Use three terms of the expansion for root 1 + x to calculate the value of root 1.5772. root 1.5772 = 1.5772 _________ (Do not round until the final answer. Then round to four decimal places as needed.) Solution sqrt[1 + x] = 1 + x/2 + (1/2)(-1/2)x*x/2! x = 0.5772 1 + 0.2886 - 0.041644 1.2469.

1. Use |v * w| = ||v||||w||| to prove the triangle inequality ||v + w|| ||v|| + ||w||
Solution
Statement of the inequality The Cauchy–Schwarz inequality states that for all
vectors x and y of an inner product space, |langle x,y angle| ^2 leq langle x,x angle cdot
langle y,y angle where langlecdot,cdot angle is the inner product. Equivalently, by taking
the square root of both sides, and referring to the norms of the vectors, the inequality is written as
|langle x,y angle| leq |x| cdot |y| Moreover, the two sides are equal if and only if x and
y are linearly dependent (or, in a geometrical sense, they are parallel or one of the vectors is
equal to zero). If scriptstyle x_1,, ldots,, x_n,in,mathbb C and scriptstyle y_1,,
ldots,, y_ninmathbb C are any complex numbers and the inner product is the standard inner
product then the inequality may be restated in a more explicit way as follows: |x_1 bar{y}_1 +
cdots + x_n bar{y}_n|^2 leq left(|x_1|^2 + cdots + |x_n|^2 ight) left(|y_1|^2 + cdots +
|y_n|^2 ight) When viewed in this way the numbers x1, …, xn, and y1, …, yn are the
components of x and y with respect to an orthonormal basis of V. Even more compactly written:
left|sum_{i=1}^n x_i bar{y}_i ight|^2 leq left(sum_{j=1}^n |x_j|^2 ight)
left(sum_{k=1}^n |y_k|^2 ight) Equality holds if and only if x and y are linearly dependent,
that is, one is a scalar multiple of the other (which includes the case when one or both are zero).
The finite-dimensional case of this inequality for real vectors was proved by Cauchy in 1821,
and in 1859 Cauchy's student Bunyakovsky noted that by taking limits one can obtain an
integral form of Cauchy's inequality. The general result for an inner product space was obtained
by Schwarz in the year 1885. [edit] Proof Let u, v be arbitrary vectors in a vector space V over F
with an inner product, where F is the field of real or complex numbers. We prove the inequality
big| langle u,v angle big| leq left|u ight| left|v ight| This inequality is trivial in
the case v = 0, so we may assume from here on that v is nonzero. In fact, as both sides of the
inequality clearly multiply by the same factor ? when v is multiplied by a positive scaling factor
?, it suffices to consider only the case where v is normalized to have magnitude 1, as we shall
assume for convenience in the rest of this section. Any vector can be decomposed into a sum of
components parallel and perpendicular to scriptstyle v; in particular, scriptstyle u can be
decomposed into scriptstyle langle u,, v angle v ,+, z, where scriptstyle z is a vector
orthogonal to scriptstyle v (this orthogonality can be seen by noting that scriptstyle langle
u,, v angle ;=; langle langle u,, v angle v ,+, z,, v angle ;=; langle u,, v
angle ,+, langle z,, v angle, so that scriptstyle langle z,, v angle ;=; 0).
Accordingly, by the Pythagorean theorem, which is to say, by simply expanding out the
calculation of scriptstyle langle u,, u angle, we find that scriptstyle left|u ight|^2 ;=;
|langle u,, v angle|^2 ,+, left|z ight|^2 ;geq; |langle u,, v angle|^2, with

2. equality if and only if scriptstyle z ;=; 0 (i.e., in the case where scriptstyle u is a multiple of
scriptstyle v). This establishes the theorem or
http://en.wikipedia.org/wiki/Cauchy%E2%80%93Schwarz_inequality