

Apply the fraction rule:

\frac{a \pm b}{c} = \frac{a}{c} \pm \frac{b}{c}

\frac{3 π ^{2} x ^{2} - π ^{3} x}{2} = \frac{3 π ^{2} x ^{2}}{2} - \frac{π ^{3} x}{2}

= x^{4} - 2 π x^{3} + \frac{3 π ^{2} x ^{2}}{2} - \frac{π ^{3} x}{2} + \frac{π ^{4}}{16}

\frac{( s + 10 ) ( s + 1 )}{( s + 100 ) ( s + 2 ) ^{3}}

x (5 x^{\frac{1}{3}} + 4 x^{\frac{1}{4}})

12 (e^{- x} - e^{- x} x)

e^{2 x} \cdot (- lo g_{n} (2 x + 1) - 2 x)

6 (x + 7)

What is expand (x-pi/2)^4 ?
The solution to expand (x-pi/2)^4 is x^4-2pix^3+(3pi^2x^2)/2-(pi^3x)/2+(pi^4)/(16)