A calculus problem by Aditya Narayan Sharma

Calculus Level 4

If $f(x)$ is a polynomial satisfying $f'(x)+2f(x)=x+1$ , where $f:\mathbb{R} \to \mathbb{R}$ . Find $f(21.5)$

The answer is 11.

2 solutions

Aditya Narayan Sharma
Jun 8, 2016

A $\text{Linear Operator}$ $L$ is defined by the following two conditions :

$\displaystyle \begin{cases} L(u+v) = L(u)+L(v) \\ L(cu)=cL(u) \end{cases}$

Naturally the differential operator satisfies both the $\text{Additivity}$ & $\text{Homogeneity}$ together called $\text{Linearity}$

So Naturally we could frame it as $\displaystyle (\frac{d}{dx} +2)f=x+1$ where $f$ is a quadratic expression.

$\displaystyle (\frac{d}{dx} +2)f=\frac{d(ax^2+bx+c)}{dx}+2ax^2+2bx+2c = 2ax^2 + (2a+2b)x + (b+2c)$

Conveniently $\displaystyle [\begin{pmatrix} 2&0&0 \\ 2&2&0 \\ 0&1&2 \end{pmatrix}\begin{pmatrix}a\\b\\c \end{pmatrix}]_{B}$

where $\displaystyle ax^2+bx+c = [\begin{pmatrix}a\\b\\c \end{pmatrix}]_{B}$

So we get a equation as follows : $\displaystyle \begin{pmatrix} 2&0&0 \\ 2&2&0 \\ 0&1&2 \end{pmatrix}\begin{pmatrix}a\\b\\c \end{pmatrix} = \begin{pmatrix} 0\\1\\1\end{pmatrix}$

Solving we get the solution vector $\displaystyle \begin{pmatrix} 0 \\ \frac{1}{2} \\ \frac{1}{4}\end{pmatrix}$

So $\displaystyle f(x) = \frac{2x+1}{4} \implies f(21.5)=11$

You are using a huge theorem in analysis in order to apply your "linear operator" proof. In fact, you are also making a huge assumption that your function must be a smooth function .

IE What is the domain and range space of your linear operator $L$ ? To justify that those are correct, you need some kind of representation theorem.

Hence I object to "Note the topic is algebra. Calculus solutions will be frowned.". I have changed the topic to calculus.

Calvin Lin Staff - 5 years ago

Ok sir. In that sense if i was to make it correct I have to derive all the properties of linear algebra. So its better to be a calculus sum

Aditya Narayan Sharma - 5 years ago

It's not just the "properties of linear algebra". In fact, that is the trivial part of doing the calculations.

To start off, if the domain space of $L$ is "all differentiable functions", then what do you think is the range space of $L$ ? Note that it is possible for the derivative of a differentiable function to have an essential discontinuity , and so the function isn't just
L: {differentiable functions} to {polynomials}.

That is where the "theorem in analysis" comes in, which is that what you are currently saying is applying the stone-weierstrass theorem to conclude that the set of polynomials are dense in the differentiable functions, which is why the linear operator interpretation works with a basis of $\{ x^n \}$ . We then need to do more work, to get
L: {smooth functions} to {smooth functions}, where we use a basis of $\{ x^n\}$ to represent the space of smooth functions.

It would be a purely algebra question if you stated that $f(x)$ is a polynomial function and allowed the algebraic use of $(x^n) ' = n x^{n-1}$ along with the linearity of derivatives. This gets past using the theorem that I stated above.

Calvin Lin Staff - 5 years ago

@Calvin Lin – Thank you for ntroducing me to the that theorem, and the use of the derivative however brings the use of calculus. As we need to use the derivative of $x^n$ , we are introducing calculus there. So there isnt a pure algebraic method for these maybe.

Aditya Narayan Sharma - 5 years ago

Let $\displaystyle S= \lim_{x\to\infty} e^{-x} \int_{0}^{x} \int_{0}^{x} \frac{e^u-e^v}{u-v}du dv$

$\displaystyle S = \lim_{x\to\infty} e^{-x} \int_{0}^{x}\int_{0}^{x} e^v \frac{e^{u-v}-1}{u-v} du dv$

$\displaystyle S = \lim_{x\to\infty} e^{-x} \sum_{k=1}^{\infty}\frac{1}{k!} \int_{0}^{x}\int_{0}^{x} e^v (u-v)^{k-1}du dv$

$\displaystyle S = \lim_{x\to\infty} e^{-x} \sum_{k=1}^{\infty} \frac{1}{k.k!} \int_{0}^{x} e^v [(x-v)^k-(-v)^k] dv$

$\displaystyle S = \lim_{x\to\infty} e^{-x} \sum_{k=1}^{\infty} \frac{1}{k.k!} \int_{0}^{x} e^v (x-v)^k dv - \sum_{k=1}^{\infty} \frac{1}{k.k!} e^{-\infty} \int_{0}^{\infty} e^v (-v)^k dv$

$\displaystyle S = \lim_{x\to\infty} e^{-x} \sum_{k=1}^{\infty} \frac{1}{k.k!} \int_{0}^{x} e^x.e^{-v} v^k dv$

$\displaystyle S = \sum_{k=1}^{\infty} \frac{1}{k.k!} \int_{0}^{\infty} e^{-v} v^k dv$

$\displaystyle S = \sum_{k=1}^{\infty} \frac{\Gamma(k+1)}{k.k!} = \sum_{k \ge 1}^{}\frac{1}{k}=+\infty$

Aditya Narayan Sharma - 4 years, 11 months ago

Oli Hohman
Jun 9, 2016

This is a first order linear differential equation in standard form. The standard method for solving these is using the integrating factor method, where you multiply both sides of the equation y'+p(x)*y=q(x) by e^(integral of p(x)dx) in order for the LHS to look like the derivative of the product of two functions.

So, e^(2x) will be our integrating factor. After integrating the derivative product on the LHS, you're left with e^(2x) y = the integral of (e^(2x) (x+1))dx. Split the RHS integral into two parts. The integral of e^(2x)*x dx can be evaluated using integration by parts and letting u=x and dv = e^(2x)dx.

e^(2x) y= (1/2) x e^(2x) - integral ((1/2)e^(2x)dx) + (1/2)e^(2x) e^(2x) y = (1/2) x e^(2x)-(1/4)e^(2x)+(1/2)e^(2x) + C multiplying both sides by e^(-2x) yields: y= (1/2) x - (1/4) + (1/2) + C y= (1/2) x + (1/4)

Now, evaluating y(21.5)= (1/2)(43/2)+(1/4) = 11

Moderator note:

Simple standard approach. This is how we avoid the assumption that the function is a polynomial.

Yes, it's like this $\uparrow$ , but I think there is a small typo in the penultimate row, I think it should be $y = (1/2)x - (1/4) + (1/2) + C \cdot e^{-2x}$ . And, since y is a polynomial it must be $C = 0$ . And your reply is very good too... If someone wants more details he can visit Integrating factors in diferential equations and Differential equations with constant coefficients ... @Aditya Sharma , would you like to create with me a wiki about variable changes in differential equations, dealing for example Bernouilli equation, Riccati equation, and Euler equation and others strange changes? I start and you later can help me...

Guillermo Templado - 5 years ago

Yes obviously, Please start creating one. The intermix of the topics make this hugely interesting. I will contribute to it surely. :)

Aditya Narayan Sharma - 5 years ago

Thank you very much, I'll start very soon... When I think you can deal, for example, Bernouilli equation I can name you... If you don't agree, or you want to ask me some doubt or something, here is my adress gtemplado@hotmail.es, cheers and thank you, see you soon...

Guillermo Templado - 5 years ago

Guillermo, thank you! You're correct. I will try to contribute to the wiki! :)

Oli Hohman - 5 years ago

Thank you very much!...you and anyone are welcome if they want to contribute to the wiki . I'm going to try to start it today...

Guillermo Templado - 5 years ago

Another method could be to write y'+2y=x+1 and then solve for its characteristic equation, i.e. when the equation is homogeneous.

the characteristic equation is:

r+2=0 so the complementary solution is y=c1*e^(-2x), but the complementary solution is of the least importance because we need a particular solution.

The best guess for a particular solution is Ax+B. The particular solution's derivative is y_p' = A

Plugging these back into the original yields

Setting coefficients equal A+2(Ax+B) = x+1 A+2B=1 2A=1 A=1/2 B=1/4

So you have y_p = .5*x+.25

Oli Hohman - 5 years ago

A calculus problem by Aditya Narayan Sharma

The answer is 11.

2 solutions

Moderator note:

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