Question

Show that the swap rate for a standard swap with level notional amount is the same as the at-par yield rate for a bond with same time to maturity as the length of the swap. At-par yield was defined in Chapter 6 as coupon rate r at which a bond would have a yield to maturity of r under the current term structure.

Answer 1

Answer:

I don't understand

Step-by-step explanation:

but please mark me as a brainlist

Answer 2

Step-by-step explanation:

$\orange{\bold{\underbrace{\overbrace{❥Question᎓}}}}$

Integrate the function

$\huge\green\tt\frac{ \sqrt{tanx} }{sinxcosx}}$

⇛$\huge\tt\frac{ \sqrt{tanx} }{sinxcosx}$

ㅤ ㅤ ㅤ ㅤ ㅤ

⇛$\huge\tt \frac{ \sqrt{tanx} }{sinxcosx \times \frac{cosx}{cosx}}$

ㅤ ㅤ ㅤ ㅤ ㅤ

⇛$\huge\tt \frac{ \sqrt{tanx} }{sinx \times \frac{ {cos}^{2} x}{cosx}}$ ㅤ ㅤ ㅤ

⇛ $\huge\tt\frac{ \sqrt{tanx} }{ {cos}^{2} x \times \frac{sinx}{cosx} }$

ㅤ ㅤ ㅤ ㅤ ㅤ

⇛$\huge\tt\frac{ \sqrt{tanx} }{ {cos}^{2}x \times tanx }$

⇛$\huge\tt {tan}^{ \frac{1}{2} - 1 } \times \frac{1}{ {cos}^{2} x}$ㅤ ㅤ ㅤ ㅤ ㅤ

⇛$\huge\tt {(tan)}^{ - \frac{ 1}{2} } \times \frac{1}{ {cos}^{2}x } = {(tanx)}^{ - \frac{1}{2} } \times {sec}^{2} x⇛(tan)$

ㅤ ㅤ ㅤ ㅤ ㅤ

⇛$\huge\tt {(tan)}^{ - \frac{ 1}{2} } \times \frac{1}{ {cos}^{2}x } = ∫ {(tanx)}^{ - \frac{1}{2} } \times {sec}^{2} x \times dx⇛(tan)$

ㅤ ㅤ ㅤ ㅤ ㅤ

$\bold\blue{☛\: Let tanx=t}$

$\bold\blue{☛ \:Differentiating \: both \: sides \: w.r.t.x}$

ㅤ ㅤ ㅤ ㅤ ㅤ

⇛$\huge\tt {sec}^{2} x = \frac{dt}{dx}$

⇛$\huge\tt{dx \frac{dt}{ {sec}^{2}x }}$

ㅤ ㅤ ㅤ ㅤ ㅤ

⇛$\huge\tt∴∫ {(tanx)}^{ - \frac{1}{2} } \times {sec}^{2} x \times dx$

⇛$\huge\tt ∫ {(t)}^{ - \frac{1}{2} } \times {sec}^{2} x \times \frac{dt}{ {sec}^{2}x }$

⇛$\huge\tt ∫ {t}^{ - \frac{1}{2} }$ㅤ ㅤ

⇛ $\huge\tt\frac{ {t}^{ - \frac{1}{2} + 1} }{ - \frac{1}{2} + 1 }$

⇛ $\huge\tt \frac{ {t}^{ \frac{1}{2} } }{ \frac{1}{2} } + c = 2 {t}^{ \frac{1}{2} } + c = 2 \sqrt{t}$

⇛$\huge2 \sqrt{t} + c = 2 \sqrt{tanx}$

╚════════════════════════