# Tag Info

5

If $A$ is a set, then $\mathbb{1}_A$ is the indicator function of the set. That is, $\mathbb{1}_A(t) = 1$ if $t \in A$ and $\mathbb{1}_A(t)=0$ if $t \notin A$.

4

Expand: $$(1+j)^2=1+2.1j$$ $$1+2j+j^2=1+2.1j$$ Take everything to one side: $$j^2-0.1j=0$$ Factorize: $$j(j-0.1)=0$$ Hence: $j=0$ or $j=0.1$ In the scenario given there must be a reason for $j\neq0$ hence it is excluded.

3

Inspired by this question of mine, we can approximate the solution "quite" easily using Padé approximants. Let the equation be $$\frac{(1+i)^{13}\cdot i}{(1+i)^{13}-1}\cdot \frac{1}{1+i}-r=0$$ Building the simplest $[1,1]$ Padé approximant around $i=0$, we have $$0=\frac{\frac{2}{39} i (26 r+7)+\frac{1}{13} (1-13 r)}{1-\frac{4 i}{3}}$$ Canceling the ...

2

Deposit of $a_0$ must fund a stream of withdrawals $d$ in the end of year $i\in [1,N]$ whose present value is $d(1+r)^{-i}$. Summing that stream in present terms we get: $$a_0=d(1-r)^{-1}\frac {1-(1+r)^{-N}}{1-(1+r)^{-1}}$$ hence $$d=\frac {a_0r} {1-(1+r)^{-N}}\approx \frac {a_0r}{1-e^{-Nr}}$$ which tells you by how much you can increase your withdrawal ...

1

There is no implicit differentiation necessary. First you have to consider that $D_A$ and $D_B$ are the quantity units x and y respectively. The profit function becomes: $P(x,y)=30x-x^2-x^{2}+25y-y^2-y^{2}-2xy-10$ $=30x-2x^2+25y-2y^2-2xy-10$ Differenting w.r.t $x$ and $y$ $\frac{\partial P}{\partial x}=30-4x-2y=0$ $\frac{\partial P}{\partial ... 1 Iterate: $$x_{n+1}=\frac {-c}{a(x_n)^{12}}-\frac{b}{a}$$ $$x_0=\frac {-b}{a}$$ I will try to edit my answer and put bounds that indicate the rate of convergence 1 Your function is monotonically increasing and convex (as long as the$c_k$stay non-negative). Thus you can use any of Newton's method, regula-falsi with Illinois anti-stalling, secant method, etc. with guaranteed and fast convergence. Additionally, since the interest rate, outside of usury, is a number$r$close to$0$you get for$x=1+r$the equation $$... 1 In my opinion, the question in interesting with respect to many items : for this kind of problems related to financial mathematics, there is rarely more than one real solution typically, the solution corresponds to a rather small number (if x is an interest rate, it is "close" to 0; if x is a return on investment, it is "close" to 1 and so r=1-x ... 1 The way you have explained the problem, it is a compound interest one. The correct equation for it would thus be 100(1+r)^3 + 15(1+r)^2 + 10(1+r) = 148.176, which on solving, yields r = 0.0642608, i.e. 6.42608\% If you check with this interest rate, you will get back 148.176 1 If you expand out an equation for your example, you get:$$100x^3+15x^2+10x=148.176$$Running the equation through Wolfram Alpha gives a real solution of 1.06426, so your method of approximating the root is close. In general, expanding out will get a polynomial, where the roots may not have nice closed forms. This root can be found numerically through ... 1 I suppose that you need to use the classical formula$$A = P\frac{i(1 + i)^n}{(1 + i)^n - 1}$$where$A$is the monthly payment amount,$P$the amount of the loan,$i$the monthly interest and$n$the total number of payments. Everything is simple to compute using this formula except the interest (otherwise you would not ask the question) and, as far as I ... 1 If it is a$\text{minimum variance portfolio}$, then the equation for asset A is$x_1=\frac{\sigma_2^2-\sigma_{12}}{\sigma_1^2+\sigma_2^2-2\sigma_{12}}$where$x_1$is the portfolio weight of asset A and$\sigma_{12}$is the covariance between asset A and asset B. Similar formula for the weight of asset B: ... 1 One way to go would be to solve the integral directly and have a closed form formula for your function.$e^{-\delta t}$usually integrates very well :) (with$t e^{-\delta t}\$ usually integrating by parts!)

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