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NCERT Exemplar · Q23

Q.A function f(θ)f(\theta) is defined as: f(θ)=1−θ+θ22!−θ33!+θ44!f(\theta) = 1 - \theta + \dfrac{\theta^2}{2!} - \dfrac{\theta^3}{3!} + \dfrac{\theta^4}{4!}. Why is it necessary for θ\theta to be a dimensionless quantity?

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The function f(θ)f(\theta) is a polynomial in θ\theta where terms of different powers are added together. For such addition to be physically meaningful, every term must have the same dimension — which forces θ\theta to be dimensionless.

The key idea here is dimensional homogeneity. In any valid physical equation, you can only add or subtract quantities that have the same dimensions. If you mix metres with seconds squared, the result is nonsense — and the same principle applies here.

Look at f(θ)f(\theta):

f(θ)=1−θ+θ22!−θ33!+θ44!f(\theta) = 1 - \theta + \frac{\theta^2}{2!} - \frac{\theta^3}{3!} + \frac{\theta^4}{4!}

The first term is the number 11. A pure number has no dimension — it's dimensionless. That means every other term in the sum must also be dimensionless, otherwise you'd be adding apples to oranges.

Let's check term by term:

  1. The constant term 11 is dimensionless. So the entire expression must be dimensionless.

  2. The term −θ-\theta must have the same dimension as 11. Since 11 is dimensionless, θ\theta itself must be dimensionless. If θ\theta had dimensions (say, length or time), then −θ-\theta would have those dimensions — and you cannot add a length to a pure number.

  3. The term θ22!\frac{\theta^2}{2!} is θ\theta squared divided by 22. If θ\theta had dimensions, θ2\theta^2 would have those dimensions squared. That would be a different dimension from θ\theta itself, and certainly not dimensionless. The only way θ2\theta^2 can be dimensionless is if θ\theta itself is dimensionless.

  4. The same logic applies to θ33!\frac{\theta^3}{3!} and θ44!\frac{\theta^4}{4!}. Each successive power would introduce a new dimension (length3^3, length4^4, etc.) unless θ\theta is dimensionless. …

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